1

Unit 1

REVIEW OF PLANE TRIGONOMETRY

1. analyze a right triangle. 2. add vectors in parallelogram and polygon method. 3. use trigonometric identities. 4. analyze oblique triangles. 5. draw graphs of trigonometric functions. 6. define a complex number. 7. illustrate the graphical representation of a complex number. 8. convert complex numbers to different forms. 9. perform mathematical operations of complex numbers. 10. apply complex numbers. 11.

LEARNING OUTCOMES

At the end of this unit, you

are expected to:

2

complex number polygon method

conjugate quadrant cosine law rectangular form

exponential form right triangle

oblique triangle sine law parallelogram trigonometric form

polar form vector addition

1.1 The Right Triangle

𝑠𝑖𝑛 𝐴 =𝑎

𝑐 𝑐𝑜𝑠 𝐴 =

𝑏

𝑐 𝑡𝑎𝑛 𝐴 =

𝑎

𝑏

𝑐𝑠𝑐 𝐴 =𝑐

𝑎 𝑠𝑒𝑐 𝐴 =

𝑐

𝑏 𝑐𝑜𝑡 𝐴 =

𝑏

𝑎

1.1.1 Solution of a Right Triangle

Referring to Figure 1.1, the relationships of the sides a, b, and c can be expressed as

𝑐2 = 𝑎2 + b2 (Pythagorean Theorem)

To find c, a, and b

𝑐 = 𝑎2 + b2 a = 𝑐2 − b2

b = 𝑐2 − a2

𝐵

𝑐 𝑎

𝐴 𝑏 𝐶

Important Terms

3

Example 1.1. Solve for b and the angles A and B.

Given: 𝑎 = 12 𝑢𝑛𝑖𝑡𝑠

𝑐 = 13 𝑢𝑛𝑖𝑡𝑠

Find: side b and angle B

Known: 𝑏 = 𝑐2 − a2

𝑠𝑖𝑛 𝐴 =𝑎

𝑐 ; 𝐴 = sin−1 a

c

𝐴 + 𝐵 = 90 ; B = 90 – A

Solution: b = 132 − 122

b = 5 𝑢𝑛𝑖𝑡𝑠

𝐴 = sin−1 12

13

𝐴 = 67.38

𝐵 = 90 – 67.38

𝐵 = 22.62

Answer: 𝑏 = 5 𝑢𝑛𝑖𝑡𝑠 , 𝐵 = 22.62°

Example 1.2. Find the sides a and b.

𝐵

𝑐 = 13 𝑎 = 12

𝐴 𝑏 𝐶

40

𝐵

𝑐 = 13 𝑎

𝐴 𝑏 𝐶

40

4

Solution:

Solving for a 𝑠𝑖𝑛 𝐴 =𝑎

𝑐

𝑎 = 𝑐 sin 𝐴

= 13 sin 40

= 8.36 𝑢𝑛𝑖𝑡𝑠

To find b 𝑐𝑜𝑠 𝐴 =𝑏

𝑐

𝑏 = 𝑐 cos 𝐴

= 13 cos 40

= 9.96 𝑢𝑛𝑖𝑡𝑠

Drill Problems

1. The height of the tower is unknown. There is now no way that you can climb the tower to determine its height. The only possible way is to use a transit to find the

angle of elevation of the tower () and the distance (d) between the transit and the foot of the tower. If the height of the transit is 1 meter (m), determine the height of the tower.

Input any value of d and and click answer. Be sure you have solved the problem with the assumed

values of d and before knowing the answer.

2. Find the x and y components of the force exerted by the father in pulling the cart with his child riding.

D m ° Answer m

d

5

Input any value of F and and click answer. Be sure you have solved the problem with the assumed

values of d and before knowing the answer.

1.2 Vector Addition

1.2.1 Parallelogram Method

1.2.2 Polygon (Head to Tail method) Method

F N ° Answer Fx Fy N

y

F1 FT

x F2

F

6

Note: F1 and F2 are the given vectors, and FT is the equivalent (total or resultant) vector.

Example 1.3. Determine graphically the resultant of the following forces using:

(a) parallelogram method. (b) polygon method.

Solution:

(a) By parallelogram method, get first the resultant of F1 and F2 and call it FA. Then use again

parallelogram method to determine the resultant vector of FA and F3 to get the total or

equivalent force.

y

FT

F1

F2

F1

F 2

F3

7

(b) Using polygon method

FA

F1

FT

F 2

F3

F2

F1 F3

F T

8

1.3 Formulas Involving Addition or Subtraction of Functions

sin2 + cos2 = 1

sec2 = 1 + tan2

csc2 = 1 + cot2

1.4 Sine and Cosine of the Sum or Difference of Two Angles

sin ( ) = sin cos cos sin

cos ( +) = cos cos - cos sin

cos ( - ) = cos cos + cos sin

sin (-A) = -sin A

cos (-A) = cos A

sin (x + 90) = cos x

cos (x + 90) = - sin x

sin (x – 90) = - cos x

sin (180 - x) = sin x

cos (180 - x) = - cos x

1.5 Double Angle Formulas

sin 2 = 2 sin cos

cos 2 = cos2 – sin2 = 2cos2 – 1 = 1 – 2 sin2

tan 2 = 2 ta n

1−𝑡𝑎𝑛 2

cot 2 = 𝑐𝑜𝑡2 − 1

2 cot

𝑐𝑜𝑠2 = 1 + 𝑐𝑜𝑠2

2

𝑠𝑖𝑛2 = 1 − 𝑐𝑜𝑠2

2

1.6 Oblique Triangles

b

9

1.6.1 Sine Law

𝑎

sin𝐴 =

𝑏

sin𝐵 =

𝑐

sin 𝐶

1.6.2 Cosine Law

𝑎2 = 𝑏2 + 2− 2bc cos A

Example 1.4. Find the x and y components of F1.

c a

A b

F1 = 18 N FT = 30 N

F2= 20 N

B

c a

A b C

10

Solution:

can be found by applying cosine law:

FT2 = F1

2 + F22 - 2 F1 F2 cos

302 = 182 + 202 – 2(18)(20) cos

= 104.15

Then = 180 - = 180 - 104.15 = 75.85

Considering F1

F1x = F1 cos

= 18 cos 75.85

= 4.4 N

F1y = F1 cos

= 18 sin 75.85

= 17.45 N

F1 = 18 N FT = 30 N

F2= 20 N

This is also

equal to F1

F1 = 18 N F1y

F1x

11

1.7 Graphs of Trigonometric Functions

Note: It is assumed that the amplitude of these waves is 1.

Drill Problem

1. Two horses are pulling a barge along a canal as shown in the figure. The cable connected to

the first horse makes an angle of 1 with respect to the direction of the canal (x-axis), while

the cable attached to the second horse makes an angle of 2 . The first horse exerts a force F1

and the second horse exerts a force F2. Find the total force exerted by the two horses.

y = sin x

y = cos x

(a)

(b)

12

Input any value of F1 and F2 . Assume values of not less than 600 N. Also input values of 1 and 2

that are less than 90°. Solve first the problem with the assumed values before clicking the answer.

2. A school bus leaves a school at 1 PM traveling at a speed of v1 along a straight road. At 2 PM

it turns right at an angle of with respect with the first road and runs at a speed of v2. What

will be its distance from the school at 2:30 PM?

Input any value of v1 and v2 and . Solve first the problem using the assumed values before clicking

the answer.

F1 N F2 N 1 ° 1 ° Ans N

v1 kph v2 kph ° Answer N

13

1.8 The Quadrants

Counterclockwise rotation Clockwise rotation

270

3

2

0 or 360

(0 or 2)

180

(/2)

90

(/2)

II I

IV

III

-270

−3

2

0 or 360

(0 or 2)

-180

(-/2)

-90

(-/2)

II I

IV

III

I II III IV

0 90 180 270 360

0 ð

2

3

2 2

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Problem Set No. 1

REVIEW OF PLANE TRIGONOMETRY

1. A ladder leans against a wall with its foot 10 ft from the wall. If the ladder makes an angle of

60 with the ground, how long is the ladder. a. 20 ft b. 15 ft c. 10 ft d. 5 ft

2. In no. 1, how far above the ground is the top of the ladder. a. 12.73 ft b. 13.72 ft c. 17.32 ft d. 19.73 ft

3. A tower 160 ft casts a shadow 92.4 ft long on the ground. Find the angle of elevation of the

sun. (The angle of elevation is the angle with its rays make with level ground.) a. 35° b. 60° c. 45° d. 75°

4. From the top of a tower at C the angle of depression of point A is 56. If the distance between point A and point B (the foot of the tower) is 180 meters, how high is the tower? a. 168.94 m b. 178.5 m c. 230.89 m d. 266.86 m

5. A boy pushes a force of 15 lb toward north. Another boy pushes the same box toward east

with a force of 20 lb. Find the resultant force. a. 25 lb b. 30 lb c. 35 lb d. 40 lb

6. One of the diagonals of a parallelogram makes angles of 2235’ and 4848’, respectively, with two adjacent sides. If the diagonal is 18.54 ft long, find the lengths of the sides of the parallelogram. a. 7,25 ft, 13.73 ft b. 7.51 ft, 14.72 ft c. 8.34 ft, 15.83 ft d. 9.35 ft, 16.72 ft

7. In the figure shown below, find the magnitude of F1 if F2 = 100 N and makes an angle of 30 with the respect to the x axis and the resultant of the two forces is 167 N and makes an angle

of 50 with respect to the x axis.

15

a. 80.64 ft

b. 82.64 ft

c. 89.49 ft

d. 90.29 ft

F1

F2

16

1.9 Complex Numbers y (imaginary axis)

+ −1 = + jA

-A A x (real axis)

− −1 = - j

A complex plane.

A complex number is one of the form

x j y

that is, it is the sum of a real number and an imaginary one.

A complex number is the conjugate of another if their imaginary parts are of opposite signs.

For example, -5 + j6 is the conjugate of –5 – j6.

1.9.1 Graphical Representation of Complex Numbers

Let z be the vector connecting origin and point P whose coordinates are x and y. Then, in complex

form

z = x + jy

r

y

x

P (x, y)

17

To find r and in terms of x and y:

r = 𝑥2 + 𝑦2 Note that r is the magnitude of z

= tan-1 𝑦

𝑥

To find x and y in terms of r and :

x = r cos

y = r sin

1.9.2 Forms of a Complex Number

Rectangular Form: z = x + jy

Trigonometric Form: z = r cos + j r sin

= r (cos + j sin )

Polar Form: z = r cjs = r

Exponential Form: z = rej , where is in radians

1.9.3 Addition and Subtraction of Complex Number

1.9.3.1 Algebraic Method

(a) Addition. To add two or more complex numbers, add the real parts and then the pure imaginary parts.

Example 1.5. Add z1 = 5 + j3 and z2 = -2 – j5

Solution: z1 + z2 = (5 + j3) + (-2 – j5)

= 3 - j2

(b) Subtraction. To subtract one complex number from another, subtract the real parts and then subtract the pure imaginary parts.

Example 1.6. Subtract z1 = 5 + j3 by z2 = -2 – j5

Solution: z1 - z2 = (5 + j3) - (-2 – j5)

= 7 + j8

1.9.3.2 Graphical Method

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(a) Addition. Use parallelogram method to find the sum of two or more complex

numbers.

Example 1.7. Add z1 = 5 + j3 and z2 = -2 – j5

Solution:

First plot z1 and z2.

1. Use parallelogram method to determine the sum.

2. Count the x and y components of the sum.

As you can see the sum is 3 - j2

(b) Subtraction.

Example 1.8. Subtract z1 = 5 + j3 by z2 = -2 – j5

Solution:

1. First plot z1 and z2.

2. Take the negative of z2. This is the opposite of z2.

3. Find the difference using parallelogram method. This is similar to addition.

4. Counting the x and y components, the result is 7 + j8.

0

8

7

6

5

4

3

2

1

-2

-3

-4

-5

-6

-7

-8

-9

-9 -8 -7 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7 8 9

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1.9.3 Transformation of a Complex Number from One Form to Another

Example 1.9. Express each of the following in rectangular, polar and exponential forms.

0

8

7

6

5

4

3

2

1

-2

-3

-4

-5

-6

-7

-8

-9

-9 -8 -7 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7 8 9

3

4

4

2

1

r4

r3

r2

r1 z2

0

8

7

6

5

4

3

2

1

-2

-3

-4

-5

-6

-7

-8

-9

-9 -8 -7 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7 8 9

z1

z3

z4

20

Solution:

Note: Always use the smaller angle between the positive real axis (the reference) and the vector.

For z1: r1 = 42 + 52 = 6.4 For z2: r2 = 62 + 32 = 6.71

1 = tan-1 5

4 = 51.34 1 = 180 - tan-1

3

6

= 153.43

For z3: r3 = 42 + 52 = 6.4 For z4: r4 = 62 + 42 = 7.2

3 = 180 - tan-1 5

4 = 128.66 4 = tan-1

4

6 = 33.69

Answers

Rectangular Form Polar Form Exponential Form

z1 4 + j5 6.4 51.34 6.4 ej0.8961

z2 -6 + j3 6.71 153.43 6.71 ej2.6779

z3 -4 - j5 6.4 -128.66 6.4 e–j2.2455

z4 6 – j4 7.2 -33.69 7.2 e-j0.588

1.9.4 Multiplication and Division of Complex Numbers

(a) Multiplication of Complex Numbers in Rectangular Form

Example 1.10. Multiply z1 = 5 + j2 by z2 = 3 – j4

Solution:

z1 z2 = (5 + j2)(3 – j4)

= (5)(2) + (j2)(3) + (5)(-j4) + (j2)(-j4)

= 10 + j6 – j20 – j2 8 where j2 = -1

= 10 + j6 – j20 – (-1)(8)

= 18 – j14

21

(b) Multiplication of Complex Numbers in Polar Form

Let z1 = r11 z2 = r22

z1z2 = r11 r22 = r1 r21 + 2

Example 1.11. Multiply z1 = 2045 by z2 = 2-30

Solution: z1 z2 = (2045 )( 2-30)

= 20245+(-30)

= 4015

(c) Division of Complex Numbers in Rectangular Form

Example 1.12. Divide z1 = 5 + j2 by z2 = 3 – j4

Solution:

𝒛𝟏

𝒛𝟐 =

5 + j2

3 – j4

Rationalize, i.e., multiply the numerator and denominator by the conjugate of the denominator. The conjugate of z2 = 3 – j4 is 𝐳𝟐 = 3 + j4 𝒛𝟏

𝒛𝟐 =

5 + j2

3 – j4

3 + j4

3 + j4

𝒛𝟏

𝒛𝟐 =

7 + 𝑗26

32+ 42 = 7 + 𝑗26

25 = 0.28 + j1.04

(d) Division of Complex Numbers in Polar Form

Let z1 = r11 z2 = r22 𝒛𝟏

𝒛𝟐 =

𝑟11

𝑟22 =

𝑟1

𝑟2 1+ 2

Example 1.13. Divide z1 = 2045 by z2 = 2-30

Solution: 𝒛𝟏

𝒛𝟐 =

2045

2−30 =

20

2 45 − (−30) = 1075

22

Problem Set No. 2

COMPLEX NUMBERS

1. Add the following using algebraic and graphical methods. (a) 8 + j4, 4 – j6 (b) -7 – j7, 8 + j3, 5 – j3

2. Perform the following using algebraic and graphical methods. (a) (8 + j4) – (4 – j6) (b) (-7 – j7) + ( 8 + j3) – (5 – j3)

3. Convert the following to polar and exponential forms: (a) (8 + j4) (b) (-4 + j6) (c) (-7 – j7) (d) ( 8 - j3)

4. Convert the following to rectangular and exponential forms.

(a) 660 (b) 12-300 (c) 10120 (d) 9-120

5. Convert the following to rectangular and polar forms.

(a) 5ej1.45 (b) 9e-j0.9823 (c) 12ej/3 (d) 7e-j

6. Perform the following. Express answers in polar form.

(a) 3045 4-30

(b) 30ej/4 4-30

(c) 3045 4−30

3045 + 4−30

(d) (515 + 4e-j/3)(30 ej/4 - 490)

8. Solve this problem using complex numbers: In the figure shown below, find the magnitude of F1, its anlge with respect to the + x aixs, its x and y components , if F2 = 100 N and makes an

angle of 30 with the respect to the x axis and the resultant of the two forces is 167 N and

makes an angle of 50 with respect to the x axis.

F1

F2

23

Unit 2

INTRODUCTION TO ELECTRICITY

1. define electricity. 2. name some scientists who contributed to the development of

electricity and electronics. 3. discuss the scientist’s contributions to electricity and electronics. 4. quote some applications of electricity and electronics. 5. identify various electrical components. 6. use metric prefixes in simplifying large and small numbers. 7. perform mathematical operations involving powers of ten and

metric prefixes. 8. discuss the difference between direct current and alternating

current.

LEARNING OUTCOMES

At the end of this unit, you

are expected to:

24

Electricity

semiconductor

static electricity

active element

dynamic electricity

passive element

Resistor

electrical quantities

Resistance

metric prefixes

Inductor

direct current

Inductance

alternating current

Transformer

2.1 Definition of Electricity

Electricity is a physical phenomenon arising from the existence and

interaction of electric charge. It is a form of energy generated by friction,

heat, light, magnetism, chemical reaction, etc.

2.1.1 Two Types of Electricity:

a. Static electricity – electricity at rest. It cannot flow from one place to another.

b. Dynamic electricity – also known as current electricity. Electricity in motion. It can be

transmitted from one place to the other.

2.1.2 Methods of Producing Electricity

There are six methods for producing electricity:

1. Magnetism

2. Chemical action

3. Pressure

4. Heat

5. Friction

6. Light

2.1.3 Electrical Effects

With the exception of friction, electricity can be used to cause the same effects that cause it.

Important Terms

25

1. Magnetism

2. Chemical action

3. Pressure

4. Heat

5. Light

2.2 History of Electricity and Electronics

Prehistoric people experienced the properties of magnetite – permanently magnetized

pieces of ore, often called lodestones. These magnetic stones were strong enough to lift pieces of

iron.

The philosopher Thales of Miletus (640 – 546 B.C.) is thought to have been the first person who

observed the electrical properties of amber. He noted that when amber was rubbed, it acquired the

ability to pick up light objects such as straw and dry grass. He also experimented with the lodestones

and knew of its power to attract iron. By the thirteenth century, floating magnets were used for

compasses.

One of the first important discoveries about static electricity is attributed to William

Gilbert (1540-1603). Gilbert was an English Physician who, in a book published in 1600, described

how amber differs from magnetic loadstones in its attraction of certain materials. He found that

when amber was rubbed with a cloth, it attracted only lightweight materials, whereas loadstones

attracted only iron. Gilbert also discovered that other substances, such as sulfur, glass, and resin,

behave as amber does. He used the Latin word electron for amber and originated the word electrical

for the other substances that acted similar to amber. The word electricity was used for the first time

by Sir Thomas Browne (1605-82), an English physician.

Following Gilbert’s lead, Robert Boyle published his many experimental results in 1675.

Boyle was one of the early experiments with electricity in a vacuum.

Otto von Guericke (1602 – 1686) built an electrical generator and reported it in his

Experimental Nova of 1672. This device was a sulfur globe on a shaft that could be turned on its

bearing . When the shaft was turned with a dry hand held on the surface, an electrical charge

gathered on the globe’s surface. Guericke also noted small sparks when the globe was discharges.

In his studies of attraction and gravitation, Guericke devised the first electrical generator. When a

hand was held on a sulfur ball revolving in its frame, the ball attracted paper, feathers, chaff, and

other light objects.

Another Englishman, Stephen Gray (1696-1736), discovered that some substances conduct

electricity and some do not. Following Gray’s Lead, a Frenchman named Charles du Fay

experimented with the conduction of electricity. These experiments led him, to believe that there

were two kinds of electricity. He called one type vitreous electricity and the other type resinous

electricity. He found that objects having vitreous electricity repelled each other and those having

26

resinous electrify attracted each other. It is known today that two types of electrical charge exist,

positive and negative. Negative charge results from an excess of electrons in a material and positive

charge results from a deficiency of electrons.

A major advance in electrical science was made in Leyden, Holland, in 1746, when Peter van

Musschenbroek introduces a jar that served as a storage apparatus for electricity. The jar was

coated inside and out with a tinfoil, and a metallic rod was attached to the inner foil lining and

passed through the lid. Leyden jar were gathered in groups (called batteries) and arranged with

multiple connections, thereby further improving the discharge energy.

Benjamin Franklin (1706-90) conducted studies in electricity in the mid-1700s. He theorized

that electricity consisted of a single fluid, and he was the first to use the terms positive and negative.

In his famous kite experiment, Franklin showed that lightning is electricity.

Charles Augustin de Coulomb (1736-1806), a French physicist, in 1785 proposed the laws

that govern the attraction and repulsion between electrically charged bodies. Today, the unit of

electrical charge is called the coulomb.

Luigi Galvani (1737-98) experimented with current electricity in 1786. Galvani was a

professor of anatomy at the University of Bologna in Italy. Electrical current was once known as

galvanism in his honor.

In 1800, Alessandro Volta (1745-1827), an Italian professor of physics, discovered that the

chemical action between moisture and two different metals produced electricity. Volta constructed

the first battery, using copper and zinc plates separated by paper that had been moistened with a

salt solution. This battery, called the voltaic pile, was the first source of steady electric current.

Today, the unit of electrical potential energy is called the volt in honor of Professor Volta.

A Danish scientist, Hans Christian Oersted (1777-1851), is credited with the discovery of

electromagnetism, in 1820. He found that electrical current flowing through a wire caused the

needle of a compass to move. This finding showed that a ,magnetic field exists around a current-

carrying conductors and that the field is produced by the current.

The modern unit of electrical current is the ampere (also called amp) in honor of the French

physicist André Ampère (1775-1836). In 1820, Ampère measured the magnetic effect of an electrical

current. He found that two wires carrying current can attract and repel each other, just as magnets

can. By 1822, Ampère had developed the fundamental laws that are basic to the study of electricity.

One of the most well known and widely used laws in electrical circuits today is Ohm’s law. It

was formulated by Georg Simon Ohm (1787-1854), a German teacher, in 1826. Ohm’s law gives us

the relationship among the three important electrical quantities of resistance, voltage, and current.

Although it was Oersted who discovered electromagnetism, it was Michael Faraday (1791-

1867) who carried the study further. Faraday was an English physicist who believed that electricity

could produce magnetic effects, then magnetism could produce electricity. In 1831 he found that a

moving magnet cause an electric current in a coil of wire placed within the field of the magnet. This

effect, known today as electromagnetic induction, is the basic principle of electric generators ands

transformers.

27

Joseph Henry (1797-1878), an American physicist, independently discovered the same

principle in 1831, and it is in his honor that the unit of inductance is called the henry. The unit of

capacitance, the farad, is named in honor of Michael Faraday.

A paper published by James Prescott Joule in 1841 claimed the discovery of the relationship

between a current and the heat or energy produced, which today we call Joule’s law. The unit of

energy is called the joule in his honor.

In the 1860s, James Clerk Maxwell (1831-79), a Scottish Physicist, produced a set of

mathematical equations that expressed the laws governing electricity and magnetism. These

formulas are known as Maxwell’s equations. Maxwell also predicted that electromagnetic waves

(radio waves) that travel at the speed of light in space could be produced.

It was left to Heinrich Rudolph Hertz (1857-94), a German physicist, to actually produce

these waves that Maxwell predicted. Hertz performed this work in the late 1880s. Today, the unit of

frequency is called the hertz.

The Beginning of Electronics

The early experiments in electronics involved electric currents in glass tubes. One of the first to

conduct such experiments was a German named Heinrich Geissler (1814-79). Geissler found that

when he removed most of the air from a glass tube, the tube glowed when an electrical potential

was placed across it.

Around 1878, Sir William Crookes (1832-1919), a British scientist, experimented with tubes

similar to those of Geissler. In his experiments, Crookes found that the current in the tubes seemed

to consist of particles.

Thomas Edison (1847 – 1931), experimenting with the carbon-filament light bulb that he had

invented, made another important finding. He inserted a small metal plate in the bulb. When the

plate was positively charged, a current flowed from the filament to the plate. This device was the

first thermionic diode. Edison patented it but never used it.

The electron was discovered in the 1890s. The French physicist Jean Baptiste Perrin (1870 –

1942) demonstrated that the current in a vacuum tube consists of negatively charged particles.

Some of the properties of these particles were measured by Sir Joseph Thomson (1856 – 1940), a

British physicist, in experiments he performed between 1895 and 1897. These negatively charged

particles later became known as electrons. The charge on the electron was accurately measured by

an American physicist, Robert A. Millikan (1868 – 1953), in 1909. As a result of these discoveries,

electrons could be controlled, and the electronic age was ushered in.

Putting the Electron to Work

A vacuum tube that allowed electrical current in only one direction was constructed in 1904 by John

A. Fleming, a British scientist. The tube was used to detect electromagnetic waves. Called the

Fleming valve, it was the forerunner of the more recent vacuum diode tubes.

28

Major progress in electronics, however, awaited the development of a device that could

boost, or amplify, a weak electromagnetic wave or radio signal. This device was the audion,

patented in 1907 by Lee de Forest, an American. It was a triode vacuum tube capable of amplifying

small electrical signals.

Two other Americans, Harold Arnold and Irving Langmuir, made great improvements in the

triode tube between 1912 and 1914. About the same time, de Forest and Edwin Armstrong, an

electrical engineer, used the triode tube in an oscillator circuit. In 1914, the triode was incorporated

in the telephone system and made the transcontinental telephone network possible.

The tetrode tube was invented in 1916 by Walter Schottky, a German. The tetrode, along

with the pentode (invented in 1926 by Tellegen, a Dutch engineer), provided great improvements

over the triode. The first television picture tube, called the kinescope, was developed in the 1920s by

Vladimir Zworykin, an American researcher.

During World War II, several types of microwave tubes were developed that made possible

modern microwave radar and other communications systems. In 1939, the magnetron was invented

in Britain by Henry Boot and John Randall. In the same year, the klystron microwave tube was

developed by two Americans, Russell Varian and his brother Sigurd Varian. The traveling-wave tube

was invented in 1943 by Rudolf Komphner, an Austrian-American.

The Computer

The computer probably has had more impact on modern technology than any other single type of

electronic system. The first electronic digital computer was completed in 1946 at the University of

Pennsylvania. It was called the Electronic Numerical Integrator and Computer (ENIAC). One of the

most significant developments in computers was the stored program concept, developed in the

1940s by John von Neumann, an American mathematician.

Solid State Electronics

The crystal detectors used in the early radios were the forerunners of modern solid state devices.

However, the era of solid state electronics began with the invention of the transistor in 1947 at Bell

Labs. The inventors were Walter Brattain, John Bardeen, and William Shockley.

In the early 1960s, the integrated circuit was developed. It incorporated many transistors and

other components on a single small chip of semiconductor material. Integrated circuit technology

continues to be developed and improved, allowing more complex circuits to be built on smaller

chips.

The introduction of the microprocessor in the early 1970s created another electronics

revolution: the entire processing portion of a computer placed on a single, small, silicon chip.

Continued development brought about complete computers on a single chip by the late 1970s.

Major Events in Electrical Sciences and Engineering

1672 Ottoo von Guiricke published Experimenta Nova.

29

1675 Robert Boyle was published Production of Electricity. 1746 The Leyden jar was demonstrated in Holland. 1750 Benjamin Franklin invented the lightning conductor. 1767 Joseph Priestley published the Present State of Electricity. 1786 Luigi Galvani observed electrical convulsion in the legs of dead frogs. 1800 Alessandro Volta announced the voltaic pile. 1801 Henry Moyes was the first to observe an electric arc between carbon rods.

1820 Hans Oersted discovered the deflection of a magnetic needle by current on a wire.

1821 Michael Faraday produced magnetic rotation of a conductor and magnet- the first electric motor.

1825 Andre-Marie Ampere defined electrodynamics. 1828 Joseph Henry produced silk-covered wire and more powerful electromagnets. 1831 Michael Faraday discovered electromagnetic induction and carried out

experiments with an iron ring and core. He also experimented with a magnet and rotating disk.

1836 Samuel Morse devised a simple relay. 1836 Electric light from batteries was shown at Paris Opera. 1841 James Joule stated the relation between current and energy produced. 1843 Morse transmitted telegraph signals from England to France. 1850 First channel telegraph signals from Baltimore to Washington, D.C. 1858 Atlantic telegraph cable was completed and the first message sent. 1861 Western Union established telegraph service from New York to San Francisco. 1862 James Clerk Maxwell determined the ohm. 1873 Maxwell published Treatise on Electricity and Magnetism. 1874 Alexander Graham Bell invented the telephone. 1877 Thomas Edison invented the telephone. 1877 Edison Electric Light Company was formed. 1881 First hydropower station was brought into use at Niagara, New York. 1881 Edison constructed the first electric power station at Pearl Street, New York. 1883 Overhead trolley electric railways were started at Portrush and Richmond,

Virginia. 1884 Philadelphia electrical exhibition was held. 1885 The American Telephone and Telegraph Company was organized. 1886 H. Hallerith introduced his tabulating machine. 1897 J.J. Thomson discovered the electron. 1898 Guglielmo Marconi transmitted radio signals from South Foreland to Wimereux,

England. 1904 John Ambrose Fleming invented the thermionic diode.

2.3 Applications of Electricity and Electronics

30

Computers Communications

Medicine Automation

Consumer Products

2.4 Circuit Components and Measuring Instruments

Resistors

31

These can be the carbon-composition type or wound with special resistance wire. Their function is to

limit the amount of current or divide the voltage in a circuit.

Capacitors

A capacitor is constructed of two conductor plates separated by an insulator (called a dielectric). Its

basic function is to concentrate the electric field of voltage across the dielectric. As a result, the

capacitor can accumulate and store electric charge from the voltage source. Furthermore, the

dielectric can discharge the stored energy when the charging source is replaced by a conducting

path.

When ac voltage is applied, the capacitor charges and discharges as the voltage varies. The practical

application of this effect is the use of capacitors to pass an ac signal but to block a steady dc voltage.

Capacitors Inductors

Inductors

An inductor is just a coil of wire. Its basic function is to concentrate the magnetic field of electric

current in the coil. Most important, an induced voltage is generated when the current with its

associated magnetic field changes in value or direction. Inductors are often called chokes.

In the practical application of a choke, the inductor can pass a steady current better than alternating

current. The reason is that a steady current cannot produce induced voltage. Note that the effect of

a choke, passing a steady current, is the opposite of that of a coupling capacitor, which blocks dc

voltage.

Transformers

32

A transformer consists of two or more coil windings in the same magnetic field. Induced voltage is

produced when the current changing in any winding. The purpose of a transformer is to increase or

decrease the amount of ac voltage coupled between the windings. However, the transformer

operates only with alternating current.

Transformers Semiconductor Devices

Semiconductor Devices

These include diode rectifiers and transistor amplifiers, either as separate, discrete components or as

part of an IC chip. A diode has two electrodes; the transistor has three. In addition, the silicon

controlled rectifier (SCR) and triac are used for power-control circuits.

Active and Passive Elements

Active elements - are capable of delivering power to some external device.

Examples: dependent and independent voltage and current sources

Passive elements – are capable of receiving power. They are able to store to store finite amounts of

energy and then return that energy later to various external devices. Examples are resistors,

inductors, and capacitors.

Electronic Instruments

Typical instruments include the power supply, for providing voltage and current; the voltmeter, for

measuring voltage; the ammeter, for measuring current; the ohmmeter, for measuring resistance;

the wattmeter, for measuring power; and the oscilloscope for observing and measuring AC voltages.

33

2.5 Electrical Quantities and Units with SI symbols.

Quantity Symbol Unit Symbol

Capacitance C farad F

Charge Q coulomb C

Conductance G Siemen S

Current I ampere A

Energy W Joule J

Frequency F hertz Hz

Impedance Z ohm Inductance L henry H

Power P Watt W

Reactance X ohm Resistance R ohm Voltage V Volt V

34

2.6 Metric Prefixes

Power of Ten Value Metric Prefix Metric Symbol

109 one billion giga G

106 one million mega M

103 one thousand kilo k

10-3 one-thousandth milli m

10-6 one-millionth micro 10-9 one-billionth nano n

10-12 one-trillionth pico p

35

Problem Set No. 3

METRIC PREFIXES

I. Express each of the following as quantity having a metric prefix:

1) 31 𝑥 10−3𝐴

a. 0.31 mA

b. 3.1 mA

c. 31 mA

d. 310 mA

2) 5.5 𝑥 103 𝑉

a. 5.5 kV

b. 55 kV

c. 550 kV

d. 5.5 MV

3) 200 𝑥 10−12 𝐹

a. 200 pF

b. 200 nF

c. 200 µF

d. 2000 pF

4) 0.000003 𝐹

a. 3 µF

b. 30 µF

c. 3 nF

d. 30 nF

5) 3,300,000

a. 3.3 kΩ

b. 33 kΩ

c. 3.3 MΩ

d. 33 MΩ

6) 350 𝑥 10−9𝐴

a. 35 nF

b. 350 nF

c. 3.5 pF

d. 35 pF

36

II. Express each quantity as a power of ten:

7) 5 𝐴

a. 5 x 10-3 A

b. 5 x 10-6 A

c. 5 x 10-9 A

d. 50 x 10-3 A

8) 43 𝑚𝑉

a. 43 x 10-3 V

b. 43 x 10-6 V

c. 43 x 10-9 V

d. 43 x 10-12 V

9) 275 𝑘

a. 275 x 106 V

b. 275 x 103 V

c. 275 x 10-3

V

d. 275 x 10-6 V

10) 10 𝑀𝑊

a. 10 x 1012 W

b. 10 x 109 W

c. 10 x 106 W

d. 10 x 103 W

III. Add the following quantities:

11) 6 𝑚𝐴 + 3 A = _________ mA

a. 6.03

b. 60.03

c. 6.003

d. 6.3

37

12) 550 𝑚𝑉 + 3.2 V = _________ V

a. 375

b. 37.5

c. 3.75

d. 0.375

13) 12 𝑘 + 6800 = ________ k

14) 1880

15) 188

16) 18.8

17) 1.88

18) 15 𝑀𝑊 + 7500 𝑘𝑊 = __________ MW

a. 0.0225

b. 0.225

c. 2.25

d. 22.5

38

2.7 Comparison of AC and DC

Direct Current. The DC electricity, flows in one direction. The flow is said to be from negative to

positive. The normal source of a DC electricity, is the dry cell or storage battery.

Alternating Current. The AC electricity constantly reverses its direction of flow. It is generated by

machine called generator. This type of current is universally accepted because of its limited number

of applications with the following advantages.

1. It is easily produced. 2. It is cheaper to maintain. 3. It could be transformed into higher voltage. 4. It could be distributed to far distance with low voltage drop. 5. It is more efficient compared with the direct current.

Comparison of DC Voltage and AC Voltage

DC Voltage AC Voltage Fixed Polarity Reverses polarity Can be steady or vary in magnitude Varies between reversals in polarity Steady value cannot be stepped up or down by a transformer

Can be stepped up or down for electric power distribution

Easier to measure Easier to amplify Heating effect is the same for direct or alternating current

AC

+ DC Voltage AC Voltage

39

Objective Test No. 1

INTRODUCTION TO ELECTRICITY

1. Which of the following is not an electrical quantity? time

power

current

voltage

2. The unit of current is volt

watt

joule

ampere

3. The unit of voltage is ohm

volt

watt

farad

4. The unit of resistance is ohm

hertz

henry

ampere

5. 15,000 W is the same as

15 W

15 mW

15 kW

15 MW

6. The quantity 4.7 x 103 is the same as 0.0047

470

4700

47,000

7. The quantity 56 x 10-3 is the same as 0.056

0.560

560

56,000

8. The number 3,300,000 can be expressed as 3.3 x 10-6

40

3.3 x 106

3.3 x 109

3.3 x 1012

9. Ten milliamperes can be expressed as

10 A

10 mA

10 kA

10 MA

10. Five thousand volts can be expressed as 5 kV

50 MV

500 kV

5000 kV

11. Twenty million ohms can be expressed as

20

20 m

20 M

20 MW

12. Hertz is the unit of time

power

frequency

inductance

13. An oscilloscope is usually used to measure rms voltage

average voltage

maximum voltage

effective voltage

14. A step-down transformer, lowers both the voltage and current. increases both the voltage and current. lowers the voltage and increases the current. lowers the current and increases the voltage.

15. The prefix pico means

10-15 of a unit

10-12 of a unit

10-9 of a unit

10-6 of a unit

41

Objective Test No. 2

INTRODUCTION TO ELECTRICITY

Note: Answers of some items may not be found in this text. Look for answers in other references.

1. Discovered the difference between conductors and insulators in 1729. Luigi Galvani Stephen Gray Alessandro Volta Gottfried Wilhelm Leibniz

2. Discovered Galvanic action in 1780. Luigi Galvani Stephen Gray Alessandro Volta Gottfried Wilhelm Leibniz

3. Invented the electric dry cell in 1800. Luigi Galvani Stephen Gray Alessandro Volta Gottfried Wilhelm Leibniz

4. Discovered electromagnetism and invented in 1820. J W Ritter Luigi Galvani William Herschel Hans Christian Oersted

5. Discovered thermoelectricity in 1821. J W Ritter T J Seebeck Stephen Gray Hans Christian Oersted

6. Developed Fourier analysis in 1828. T J Seebeck Stephen Gray Hans Christian Oersted Jean-Baptiste-Joseph Fourier

7. Formulated Ohm’s Law in 1826. T J Seebeck George S. Ohm Hans Christian Oersted

42

Jean-Baptiste-Joseph Fourier

8. Discovered electromagnetic induction in 1831. T J Seebeck

George S. Ohm

Michael Faraday

Jean-Baptiste-Joseph Fourier

9. Invented the transformer in 1831. T J Seebeck George S. Ohm Michael Faraday

Jean-Baptiste-Joseph Fourier

10. Formulated the law of electrolysis in 1834. George S. Ohm

Charles Babbage

Michael Faraday

Jean-Baptiste-Joseph Fourier

11. Invented the electric motor in 1837. Samuel Morse Charles Babbage Michael Faraday

Thomas Davenport

12. Invented the magnetohydrodynamic battery in 1839. Samuel Morse

Charles Babbage

Michael Faraday

Thomas Davenport

13. Discovered the photovoltaic effect in 1839. Samuel Morse

Michael Faraday

Edmond Becquerel

Thomas Davenport

14. Invented the fuel cell in 1839. William Grove

Michael Faraday

Edmond Becquerel

Thomas Davenport

15. Invented the differential resistance measurer in 1843. John Herschel William Grove

43

Charles Wheatstone

Edmond Becquerel

16. Formulated KCL and KVL in 1845. John Herschel

George S. Ohm

Gustav Kirchhoff

Charles Wheatstone

17. Invented submarine cable insulation in 1847. Charles Wheatstone Gustav Kirchhoff George S. Ohm Werner Siemens

18. Discovered magnetostriction in 1847. James Prescott Joule

Gustav Kirchhoff

George S. Ohm

Werner Siemens

19. Invented the lead acid cell in 1860. Plante

Michael Faraday

George Boole

J P Reis

20. Invented the Leclanche in 1868. Plante

Georges Leclanche

James Clerk Maxwell

J P Reis

21. Invented the induction motor in 1888. Henry Hunnings

Nikola Tesla

Augustus Desire Waller

Thomas Alva Edison

22. Discovered electron in 1897. Joseph John Thomson

Guglielmo Marconi

Wilhelm Rontgen

Almon Brown Strowger

23. Invented Nickel-iron cell in 1900. Guglielmo Marconi

Thomas Alva Edison

44

Almon Brown Strowger

Joseph John Thomson

24. Invented synchronous induction motor in 1902. Ernst Danielson

Heaviside and Kennely

Thomas Alva Edison

Max Planck

25. Discovered superconductivity in 1911. Lee De Forest

Thadius Cahill

Alessandro Artom

Kamerlingh Onnes

26. Discovered neutron in 1932. James Chadwick

Tellegen and Hoist

H S Black

Julius Lilienfield

27. Conceptualized FET in 1935. James Chadwick

Westinghouse Co.

E H Armstrong

Oskar Heil

28. Invented Digital Voltmeter in 1952. M V Wilkes

J I Nishizawa

Yoshire Nakamats

Andy Kay

29. Invented LED in 1960. Leo Esaki

J I Nishizawa

Allen and Gibbons

Andy Kay

30. Invented IC in 1958. G T Wright

Johnson and deLoach

Jack Kilby

Prager, Chang and Weisbrod

45

Direction: Search for names of scientist who contributed to the development of electricity, electrical

components and instruments found in this puzzle. Encircle the name or word vertically,

horizontally, backward, upward or downward.

C A B A D E T S R E O N A I T S I R H C S N A H O Q W E H R T T U C Y R Y Q O O N P H A S D F G H J U K H W L Z

A L E S S A N D R O V O L T A Z X C V B N M Q V M W E R

R L L Y V P M U I P A S D T L U I O P T P A Q S M D F I

L E U C G A I J O S E P H H E N R Y H J K Y K L E Z T N

E W O D S C C B D F G H J K S L R E S I S T O R T A R A

S X J Z R I H E Z X C V B N O I M Q C E D R T Y E U A V

D A T I T T A N O P T G I H F J E K J Z X C N V R B N L U M T Q W O E J R Y M U B I M I O P A S D F G H J K S A

F K O Z X R L A C V H B N M I R O B E R T B O Y L E F G

A R C U M Q F M W R O Y U M L I P O T A S K A S D F O I

Y E S H V J A I K P N R E T E M T L O V K L Z A C V R G

B L E N W M R N Y O O Z J F T W A X Q E W E R T T Y M I

A C R A W B A F N U M I P O U M U T C D R E Z X C V E U

N S P Q W E D R R X I T K Y S E M I C O N D U C T O R L

W E S I R O A A M P S A S D F S T E P H E N G R A Y G H O M E L Z X Y N C V G B W A T T M E T E R B N M Q W E R

R A M L U I O K E O R B N E H C S S U M N A V R E T E P

B J A P O G A L S D O F G H J K Q K L M N I B J C K X Z

S X J C S V B I B N E P M E N O F B C K F X P L M L R G

A C X V B Q N N B M G Q D W E R T Y H U I O P A S F E R

M Y A G S D F G H M J W I L L I A M G I L B E R T Y T Y

O P O O Z I U Y N G T R E W Q J A S D F G H H J K L E M H D H G C H A R L E S A U G U S T I N D E C O U L O M B

T A S D F G H J K L A A W R I T Y U I O P P A S D F M G

R A N D R E A M P E R E H G H J Z D Y Q W C E D R T A E

I B K L C M N B V E K C I R I U G N O V O T T O B V C X

S A H E I N R I C H R U D O L P H H E R T Z A Z X C V B

How Much Have You Learned?

46

Unit 3

INTRODUCTION TO ALTERNATING CURRENT AND VOLTAGE

1. identify sine waves and measure their characteristics. 2. explain how frequency and period are related. 3. measure the following voltage or current values of a sine wave:

instantaneous, peak, peak-to-peak, rms, and average. 4. define a form factor. 5. describe how sine waves are generated. 6. measure points on a sine wave in terms of angular units. 7. determine the phase angle lead and phase lag. 8. express sine waves with a mathematical formula. 9. discuss phasors and how they can be used to represent sine

waves. 10. Apply Ohm’s law and Kirchhoff’s laws to ac circuits as well as to dc

circuits.

LEARNING OUTCOMES

At the end of this unit, you

are expected to:

47

alternating current

peak value power transmission

peak-to-peak value

electricity

rms value sine wave

average value

waveform

Faraday's law cycle

electromagnetic induction

polarity

phase alternation

phasor

period

Ohm's law frequency

Kirchhoff's law

instantaneous value

3.1 The Alternating Current

Alternating current circuits improves the versatility and usefulness of electrical

power system. Alternating current plays a vital role in today’s energy generation

Important Terms

48

Once a big controversy ensued between the proponents of the of the DC

electricity led by Thomas Edison and the advocates of the AC electricity led

by George Westinghouse. According to Thomas Edison,

“The AC electricity is dangerous, because it involves high voltage

transmission lines.”

The AC advocates on the other hand, countered that:

“The AC alternation is just like a handsaw which cuts on the upstroke and the down stroke.

The high voltage in the transmission line could be reduced to the desired voltage as it passes the

distribution line.”

3.2 The Power Transmission

Figure 3-1. Electricity leaves the power plant. (2) Its voltage is increased at a step-up transformer. (3) The

electricity travels along a transmission line to the area where power is needed. (4) There, in the substation,

voltage is decreased with the help of step-down transformer. (5) Again the transmission lines carry the

electricity. (6) Electricity reaches the final consumption points.

More than 90 per cent of the electrical energy used for commercial purposes is generated as

alternating current. This is not due primarily to any superiority of alternating over direct current so

far as applicability to industrial and domestic uses is concerned. In fact, there are many instances

49

where direct current is absolutely necessary for industrial purposes, such as municipal traction,

electrolytic processes, and certain types of arc lamps; also, direct current motors are superior for

elevators, printing presses, and many variable-speed drives. However, for these various purposes

the energy is generated and transmitted almost always as alternating current and then converted to

direct current.

Some of the reasons for generating electrical energy as alternating current are the following:

Alternating current can be generated at comparatively high voltages, and these voltages can be raised and lowered readily by means of static transformers.

For constant-speed work, the alternating-current induction motor is more efficient than the direct-current motor and is loess in first cost and in maintenance, owing in part to the fact that the induction motor has no commutator.

A circuit operating at increased voltage, has a lower power loss, power voltage drop, and economically constructed for using smaller copper wires. On transmission and distribution line, power loss is the most important problem to resolve. This is the main reason why Alternating Current (AC) gained more favor and acceptance during the middle part of the 19th century. In the USA, an ordinary house current is described as 120 volts 60 Hertz.

3.3 The Sine Wave

Waveforms are a graphical representation of how voltage or current varies with time. The most

common type of a waveform is the sine wave.

The sine wave is one very common type of alternating (ac) and alternating voltage. It is also referred

to as a sinusoidal wave or, simply, sinusoid. The electrical service provided by the power companies

is in the form of sinusoidal voltage and current.

Sine waves are produced by two types of sources: rotating electrical machines (ac generators) and

electronic oscillator circuits, which are in instruments known as electronic signal generators.

Figure 3.2. The Sine Wave

-15

-10

-5

0

5

10

15

0 1 2 3 4 5 6 7

v

o

l

t

a

g

e

o

r

c

u

r

r

e

n

ttime (sec)

50

Figure 3-3. Parts of a Sine Wave

3.3.1 Cycle

A cycle is a complete set of positive and negative values.

3.3.2 The Polarity of a Sine Wave

(a) Positive voltage: current direction as shown.

(b) Negative voltage: current reverses direction

Figure 3-4. Alternating current and voltage.

Positive

alternation

Vs

0 t

+

Vs

_

+

Vs

_

_

Vs

+

VS

Negative

alternation

Voltage (V) or

Current (I) Peak (maximum) value

Negative-going zero-

crossing

Positive-going

zero crossing

51

3.3.3 The Period of a Sine Wave

The period (T) of a sine wave is the time required to complete one cycle. 3.3.4 The Frequency of a Sine Wave Frequency is the number of cycles that a sine wave completes in 1 sec. The Unit of Frequency Frequency (f) is measured in units of hertz, abbreviated Hz. One hertz is equivalent to one cycle per second; 60 Hz is 60 cycles per second; and so on. 3.3.5 Relationship of Frequency and Period The relationship between frequency and period is very important. The formulas for this relationship are as follows:

f = 1

𝑇 ; T =

1

𝑓

Example 3.1. The period of a certain sine wave is 10 ms. What is the frequency? Solution:

f = 1

𝑇 =

1

10 ms = 100 Hz

Example 3.2. The frequency of a sine wave is 60 Hz. What is the period?

Solution:

T = 1

𝑓 =

1

160 Hz = 16.67 ms

.

52

Problem Set No. 4

SINE WAVES

1. Describe one cycle of a sine wave.

2. At what point does a sine wave change polarity? 3. How many maximum points does a sine wave have during one cycle?

4. How is the period of a sine wave measured?

5. Define frequency, and state its unit.

6. Determine f when T = 5 s. a. 200 kHz

b. 20 kHz

c. 2 kHz

d. 200 Hz

7. Determine T when f = 120 Hz. a. 8.33 ms b. 6.33 ms c. 4.25 ms d. 2.45 ms

8. A sine wave goes through 5 cycles in 10 s. What is its period?

a. 1 s

b. 2 s

c. 2 s

d. 2 s

9. A sine wave has a frequency of 50 kHz. How many cycles does it complete in 10 ms? a. 200 cycles b. 400 cycles c. 500 cycles d. 600 cycles

53

3.4 THE 60-Hz AC POWER LINE

Practically all homes in the Unites States are supplied alternating voltage between 115 and 125 V

rms, at a frequency of 60 Hz. This is a sine-wave voltage produced by a rotary generator. The

electricity is distributed by high voltage power lines from the generating station and reduced to the

lower voltages used in the home. Here the incoming voltage is wired to all the wall outlets and

electrical equipment in parallel. The 120-V source of commercial electricity is the 60-Hz power line

or the mains, indicating it is the main line for all the parallel branches.

Advantages. The incoming electric service to residences is normally given as 120 V rms. With an rms

value of 120 V, the ac power is equivalent to 120-V dc power in heating effect. If the value were

higher, there would be more danger of a fatal electric shock. Lower voltages would be less efficient

in supplying power.

Higher voltage can supply electric power with less I2R; sine the same power is produced with less

I. Note that the I2R power loss increases as the square of the current. For applications where large

amounts of power are used such as central air-conditioners and clothes dryers, a line voltage of 240

V is often used.

The advantage of ac over dc power is greater efficiency in distribution from the generating

station. Alternating voltages can easily be stepped up by means of a transformer, with very little

loss, but a transformer cannot operate on direct current. The reason is that a transformer needs the

varying magnetic field produced by an ac voltage.

Using a transformer, the alternating voltage at the generating station can be stepped up to

values as high as 500 kV for high-voltage distribution lines. These high-voltage lines supply large

amounts of power with much less current and less I2R loss, compared with a 120-V line. At the

home, the lower voltage required is supplied by a step-down transformer. The step-up and step-

down characteristics of a transformer refer to the ration of voltages across the input and output

connections.

The frequency of 60 Hz is convenient for commercial ac power. Much lower frequencies would

require much bigger transformers because larger windings would be necessary. Also, too low a

frequency for alternating current in a lamp would cause the light to flicker. For the opposite case,

too high a frequency results in excessive iron-core heating in the transformers because of eddy

currents and hysteresis losses. Based on these factors, 60 Hz is the frequency of the ac power line in

54

the United States. It should be noted that the frequency of the ac power mains in England and most

European counters is 50 Hz.

3.5 Voltage and Current Values of a Sine Wave

3.5.1 Instantaneous Value

Figure 3-5 illustrates that any point in time on a sine wave, the voltage (or current) has an

instantaneous value. This instantaneous value is different at different point in the curve.

Instantaneous values are positive during the positive alternation and negative during the negative

alternation. Instantaneous values of voltage or current are symbolized by lower case v and i,

respectively.

Figure 3-5. Example of instantaneous value of a sine voltage.

3.5.2 Peak Value

The peak value of a sine wave is the value of voltage (or current) at the positive or the negative maximum (peaks) with respect to zero. Since the peaks are equal in magnitude, a sine wave is characterized by a single peak value, as is illustrated in Figure 3-6. For a given sine wave, the peak value is constant and is represented by VP or Vm for voltage and IP or Im for current.

-10

-8

-6

-4

-2

0

2

4

6

8

10

0 1 2 3 4 5 6 7

V

v

o

l

t

s

time (ms)

v2

v1

t1 t2

55

Figure 3-6. Example of a peak value of a sine wave.

3.5.3 Peak-to-Peak Value

The peak-to-peak value of a sine, as illustrated in Figure 3-6, is the voltage (or current) form the

positive peak to negative peak. Of course it is always twice the peak value as expressed in the

following equations:

VPP = 2VP

IPP = 2IP

3.5.4 rms Value

The term rms stands for root mean square. It refers to the mathematical procedure which this value

is derived. The rms value is also referred to as the effective value. Most ac voltmeters display rms

voltage. The 220 V at your wall outlet is an rms value.

The rms value of a sine wave is actually a measure of the heating effect of a sine wave. For

an example, when a resistor is connected across an ac (sine wave) source, as shown in Figure 3-7(a),

a certain amount of heat is generated by the resistor. Part (b) shows the same resistor connected

across a dc voltage source.

-15

-10

-5

0

5

10

15

0 1 2 3 4 5 6 7

v

o

l

t

a

g

e

o

r

c

u

r

r

e

n

ttime (sec)

Vp

Vp-p

56

Figure 3-7. When the same amount of heat is being produced in both cases, the sine wave has an rms value

equal to the DC voltage.

The value of the AC voltage can be adjusted so that resistor gives off the same amount of

heat as it does when connected to the DC source.

The rms value of a sine wave is equal to the dc voltage that produces the same

amount of heat in a resistance as the sinusoidal voltage.

The peak value of a sine wave can be converted to the corresponding rms value using the following

relationships for either voltage or current:

Vrms = Vp

2

Irms = Ip

2

Using these formulas, we can also determine the peak value knowing the rms value, as

follows:

VP = 2 Vrms

Ip = 2 Irms

V

12 V

R30Ω

XWM1

V I

R30Ω

XWM1

V I

E

12 Vrms 60 Hz 0°

(a)

(b)

57

3.5.5 Average Value

The average value of a sine wave when taken over one complete cycle is always zero, because the

positive values (above the zero crossing) offset the negative values (below the zero crossing).

To be useful for comparisons, the average value of a sine wave is defined over a half-cycle

rather than over a full cycle. The average value is the total area under the half-cycle divided by the

distance of the curve along the horizontal axis. The result is expressed in terms of the peak value as

follows for both voltage and current sine waves:

Vavg = 2

V𝑃

Iavg = 2

𝐼𝑃

58

Problem Set No. 5

VALUES OF SINE WAVE VOLTAGE AND CURRENT

1. Derive the formula for the rms value of current. 2. Derive the formula for the average value of a sine wave in half-cycle. 3. A sine wave has a peak value of 12 V. Determine the following values: (a) rms (b) peak-to-peak (c) half-cycle average

4. A sinusoidal current has an rms value of 5 mA. Determine the following value: (a) peak (b) half-cycle average (c) peak-to-peak

5. A direct current of 12.5 A flows in a 25-ohm noninductive resistance. Determine maximum value of an alternating current that will produce heat at the same rate in this resistance. a. 2 A b. 12.5 A c. 17.68 A d. 31.25 A

6. Number 6 AWG underground cable, which supplies a series incandescent lamp system with alternating current is guaranteed to operate safely with 5,000 volts (rms) alternating. If the system were changed to direct current, at what voltage would it be safe to operate the system? a. 3535.53 V b. 5,000 V c. 7071.07 V d. 10,000 V

59

3.6 Sine Wave Voltage Sources

A. An AC Generator

An AC generator is a rotating electrical machine that uses the principle of electromagnetic

induction. It converts mechanical energy into electrical energy.

Figure 3-8. An elementary generator.

3.6.1 Parts of an Elementary Generator

1. Magnetic Poles – provide the magnetic field

2. Loop of wire – cuts the magnetic field so that emf will be induced.

3. Slip rings – connected to the loop of wire.

4. Brushes – slide with the slip rings to collect the current from the loop of wire.

5. Terminals - connected to the load

3.6.2 Generation of Alternating EMFs.

1 2

3

4

5

60

The generation of emf started with the discovery of electromagnetism by Hans Christian Oersted in

1820. He found that when current flows through a coil a magnetic field is established around it as

shown in Figure 3-9.

Figure 3-9. Magnetic lines of force around a current-carrying conductor.

After the discovery (by Oersted) that electric current produces a magnetic field, scientists began to search for the converse phenomenon from about 1821 onwards. The problem they put to themselves was how to ‘convert’ magnetism into electricity. It is recorded that Michael Faraday was in the habit of walking about with magnets in his pockets so as to constantly remind him of the problem. After nine years of continuous research and experimentation, he succeeded in producing electricity by ‘converting magnetism’. In 1831, he formulated basic laws underlying the phenomenon of electromagnetic induction (known after his name), upon which is based the operation of most of the commercial apparatus like motors, generators and transformers etc.

3.6.3 Faraday’s Laws of Electromagnetic Induction Faraday summed up the above facts into two laws known as Faraday’s Laws of Electromagnetic Induction. First Law. It states that whenever the magnetic flux linked with a circuit changes, an e.m.f. is always induced in it. or whenever a conductor cuts magnetic flux, an e.m.f. is induced in that conductor. Second Law. It states that the magnitude of the induced e.m.f. is equal to the rate of change of flux-linkages.

61

Explanation. Suppose a coil has N turns and flux through it changes from an initial value of Φ1 webers to the final value of Φ2 webers in time t seconds. Then, remembering that by flux-linkages mean the product of number of turns and the flux linked with the coil, we have Initial flux linkages = NΦ1, and final flux linkages = NΦ2

induced e.m.f. e = NΦ2 − NΦ1

𝑡2− 𝑡1 Wb/s or volt e =

N(Φ2 − Φ1)

𝑡2− 𝑡1 volt

Putting the above expression in its differential form, we get

e = d(NΦ)

dt =

Nd Φ

dt volt

Usually, a minus sign is given to the right-hand side expression to signify the fact that the induced e.m.f. sets up current in such a direction that magnetic effect produced by it opposes the very cause producing it . (Lenz’s Law)

e = − Nd Φ

dt volt

3.6.4 Three Ways of the Generation of Voltage.

A voltage can be developed in a coil of wire in one of three ways; these are (1) by changing

the flux through the coil as shown in Figure 3-10, (2) by moving the coil through a magnetic field so

that flux cutting results as in Figure 3-11, and (3) by altering the direction of the flux with respect to

the coil as shown in Figure 3-12. In the first of these the voltage is said to be an induced emf and, in

accordance with Faraday’s law, its magnitude at any instant of time is given by the equation

In CGS

e = N𝑑

𝑑𝑡 𝑥 10−8 𝑣𝑜𝑙𝑡𝑠

where N = number of turns in the coil

= flux in maxwells

t = time in seconds

𝑑

𝑑𝑡 = rate at which flux, in maxwells, changes through the coil

In SI

e = N𝑑

𝑑𝑡 volts

where: = the flux in Weber (Wb)

62

Figure 3-10. Induction of voltage due to the change of flux which results from the changing current.

Figure 3-11. Electromagnetic Induction in a Coil

e

63

Figure 3-12. Electromagnetic induction in a coil with magnetic field moving.

Note particularly that, by this method (the fist method) of developing an emf, there is no physical

motion of coil or magnet; the current though the exciting coil that is responsible for the magnetism is

altered to change the flux through the coil in which the voltage is induced. By the second or third

method there is actual physical motion of coil or magnet, and in altered positions of coil or magnet

flux through the coil changes. A voltage developed in either of these ways is said to be a generated

emf and is given by the equation

e = Blv x 10-8 volts

64

where: B = flux density, lines per square inch

l = length of the wire, in., that is moved relative to the flux

v = velocity of the wire, in. per second, with respect to the flux

In SI:

e = Blv volts

where: B = flux density in Tesla or Wb/m2

l = length of the wire that is moved relative to the flux, in meters

v = velocity of the wire with respect to the flux, in meters per second

In CGS:

e = Blv x 10-8 volts

where: B = flux density in Gauss

l = length of the wire that is moved relative to the flux, in centimeters

v = velocity of the wire with respect to the flux, in centimeters per second

Note that formulas above are applied to a length of conductor as shown in Figure 3.11. If there N

number of conductors they are just multiplied by N. Also take note that B, l, and v are mutually

perpendicular so that

e = N Blv volts

Example 3.3. A coil has 500 turns in which a current rises a current from 0 to 0.25 A in 0.1 s. As the

current rises linearly, it produces a flux of 2.5 Wb. What is the induced emf in the coil?

Solution: e = N(Φ2 − Φ1)

𝑡2− 𝑡1 =

500(2.5 Wb − 0)

0.1 − 0 = 0.0125 𝑉

Example 3.4. A piece of conductor 10 cm long moves across a magnetic field of 10,000 gauss at a

velocity of 120 cm/sec. What is the voltage across the conductor?

Solution: e = Blv x 10-8 = (10,000)(10)(120) x 10-8 = 0.12 V

3.7 Generation of a Sine Wave Voltage

There are actually many turns of wire in the coil of a generator but in Figure 3-13 shows only a single

turn. Considering one side of the single-turn coil, as this cuts magnetic field a voltage is induced

across it. Figure 3-13 shows the direction of flux, induced emf, and the movement of conductor.

These directions are determined by following the right-hand rule.

65

Figure 3-13. Generation of emf in a conductor.

Now, the cross-section of the conductor is being considered as shown in Figure 3.14. As the

conductor rotates around the magnetic field flux cutting results and an emf is induced across it. As

this continually rotated a sine wave voltage is produced.

Figure 3-14. Sine wave voltage produced by the rotating conductor.

3.7.1 Reasons for using alternating current (or voltage) of sinusoidal form:

An alternating current (or voltage) sinusoidal form is normally used because of the following

reasons:

1. Mathematically, it is quite simple.

66

2. Its integrals and differentials both are sinusoidal.

3. It lends itself to vector representation.

4. A complex waveform can be analyzed into a series of sine waves of various frequencies,

and component can be dealt with separately.

5. This waveform is desirable for power generation, transmission and utilization.

3.7.2 Equations of the Alternating Voltages and Currents

Figure 3.15. Conductor that rotates at an angle of .

In a single-coil generator the e.m.f. generated in one side of the coil which contains N conductors, is given by,

e = N Bl v

Using SI units, B = flux density in Wb/m2 or Tesla, N = number of turns, l length of the wire that is

moved relative to the flux, in meters, and v = velocity of the wire with respect to the flux, in meters

per second

67

Total e.m.f. generated in both sides of the coil is

e = 2BNl v sin θ volt

From Figure 3.13, getting the v sin -component because that is the one which is perpendicular to the field, the equation becomes,

e = 2BNl v sin θ volt

Now, e has maximum value of Em (say) when θ = 90. Hence, Em = 2BNl v Therefore e = Em sin θ If b = width of the coil in meters ; f = frequency of rotation of coil in Hz, then v = π bf

Em = 2 B N l ( π b f) = 2 π f N B A volts

Example 3.5. A square coil of 10 cm side and 100 turns is rotated at a uniform speed of 1000 revolutions per minute, about an axis at right angles to a uniform magnetic field of 0.5 Wb/m2. Calculate the instantaneous value of the induced electromotive force, when the plane of the coil is (i) at right angles to the field (b) in the plane of the field. Solution: Let the magnetic field lie in the vertical plane and the coil in the horizontal plane. Also, let the angle θ be measured from X-axis. Maximum value of the induced e.m.f., Em = 2 π f N Bm A volt. Instantaneous value of the induced e.m.f. e = Em sin θ Now f = 100/60 = (50/3) rps, N = 100, Bm = 0.5 Wb/m2, A = 10−2 m2

(a) In this case, θ = 0

e = 0

(b) Here θ = 90, e = Em sin 90 = Em

Substituting the given values, we get e = 2π x (50/3) x 100 x 0.5 x 10−2 = 52.3 V Answer

3.7.3 Factors That Affect Frequency in ac Generator (a) rate of rotation of the loop (b) magnetic poles

60 Hz 120 Hz

68

f = number of pole pairs x revolution per second

Example 3.6 A four-pole generator has a rotation speed of 100 rev/sec. Determine the frequency

of the output voltage.

Solution: f = number of pole pairs x revolution per second

𝑓 =4

2 (100) = 200 cycles per second or 200 Hertz (Hz)

3.7.4 Factors That Affect Amplitude in ac Generator

Number of turns (N) Rate of change with respect to the magnetic field

B. Signal Generator- a source of sine wave and other waveforms which uses electronic oscillator circuits.

3.7.5 Derivation of the RMS Value of Voltage or Current

Mid-ordinate Method

In Fig. 3-16 are shown the positive half cycles for both symmetrical sinusoidal and non-sinusoidal alternating currents. Divide time base ‘t’ into n equal intervals of time each of duration t/n seconds. Let the average values of instantaneous currents during these intervals be respectively i1, i2, i3 .... in (i.e. mid-ordinates in Fig. 3-16). Suppose that this alternating current is passed through a circuit of resistance R ohms. Then,

N

S S

N

1 revolution = 2 cycles

69

Figure 3-16.

Heat produced in 1st interval = 0.24 x 10-3 𝑖12 Rt/n kcal (where 1/J = 1/4200 = 0.24 x 10-3

Heat produced in 2nd interval = 0.24 x 10-3 𝑖22 Rt/n kcal

: : : :

: : : :

Heat produced in nth interval = 0.24 x 10-3 𝑖𝑛2 Rt/n kcal

Total heat produced in t seconds is = 0.24 x 10-3 Rt 𝑖1

2+ 𝑖22+ …+𝑖𝑛

2

𝑛

Now, suppose that a direct current of value I produces the same heat through the same resistance during the same time t. Heat produced by it is = 0.24 x 10−3 I2Rt kcal. By definition, the two amounts of heat produced should be equal.

0.24 x 10-3 I2 Rt = 0.24 x 10-3 Rt 𝑖1

2+ 𝑖22+ …+𝑖𝑛

2

𝑛

I2 = 𝑖1

2+ 𝑖22+ …+𝑖𝑛

2

𝑛 and I =

𝑖12+ 𝑖2

2+ …+𝑖𝑛2

𝑛

= square root of the mean of the squares of the instantaneous currents

Similarly, the r.m.s. value of alternating voltage is given by the expression

V = 𝑣1

2+ 𝑣22+ …+𝑣𝑛

2

𝑛

t/n

in

i4 i3

i2

i1 in

i3

i2

i1

Current Current i4

t

t

70

Analytical Method

The standard form of a sinusoidal alternating current is i = Im sin ωt = Im sin θ. The mean of the squares of the instantaneous values of current over one complete cycle is (even the value over half a cycle will do).

= 𝑖2𝑑

(2 −0)

2

0

The square root of this value is = 𝑖2𝑑

2

2

0 put i = Im sin θ and integrate

Hence, the r.m.s. value of the alternating current is

I = 𝐼𝑚

2 = 0.707 Im

Hence, we find that for a symmetrical sinusoidal current r.m.s. value of current = 0.707 . max. value of current

The r.m.s. value of an alternating current is of considerable importance in practice, because the ammeters and voltmeters record the r.m.s. value of alternating current and voltage respectively. In electrical engineering work, unless indicated otherwise, the values of the given current and voltage are always the r.m.s. values. It should be noted that the average heating effect produced during one cycle is

= I2R = ( Im/ 2)2 R = ½ Im2 R

Average Value The average value Ia of an alternating current is expressed by that steady current which transfers across any circuit the same charge as is transferred by that alternating current during the same time . In the case of a symmetrical alternating current (i.e. one whose two half-cycles are exactly similar, whether sinusoidal or non-sinusoidal), the average value over a complete cycle is zero. Hence, in their case, the average value is obtained by adding or integrating the instantaneous values of current over one half-cycle only. But in the case of an unsymmetrical alternating current (like half-wave rectified current) the average value must always be taken over the whole cycle.

(i) Mid-ordinate Method

With reference to Fig. 11.16, Iav = 𝑖1+ 琠2 + …+ 𝑖𝑛

𝑛

This method may be used both for sinusoidal and non-sinusoidal waves, although it is specially convenient for the latter.

71

(ii) Analytical Method

The standard equation of an alternating current is, i = Im sin θ

Iav = 𝑖𝑑

(2 −0)

2

0 putting the value of i and integrate

Iav = 2

Im = 0.637 Im

Form Factor

It is defined as the ratio, Kf = 𝑟𝑚𝑠 𝑣𝑎𝑙𝑢𝑒

𝑎𝑣𝑒𝑟𝑎𝑔 𝑒 𝑣𝑎𝑙𝑢𝑒 =

0.707𝐼𝑚

0.637 𝐼𝑚 = 1.1. (for sinusoidal alternating currents only)

In the case of sinusoidal alternating voltage also, Kf = 𝑟𝑚𝑠 𝑣𝑎𝑙𝑢𝑒

𝑎𝑣𝑒𝑟𝑎𝑔 𝑒 𝑣𝑎𝑙𝑢𝑒 =

0.707𝐸𝑚

0.637 𝐸𝑚 = 1.11

As is clear, the knowledge of form factor will enable the r.m.s. value to be found from the arithmetic mean value and vice-versa. Crest or Peak or Amplitude Factor

It is defined as the ratio Ka = 𝑚𝑎𝑥𝑖𝑚𝑢 𝑚 𝑣𝑎𝑙𝑢𝑒

𝑟𝑚𝑠 𝑣𝑎𝑙𝑢𝑒 =

𝐼𝑚

𝐼𝑚 / 2 = 2 = 1.414 (for sinusoidal AC only)

For sinusoidal alternating voltage also, Ka = 𝐸𝑚

𝐸𝑚 / 2 = 1.414

Knowledge of this factor is of importance in dielectric insulation testing, because the dielectric stress to which the insulation is subjected, is proportional to the maximum or peak value of the applied voltage. The knowledge is also necessary when measuring iron losses, because the iron loss depends on the value of maximum flux.

3.8 Angular Relationship of a Sine Wave

270

3

2

0 or 360

(0 or 2)

180

(/2)

90

(/2)

II I

IV

III

72

3.8.1 Radian-Degree Conversion

rad = 𝜋 𝑟𝑎𝑑

180 x degrees

degrees = 180

𝜋 𝑟𝑎𝑑 x radians

3.8.2 Sine Wave Angles

3.9 Phase of a Sine Wave

The phase of a sine wave is an angular measurement that specifies the position of the sine wave

relative to a reference.

(a) In Phase

0 90 180 270 360

0 ð

2

3

2 2

73

Two or more waveforms are said to be in phase if they reach their zero and maximum values at the same time.

(b) Out-of-Phase – one wave either leads or lags the other.

o Lagging Phase Difference o Leading Phase Difference

In these two waveforms, sine wave A leads sine wave B by 90 or sine wave B lags sine wave A by 90 Note: To get the phase angle always consider the positive-going zero crossing nearest to the reference or to other wave. Example 3.7 What is the phase angle between the two sine waves in the figures below? Which one is lagging? Which one is leading?

-30

-20

-10

0

10

20

30

0 1 2 3 4 5 6 7

curr

en

t

time (sec)

i1

i2

I1m

I2m

A B

0 45

Solution:

The phase angle between the

two sine waves is 45 ( = 45 -

0). Sine wave A leads B by 45

B

A

74

3.10 The Sine Wave Equation

Figure 3-17. Sine wave with phase angle of 0.

The equation of a sine wave is

y = A sin ( )

where: y = ordinate; the value of the sine wave at an angle .

A = amplitude ( the peak or maximum value)

= the phase angle which is the angle between the reference (the origin) and the given sine

wave.

In the sine wave of Figure 3.17, the phase angle is 0 since its positive-going zero-crossing is at 0

(or at the origin), so that the equation becomes,

y = A sin

0 30 90

Solution:

The phase angle between the

two sine waves is 60( = 90 -

30). Sine wave A lags B by 60

or sine wave B leads sine wave

A by 60.

B

A

A

y

75

Example 3.8 Assuming that the sine wave below is a voltage, find its value at 30 and 60.

Solution: The equation is v = Vm sin

At = 30, v = 90 sin 30 = 45 V

At = 60, v = 90 sin 60 = 77.94 V

3.11 Expressions for Phase-Shifted Sine Waves

Phase-shifted to the right Phase-shifted to the left

y = A sin ( - ) y = A sin ( + )

Example 3.9 Determine the instantaneous value at the 90 reference point on the horizontal axis

of each sine wave current.

V

90 V

30 60

76

Solution:

The equation of sine wave current a is, a = A sin ( - 60); where A = 10 A

so that a = 10 sin ( - 60) A;

When = 90

a = 10 sin ( - 60) = 10 sin (90 - 60) = 5 A

The equation of sine wave current b is b = B sin ( + 30) A; where B = 8 A

so that b = 8 sin ( + 30)

When = 90

b = 8 sin ( + 30) = 8 sin (90 + 30) = 6.93 A

The equation of sine wave current c is c = C sin ( + 0); where C= 5 A

so that b = 5 sin A

When = 90

c = 5 sin = 5 sin 90 = 5A

3.12 Expression of Voltage/Current in terms of Time

-15

-10

-5

0

5

10

15

-1 0 1 2 3 4 5 6 7

c

u

r

r

e

n

t

(degrees)

c

b

a

AB

C

E1m

E2m

30° 0° 60°

77

In Figure 3.13,

= 2 𝑟𝑎𝑑𝑖 牡𝑛𝑠

𝑐𝑦𝑐𝑙𝑒 x f (cycle/second) x t ( time in sec)

= 2ft radians

= 360 ft degrees

where 2f = (angular velocity in radians/second)

or = 360 f (degrees per second)

and so from e = Em sin, this equation becomes

e = Em sint This equation can also be written as

e(t) = Em sint meaning e is a function of t. and for current

i = Im sint

Example 3.10 A 60-Hz alternating current has an rms value of 42.42 A making its maximum value

of 60 A. Draw current wave to scale. Find the time corresponding to an angle of 45, 90, 135 ,

225 , and 270 . Solution

60 A

45 90 135 225 270

Vm or Im

v or i

(degrees or radians) t t (seconds)

78

Use the formula = 360 ft

t =

360 f =

360 (60)

when = 45, t =

360 (60) =

45

360 (60) = 2.08 ms

when = 90, t =

360 (60) =

90

360 (60) = 3.85 ms

when = 135, t =

360 (60) =

135

360 (60) = 6.25 ms

when = 225, t =

360 (60) =

225

360 (60) = 10.42 ms

when = 270, t =

360 (60) =

270

360 (60) = 12.5 ms

Example 3.11 A sine-wave alternating voltage has a maximum value of 170 volts and a frequency of 25 Hz. Determine (a) value of voltage 0.001, 0.004, 0.01 sec after crossing zero axis in a positive direction; (b) angles corresponding to each value of time; (c) rms and average (for a half-cycle) values. Solution

(a) The equation of the voltage is, v = 170 sin 225t V

170 V

T = 40 ms

79

(1) when t = 0.001 sec; v = 170 sin [225(0.001)] = 26.59 V

(2) when t = 0.004 sec; v = 170 sin [225(0.004)] = 99.92 V

(3) when t = 0.01 sec, v = 170 sin [225(0.01)] = 170 V

(b) The formula is = 360ft degrees = 360(25)t

(1) when t = 0.001 sec; = 360 x 25 𝑐𝑦𝑐𝑙𝑒𝑠

𝑠𝑒𝑐 x 0.001 sec = 9

(2) when t = 0.004 sec; = 360x 25 𝑐𝑦𝑐𝑙𝑒 尠

𝑠𝑒𝑐 x 0.004sec = 36

(3) when t = 0.01 sec; = 360x 25 𝑐𝑦𝑐𝑙𝑒𝑠

𝑠𝑒𝑐 x 0.01 sec = 90

(c) Vrms = 170

2 = 120.21 V

Vavg = 2

(170) = 108. 23 V

Problem Set No. 6

EXPRESSION AND PHASE RELATIONSHIP OF SINE WAVES

Solve the following problems. Draw the corresponding sine waves.

1. Sine wave A has a positive-going zero crossing at 30. Sine wave B has a positive-going zero

crossing at 45. Determine the phase angle between the two signals. Which signal leads?

2. One sine wave has a positive peak at 75, and another has a positive peak at 100. How much

is each sine wave shifted in phase from the 0 reference? What is the phase angle between them?

3. Make a sketch of two sine waves as follows: Sine wave A is the reference, and sine wave B lags

A by 90. Both have equal magnitude.

4. Convert the following angular values from degrees to radians:

(a) 30 (b) 45 (c) 135 (d) 300

5. Convert the Following angular value from radians to degrees:

(a) /8 (b) /2 (c) 3/5 (d) 6/5

6. A certain sine wave has a positive-going zero crossing at 0 and an rms value of 20 A. Calculate its instantaneous value at each of the following angles:

(a) 15 (b) 50 (c) 135 (d) 300

7. Sine wave A lags sine wave B by 30. Both have peak values of 15 V. Sine wave A is the

reference with a positive-going zero crossing at 0. Determine the instantaneous value of sine

wave B at 30, 45, 90, 180, 200, and 300.

80

8. A 50-Hz alternating current has a maximum instantaneous value of 42.42 A. It crosses the zero

axis in a positive direction when time is zero. Determine (a) time when current first reaches a value of 30 A; (b) time when current, after having gone through its maximum positive value, reaches a value of 36.7 A; (c) value of current when the time is 1/120 sec; (d) value of time when current first reaches a negative value of 21.21 A.

9. A current is given by i = 22.62 sin 377t. Determine (a) maximum value; (b) rms value; (c) frequency; (d) radians through which its vector has gone when t = 0.01 sec; (e) number of degrees in (d); (f) value of current at instant in (d).

10. A 25-Hz emf has an rms value of 250 volts, is zero and increasing positively when t = 0. Determine (a) maximum value; (b) equation; (c) radians at t = 1/75 sec; (d) degrees in (c); (e) emf at time t in (c).

3.13 Addition of Sine Waves

(a) In Phase Sine Waves

The equation of i1 = Im1 sin t and that of i2 = Im2 sin t

The sum (total, resultant, or equivalent) of the two waveforms is

iT = (Im1 + Im2) sin t

-25

-20

-15

-10

-5

0

5

10

15

20

25

0 1 2 3 4 5 6 7

curr

ent

time (sec)

i1

i2

I1m

I2m

81

Example 3.12 If the maximum values of i1 and i2 are 15 A and 20 A respectively, and the frequency

is 60 Hz, find the equation of the resultant of the two waves.

Solution

The equation of i1 = 15 sin 260t A and that of i2 = 20 sin 260t A

The resultant equation is

iT = i1 + i2

= 15 sin 260t A + 20 sin 260t A

iT = 35 sin 260t A Answer

The figure below shows the waveform of the resultant current

Example 3.13 Find the equation of the resultant of the two 60-Hz sine waves below.

-40

-30

-20

-10

0

10

20

30

40

0 1 2 3 4 5 6 7curr

en

t

time (sec)

i1

i2

iT

82

The equation of e1 = 90 sin (260t – 30) V and that of e2 = 70 sin (260t - 30) V

or converting 30 to radians we have, /6 radians

So that e1 = 90 sin (260t – /6) V and e2 = 70 sin (260t - /6) V

The resultant equation is

eT = 160 sin (260t - /6) V Answer

The figure below shows the resultant voltage.

(b) Waves Differing in Phase by 90 degrees

-100

-80

-60

-40

-20

0

20

40

60

80

100

0 1 2 3 4 5 6 7 8

volt

age

(degrees)

e1

e2

E1m

E2m

30°

-200

-150

-100

-50

0

50

100

150

200

0 1 2 3 4 5 6 7 8

volt

age

(degrees)

e1

e2

eT

ETm

30°

83

Let A = be the maximum value of sine wave a

B = the maximum value of sine wave b

C = the maximum value of the sum or resultant of the two sine waves

= the phase angle of the resultant.

The equation of sine wave a = A sin t and b = B sin (t + 90) = B cos t and

c = C sin(t + )

Now, find C and in terms of A and B

a + b = c

A sin t + B cos t = C sin(t + )

Expanding sin(t + ) = sint cos + cos t sin

A sin t + B cos t = C (sint cos + cos t sin )

A sin t + B cos t = C sint cos + Ccos t sin

Equating coefficients of sin t,

A = C cos (1)

Equating coefficients of cos t,

-20

-15

-10

-5

0

5

10

15

20

-2 -1 0 1 2 3 4 5 6 7

a

b

c90° 0° 90° 180° 270° 360°

CB

A

84

B = C sin (2)

To find C, square equations 1 and 2 then add,

A2 + B2 = C2 (cos2 + sin2 )

A2 + B2 = C2

C = A2 + B2

To find , divide equation 2 by equation 1

B

A =

C si n

C co s = tan ; = tan-1

B

A

Hence,

A sin t + B cos t = c = A2 + B2 sin (t + tan-1 B

A)

Example 3.14 A 60-Hz current i1 = 9 sint A is added to 60-Hz current i2 = 15cost, where =

260. Determine (a) the equation of the resultant current; (b) time at which the two currents are

equal.

Solution

iT = I𝑚1

2 + I𝑚2

2 sin (t + tan-1 I𝑚 2

I𝑚 1

)

-20

-15

-10

-5

0

5

10

15

20

-2 -1 0 1 2 3 4 5 6 7

curr

ent

(degrees)

i1

i2

iT

90° 0° 90° 180° 270° 360°

ITm

I2m

I1m

85

where Im1 = 9 A , Im2 = 15 A , = 260 rad/sec.

iT = 92 + 152 sin (260 t + tan-1

15

9)

so that , iT = 17.49 sin (260t + 59.04) or iT = 12.04 sin (260t + 1.03)

(c) Waves Differing in Phase Other than 90 degrees

Example 3.15 Two 25-Hz emfs differing in phase by 105 are given by

e1 = 120 sin (t + 60) V and e2 = 60 sin (t - 45) V

Solution eT = e1 + e2

= 120 sin (t + 60) + 60 sin (t - 45)

= 120 (sin tcos60 + cost sin 60) + 60 (sin t cos 45 - cos t sin 45)

= 60sint + 103.92 cost + 42.43sin t - 42.43cos t

= 102.43 sint + 61.49 cost

Following the formula

c = A sin t + B cos t = A2 + B2 sin (t + tan-1 B

A)

-150

-100

-50

0

50

100

150

-2 -1 0 1 2 3 4 5 6 7 8

volt

age

(degrees)

e1

e2

eT

60° 0° 45°

86

where A = 102.43, B = 61.49

and let c = be the eT

Therefore,

eT = (102.43)2 + (61.49)2 sin (t + tan-1 61.49

102.43)

eT = 119.47 sin (t + 30.98) V

In this equation, ITm = 119.47 V and = 30.98

Problem Set No. 7

ADDITION OF SINE WAVES

1. An emf e1 = 100 sin 260t is in series with an emf e2 = 120 sin 260t. Determine (a) their resultant e3; (b) angle between e1 and e3.

2. Two currents i1 = 12 sin 260t and i2 = 9 cos 260t flow in a wire. (a) Determine the equation of

resultant current i3 and angle between i1 and i3 ; (b) Determine rms value of i3.

3. Two 50-Hz currents i1 = 2.5 sin (ωt - 15) and i2 = 3.5 sin (ωt - 75) flow in a common wire. Determine (a) their resultant i3; (b) angle between i1 and i3 .

4. Two 50-Hz emf having rms values of 22.6 and 33.9 volts differ in phase by an angle of 60, the

latter current lagging and its positive-going zero crossing is at 0. Determine the resultant emf if they are connected in series.

87

3.14 Introduction to Phasors

Phasors can be used to represent time-varying quantities, such as sine waves, in terms of their

magnitude and angular position (phase angle)

Sine Waves Phasor Diagram

i = 60 sin (t + 0) A 𝐈 = 42.43 A0

𝐈

60 A i

88

e = 120 sin (t - 60) 𝐄 = 84.85 V -60

e = 120 sin (t - 60) 𝐄 = 84.85 V -60

Note: For the magnitude of the phasors always use the rms value of the sine wave voltage and

current.

3.15 Addition of Sine Waves and Phasors

Example 3.16 Find the total current if the maximum values of i1 and i2 are 20 and 15 amp,

respectively.

120 V

e

60

=

60

E

60

120 V

e

=

60

60

E

89

Solution For the maximum values of the currents:

I1 = 14.14 A

I2 = 10.61 A

𝐈𝐓 = 𝐈𝟏 + 𝐈𝟐

= 14.140 + 10.610

= 24.75 A0

To convert this to a sine wave equation, the maximum value of the total current

ITm =2( 24.75) = 35 A. The equation of the total current is

iT = 35 sin(t + 0)

Example 3.17

-40

-30

-20

-10

0

10

20

30

40

0 1 2 3 4 5 6 7curr

ent

time (sec)

i1

i2

iT

I1 = 14.14 A I2 = 10.61 A

24.75 A

24.75 A

90

Solution E1 = 90 𝑉

2 = 63.64 A

2

E2 = 70 𝑉

2 = 49.5 A

𝐄𝐓

= 𝐄𝟏

+ 𝐄𝟐

= 63.64 V-30 + 49.5 V-30

= 113.14 V-30

To convert this to a sine wave equation, the maximum value of the total current

ETm = 2 ( 113.14) = 160 A. The equation of the total current is

eT = 160 sin(t - 30)

Example 3.18 A 60-Hz current i1 = 9 sint A is added to 60-Hz current i2 = 15cost, where =

260. Determine (a) the equation of the resultant current; (b) time at which the two currents are

equal.

Solution

-200

-150

-100

-50

0

50

100

150

200

0 1 2 3 4 5 6 7 8

volt

age

(degrees)

e1

e2

eT

ETm

E1m

E2m

30°

E2 = 49.5 V

ET = 113.14 V

E1 = 63.64 V

30

91

The resultant equation of the two waves is,

iT = 17.49 sin (260t + 59.04) A

Converting these sine waves as phasors,

Adding the phasors,

𝑰𝑻 = 𝑰𝟏 + 𝑰𝟐

= 6.36 A0 + 10.61 A 90

= (6.36 + j0) + (j10.61)

= 6.36 + j 10.61

= 12.37 A 59.04

-20

-15

-10

-5

0

5

10

15

20

-2 -1 0 1 2 3 4 5 6 7

curr

en

t

(degrees)

i1

i2

iT

90° 0° 90° 180° 270° 360°

ITm

I2m

I1m

I2 = IT = 12.37 A

10.61 A

= 59.04

I1 = 6.36 A

92

Problem Set No. 8

PHASOR ANALYSIS

1. Sketch the sine waves represented by the phasor diagram shown. Find also the rms value of the resultant.

E1 = 25 V

45 E2 = 15 V

60

93

2. An emf e1 = 100 sin 260t is in series with an emf e2 = 120 sin 260t. Determine (a) their

resultant 𝐸3 in polar form (b) angle between 𝐸1

and 𝐸3 .

3. Two currents i1 = 12 sin 260t and i2 = 9 cos 260t flow in a wire. (a) Determine theresultant

current 𝐼3

in polar form ; (b) Determine rms value of 𝐼3 .

4. Two 50-Hz currents i1 = 2.5 sin (ωt - 15) and i2 = 3.5 sin (ωt - 75) flow in a common wire.

Determine (a) their resultant 𝐼3 (b) angle between 𝐼1 and 𝐼3

.

5. Two 50-Hz emf having rms values of 22.6 and 33.9 volts differ in phase by an angle of 60, the

latter current lagging and its positive-going zero crossing is at 0. Determine the resultant emf if they are connected in series.

3.16 Ohm’s Law and Kirchhoff’s Laws in AC Circuits

When time-varying ac voltages such as the sine wave are applied to resistive circuits, the

circuit laws that you studies earlier still apply. Ohm’s law applies to resistive ac circuits in the same

way that it applies to dc circuits. If a sine wave voltage is applied across a resistor as shown in the

figure below, a sine wave current flows. The current is zero when the voltage is zero and is maximum

when the voltage is maximum. When the voltage changes polarity, the current reverses direction.

As a result, the voltage and current are said to be in phase with each other.

94

When using Ohm;s law in ac circuits, remember that both the voltage and the current must

be expressed consistently, that is, both as peak values, both as rms values, both as average values,

and so on.

Sine wave voltage produces a sine wave current.

Note: A quantity with no rms subscript is understood as rms quantity. For instance, in 110

V (rms) the rms can be deleted and it is still understood as an rms quantity; in I(rms) = 110 V or I =

110 V (rms) can be written as I = 110 V.

The Power formulas P = I2R = 𝑉2

𝑅 = IV are applicable to both DC and AC circuits,

Where I and V = are either dc values and rms values

v

i

VACR

I

R1 = 1 kohm110 V dc 𝑉 = 𝑅𝐼

In DC circuit,

I = V

R

R = V

I

R1 = 1 kohmV ac

In AC circuit,

Vrms = RIrms or 𝑉 = 𝑅𝐼

Irms = 𝑉𝑟𝑚𝑠

R or I = V

R

95

P = power in watts

R = resistance

Example 3.19 Determine the rms voltage across each resistor and the rms current. The source

voltage is given as rms value.

Solution

The total resistance of the circuit is

RT = R1 + R2 = 1 k + 560 = 1.56 k

Use Ohm’s law to find the rms current,

Irms = Vs (rm s)

RT =

110 𝑉

1.56 𝑘Ω = 70.5 mA

The rms voltage drop across each resistor is

V1(rms) = Irms R1 = (70.5 mA)(1 k) = 70.5 V

V2(rms) = Irms R2 = (70.5 mA)(560 ) = 39.5 V

Problem Set No. 9

OHM’S LAWS AND KIRCHHOFF’S LAWS IN AC CIRCUITS

1. Three resistors are connected in series across a 120-V (rms) source. The voltage drop across R1 is 60 V (rms) and across R2 is 40 V (rms). Find the peak voltage across R3.

2. Three resistors that are connected in parallel have currents of 6 A, 5 A, and 8 A. Find the total rms current. All values are given in rms.

Vs = 110 V

R1 = 1 kohm

R2 = 560 ohms

96

3. A 117- V 60-Hz source is connected to a series circuit consisting of three resistors. If the ohmic

values of the latter are 20, 30, and 40 ohms, respectively, calculate the current through the circuit and the voltage drop across each resistor.

4. Three incandescent lamps (resistors) are connected in parallel, and to a 115-volt 60-Hz source. If the lamp ratings are 75, 100, and 150 watts, (a) calculate the rms value of the resultant current, (b) write the equation for the resultant current.

5. A resistance load of 4 ohms is connected to a 220-V 60-Hz line which leads back to the source through a pair of wires, the resistance of each of which is 0.08 ohm. What is the voltage at the source?

Objective Test No. 3

INTRODUCTION TO ALTERNATING CURRENT AND VOLTAGE

1. The difference between alternating current (AC) and direct current (DC) is

a. AC changes value and DC does not. b. DC changes direction and AC does not. c. DC changes value and AC does not.

2. When using circuit laws and rules we must use

97

a. maximum value.

b. effective value

c. average value

d. peak-to-peak value

3. If the peak value of a sine wave is 20 V, the rms value is

a. 14.14 V

b. 6.37 V

c. 7.07 V

d. 0.707 V

4. The average value of a 10-V peak sine wave over one complete cycle is

a. 0 V

b. 6.37 V

c. 7.07 V

d. 5 V

5. The instantaneous value of a 15-A peak sine wave at a point 32º from its positive-going

zero crossing is

a. 7.95 A

b. 7.5 A

c. 2.13 A

d. 7.95 V

6. The value of ac that would have the same effect in power produced as a similar value

of DC is known as

a. peak value.

b. rms value

c. average value

d. peak-to-peak value

7. If e1 = A sin t and e2 = B sin (t - ) then

a. e1 lags e2 by

b. e2 lags e1 by

c. e2 leads e1 by

8. The equation for 25 cycles current sine wave having rms value 30 amperes will be

a. 30 sin 25t

b. 30 sin 50t

c. 42.4 sin 25t

d. 42.4 sin 50t

9. A phasor is a

98

a. line representing the magnitude and direction of an alternating current.

b. line which represents the magnitude and phase of an alternating current.

c. color band for distinguishing between different phases of a 3-phase supply.

d. Instrument used for measuring phases of an unbalanced 3-phase load.

10. A 60-Hz frequency would cause an electric light to

a. turn on and off 120 times per second.

b. flicker noticeable.

c. turn and off 180 times per second.

d. turn and off 60 times per second.

11. The difference between the positive peak value and the negative peak value of a sine

wave is called the

a. maximum value

b. average value

c. effective value

d. peak-to-peak value

12. The relationship between frequency f, number of revolutions per second n and pair of poles p

is given by

a. f = n/p

b. f = n/2p

c. f = np

d. f = 2np

13. The root-mean-square (rms) value of ac is the same as

a. instantaneous value

b. effective value

c. average value

d. maximum value

14. The rms value of sine wave is equal to

a. 0.637 max. value

b. 0.707 max. value

c. 0.506 max. value

d. 1.414 max. value

15. Form factor is defined as

a. rms value/peak value

b. max. value/rms value

c. rms value/average value

d. effective value/rms value

16. The value of form factor for a pure sine wave is

99

a. 1.414

b. 0.707

c. 0.637

d. 1.11

17. The value of peak factor for a pure sine wave

a. 1.414

b. 0.707

c. 0.637

d. 1.11

18. Which of the following statement concerning the sinusoidal waveform, is most correct?

a. it represents AC

b. it represents DC

c. it represents half-wave rectified AC

d. it represents sum of AC and DC

19. The average value of a sine wave is 2 times the maximum value.

a. true

b. false

20. The form factor of dc supply voltage is always

a. infinite

b. zero

c. 0.5

d. unity

21. The rms value of a sinusoidal AC current is equal to its value at angle of ______ degrees.

a. 90

b. 60

c. 45

d. 30

Directions: Solve the crossword puzzle. Use the given clues to arrive at the right answer.

How Much Have You Learned?

100

1 2 3C

R V Y

E 2 4 C

Q R L

U A 3

E G

N E 5 H A S O R

C V

y A 4

L

U 6

E 6

7

7 8

9 8

DOWN ACROSS 1 number of cycle per second 1 electromagnetic induction 2 0.637 x peak value 2 maximum value 3 a set of positive and negative values 3 effective value 4 Gustav _____________ 4 V = IR 5 positive and negative 5 a line representing an AC voltage or current 6 common waveform 6 symbol T 7 relative time difference 7 flow of electrons 8 alternating current 8 saw tooth, sine wave, triangular wave 9 value of voltage or current at any instant

Unit 4

INDUCTOR AND INDUCTANCE

101

1. describe the basic construction of an inductor. 2. define inductance. 3. explain how an inductor stores energy. 4. relate various physical parameters to inductance value. 5. explain why practical inductors have both resistance and

capacitance. 6. state Lenz’s law and Faraday’s law. 7. identify various types of inductors. 8. determine total series inductance. 9. determine total parallel inductance. 10. define time constant as related to an inductive circuit. 11. describe some common inductor applications. 12. check out an inductor with an ohmmeter.

LEARNING OUTCOMES

At the end of this unit, you

are expected to:

Important Terms

102

inductor

winding resistance electromagnetic field

winding capacitance

self-inductance Faraday’s law henry

Lenz’s law

induced voltage

permeability energy storage

time constant

core

4.1 The Inductor

When a length of wire is formed into a coil, as shown in Figure 4-1, it becomes a basic inductor.

Current through coil produces an electromagnetic field. The magnetic lines of force form a strong

magnetic field within and around a coil.. The net direction of the total magnetic field creates a north

and a south pole, as indicated.

Figure 4-1. A coil of wire forms an inductor. When current flows through it, a three-dimensional

electromagnetic field is created, surrounding the coil in all directions.

4.2 Self-Inductance

When there is current through an inductor, an electromagnetic field is established. When the

current changes, the electromagnetic field also changes. An increase in current expands the field,

and a decrease in current reduces it. Therefore, a changing current produces a changing

electromagnetic field around the inductor (coil). In turn, the changing electromagnetic field

produces a voltage across the coil in a direction to oppose the change in current. This property is

called self-inductance, but it is usually referred to as simply inductance. Inductance is symbolized by

L.

Inductance is a measure of a coil’s ability to establish an induced voltage as a result of a

change in its current and that induced voltage is in direction to oppose that change in current.

103

4.2.1 The Unit of Inductance

The henry, symbolized by H, is the basic unit of inductance. By definition, the inductance is one

henry when current through the coil, changing at the rate of one ampere per second, induces one volt

across the coil. In many practical application, millihenries (mH) and microhenries (H) are the most

common units. A common schematic symbol for the inductor is shown in Figure 4-2.

Figure 4.2. Symbol of inductor

4.3 The Induced Voltage in an Inductor

A changing current in an inductor causes a changing magnetic field though it. Since according to

Faraday’s law a changing magnetic field results to the induction of voltage across the inductor.

The formula for the induced emf (or voltage) across the coil or inductor is,

𝑒𝐿 = 𝐿𝑑𝑖

𝑑𝑡

where eL = the induced emf across a coil or inductor in volts (V)

L = the inductance in henry (H)

di/dt = rate of change of current in amp/sec.

Example 4.1 A group of electromagnets that create a flux in a dc generator – the field circuit – has

an inductance of 15 henrys. If the 2.6 amp excitation is interrupted in 0.04 sec by the opening of the

field switch, what average voltage is induced in the winding?

Solution eL = 15 x 2.6

0.04 = 975 volts

4.4 Energy Storage

An inductor stores energy in the magnetic field created by the current. The energy stored is

expressed as follows

W = ½ LI2

As you can see, the energy stored is proportional to the inductance and the square of the current.

When I is in amperes and L is in henries, the energy is in joules.

L

104

4.5 Physical Characteristics

The following characteristics are important in establishing the inductance of a coil, the core material,

the number of turns of wire, the length, and the cross-sectional area.

4.5.1 Core Material

As discussed earlier, an inductor is basically a coil of wire. The material around which the coil is

formed is called the core. Coils are wound on either nonmagnetic or magnetic materials. Examples

of nonmagnetic materials are air, wood, copper, plastic, and glass. The permeabilities of these

material are the same as for a vacuum. Examples of magnetic materials are iron, nickel, steel, cobalt,

or alloys. These materials have permeabilities that are hundreds or thousands of times greater than

that of a vacuum and are classified as ferromagnetic. A ferromagnetic core provides a better path for

the magnetic lines of force and thus permits a stronger magnetic field.

The permeability () of the core material determines how easily a magnetic field can be

established. The inductance is directly proportional to the permeability of the core material.

4.5.2 Parameters As indicated in the Figure 4-3, the number of turns of wire, the length, and the cross sectional area of

the core are factors in setting the value of inductance. The inductance is inversely proportional to

the length of the core and directly proportional to the cross-sectional area. Also, the inductance is

directly related to the number of turns squared.

Figure 4-3. Factors that determine the inductance of a coil.

This relationship is as follows:

L = 𝑵𝟐 𝑨

𝒍

where L is the inductance in henries, N is the number of turns, is the permeability, A is the cross-

sectional area in meters squared, and l is the core length in meters.

105

Example 4.2 Determine the inductance of the coil below. The permeability of the core is 0.25 x 10-3.

Solution

L = 𝑵𝟐 𝑨

𝒍 =

42(0.25 x 10−3)(0.1)

0.01 = 40 mH

4.6 Winding Resistance

When a coil is made of a certain material, for example, insulated copper wire, that wire has a certain resistance per unit of length. When many turns of wire are used to construct a coil, the total resistance may be significant. This inherent resistance is called the dc resistance of the winding resistance (Rw). Although this resistance is distributed along the length of the wire, it effectively appears in series with the inductance of the coil, as shown in Figure 4-4. In many applications, the winding resistance can be ignored and the coil considered as an ideal inductor. In other cases, the resistance must be considered.

(a) The wire has resistance (b) Equivalent circuit

Figure 4-4. Winding resistance of a coil.

4.7 Winding Capacitance When two conductors are placed side by side, there is always some capacitance between them. Thus, when many turns of wire are placed close together in a coil, a certain amount of stray capacitance is a natural side effect. In many applications, this stray capacitance is very small and has no significant effect. In other cases, particularly at high frequencies, ti may become quite important. The equivalent circuit for an inductor with both its winding resistance (Rw) and its winding capacitance (Cw) is shown in Figure 4-5. The capacitance effectively acts in parallel.

0.01 m 0.1 m

2

N = 4

Rw L

Cw

Rw L

106

(a) Stray capacitance between each (b) Equivalent circuit loop appears as a total parallel capacitance

Figure 4-5. Winding capacitance of a coil.

4.8 Faraday’s Law

Faraday found that by moving a magnet through a coil of wire, a voltage was introduced across the

coil, and that when a complete path was provided, the induced voltage an induced current.

The amount of induced voltage is directly proportional to the rate of change of the

magnetic field with respect to the coil.

This principle is illustrated in the Figure 4-6, where a bar magnet is moved through a coil of wire. An

induced voltage is indicated by the voltmeter connected across the coil. The faster the magnet is

moved, the greater is the induced voltage.

When a wire is formed into a certain number of loops or turns and is exposed to a changing

magnetic field, a voltage is induced across the coil. The induced voltage is proportional to the

number of turns of wire in the coil, N, and to the rate at which the magnetic field changes.

Figure 4-6. Induced voltage is created by a changing magnetic field.

4.9 Lenz’s Law

Lenz’s law adds to Faraday’s law by defining the direction of induced voltage as follows:

When the current through a coil changes and an induced voltage is created as a result of

the changing magnetic field, the direction of the induced voltage is such that it always opposes the

change in current.

In Figure 4-7 (a), the current is constant and is limited by R1. There is no induced voltage

because the magnetic field is unchanging. In part (b), the switch suddenly is closed, placing R2 in

parallel with R1 and thus reducing the resistance. Naturally, the current tries to increases and the

107

magnetic field begins to expand, but the induced voltage opposes this attempted increases in current

for an instant.

In part (c), the induced voltage gradually decreases, allowing the current to increase. In part

(d), the current has reached a constant value as determined by the parallel resistors, and the induced

voltage is zero. In part (e), the switch has been suddenly opened, and, for an instant, the induced

voltage prevents any decreases in current. In part (f), the induced voltage gradually decreases,

allowing the current to decreases back to a value determined by R1. Notice that the induced voltage

has a polarity that opposes any current change. The polarity of the induced voltage is opposite that

of the battery voltage for an increases in current and aids the battery voltage for a decreases in

current.

(b) At instant of switch closure:

Expanding magnetic field

induces voltage, which prevents

increase in total current.

(a) Switch open: Constant current

and constant magnetic field;

no induced voltage.

108

Figure 4-7. Demonstration of Lenz’s law: When the current tries to change suddenly, the

electromagnetic field changes and induces a voltage in a direction that opposes that change in

current.

4.10 Classifications of Inductor

Inductors are made in a variety of shapes and sizes. Basically, they fall into two general categories:

fixed and variable.

(a) Fixed (b) Variable

(c) Right after switch closure: The rate

of expansion of the magnetic field

decreases, allowing the current to

increase as induced voltage

decreases.

(d) Switch remains closed: Current

and magnetic field reach

constant value.

(e) At instant of switch opening:

Magnetic field begins to collapse,

creating an induced voltage, which

prevents decrease in current.

(f) After switch opening: Rate of

collapse of magnetic field decrease,

allowing current to decrease back to

original value.

109

Both fixed and variable inductors can be classified according to the type of core material. Three

common types are the air core, the iron core, and the ferrite core. Each has a unique symbol, as

shown.

Adjustable (variable) inductors usually have a screw-type adjustment that moves a sliding

core in and out, thus changing the changing the inductance.

4.11 Types of Inductor:

(a) fixed molded inductors (b) variable coils (c) toroid inductor

4.12 Inductors in Series and Parallel

When inductors are connected in series, the total inductance, LT, is the sum of the individual

inductances. The formula for LT is expressed in the following equation for the general case of n

inductors in series:

LT = L1 + L2 + L3 + . . . + Ln

Notice that the formula for inductance in series is similar to the formula for resistance in series.

Figure 4-8 Inductors in Series

When inductors are connected in parallel, the total inductance is less than the smallest inductance.

The formula for total inductance in parallel is similar to that for total parallel resistance.

1

𝐿𝑇=

1

𝐿1+

1

𝐿2+

1

𝐿3+ … +

1

𝐿𝑛

L115H

L220H

L330H

Ln60H

(a) Air core (b) Iron core (c) Ferrite core

L1 L2 L3 Ln

110

Figure 4-9 Inductors in parallel

The general formula states that the reciprocal of the total inductance is equal to the sum of the

reciprocals of the individual inductances. LT can be found by taking the reciprocals of both sides of

the equation.

LT = 𝟏

𝟏

𝑳𝟏+

𝟏

𝑳𝟐+

𝟏

𝑳𝟑+ …+

𝟏

𝑳𝒏

or LT = (L1-1 + L2

-1 + L3-1 + . . . + Ln

-1)-1

4.13 Inductors in DC Circuits

When there is constant direct current in an inductor, there is no induced voltage. There is, however,

a voltage drop due to the winding resistance of the coil. The inductance itself appears as a short to

DC. Energy is stored in the magnetic field according to the formula W = ½ LI2. The only energy loss

occurs in the winding resistance (P = I2Rw).

This condition is illustrated in Figure 4-10.

Figure 4-10. Energy storage and loss in an inductor. The only dc voltage drop across the coil is due

to the winding resistance.

4.14 Time Constant

Because the inductor’s basic action is to oppose a change in its current, it follows that current cannot

change instantaneously in an inductor. A certain time is required for the current to make a change

from one value to another. The rate at which the current changes is determined by the time

constant. The time constant for a series RL circuit is

= 𝐿

𝑅

where is in seconds when L is in henries and R is in ohms.

111

Example 4.3 A series RL circuit has a resistance of 1 k and an inductance of 1 mH. What is the

time constant?

= 𝐿

𝑅 =

1 𝑚𝐻

1 𝑘Ω = 1 s

4.15 Energizing Current in an Inductor

In a series RL circuit, the current will increase to 63% of its value in one time constant interval after

the switch is closed. The buildup of current is analogous to the buildup of capacitor voltage during

the charging in an RC circuit; they both follow an exponential curve and reach the approximate

percentages of final value as indicated in the Figure 4-11.

Figure 4-11. Energizing current in an inductor

The change in current over five time constant intervals is illustrated in Figure 4-12. When the

current reaches its final value at approximately 5, it ceases to change. At this time, the inductor acts

as a short (except for winding resistance) to the constant current. The final value of the current is

Vs/Rw = 10 V/10 = 1 A.

0 1 2 3 4 5

63%

86%

95%

98% 99% Considered 100%

Final current

112

Figure 4-12. Illustration of the exponential buildup of current in an inductor. The current

increases another 63% during each time constant interval. A winding resistance of 10 is

assumed. A voltage (VL) is induced in the coil that tends to oppose the increase in current.

Example 4.4 Calculate the time constant for the circuit shown below. Then determine the current

and the time at each time constant interval, measured from the instant the switch is closed.

20 V

R

100 ohms

L

50 mH

113

Solution

Ifinal = 𝑉

𝑅 =

20 𝑉

100 Ω = 0.2 A

= 𝐿

𝑅 =

50 𝑚𝐻

100 Ω = 0.5 ms

At 1 = 0.5 ms: i = 0.63(0.2 A) = 0. 126 A

At 2 = 1.0 ms: i = 0.86(0.2 A) = 0. 172 A

At 3 = 1.5 ms: i = 0.95(0.2 A) = 0. 190 A

At 4 = 2.0 ms: i = 0.98(0.2 A) = 0. 196 A

At 5 = 2.5 ms: i = 0.99(0.2 A) = 0. 198 A 0.2 A

4.16 Inductor Applications Power Supply Filter rf Choke Tuned Circuits

4.17 Testing Inductors The most common failure in an inductor is an open coil. To check for an open, remove the coil from the circuit. If there is an open, an ohmmeter check will indicate infinite resistance, as shown in Figure 4-13. If the coil is good, the ohmmeter will show the winding resistance. The value of winding resistance depends on the wire size and length of the coil. It can be anywhere from one ohm to several hundred ohms. Occasionally, when an inductor is overheated with excessive current, the wire insulation will melt, and some coil will short together. This produces a reduction in the inductance by reducing the effective number of runs and a corresponding reduction in winding resistance.

Figure 4-13. Checking a coil by measuring the resistance.

114

Problem Set No. 10

INDUCTOR AND INDUCTANCE 1. Derive the formula for the energy stored in an inductor. 2. Derive the formula for the inductance. 3. Derive the formula for the total inductance of series inductors. 4. Derive the formula for the total inductance of series inductors. 5. Convert the following to millihenries:

a. 1 H b. 250 H c. 10 H d. 0.0005 H

6. How many turns are required to produce 30 mH with a coil wound on a cylindrical coil having a cross-sectional area of 10 x 10-5 m2 and a length of 0.05 m? The core has a permeability of 1.2 x 10-6? a. 3536 turns b. 3679 turns c. 3987 turns d. 4502 turns

7. A 12-V battery is connected across a coil with a winding resistance of 12 . How much current is there in the coil? a. 1 A b. 2 A c. 3 A d. 4 A

8. How much energy is stored by a 100-mH inductor with a current of 1 A?

a. 0.02 J b. 0.04 J c. 0.05 J d. 0.09 J

9. The current through a 100-mH coil is changing at a rate of 200 mA/s. How much voltage is

induced across the coil? a. 0.02 V b. 0.04 V c. 0.01 V d. 0.08 V

10. Suppose that you require a total inductance of 50 mH. You have available a 10-mH coil and a 22-

mH coil. How much additional inductance do you need? a. 16 mH b. 18 mH c. 20 mH d. 36 mH

115

11. Determine the total parallel inductance for the following coils in parallel: 75 H, 50 H, 25 H,

and 15 H. a. 7.14 µH b. 8.90 µH c. 9.28 µH d. 10.67 µH

12. You have a 12-mH inductor, and it is your smallest value. You need an inductance of 8 mH.

What value can you use in parallel with the 12-mH to obtain 8 mH? a. 12 mH b. 18 mH c. 20 mH d. 24 mH

116

Objective Test No. 4

INDUCTOR AND INDUCTANCE

1. When the current though an inductor increases, the amount of energy stored in the electromagnetic field

Decreases remains constant increases doubles

2. When the current though an inductor doubles, the stored energy

Doubles Quadruples is halved does not change

3. The winding resistance of a coil can be decreased by

reducing the number of turns using a larger wire changing the core material

4. The inductance of an iron-core coil increases if

the number of turns is increased the iron core is removed the length of the core is increased larger wire is used

5. An inductor, a resistor, and a switch are connected in series to a 12-V battery. At the instant the

switch is closed, the inductor voltage is 0 V 12 V 6 V 4 V

6. An ohmmeter is connected across an inductor and the pointer indicates an infinite value. The

inductor is Good Open shorted resistive

7. The property that opposes any change in current

mutual inductance

friction

self-inductance

losses

117

8. An open coil has

infinite resistance and inductance zero resistance and inductance zero resistance and infinite inductance infinite resistance and zero inductance

9. If the number of turns in an inductor is increased, its inductance will Vary

Decrease

increase

remain the same

10. At DC steady state, an inductor acts like ___________. an open circuit

a short circuit

a capacitor

an insulator

11. Unit of inductance Farad Ohm Henry siemen

118

Unit 5

CAPACITOR AND CAPACITANCE

1. describe the basic construction of a capacitor. 2. define capacitance and tell how it is measured. 3. explain how a capacitor stores energy. 4. state Coulomb’s law and discuss how it relates to an electric field and

the storage of energy. 5. illustrate the charging and discharging of a capacitor. 6. relate various physical parameters to capacitance value. 7. determine total series capacitance. 8. determine total parallel capacitance. 9. define time constant as related to a capacitive circuit. 10. relate the charging and discharging of a capacitor to the time

constant. 11. explain why a capacitor blocks dc. 12. explain why a capacitor produces no energy loss. 13. explain the significance of reactive power in a capacitive circuit. 14. describe some common capacitor applications. 15. check out a capacitor with an ohmmeter.

LEARNING OUTCOMES

At the end of this unit, you

are expected to:

119

dielectric

voltage rating

charging

dielectric strength

discharging temperature coefficient

capacitance

leakage

farad

dielectric constant

Coulomb’s law

time constant

energy storage

permittivity

5.1 The Capacitor: Basic Construction

Figure 5-1. Basic Construction of Capacitor

In its simplest form, a capacitor is an electrical device constructed of two parallel conductive plates

separated by an insulating material called the dielectric.

dielectric

Parallel metal plates

terminal

Important Terms

120

5.2 Charge Storage of Capacitor

In the neutral state, both plates of a capacitor have an equal number of free electrons, as indicated in

Figure 5-2. When the capacitor is connected to a voltage source through a resistor, electrons

(negative charge) are removed from plate A, and an equal number are deposited on plate B. As plate

A loses electrons , plate B gains electrons, plate A becomes positive with respect to plate B. During

the charging process, electrons flow only through the connecting leads and the source. No electrons

flow through the dielectric of the capacitor because it is an insulator. The movement of electrons

ceases when the voltage across the capacitor equals the source voltage. If the capacitor is

disconnected from the source retains the stored charge for a long period of time (the length depends

upon the type of capacitor) and still has the voltage across it. Actually charged capacitor can be

considered as a temporary battery.

Figure 5-2. Illustration of a capacitor storing charge.

5.3 Capacitance

The amount of charge per unit of voltage that a capacitor can store is its capacitance, designated C.

That is, capacitance is a measure of capacitor’s ability to store charges. The more charge per unit of

voltage that a capacitor can store, the greater its capacitance, as expressed by the following formula:

121

C = 𝑄

𝑉

where C is the capacitance, Q is charge, and V is voltage

The Unit of Capacitance

The farad (F) is the basic unit of capacitance. By definition,

One farad is the amount of capacitance when one coulomb of charge is stored with one

volt across the plates.

Most capacitors that you will use in electronics work have capacitance values in microfarads (F) and

picrofarads (pF).

5.4 How a Capacitor Stores Energy

A capacitor stores energy in the form of an electric field that is established by the opposite charges

on the two plates. The electric field is represented by lines of force between the positive and

negative charges and concentrated within the dielectric.

Electric field exists between the plates of a charged capacitor.

A force exists between a charged bodies.

Lines of force

Q1 Q2

d

d

Q1 F Q2

d

122

5.5 Coulomb’s law

A force exists between two charged bodies that is directly proportional to the product of

the twp charges and inversely proportional to the square of the distance between the bodies.

This relationship is expressed as

F = 𝑘𝑄1𝑄2

𝑑2

where F is the force in newtons, Q1 and Q2 are the charges in coulombs, d is the distance between

the charges in meters, and k is a proportionality constant equal to 9 x 109.

5.6 The Energy Stored in a Capacitor

The formula for the energy stored by a capacitor is as follows:

W = ½ CV2

where the energy, W, is in joules when C is in farads and V is in volts.

5.7 Voltage Rating

Every capacitor has a limit on the amount of voltage that it can withstand across its plates. The

voltage rating specifies the maximum dc voltage that can be applied without risk of damage to the

device. If this maximum voltage, commonly called the breakdown voltage or working voltage, is

exceeded, permanent damage to the capacitor can result.

Both the capacitance and the voltage rating must be taken into consideration before a

capacitor is used in a circuit application. The choice of capacitance value is based on particular circuit

requirements (and or factors that are studied later). The voltage rating should always be well above

the maximum voltage expected in a particular application.

5.8 Dielectric Strength

The breakdown voltage of a capacitor is determined by a dielectric strength of the dielectric material

used. The dielectric strength is expressed in volts/mil (1 mil = 0.001 in.) Table 5-1 below typical

values for several materials. Exact values vary depending on the specific composition of the material.

123

Table 5-1. Some Common Dielectric Materials and their Dielectric Strengths.

Material Dielectric Strength

(volts/mil)

Air 80

Oil 375

Ceramic 1000

Paper 1200

Teflon 1500

Mica 1500

Glass 2000

5.9 Temperature Coefficient

The temperature coefficient indicates the amount and direction of a change in capacitance value

with temperature. A positive temperature coefficient means that the capacitance increases with an

increase in temperature or decreases in temperature. A negative coefficient means that the

capacitance decreases with an increases in temperature or increases with a decreases in

temperature.

Temperature coefficients typically are specified in parts per million per degree Celsius

(ppm/C). For example, a negative temperature coefficient of 150 ppm/C for a 1-F capacitor

means that for every degree rise in temperature, the capacitance decreases by 150 pF (there are one

million picofarads in one microfarad).

5.10 Leakage

No insulating material is perfect. The dielectric of any of any capacitor will conduct some very small

amount of current. Thus, the charge on a capacitor will eventually leak off. Some types of

capacitors have higher leakages than others. An equivalent circuit for a nonideal capacitor is shown

in Figure 5-3. The parallel resistor represents the extremely high resistance of the dielectric material

through which leakage current flows.

124

Figure 5-3. Equivalent Circuit for a Nonideal Capacitor

5.11 Physical Characteristics of a Capacitor The following parameters are important in establishing the capacitance and the voltage rating of a

capacitor: plate area, plate separation, and dielectric constant.

5.11.1 Plate Area

Capacitance is directly proportional to the physical size of the plates as determined by the plate area.

5.11.2 Plate Separation

Capacitance is inversely proportional to the distance between the plates.

5.11.3 Dielectric Constant

As you know, the insulating material between the plates of a capacitor is called the dielectric. Every

dielectric material has the ability to concentrate the lines of force of the electric field existing

between the oppositely charged plates of a capacitor and thus increase the capacity for energy

storage. The measure of a material’s ability to establish an electric field is called the dielectric

constant or relative permittivity , symbolized by r (the Greek letter epsilon).

Capacitance is directly proportional to the dielectric constant. The dielectric constant

(relative permittivity) is dimensionless, because it is a relative measure and is a ration of the

absolutely permittivity, , of a material to the absolute permittivity, o, of a vacuum, as expressed by

the formula:

r =

𝑜

The value of o is 8.85 x 10-12 (farads per meter).

C R leak

125

Table 5-2. Some Common Dielectric Materials and their Dielectric Strengths.

Material Typical r values

Air (vacuum) 1.0

Teflon 2.0

Paper (paraffined) 2.5

Oil 4.0

Mica 5.0

Glass 7.5

Ceramic 1200

5.12 Formula for Capacitance in Terms of physical Parameters

An exact formula for calculating the capacitance in terms of the three quantities mentioned is as

follows:

C = A r (8.85 x 10−12 F/m)

d

where A is in square meters (m2), d is in meters (m), C is in farads (F) and r is the relative permittivity.

Example 5.1 Determine the capacitance of a parallel plate capacitor having a plate area of 0.01

m2 and a plate separation of 0.02 m. The dielectric is mica, which has a dielectric constant of 5.0.

Solution

C = A r (8.85 x 10−12 F/m)

d

= (0.01 𝑚2) 5.0 (8.85 𝑥 10−12𝐹/𝑚 )

0.02 𝑚

d = 22.13 pF

5.13 Types of Capacitors

Mica Capacitors

126

Ceramic Capacitors

Paper/Plastic capacitors

Electrolytic Capacitors

Variable Capacitors

Air Capacitors

Trimmers and Padders

Varactors

5.14 Series Capacitors

While charging, I = Q/t is the same at all points so that all capacitors store the same amount of

charge (QT = Q1 = Q2 = Q3)

By Kirchhoff’s voltage law, Vs = V1 + V2 + V3

Using the fact that V = Q/C, we can substitute into the formula for Kirchhoff’s law and get the

following relationship (where Q = QT = Q1 = Q2 = Q3):

𝑄

𝐶𝑇=

𝑄

𝐶1+

𝑄

𝐶2+

𝑄

𝐶3

Canceling out Q we have,

1

𝐶𝑇=

1

𝐶1+

1

𝐶2+

1

𝐶3

Taking the reciprocal of both sides gives the formula for the total capacitance:

CT = 1

1

𝐶1+

1

𝐶2+

1

𝐶3

Voltage Division in Series Capacitors

Vx = 𝐶𝑇

𝐶𝑥 x Vs

C1 C2 C3

Vs

127

where Vx is the voltage across Cx which is any capacitor, such as C1, C2, and so on.

5.15 Parallel Capacitors

The charged stored by the capacitors together equals the total charge that was delivered from the

source:

QT = Q1 + Q2 + Q3

Using the fact that Q = CV, we can substitute into the preceding formula and get the following

relationship:

CTVs = C1V1 + C2V2 + C3V3

Since Vs = V1 = V2 = V3, they can be canceled, leaving

CT = C1 + C2 + C3, etc.

5.16 Capacitors in DC Circuits

In this section, the response during charging and discharging of a simple capacitive circuit with a dc

source is examined. Figure 5-4 shows a capacitor connected in series with a resistor and a switch to

a dc voltage source. Initially, the switch is open and the capacitor is uncharged with zero volts across

its plates. At the instant the switch is closed, the current jumps to its maximum value and the

capacitor begins to charge. The current is maximum initially because the capacitor has zero volts

across it and therefore, appears as a short; thus, the current is limited only by the resistance. As time

passed and the capacitor charges, the current decreases and the voltage VC across the capacitor

increases. The resistor voltage is proportional to the current during this charging period.

C1 C2 C3Vs

128

Figure 5-4 Charging and discharging of a capacitor.

After a certain period of time, the capacitor reaches full charge. At this point, the current is zero and the capacitor voltage is equal to the dc source voltage, as shown in Figure 5-4. If the switch were opened now, the capacitor would retain its full charge (neglecting any leakage). In Figure 5-4, the voltage source has been removed. When the switch is closed, the capacitor begins to discharge. Initially, the current jumps to a maximum but in a direction opposite to its direction during charging. As time passes, the current and capacitor voltage decrease. The resistor voltage is always proportional to the current. When the capacitor has fully discharged, the current and the capacitor voltage are zero.

Remember the following about capacitors in dc circuits:

129

1. Voltage across a capacitor cannot change instantaneously. 2. Current in a capacitive circuit can change instantaneously. 3. A fully charged capacitor appears as an open to nonchanging current. 4. An uncharged capacitor appears as a short to an instantaneous change in current.

5.17 The RC Time Constant As you have seen, when a capacitor charges or discharges through a resistance, a certain time is required for the capacitor to charge fully or discharge fully. The voltage across a capacitor cannot change instantaneously, because a finite time is required to move charge from one point to another. The arte at which the capacitor charges or discharges is determined by the time constant of the circuit. The time constant of a series RC circuit is a time interval that equals the product of the resistance and the capacitance.

The time constant is symbolized by , and the formula is as follows

= RC Recall that I = Q/t. The current is the amount of charge moved in a given time. When the resistance is increased, the charging current is reduced, thus increasing the charging time of the capacitor. When the capacitance is increased, the amount of charge increases; thus, for the same current, more time is required to charge the capacitor.

Example 5.2 A series RC circuit has a resistance of 1 M and a capacitance of 5 F. What is the time constant?

Solution = RC = (1 M)(5 F) = 5s

5.18 The Charging Curve

ic

Initial value 100%

37%

14%

5%

2%

1%

0 1 2 3 4 5

130

Voltage and Current in a Capacitor during charging

Voltage across the

capacitor

Current through the capacitor

time

time

131

Problem Set No. 11

CAPACITOR AND CAPACITANCE

1. Two capacitors connected in parallel across a 250-V mains have charges of 3,000 µC and 5,000

C, respectively. Find the total capacitance of the combination.

a. 32 µF

b. 45 µF

c. 56 µF

d. 76 µF

2. A 0.4-F capacitor has a charge of 20 C. How much is the voltage across it?

a. 20 V

b. 30 V

c. 45 V

d. 50 V

3. The equivalent capacitance of two capacitors in series is 2.4 F. If one of the capacitors has a

capacitance of 4 F, what is the capacitance of the other?

a. 2 µF

b. 4 µF

c. 6 µF

d. 8µF

4. Three capacitors having capacitance of 4 F , 6 F and 8 F respectively are connected in series.

Find the equivalent capacitance of the combination.

a. 1.08 µF

b. 2.84 µF

c. 1.84 µF

d. 4.84 µF

5. The energy stored in a 0.125 F capacitor is 50 J, solve for the charge accumulated.

a. 3.54 mC

b. 5.34 mC

c. 6.23 mC

d. 8,45 mC

6. A certain capacitor is charged at 48 volts after its stored energy is 5.76 x 10-2 joules. What is the

capacitance of the capacitor?

a. 25 µF

b. 50 µF

c. 75 µF

132

d. 89 µF

7. Calculate the capacitance between two plates each of which is 100 cm2 and 2 mm apart in air.

a. 44.27 nF

b. 48.90 nF

c. 56.84 nF

d. 76.43 nF

8. A capacitor whose plates is 20 cm x 3.0 cm and is separated by a 1.0-mm air gap is connected

across a 12-V battery. Determine the charge accumulated on each plate after a long time.

a. 43.89 nC

b. 56.32 nC

c. 63.75 nC

d. 89.56 nC

9. Three capacitors A, B, and C are charged as follows: A: 10F, 100 volts; B: 15 F, 150 volts; C:

25 F, 200 volts. They are then connected in parallel with terminals of like polarity together. What is the voltage across the combination? a. 165 V

b. 175 V

c. 185 V

d. 195 V

10. A given capacitor has a capacitance of 100 F. Calculate its elastance. a. 1,000 D

b. 10,000 D

c. 100,000 D

d. 1,000,000 D

11. Three capacitors of 5 F , 10 F and 15 F respectively are connected in series across a 100-V

supply. Solve for the voltage across the 15-F capacitor.

a. 18.2 V

b. 19.6 V

c. 20.7 V

d. 25.4 V

133

Objective Test No. 5

CAPACITOR AND CAPACITANCE

1. Which of the following statement(s) accurately describes a capacitor?

The plates are conductive. The dielectric is an insulator between the plates. Constant dc flows through a fully charged capacitor. A practical capacitor stores charge indefinitely when disconnected from the source.

2. The capacity of a condenser is proportional to ___________.

area of its plates

volume of its plates

the specific resistance of the plate material

the temperature coefficient of the plate material

3. The capacity of the capacitor is inversely proportional to ________.

the temperature of the dielectric

the material of the dielectric

the thickness of the dielectric

the permeability of the material and inductance

4. The capacitors in series have the same _______.

voltage

capacity

charge

energy loss

5. The capacitors are named according to the _______ used.

material of the plate

dielectric used

enclosures used

voltage

6. When one of the following statements is true?

There is current through the dielectric of a charging capacitor.

When a capacitor is connected to a DC voltage source, it will charge to the value of the

source.

An ideal capacitor can be discharged by disconnecting it from the voltage source.

7. A capacitance of 0.01 F is larger than 0.00001 F

100,000 pF

134

1000 pF

all of the above

8. When the voltage across a capacitor is increased, the stored charge

increases

decreases

remains constant

fluctuates

9. When the voltage across a capacitor is doubled, the stored charge

stays the same

is halved

increases by four

doubles

10. The voltage rating of a capacitor is increased by

decreasing plate area

increasing plate separation

increasing the plate area

a and b

11. The capacitance value is increased by decreasing plate area

increasing plate separation

decreasing plate separation

increasing plate area

12. An uncharged capacitor and a resistor are connected in series with a switch and a 12-V battery. At the instant the switch is closed, the voltage across the capacitor is 12 V

6 V

24 V

0 V

13. In Question 12, the voltage across the capacitor when it is fully charged is 12 V

6 V

24 V

-6 V

14. An ohmmeter is connected across a discharged capacitor and the needle stabilizes at

approximately 50 k. The capacitor is

good

charged

too large

135

leaky

15. A good capacitor has a __________ resistance.

negligible

very high

negative

none of these

16. A capacitor opposes any change in __________.

current

voltage

Resistance

flux

17. The capacitance of a capacitor is directly proportional to

area of its plate

thickness of dielectric

18. Capacitors are used to

filter AC currents and pass DC currents

filter AC and DC currents

filter DC currents and pass AC currents

pass AC and DC currents

19. A capacitor consists of two _________

insulators separated by a conductor.

conductors separated by an insulator

conductors

insulators

20. Capacitors designed to be used in places where a high dielectric breakdown voltage is important

.

paper capacitors

ceramic capacitors

electrolytic capacitors

mica capacitors

21. Reciprocal of capacitance.

Inductance

Elastance

Reluctance

Daraf

22. The capacitance of a capacitor is NOT affected by _____.

136

type of dielectric material

distance between plates

area of the plates

type of material used in the plates

23. The capacitor stores the electricity in the shape of _____.

dynamic charge

static charge

current electricity

molecules

24. Capacitors are used in electric circuits to ____________.

store energy

introduce a voltage drop

produce a low opposition path to high frequencies

all of these

137

Unit 6

ALTERNATING CURRENT CIRCUITS

1. Apply Ohm’s law and Kirchhoff’s laws to ac circuits as well as to dc circuits. 2. explain the behavior of a resistor and other purely resistive load in ac circuit. 3. explain the behavior of an inductor in ac circuit. 4. discuss the behavior of a capacitor in ac circuit. 5. analyze the response of an RL, RC, and RLC circuit with a sine wave input. 6. define impedance. 7. determine the impedance of both series and parallel RL, RC, and RLC circuit. 8. convert a parallel circuit to an equivalent series circuit. 9. determine the effects of frequency on an RL, RC, and RLC circuits in terms of

changes in impedance and phase angle. 10. define conductance, capacitance, capacitive and inductive susceptance, and

admittance. 11. Apply Ohm’s law to RC circuits in order to find current and voltage value. 12. determine the true power, reactive power, and apparent power. 13. define power factor and explain its significance. 14. explain how RL, RC, and RLC circuits are used as phase-shifting circuits (lead

and lag network).

LEARNING OUTCOMES

At the end of the lesson, you are expected to:

138

6.1 AC Circuit with Resistance Only This is a circuit in which the applied voltage is ac connected to any resistive load. What is a resistive load? Any component which convert electrical energy to heat and light energy such as lamp, flat iron, electric stove, etc. These components are made up of metals which posses certain amount of resistance. The most common type of resistive load is resistor in which the common function is to resist and limit the amount of current. Heat is also produced by the resistor when current flows through it. Motor also have resistance because their windings are made up of copper wire.

.

Let VR be the voltage across the resistor (or any resistive load) in rms value

Vs be the source voltage in rms value.

Note:

The load here represents any purely resistive load, that is, a load which has resistance only. rms value is used because it is the effective value in the circuit. VR = Vs because there is only component in the circuit.

If the sine wave equation of the of Vs is vs = Vsm sint the equation of VR is definitely vR = VRm sint.

To find the current through the resistor using Ohms’ law,

𝑖 = 𝑣R

R =

VR m si n t

R

Let 𝐼𝑅𝑚 =

VR m

R

so that the equation of the current i = IRm sint. where i = is the instantaneous value of the current VS and Vr = the instantaneous value of voltage source and the voltage across the purely load, respectively.

IRm , VSm, VRm = the maximum value of the current through the load, maximum value of the

voltage source, and the maximum value of the voltage across the resistor.

RVs

139

As you can see, the phase angle of current and voltage are both equal to zero. 6.1.1 The Phase Relationship of Current and Voltage

When the equation of the voltage across the resistor is vR = VRm sint the current is i = IRm sint,

therefore,

In a purely resistive load the current is in phase with the voltage.

This phase relationship is illustrated below where the phase angles of the voltage and current are

both zero meaning one of the positive-going zero crossings is at 0.

If in case the phase angle of voltage is 30, then the phase angle of current is also 30

Note: The phase angle that is being referred here in the angle of positive-going zero crossing of a

sine wave nearest to the origin.

-80

-60

-40

-20

0

20

40

60

80

0 1 2 3 4 5 6 7

current through the resistor

voltage across the resistor

0° 90° 180° 270° 360°

IR VR

Phasor Diagram

140

If the equation of the voltage across the resistor is v = VSm sin (t + 30) then the current is

i = IRm sin (t + 30).

From IRm = VR m

R since, IRm = 2 IR rms and VRm = 2 VR rms

2 IR rms = 2 VR rm s

𝑅

Canceling 2 , we have,

IR rms = VR rm s

𝑅 or simply IR =

VR

𝑅

Also VR = IR R

R = VR

I

Note: These formulas are the same as in DC circuit and these are applicable regardless of the frequency of the source because frequency has no effect on the resistance.

-80

-60

-40

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0

20

40

60

80

-1 0 1 2 3 4 5 6 7

iR

vR

VRm

IRm

30°

VR

IR 30

Phasor Diagram

141

Example 6.1 In the circuit below, find the sine wave equation of the current if the source is 120 V

(rms).

Solution

The maximum value of the source is 2 (120) = 169.71 V. Assuming that its positive-going

zero-crossing is at 0, therefore its equation is vS = 169.70 sin 260t and the voltage across the

resistor is vR= 169.70 sin 260t.

Using the formula IR = VR

𝑅 =

169.70 sin 260t V

100 , so that the equation of the current is

iR = 1.697 sin 260t A answer

Another solution is simply using the formula IR = VR

𝑅

where VR = Vs = 120 V

IR = 120 V

100 Ω = 1.2 V the rms value of which is 2 (1.2) = 1.6971 A,

The equation is also

iR = 1.697 sin 260t A answer

Example 6.2 In the circuit below, find the value of the resistance if the vR = 127.28 sin (t - 60)

V and iR = 14.14 sin(t - 60) A.

Solution

R = 𝑣𝑅

𝑖𝑅 =

127.28 sin (t − 60) V

14.14 si n(t − 60) A = 9

Example 6.3 Find the rms value of the current in the circuit shown below

R

100 ohms

Vs

120 V

60 Hz

RVs

142

Solution

IR = VR

𝑅 =

100 V

10 Ω = 10

6.1.2 The Power in a Circuit with Resistance Only

-20

-10

0

10

20

30

40

50

60

70

80

0 1 2 3 4 5 6 7

new axis iR

vR

P

0° 90° 180° 270° 360°

Pave

R

10 ohmsVs

100 V

RE

143

Let p = be the instantaneous power absorbed by the resistor, where p = vRiR

In the sine waves below iR = IRm sin t and vR = VRm sin t, so that

p = (IRm sin t )(VRm sin t)

= IRm VRm sin2 t where sin2t =

1 − cos 2t

2

p = IRm VRm 1 − cos 2t

2

The average value of the power wave is IR m VR m

2

P = IR m VR m

2

where P = average value of the power. This average value is also known as true or active power. 6.1.3 What is meant by true power? This is the rate at which energy is being dissipated by the resistor or any resistive load. 6.1.4 The Formula of True Power

From P = IR m VR m

2 , IRm = 2 IRrms or 2 IR and VRm = 2 VRrms or 2 VR

Note: Eliminating rms does not change the meaning of the voltage and current; they are still rms values. Then the power formulas are

P = IRVR = I2R = 𝑉𝑅

2

𝑅

where P = power in watts IR = rms value in case of ac circuit and dc value in case of dc circuit. VR = rms value in ac circuit and dc value in dc circuit R = resistance in ohms Example 6.4 The emf and current waves of a circuit having resistance only are

e = 170 sin 260t and i = 14.14 sin 260t. Determine (a) equation of power wave; (b) frequency of power wave; (c) maximum value of power wave; (d) average power; (e) power when t = 1/480 sec, 1/240 sec, 1/90 sec. (f) Draw voltage, current, and power waves. Solution

144

(a) p = ei

= (170 sin 260t)(14.14 sin 260t)

= 2403.8 sin2 260t

= 2403.8 1 − cos 2260t

2

= 2403.8

2 (1 - cos2260t)

= 1201.9 (1 - cos2260t) watts

= 1201.9 (1 - cos2120t) watts (d) f = 120 Hz (e) Pmax = 2403.8 watts (f) Pave = 1201.9 watts

(g) From the power wave equation p = 1201.9 (1 - cos2120t) watts

when t = 1/480 sec ; p = 1201.9 [1 - cos2120(1/480)] = 1201.9 watts

when t = 1/240 sec p = 1201.9 [ 1 - cos2120(1/240)] = 2403.8 watts

when t = 1/90 sec ; p = 1201.9 [ 1 - cos2120(1/90)] = 1802.85 watts

Example 6.5 A lamp load consist of thirty 100-watt lamps each taking rated power from a 120-V

supply. Determine (a) power when supply is DC; (b) power when supply is 60-Hz sinusoidal AC; (c)

equation of AC voltage wave if voltage is zero and increasing when t = 0; (d) equation of current

wave; (e) equation of power wave.

Solution

(a) P = 30 lamps x 100 watts/lamp = 3,000 watts

-500

0

500

1000

1500

2000

2500

3000

0 1 2 3 4 5 6 7

iR

vR

p

145

(b) 3,000 watts also because the frequency has no effect on the resistance.

(c) e = vR = 2(120) sin 260t = 169.71 sin 260t V

(d) I = 𝑃

𝐸 =

3,000 𝑊

120 𝑉 = 25 A

Im = 2 (25) = 35.36 A

The equation of the current wave is i = 35.36 sin 260t A

(e) p = ei

= (120 sin 260t)(35.36 sin 260t)

= 4243.2 sin2 260t

= 4243.2 1 − cos 2260t

2

= 4243.2

2 (1 - cos2260t)

= 2121.6 (1 - cos2260t) watts

= 2121.6 (1 - cos2120t) watts

6.2 AC Circuit with Inductance Only

Practically, all inductors have resistance in them because they are made of metals and metals have

resistance. In this lesson inductors are considered as ideal, meaning the resistance is zero or

negligible.

6.2.1 What are some examples of inductive loads? Inductors, motors, and everything that is made of coils of conductors .

(a)

Vs L

146

(b)

Figure 6-1. (a) An inductor connected across an ac source; (b) The waveforms of voltage and

current in the inductor.

Note: L represents the inductance of a load which is considered as purely inductive.

6.2.2 Phase relationship of Current and Voltage in a Purely Inductive Load

In the waveforms shown in Figure 6-1(b) the current lags the voltage by 90 or the voltage leads the

current by 90. The equation of voltage across the capacitor is vL = VLmsin t V and the current is iL =

ILm sin (t - 90) A. Just subtract 90 from the phase angle of the voltage (that is 0 - 90 = -90).

In case the phase angle of the voltage is other than 0, say +30, the phase angle of the current

would be 30 - 90 = -60. The equation of the voltage would be vL = VLm sin (t + 30) V and that of

the current is iL = ILm sin (t - 60) A.

If the equation of current is iL = ILm sin (t + 30) A, the equation of the voltage is vC = VCmsin (t +

120) V. Add 90 from 30 since the voltage is leading the current by 90.

Example 6.6 The waveform below is the voltage across an ideal inductor. Draw the waveform for its current if its maximum value is 5 A. Write also their sine wave equations.

-25

-20

-15

-10

-5

0

5

10

15

20

25

0 1 2 3 4 5 6 7 8 9

VLm

ILm current through the inductor

voltage induced by the inductor

0° 90° 180° 270° 360°

147

Solution

The equation of the voltage is vL = 120 sin (t + 60) V and that of the current is iL = 5 sin (t - 30)

A.

6.2.3 Why the Current Lags the Inductor Voltage by 90

As you know, a sine wave voltage has a maximum rate of change at its zero crossings and a zero rate

of change at the peaks. From Faraday’s Law you know that the amount of voltage induced across a

coil is directly proportional to the rate at which the current is changing. Therefore, the coil voltage is

maximum at the zero crossings of the current where the rate of change of the current is the greatest.

Also, the amount of voltage is zero at the peaks of the current where its rate of change is zero. This

-150

-100

-50

0

50

100

150

-2 -1 0 1 2 3 4 5 6 7 8

VLm = 120 V

60°

-150

-100

-50

0

50

100

150

-2 -1 0 1 2 3 4 5 6 7 860° 30°

VLm = 120 V

ILm = 5A

148

relationship is shown in Figure 6-1(b) . As you can see, the current peaks occur a quarter cycle after

the voltage peaks. Thus, the current lags the voltage by 90.

6.2.4 The Inductive Reactance, XL

In Figure 6-2, an inductor is connected to a sine wave source. Note that when the source voltage is

held at a constant amplitude value and its frequency is increased, the amplitude of the current

decreases. Also, when the frequency of the source is decreased the current amplitude increases.

The reason is as follows: when the frequency of the applied voltage increases, its rate of change also

increases, as you already. Now if the frequency of the applied voltage is increased, the frequency of

the current also increases. According to Faraday’s law and Lenz’s law, this increase in frequency

induces more voltage across the inductor in a direction to oppose the current and cause it to

decrease in amplitude. Similarly, a decrease in frequency will cause an increase in current.

A decrease in the amount of current for a fixed amount of voltage indicates that opposition

to the current has increased. Thus, the inductor offers opposition to current which varies directly

with frequency.

Figure 6-2. The current in an inductive circuit varies inversely with the frequency of the applied

voltage.

The opposition to sinusoidal current in an inductor is called inductive reactance.

The symbol for inductive reactance is XL, and its unit is the ohm ()

Formula: XL = 2fL

where XL = the inductive reactance in ohms ()

f = frequency in hertz (Hz)

L = the inductance in henry (H)

149

XL is proportional to the frequency and inductance

Example 6.7 A sinusoidal voltage is applied to the circuit in Figure. The frequency is 1 kHz.

Determine the inductive reactance.

Solution XL = 2fL

= 2(1 kHz)(5 mH) = 31.4

6.2.5 Ohm’s law in Inductive Circuits

The reactance of an inductor is analogous to the resistance of a resistor. In fact, XL, just like R, is

expressed in ohms. Since inductive reactance is a form of opposition to current, Ohm’s law applies

to inductive circuits as well as to resistive circuits and capacitive circuits, and it is stated as follows:

VL = ILXL

where VL = the voltage across the inductor in volts (V). IL = the current through the inductor (A)

XL = the inductive reactance in ohms ()

Example 6.8 Determine the rms current.

Solution

First calculate XL:

XL = 2fL = 2(10 kHz)(100 mH) = 6283

Using Ohm’s law, we obtain

Vs L = 5 mH

5 V

10 kHz L = 100 mH

150

IL = VL

XL =

5 V

6283 = 795.8 A

6.2.6 Power in an Inductor

As discussed earlier, an inductor stores energy in its magnetic field when there is current through it.

An ideal inductor (assuming no winding resistance) does not dissipate energy; it only stores it. When

an ac voltage is applied to an inductor, energy is stored by the inductor during a portion of the cycle;

then the stored energy is returned to the source during another portion of thee cycle. There is no net

energy loss. Figure above shows the power curve that results from one cycle of inductor current

and voltage.

6.2.7 Instantaneous power (p)

The product of instantaneous voltage, vL, and instantaneous current, IL, gives instantaneous power, p.

At points where vL and iL is zero. When both vL and iL are positive, p is also positive. When either vL

or iL is positive and the other negative, p is negative. When both vL and iL are negative, p is positive.

As you can see, the power follows a sinusoidal-type curve. Power values of power indicates that

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0

20

40

60

0 1 2 3 4 5 6 7 8

vL

iL

p0° 90° 180° 270° 360°

Vs L

151

energy is stored by the inductor. Negative values of power indicate that energy is returned from the

inductor to the source. Note that the power fluctuates at a frequency twice that of the voltage or

current, as energy is alternately stored and returned to the source

6.2.8 True Power (Ptrue or simply P)

Ideally, all of the energy stored by the inductor during the positive portion of the power cycle is

returned to the source during the negative portion. No net energy is consumed in the inductance, so

the power is zero. Actually, because of winding resistance in a practical inductor, some power is

always dissipated. But since in this lesson, the inductor is considered ideal so its resistance is

negligible so its power dissipation or true power is zero.

6.2.9 Inductive Reactive Power (QL)

The rate at which an inductor stores or returns energy is called its reactive power, QL. The inductive

reactive power (so called because this is the reactive power of the inductor) is nonzero quantity,

because at any instant in time, the inductor actually taking energy from the source or returning

energy to it. Reactive power does not represent an energy loss. The following formula apply:

QL = VL IL

QL = VL

2

XL

QL = IL2 XL

where QL = the inductive reactive power in volt-ampere reactive (VAR or var)

VL = the rms voltage across the inductor in volts (V)

IL = the rms current through the inductor in amperes (A)

XL = the inductive reactive in ohm ()

Example 6.9 A 10-V rms signal with a frequency of 1 kHz is applied to a 10-mH coil with a

negligible resistance. Determine the reactive power (QL) and the true power (P).

Solution

XL = 2fL = 2(1 kHz)(10 mH) = 62.83

I = VL

XL =

10 V

62.83 Ω = 159.16 mA

We can find the inductive reactive using the three formulas:

QL = VL IL = (10 V)(159.16 mA) = 1.59 VAR

152

QL = VL

2

XL =

(10 V)2

62.83 Ω = 1.59 VAR

QL = IL2 XL = (159.16 mA)2(62.83) = 1.59 VAR

For the true power, P = 0 because there is no resistance in the inductor.

6.3 AC Circuit with Capacitance Only

Note: The capacitor in the circuit above is considered ideal, meaning the resistance is negligible.

What are some of the components that are considered capacitive?

Basically, all capacitors and synchronous motors.

6.3.1 The Phase Relationship of Current and Voltage in a Purely Capacitive Load

The current leads the voltage by 90, or the voltage lags the current by 90.

-60

-40

-20

0

20

40

60

-4 -2 0 2 4 6 8 10

iC

vC

Vcm

Icm

90° 0° 90° 180° 270° 360°

Vs C

153

In the waveforms above the equation of voltage across the capacitor is vC = VCmsin t V and the

current is iC = ICm sin (t + 90) A. Just add 90 from the phase angle of the voltage (that is 0 + 90 =

90).

In case the phase angle of the voltage is other than 0, say +30, the phase angle of the current

would be 30 + 90 = 120. The equation of the voltage would be vC = VCm sin (t + 30) V and that of

the current is iC = ICm sin (t + 120) A.

If the equation of current is iC = ICm sin (t + 30) A, the equation of the voltage is vC = VCmsin (t -

60) V. Subtract 90 from 30 since the voltage is lagging the current by 90.

Example 6.10 The waveform below is the voltage across an ideal capacitor, draw the waveform for

its current if its maximum value is 5 A. Write also their sine wave equations.

Solution

The equation of the voltage is vC = 120 sin (t + 60) V and that of the current is iC = 5 sin (t +

150) A

6.3.2 Why the current leads the voltage by 90?

60

60

120 V

60

154

A sine wave voltage is shown in the figure below. Notice that the rate at which the voltage is

changing varies along the sine wave curve, as indicated by the “steepness” of the curve. At the zero

crossings, the curve is changing at a faster rate than anywhere else along the curve. At the peaks,

the curve has a zero rate of change because it has just reached its maximum and is at the point of

changing direction.

The amount of charge stored by a capacitor determines the voltage across it. Therefore, the rate

at which the charge is moved (Q/t = current) from one plate to the other determines the rate at

which the voltage changes. When the current is changing at its maximum rate ( at the zero

crossings), the voltage is at its maximum value (peak). When the current is changing at its

maximum rate (zero at the peaks), the voltage is at its minimum value (zero). This relationship is

illustrated in the figure. As you can see, the current peaks occur a quarter of a cycle before the

voltage peaks. Thus the current leads the voltage by 90.

6.3.3 Capacitive Reactance, XC

Capacitive reactance is the opposition to sinusoidal current, expressed in ohms and

symbolized by XC.

In Figure, a capacitor is connected to a sine wave voltage source. Note that when the source

voltage is held at a constant amplitude value and its frequency is increased, the amplitude of the

current increases. Also, when the frequency of the source is decreased, the current amplitude

Zero rate of change

Zero rate of change

Maximum negative

rate of change

(steepest)

Maximum negative

rate of change

(steepest)

155

decreases. The reason is as follows: When the frequency of the voltage increases, its rate of

change also increases. This relationship in the figure, where the frequency is doubled.

The current in a capacitive circuit varies directly with the frequency of the applied voltage.

Rate of change increases with the frequency.

Now, if the rate at which the voltage is changing increase, the amount of charge moving though the

circuit in a given period of time must also increase. More charge in a given period of time means

more current. For example, a tenfold increase in frequency means that the capacitor is charging and

discharging 10 times a much in a given time interval. Therefore, since the rate of charge movement

has increased 10 times, the recurrent must increases by 10 because I = Q/t.

An increase in the amount of current for a fixed amount of voltage indicates that opposition

to the current has decreased. Therefore, the capacitor offers opposition to current, which varies

inversely with frequency. The opposition to sinusoidal current called capacitive reactance. The

symbol for capacitive reactance is XC, and its unit is the ohm ().

You have just seen how frequency affects the opposition to current (capacitive reactance) in

a capacitor. Now let’s see how the capacitance affects the reactance. Figure shows that when a sine

wave voltage with a fixed amplitude and frequency is applied to a 1-F capacitor, a certain amount of

current flows. When the capacitance value is increased to 2 F, the current increases. Thus, when

the capacitance increases, the opposition to current (capacitive reactance) decreases. Therefore, not

B A

B has the

greater rate of

change (steeper

slope, more

cycles)

156

only is the capacitive reactance inversely proportional to frequency, but it is also inversely

proportional to capacitance:

XC is proportional to 1

fC

It can be proven that the constant of proportionality is 1/2. Therefore, the formula for XC is

XC = 1

2fC

where XC = the capacitive reactance in ohms ()

f = frequency in hertz (Hz)

C = capacitance in farad (F)

The 2 term comes from the fact that a sine wave can be described in terms of rotational motion,

and one revolution contains 2 radians.

Example 6.11 A sinusoidal voltage is applied to a capacitor. The frequency of the sine wave is 1

kHz. Determine the capacitive reactance.

Solution

XC = 1

2fC =

1

2(1 kHz )(0.005 F) = 31.83 k

6.3.4 Ohm’s Law in Capacitive Circuits

The reactance of a capacitor is analogous to the resistance of a resistor. In fact, both are expressed

in ohms. Since both are forms of opposition to current, Ohm’s law apply to capacitive circuits as well

as to resistive circuits and is stated as follows for capacitive reactance:

VC = IC XC

where VC = the rms voltage across the capacitor in volts (V)

IC = the rms current through an inductor in amperes (A)

Vs 0.005 µF

157

XC = the capacitive reactance in ohms ()

Example 6.12 Determine the rms current through the circuit shown below.

Solution

XC = 1

2fC =

1

2(10 kHz)(0.005 F) = 3.183 k

Applying Ohm’s law:

IC = VC (rm s)

XC =

5 V

3.183 k = 1.57 mA

6.3.5 Power in a Capacitor

As discussed earlier, a charged capacitor stores energy in the electric field within the dielectric. An

ideal capacitor does not dissipate energy; it only stores it. When an ac voltage is applied to a

capacitor, energy is stored by a capacitor during a portion of the voltage cycle. There is no net

energy loss. Figure below shows the power curve that results from one cycle of capacitor voltage

and current.

Vs = 5 V

10 kHz0.005 µF

Vs C

158

6.3.6 Instantaneous Power (p)

The product of instantaneous voltage, VC, and instantaneous current, IC, gives instantaneous power,

p. At point where vC or IC is zero, p is also zero. When both vC or IC is positive and the other negative,

p is negative. When both vC and IC are negative, p is positive. As you can see, the power follows a

sinusoidal-type curve. Positive values of power indicate that energy is stored by the capacitor.

Negative values of power indicate that energy is returned from the capacitor to the source. Note

that the power fluctuates at a frequency twice that of the voltage or current, as energy is alternately

stored and returned to the source.

6.3.7 True Power (P)

Ideally, all of the energy stored by a capacitor during the positive portion of the power cycle is

returned to the source during the negative portion. No net energy is consumed in the capacitor, so

the true power is zero. Actually, because of leakage and foil resistance in a practical capacitor, a

small percentage of the total power is dissipated. But in this lesson, the capacitor is considered as

ideal (meaning the capacitor has infinite resistance) so the true power is zero.

6.3.8 Capacitive Reactive Power (QC)

The rate at which a capacitor stores or returns energy is called its capacitive reactive power, QC. The

capacitive reactive power is a nonzero quantity, because at any instant in time, the capacitor is

actually taking energy from the source or returning energy to it. Reactive power does not represent

an energy loss. The following formulas apply:

QC = VC IC

QC = VC

2

XC

QC = IC2XC

where QC = capacitive reactive power in volt-ampere reactive (VAR or var)

-200

-150

-100

-50

0

50

100

150

200

-2 -1 0 1 2 3 4 5 6 7 8

iC

vC

p

Pm

90° 0° 90° 180° 270° 360°

VCm

Icm

159

IC = current through the capacitor

VC = rms voltage across the capacitor

XC = capacitive reactance

Example 6.13 Determine the true power and the reactive power in the circuit shown.

Solution

P = 0

XC = 1

2fC =

1

2(60)(110F) = 24.114

QC = VC

2

XC =

(110 V)2

24.114 = 501.78 VARS

110 V

60 Hz110 µF

160

Problem Set No. 12

AC CIRCUIT WITH RESISTANCE, INDUCTANCE, AND CAPACITANCE ONLY

1. Prove that the average value of power in a purely resistive load is P = VRIR. 2. An electric flatiron whose heating element is practically a pure resistance takes 480 watts when

connected across 115-V DC mains. Determine (a) power that it takes from 120-V 60-Hz mains; (b) its resistance; (c) equation of AC voltage, current, and power waves (draw waves), zero time being when voltage is going through zero and increasing positively; (d) maximum value of power; (e) instantaneous power when t is 1/480 sec. (f) Show graphically that average power over 1 cycle is product of rms volts and amperes.

3. A large room is illuminated by twenty 150-watt lamps and thirty 100-wattt lamps. If the circuit

voltage is 116, calculate the total current. 4. How much power is represented by a circuit in which the voltage and current equations are e =

160 sin314t and i = 42.5 sin 314t?

5. By applying integration, prove that the average power or true power in a purely inductive load or ideal inductor is equal to zero.

6. Derive XL = 2fL. 7. Calculate XL for f = 5 kHz and L = 100 mH. (Answer: 3141.59 Ω)

8. At what frequency is the reactance of a 50-H inductor equal to 800 ? (Answer: 2.54 kHz) 9. A 50-mH inductor is connected to a 12-V rms source. What is the true power? What is the

reactive power at a frequency of 1 kHz? (Answers: 0 W, 458.366 mH) 10. Calculate the rms current in the figure below. (Answer: 0.6366 A)

11. The emf and current waves of a circuit having inductance only are e = 141.4 sin 225t and i = -17

cos 225t. Determine (a) equation of power wave; (b) frequency of power wave; (c) maximum value of power wave; (d) average power; (e) power when t = 1/200 sec, 1/100 sec, 1/80 sec. (f)

Draw voltage, current, and power waves. (Answers: a. p = -1201.9 sin425t W, b. 50 Hz, c. 2403.8 W, d. 0 W, e. -1201.9 W, 0 W, 849.87 W)

E120 Vpk

60 Hz

0°

L500mH1

2

120 V

60 Hz

161

12. A pure inductance takes 4 amp from 120-V(rms) 60-Hz mains. Determine (a) equation of voltage

and current waves, zero time being when current is going through zero and increasing positively; (b) equation of power wave; (c) maximum instantaneous power; (d) average power; (e) maximum energy stored in inductance; (f) rate at which emf of self-induction is changing when t

= 1/240 sec. (g) Plot all three waves. (Answers: a. 169.71 sin (260 t + /2), 5.66 sin 260t, b.

480.28 sin 460t, (c) 960.56 W, (d) de/dt = L 13. A reactance coil whose resistance is negligible takes 1.060 amp from 120-V 60-Hz mains.

Determine (a) inductance; (b) current when connected to 220-V 60-Hz mains. 14. By applying integration prove that the average power or true power of an ideal capacitor is equal

to zero.

15. Prove that XC = 1

2fC

16. Calculate XC for f = 5 kHz and C = 50 pF. (Answer: 636.62 kΩ)

17. At what frequency is the reactance of a 0.1-F capacitor equal to 2 k? (Answer: 796 Hz) 18. Calculate the rms current for the circuit shown below. (Answer: 0.6283 A)

19. A 1-F capacitor is connected to an ac voltage source of 12 V rms. What is the true power? (Answer: 0)

20. In problem 19, determine reactive power at a frequency of 500 Hz. (Answer: 0.4524 VAR)

21. A capacitance of 4 F is connected across a 40-V (rms) 1,000 Hz power supply. Determine (a) rms current; (b) maximum instantaneous current; (c) equations of current and emf waves, zero time being when current is crossing zero axis in a positive direction; (d) maximum rate of

change of current and of emf. (Answer: a. 1.01 A b. 1.43 A, c. 1.43 sin 21000t, 56.57 sin

(21000t - /2), d.

22. The emf and current waves in a circuit having capacitance only are e = 311 sin 250t and i = 5.65

cos 250t. Determine (a) equation of power wave; (b) frequency of power wave; (c) maximum value of power wave; (d) average power; (e) power when t = 1/800 sec, 1/400 sec, 1/300 sec. (f)

Draw the waves. (Answers: a. 878.58 sin 450t, b. 100 Hz, c. 1757.15 W, d. 0, e. 621.25 W, 878.58 W, 760.87 W)

1 V

1 MHz0.1 µF

162

23. A 4-F capacitor is connected across an emf of 50 volts (rms), 1,000 Hz. Determine (a) equation of emf and current, zero time being when emf is crossing axis in a positive direction; (b) equation of power wave; (c) maximum instantaneous power; (d) average power; (e) maximum energy stored in capacitor; (f) maximum rate of change of current. (g) Plot all three waves.

24. In a power circuit it is desired to obtain a 90 leading current of 60 amp by the use of capacitors, the voltage supply being 600 volts, 60 Hz. (Capacitors are used on power systems to correct power factor.) Determine (a) required capacitance; (b) current that capacitance would take at 440 volts, 120 Hz.

25. It is desired to obtain 43.5 A at 2,300 volts, 60-Hz, by means of a capacitors. Determine (a)

necessary capacitance in microfarads; (b) kVA rating of capacitor.

163

6.4 Power Factor

The Power factor is the ratio of true power and apparent power. The power factor ranges from 0 to

1.

pf = P

S

where pf – is the power factor which can expressed as a decimal or percentage.

P – true power in watts (W)

S – apparent power in volt-amperes (VA). The apparent power is just the product of the total

voltage and the total current in an ac circuit.

From the power triangle below which will be learned later, it can be seen that P/S is cos . Therefore

pf = P

S = cos

6.4.1 Types of Power Factor

Unity pf (pf = 1) - the voltage and current are in phase. A purely resistive load has a unity power factor since the voltage and current are in phase. Any circuit whose total inductive reactive power is equal to the total capacitive reactive power will have a unity power factor.

Lagging pf (pf is between 0 to 1) – the current lags the voltage by an acute angle .

Leading pf (pf factor is between 0 to 1) – the current leads the voltage by an acute angle .

Zero pf – if the voltage and current are out of phase by exactly 90. Any circuit which has a zero true power has zero pf. The purely inductive or capacitive load would have a zero pf

since the true power is zero and the voltage and current are out of phase by exactly 90.

S Q

P

P

P – true power

Q – reactive power

S – apparent power

- the phase angle between the total

current and the total voltage or

power factor angle.

164

6.5 Series Resistance and Inductance

Characteristics:

1. There is a common current in the circuit, that is, IT = IR = IL. 2. The total voltage is the algebraic sum of the voltage across the resistor and across the

inductor. 3. the apparent power is the algebraic sum of the true power (power taken by the resistive

load) and the inductive reactive power in the inductor.

Omit the current wave, and drawing the total (resultant) voltage. In this case, the total or resultant

voltage is the source or applied voltage.

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iT

vL

vR90° 0° 90° 180° 270° 360°

VLm

VRm

ITm

Vs

R

L

165

The Phasor Diagram of a Series RL Circuit.

Note: The angle of the reference is not always zero. It can be any angle other than 0. The figure

below shows an example. This applies to the succeeding circuits.

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vR

vL

vT

VTm

VLm

VRm

90° 0° 90° 180° 270° 360°

VL VS

IT

VR (ref)

VL VS

VR

IT (ref)

166

VT = VR + jVL volts total voltage in rectangular form

VR = VT cos VL = VT sin

VT = VT volts total voltage in polar form

where VT - is the magnitude of the total voltage

VR – the voltage across the resistance of the load

VL - the voltage across the inductance of the load.

- the phase angle, the angle between the total voltage and the total current.

The magnitude of the total voltage determined by using Pythagorean Theorem.

VT = VR2 + VL

2

and

= tan-1 VL

VR

From the rectangular form of the total voltage, divide each parameter by I since it is common in the

circuit.

VT

IT =

VR

IR+ j

VL

IL

But, VT

IT = Z ,

VR

IR = R ,

VL

IL = XL

so that,

Z = R + jXL the rectangular form of the impedance

where 𝑍 – impedance in ohms ()

𝑅 - resistance in ohms ()

𝑋 – inductive reactance in ohms ()

6.5.1 What is impedance?

The impedance is the joint effect of combining resistive and reactance (inductive or

capacitive) in an AC circuit. Impedance is also defined as the total opposition to alternating current.

The figure below shows the impedance triangle of a series RL circuit.

167

The impedance triangle of a series RL circuit.

𝒁 = 𝑍 polar form of the impedance

where Z – the magnitude of the impedance of the circuit.

- the phase angle.

By Pythagorean Theorem, the magnitude of the impedance is,

𝑍 = R2 + XL2

And

= tan-1 XL

R which is the same as in the phasor diagram

now for the Power Triangle,

The Power Triangle in Series RL Circuit.

XL

R

XL

Z

R

VL VS

IT

VR (ref)

S

QL

P

168

Multiply the voltages by the current,

𝑉𝐿 𝑥 𝐼𝐿 = 𝑄𝐿

𝑉𝑅 𝑥 𝐼𝑅 = 𝑃

𝑉𝑇 𝑥 𝐼𝑇 = 𝑆

𝑺 = 𝑃 + 𝑗𝑄𝐿 the rectangular form of the apparent

power

𝑺 = 𝑆 the polar form of the apparent power

where 𝑆 = the magnitude of the apparent power

= power factor angle which is similar to the phasor diagram and impedance triangle.

The magnitude of the apparent power can be found by Pythagorean Theorem.

𝑆 = P2 + QL2

and = tan-1 QL

P which is the same as in the phasor diagram

Also,

𝑆 = P

cos =

P

pf

P = S cos = ITVTcos

Example 6.14 In the circuit below, determine the magnitudes of the following:

a. the impedance b. total current c. voltage across the resistor and inductor d. power, reactive power, and apparent power e. angle between the total current and the total voltage f. power factor

R = 75

Vs= 24 V

60 Hz

L = 290 mH

169

Solution

(a) XL = 2fL = 2(60 Hz)(290 mH) = 109.33

Z = (75)2 + (109.33)2 = 132.58

(b) IT = VT

Z =

24 V

132.58 = 181.02 mA = IR = IL

(c) VR = IRR = (181.02 mA)(75 ) = 13.58 V

VL = ILXL = (181.02 mA)(109.33 ) = 19.79 V

(d) P = IR2R = (181.02 mA)2 (75) = 2.46 W

or P = IRVR = (181.02 mA) (13.58) = 2.46 W

QL = IL2XL = (181.02 mA)2 (109.33) = 3.58 VAR

or QL = ILVL = (181.02 mA)(19.79) = 3.58 VAR

S = P2 + QL2 = (2.46)2 + (3.58)2 = 4.34 VA

(e)

= tan-1 VL

VR = tan-1 19.79

13.58 = 55.54

= tan-1 XL

R = tan-1 109.33

75 = 55.54

(f) pf = P

S =

2.46

4.34 = 0.567 or 56.7% lagging (because the current is lagging)

or cos = cos 55.55 = 0.567 lagging

Example 6.15 A 6-ohm resistor and an 8-ohm inductive reactance when connected in series across

a 60-Hz supply take 12 amp. Determine (a) impedance of circuit; (b) voltage across resistor; (c)

VL = 19.79 V

VS = 24 V

= 55.54

VR = 13.58 V IT (ref)

170

voltage across reactance; (d) circuit voltage; (e) power; (f) angle between current and voltage; (g)

power factor; (h) inductance. (i) Draw phasor diagram.

Solution

(a) Z = 62 + 82 = 10

(b) VR = IRR = (12 A)(6 ) = 72 V

(c) VL = ILXL = (12 A)(8 ) = 96 V

(d) VT = VR2 + VL

2 = (72)2 + (96)2 = 120 V

(e) P = IRVR = (12 A)(72) = 864 W (f)

= tan-1 VL

VR = tan-1

96

72 = 53.13

= tan-1 XL

R = tan-1

8

6 = 53.13

(g) pf = cos 53.13 = 0.6 or 60% lagging

(h) Phasor diagram:

R =6

Vs=?

60 Hz

XL = 8

12 A

VL =

96 V VS = 120 V

= 53.13

VR = 72 V IT (ref)

171

Example 6.16 the equation of the emf on an inductive circuit is e = 400 sin 377t and the current is i

= 40 sin (377t - 60). What is the inductance?

Solution:

𝑉𝑇 = 282.82 0

𝐼𝑇 = 28.28 − 60

𝑍 = 𝑉 𝑇

𝐼 𝑇

= 282.82 0

28.28 −60

= 10 Ω60

In rectangular form,

𝑍 = 5 + 𝑗8.66 Ω

so that R = 5 Ω and 𝑋𝐿 = 8.66 Ω

and 𝐿 = 𝑋𝐿

=

8.66

377 = 22.97 𝑚𝐻

Example 6.17 An induction motor which is connected to 120-V 50-Hz source draws a current of 5 A.

If the power factor of the motor is 95% lagging, find (a) the apparent power, (b) the true power that

it takes.

Solution:

(a) 𝑆 = 𝐼𝑇𝑉𝑇 = (5)(120) = 600 𝑉𝐴

(b) 𝑃 = 𝐼𝑇𝑉𝑇 𝑐𝑜𝑠 = 𝐼𝑇𝑉𝑇 𝑝𝑓 = 5 120 0.95 = 570 𝑊

172

Problem Set No. 13

SERIES RESISTANCE AND INDUCTANCE

1. A 0.0159-henry inductance coil and a 4-ohm resistor are connected in series across 240-V 60-Hz mains. Determine (a) reactance; (b) impedance; (c) current; (d) true power; (e) phase angle; (f) power factor; (g) voltage across resistor; (h) voltage across inductance coil. (i) Draw phasor diagram. (Answers: a. 5.99 Ω, b. 7.2 Ω, c. 33.33 A, d. 4443.56 W, e. 56.27°, f. 0.555 lagging, g. 133.32 V, h. 199.65 V)

2. When a 12-ohm resistor and an unknown inductance coil of negligible resistance are connected

in series across a 120-V 50-Hz supply, the current is 8 amp. Determine (a) reactance; (b) inductance; (c) phase angle; (d) power; (e) power factor; (f) voltage across resistor and across inductance coil. (g) Draw phasor diagram. (Answers: a. 9 Ω, b. 28.65 mH, c. 36.87°, d. 768 W, e. 0.8, f. 96 V, 72 V)

3. The corrected readings of a voltmeter, an ammeter, and wattmeter when connected to measure

the voltage, current, and power of a circuit known to consist only of resistance and inductance coil in series are as follows: volts, 118; amperes, 3.27; power, 320 watts. The frequency is 60 Hz. Determine (a) power factor; (b) circuit phase angle; (c) resistance; (d) reactance; (e) inductance; (f) voltage across resistance and inductance coil. (g) Draw phasor diagram. (Answers: a. 0.8293, b. 33.97°, c. 29.93 Ω, d. 20.16 Ω, e. 53.48 mH, f. 97.87 V, 65.92 V)

4. The primary of a telephone induction coil has an effective resistance of 60 ohms and an

inductance of 0.154. Determine (a) impedance at 1,000 Hz; (b) current that it takes when 50 volts at 1,000 Hz is impressed across it; (c) power to coil. (Answers: a. 969.47 Ω, b. 0.05157 A, c. 0.1596 W)

5. The current in a circuit known to consist only of resistance and inductance in series is 8.31 amp

when the circuit is connected across 120-V 25-Hz mains; when connected across 120-V 60-Hz mains the current is 5.30 amp. Determine the resistance and inductance. (Answers: 51.22 mH, 12 Ω)

6. The current in a series inductive circuit is 7.5 amp at 25 Hz. The circuit takes 425, watts, and the

power factor is 0.47. Determine (a) circuit voltage; (b) series inductance; (c) resistance. (d) Draw phasor diagram. (Answers: a. 120.57 V, b. 90.33 mH, c. 7.55 Ω)

.

173

6.6 Series Resistance and Capacitance

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0 1 2 3 4 5 6 7 8 9 10

iT

vR

vC0° 90° 180° 270° 360°

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0

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0 1 2 3 4 5 6 7 8 9 10

vR

vC

vT0° 90° 180° 270° 360°

VTm

VCm

VRm

Vs

R

C

174

Phasor Diagram of Series RC Circuit

VT = VR – jVC volts total voltage in rectangular form

VR = VT cos VC = VC sin

VT = VT volts total voltage in polar form

where VT = is the magnitude of the total voltage

VR = the voltage across the resistance of the load or resistor

VC = the voltage across the capacitance of the load or capacitor

= the phase angle, the angle between the total voltage and the total current.

The magnitude of the total voltage is determined by using Pythagorean Theorem.

VT = 𝑉𝑅2 + VC

2

and = tan-1 VC

VR

From the rectangular form of the total voltage, divide each parameter by I since it is common in the

circuit.

VT

IT =

VR

IR+ j

VC

IC

But, VT

IT = Z ,

VR

IR = R ,

VC

IC = XL

So that,

Z = R - jXC the rectangular form of the impedance

where Z = impedance in ohms ()

VR IT (ref)

VT

VC

175

R = resistance in ohms ()

XL = capacitive reactance in ohms ()

The impedance triangle of a series RC circuit.

Z = Z polar form of the impedance

where Z = the magnitude of the impedance of the circuit.

= the phase angle.

By Pythagorean Theorem, the magnitude of the impedance is,

Z = R2 + XC2

and = tan-1 XC

R which is the same as in the phasor diagram

Now for the Power Triangle,

Power Triangle in Series RL Circuit.

Multiply the voltages by the current,

VC x IC = QC

R

XC

R

Z

XC

VR IT (ref)

VC

VT

P

QC

S

176

VR x IR = P

VT x IT = S

S = P + jQC the rectangular form of the apparent power

S = S the polar form of the apparent power

where S = the magnitude of the apparent power

= power factor angle which is similar to the phasor diagram and impedance triangle. The magnitude of the apparent power can be found by Pythagorean Theorem.

S = P2 + QC2

and

= tan-1 QC

P

which is the same as in the phasor diagram

also,

S = P

co s =

P

pf

Example 6.16 In the circuit below, determine the magnitudes of the following:

a. the impedance b. total current c. voltage across the resistor and capacitor d. power, reactive power, and apparent power e. angle between the total current and the total voltage f. power factor

VR = 13.85 V IT(ref)

= 54.74

VC = VS =

19.6 V 24 V

24 V

60 Hz

75 ohms

25 µF

177

Solution

(a) XL = 1

2fL =

1

2(60 Hz)(25 F) = 106.1

Z = (75)2 + (106.1)2 = 129.93

(b) IT = 𝑉𝑇

𝑍 =

24 𝑉

129.93 = 184.71 mA = IR = IC

(c) VR = IRR = (184.71 mA)(75 ) = 13.85 V

VC = ICXC = (184.71 mA)(106.1 ) = 19.6 V

(d) P = IR2R = (184.71 mA)2 (75) = 2.56 W

or P = IRVR = (184.71 mA) (13.85) = 2.56 W

QL = IL2XL = (184.71 mA)2 (106.1) = 3.62 VAR

or QL = ILVL = (184.71 mA)(19.6) = 3.62 VAR

S = P2 + QL2 = (2.56)2 + (3.62)2 = 4.43 VA

= tan-1 VC

VR = tan-1

19.60

13.85 = 54.75

= tan-1 XL

R

= tan-1

106.1

75 = 54.74

(e) pf = 𝑃

𝑆 =

2.56

4.53 = 0.578 or 57.8% leading (because the current is leading)

or cos = cos 57.8 = 57.8%

Example 6.17 A 6-ohm resistor and an 8-ohm inductive reactance when connected in series across

a 60-Hz supply take 12 amp. Determine (a) impedance of circuit; (b) voltage across resistor; (c)

voltage across reactance; (d) circuit voltage; (e) power; (f) angle between current and voltage; (g)

power factor; (h) inductance. (i) Draw phasor diagram.

Solution

178

(a) Z = 62 + 82 = 10

(b) VR = IRR = (12 A)(6 ) = 72 V

(c) VC = ICXC = (12 A)(8 ) = 96 V

(d) VT = 𝑉𝑅2 + 𝑉𝐶

2 = (72)2 + (96)2 = 120 V

(e) P = IRVR = (12 A)(72) = 864 W

(f)

= tan-1 VC

VR = tan-1 96

72 = 53.13

= tan-1 XC

R = tan-1

8

6 = 53.13

(g) pf = cos 53.13 = 0.6 or 60% leading

(h) Phasor diagram:

VR = 72 V IT (ref)

= 53.13

VS = 120 V

VL = 96 V

Vs=?

60 Hz

6 ohms

8 ohms

12 A

179

Problem Set No. 14

SERIES RESISTANCE AND CAPACITANCE

1. A 50-ohm resistance and an 80-F capacitor are connected in series across 115-V 60-Hz mains. Determine (a) current; (b) power; (c) power factor; (d) voltage across resistor; (e) voltage across capacitor. (f) Draw phasor diagram. (Answers: a. 1.92 A, b. 184.32 W, c. 0.8333, d. 96 V, e. 63.67 V)

2. A current of 2 amp at 60 Hz flows in a circuit with the resistor and a capacitor in series. The

voltage across the resistor is 60 volts, and that across the capacitor is 90.8 volts. Determine (a) circuit voltage; (b) power; (c) power factor; (d) capacitance. (e) Draw vector diagram. (Answers: a. 108.83 V, b. 120 V, c. 0.5513, d. 58.43 µF)

3. A circuit with a resistor and a capacitor in series takes 200 watts at a power factor of 0.40 from

200-Volt 50-Hz mains. Determine (a) current; (b) power-factor angle; (c) resistance; (d) impedance; (e) capacitance. (Answers: a. 2.5 A, b. 66.42°, c. 32 Ω, d. 80 Ω, e. 43.41 µF)

4. A circuit with a resistor and a capacitor in series takes 3.0 amp, 216 watts, at 0.6 power factor

from a 60-Hz supply. Determine (a) resistance; (b) circuit voltage; (c) capacitive reactance; (d) capacitance; (e) power-factor angle. (Answers: a. 24 Ω, b. 120 V, c. 32 Ω, d. 82.89 µF, e. 0.6)

5. A 33-ohm resistor is in series with a 35.3-F capacitor across a constant potential source of 100 volts. Determine (a) frequency that will give current of 2.0 amp; (b) circuit power; (c) power factor. (Answers: a. 120 Hz, b. 132 W, c. 0.66)

6. A circuit with a 50-F capacitor and an adjustable resistor in series is connected across 120-V 60-Hz mains. To what value of ohms must the resistor should be adjusted for the circuit to take 80 watts? (Two values of resistance will satisfy this condition.) (Answers: 162.7 Ω, 17.3 Ω)

180

6.7 Series Resistance, Inductance, and Capacitance

E

12 Vrms

60 Hz

0°

R1.0kΩ

L1.0mH

C1.0µF

123

4

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iT

vR

vL

vC

90° 0° 90° 180° 70° 360°

VLm

VCm

VRm

ITm

181

To find the resultant or total voltage which is also the voltage source, VS.

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VL

VR I (ref)

VC

182

The phasor diagram if VL > VC

VX = the net reactive voltage.

Formulas

VT = VR + jVL - jVC the rectangular form of the total voltage

VT = VR + j(VL - VC)

VT = VR + j VX where VX = VL - VC

VT = VT the polar form of the total voltage

where

VT = 𝑉𝑅2 + 𝑉𝑥

2

= tan-1 VX

VR

VX = net reactive voltage

From VT = VR + jVL - jVC we can divide this by the total current since it is common in the circuit , that

is, IT = IR = IL = IC

VT

IT

= VR

IR

+ j VL

IL− j

VC

IC

where

VT

IT = Z ,

VR

IR = R ,

VL

IL = XL ,

VC

IC = XC

VL

VX = VL - VC

VR I (ref)

VC

183

So that

Z = R + jXL - jXC the rectangular form of the impedance

Z = R + j(XL - XC)

Z = R + jXnet Xnet = XL - XC

Z = Z

where Z = 𝑅2 + 𝑋𝑛𝑒𝑡2

= tan-1 Xne t

R

Xnet = net reactance

For the power triangular, multiply each quantity in the phasor diagram by the each corresponding

current.

XL

Xnet = XL - XC

R I (ref)

XC

VL

VR I (ref)

VC

QL

P

QC

184

S = P + jQL - jQC rectangular form of the apparent power

S = P + j(QL - QC)

S = P + jQnet where Qnet = QL - QC = net reactive power

S = S polar form of the apparent power

where S = 𝑃2 + 𝑄𝑛𝑒𝑡2

= tan-1 Qne t

P

Now if VL < VC, the formulas above, that is for VL > VC, also apply. But the VT and VX will have a

different direction as shown below. The Power factor here is leading because the total current is

leading the total voltage by an angle .

The Phasor Diagram if VC > VL

QL

S

Qnet = QL - QC

P

QC

VL

VR I (ref)

VX = VC – VL

V

T

VC

185

Example 6.18 In the circuit below, determine the magnitudes of the following:

(a) the impedance (b) total current (c) voltage across the resistor, inductor and capacitor (d) power, reactive power (inductive and capacitive), and apparent power (e) angle between the total current and the total voltage (f) power factor

Solution

(a) impedance

By complex quantities we can get the rectangular and polar form of the impedance,

Z = 50 + j48.63 – j22.10 = 50 + j26.53

Converting this to polar form

Z = 56.6 27.95

If the required is only the magnitude, Z = 56.6

Or use the formula for the magnitude of the impedance

Z = R2 + Xne t2

where Xnet = XL - XC

Vs= 110 V

60 Hz

50 ohms

120 µF

L = 129 mH

XL = 48.63

R = 50

XC = 22.10

XL = 48.63

Xnet = Z = 56.6

26.53 = 27.95

R = 50

XC = 22.10

186

Z = R2 + (XL − XC ) 2

Z = 502 + (48.63 − 22.10) 2

= 56.6

(b) IT = VT

Z

=

110 V

56.6

= 1.94 A

(c) VR = IRR = (1.94 A)(50 ) = 97 V

VL = ILXL = (1.94 A)(48.63 ) = 94.34 V

VC = ICXC = (1.94 A)(22.10 ) = 42.87

(d) P = IR2

R = (1.94 A)2 (50 ) = 188.18 W

QL = IL2

XL = (1.94 A)2 (48.63 ) = 183.02 VARs

QC = IC2

XC = (1.94 A)2 (22.10 ) = 83.18 VARs

S = P2 + Qne t2 = P2 + (QL − QC ) 2

= (188.18)2 + (183.02 − 83.18) 2 = 213.03 VA

(e) To find , we can use any of these formulas, the value of = 27.95

= tan-1 𝑋𝐿− 𝑋𝐶

𝑅

= tan-1 VL− VC

VR

= tan-1 QL− QC

P

(f) pf = cos 27.95 = 0.883 or 88.3% lagging; since the current is lagging the total voltage as seen

in the phasor diagram.

VL = 94.34 V

VT = 110 V

Vnet =

51.47 V = 27.95

R = 50

VC = 42.87 V

I (ref)

187

Problem Set No. 15

SERIES RESISTANCE, INDUCTANCE AND CAPACITANCE

1. A series circuit with 12 ohms resistance, 32 ohms inductive reactance, and 20 ohms capacitive reactance is connected across 240-volt 60-Hz mains. Determine (a) impedance; (b) current; (c) voltage across each circuit element; (d) power; (e) power factor and a power-factor angle. (f) Draw phasor diagram. (Answers: a. 16.97 Ω, b. 14.14 A, c. 169.68 V, 452.48 V, 282.8 V, d. 2399.28 W, e. 0.7071, 45°)

2. A voltage of 220 volts at 60 Hz is impressed on a circuit having a 50-ohm resistor, 25-F capacitor, and a 0.2-henry inductor in series. Determine (a) impedance; (b) current; (c) voltage across resistor, inductor, capacitor; (d) total power; (e) power factor and power factor angle. (f) Draw phasor diagram to scale. (Answers: a. 58.67 Ω, b. 3.75 A, c. 187.5 V, 282.75 V, 397.88 V, d. 703.125 W, e. 0.8522, 31.55°)

3. A 15-ohm resistor, a 0.25-henry inductor, and a 100-F capacitor are connected in series across 200-volt 25-Hz mains. Determine (a) current; (b) power; (c) voltage across resistor, inductor, capacitor; (d) apparent power; (e) power factor and power factor angle. (f) Draw phasor diagram to scale. (Answers: a. 7 A, b. 735 W, c. 105 V, 274.89 V, 445.62 V, d. 1400 VA, e. 0.5248, 58.34°)

188

6.8 Parallel Resistance and Inductance

Let VT = VT0; since this is a parallel circuit, VR = VL = VT = VT0;

IR = 𝑉𝑅

𝑅 =

𝑉𝑇 0

𝑅

-40

-30

-20

-10

0

10

20

30

40

0 1 2 3 4 5 6 7 8

vT

iR

iL0° 90° 180° 270° 360°

VTm

ILm

IRm

VsR L

IR VT (ref) IR VT (ref)

IL

IT

189

IL = 𝑉𝐿

𝑋𝐿 =

𝑉𝑇 0

𝑋𝐿 90 =

𝑉𝑇 0

𝑗𝑋𝐿

Based on the phasor diagram above

IT = IR - jIL the rectangular form of the total voltage

IT = IT the polar form of the total voltage

where

IT = 𝐼𝑅2 + 𝐼𝐿

2

= tan-1 𝐼𝐿

𝐼𝑅

From IT = IR - jIL Divide by VT or by the corresponding voltage,

𝐼𝑇

𝑉𝑇=

𝐼𝑅

𝑉𝑅 − 𝑗

𝐼𝐿

𝑉𝐿

where 𝐼𝑇

𝑉𝑇 = admittance (Y ) in siemens (S)

𝐼𝑅

𝑉𝑅 = conductance (G ) in siemens (S)

𝐼𝐿

𝑉𝐿 = inductive susceptance (BL) in siemens (S)

Y = G - jBL

Y = Y polar form of admittance

Where Y = 𝐺2 + 𝐵𝐿2

= tan-1 𝐵𝐿

𝐺

Power Factor: pf = cos lagging (the total current is lagging the total voltage in this circuit by an

angle ).

Admittance, Conductance, and Susceptance

Admittance – the reciprocal of admittance; the property of a circuit to allow the flow of alternating

current.

Y = 1

𝑍

190

Conductance – the reciprocal of resistance; the property of a circuit to allow the flow of current.

G = 1

𝑅

Susceptance - the reciprocal of reactance.

BL – inductive susceptance; the ability of inductor to allow the flow of alternating current

BC – inductive susceptance; the ability of a capacitor to allow the flow of alternating current

BL = 1

𝑋𝐿 BC =

1

𝑋𝐶

6.9 Parallel Resistance and Capacitance

-40

-30

-20

-10

0

10

20

30

40

-2 -1 0 1 2 3 4 5 6 7 8

vT

vR

vC90° 0° 90° 180° 270° 360°

VTm

ICm

IRm

E

12 Vrms

60 Hz

0°

R1.0kΩ

C1.0µF

24

191

Let VT = VT0; since this is a parallel circuit, VR = VC = VT = VT0;

IR = 𝑽𝑹

𝑹 =

𝑉𝑇 0

𝑅

IC = 𝑽𝑹

𝑿𝑪 =

𝑉𝑇 0

𝑋𝐶−90 =

𝑉𝑇 0

−𝑗𝑋𝐶

Based on the phasor diagram above

IT = IR + jIC the rectangular form of the total voltage

IT = IT the polar form of the total voltage

where

IT = 𝐼𝑅2 + 𝐼𝐶

2

= tan-1 𝐼𝐿

𝐼𝐶

From IT = IR + jIC Divide by VT or by the corresponding voltage,

𝐼𝑇

𝑉𝑇=

𝐼𝑅

𝑉𝑅+ 𝑗

𝐼𝐶

𝑉𝐶

where 𝐼𝑇

𝑉𝑇 = admittance (Y ) in siemens (S)

𝐼𝑅

𝑉𝑅 = conductance (G ) in siemens (S)

𝐼𝐶

𝑉𝐶 = capacitive susceptance (BC) in siemens (S)

Y = G + jBC rectangular form of admittance

Y = Y polar form of admittance

where Y = 𝐺2 + 𝐵𝐶2

IC IC IT

IR VT (ref) IR VT (ref)

192

= tan-1 𝐵𝐶

𝐺

Power Factor: pf = cos leading (because the total current is leading the total voltage by an angle

)

6.10 Parallel Resistance, Inductance, and Capacitance

E

12 Vrms

60 Hz

0°

R1.0kΩ

L1.0mH C

1.0µF

24

-40

-30

-20

-10

0

10

20

30

40

-2 -1 0 1 2 3 4 5 6 7 8

vT

vR

vC

vL

90° 0° 90° 180° 270° 360°

VTm

ILm

ICm

IRm

193

Let VT = VT0; since this is a parallel circuit, VR = VL = VC = VT0;

𝑰𝑹 = 𝑽𝑹

𝑹 =

𝑉𝑇0

𝑅

𝑰𝑳 = 𝑽𝑳

𝑿𝑳 =

𝑉𝑇0

XL 90 =

𝑉𝑇0

+jXL 90

𝑰𝑪 = 𝑽𝑪

𝑿𝑪 =

𝑉𝑇0

XC 90 =

𝑉𝑇0

−jXC 90

Based on the phasor diagram above

𝑰𝑻 = 𝐼𝑅 + 𝑗𝐼𝐶 − 𝑗𝐼𝐿 the rectangular form of the total voltage

𝑰𝑻 = 𝐼𝑅 + 𝑗(𝐼𝐶 − 𝐼𝐿)

𝑰𝑻 = 𝐼𝑇 the polar form of the total voltage

where

𝐼𝑇 = 𝐼𝑅2 + 𝐼𝐶 − 𝐼𝐿 2

𝐼𝑇 = 𝐼𝑅2 + 𝐼𝑛𝑒𝑡 2 where 𝐼𝑛𝑒𝑡 (𝑛𝑒𝑡 𝑟𝑒𝑎𝑐𝑡𝑖𝑣𝑒 𝑐𝑢𝑟𝑟𝑒𝑛𝑡) = 𝐼𝐶 − 𝐼𝐿

= 𝑡𝑎𝑛−1 𝐼𝑛𝑒𝑡

𝐼𝑅

From 𝑰𝑻 = 𝐼𝑅 + 𝑗𝐼𝐶 − 𝑗𝐼𝐿 Divide by VT or by the corresponding voltage,

IC IC

IT

IR VT (ref) IR VT (ref)

IL IL

194

𝑰𝑻𝑽𝑻

= 𝐼𝑅𝑉𝑅

+ 𝑗𝐼𝐶𝑉𝐶

− 𝑗𝐼𝐿𝑉𝐿

𝒀 = 𝐺 + 𝑗𝐵𝐶 − 𝑗𝐵𝐿

where 𝑰𝑻

𝑽𝑻 = admittance (Y ) in siemens (S)

𝐼𝑅

𝑉𝑅 = conductance (G ) in siemens (S)

𝐼𝐶

𝑉𝐶 = capacitive susceptance (BC) in siemens (S)

𝐼𝐿

𝑉𝐿 = inductive susceptance (BL) in siemens (S)

𝑌 = 𝐺 + 𝑗𝐵𝐶 − 𝑗𝐵𝐿

𝑌 = 𝐺 + 𝑗(𝐵𝐶 − 𝐵𝐿)

𝑌 = 𝐺 + 𝑗𝐵𝑛𝑒𝑡 𝑤𝑕𝑒𝑟𝑒 𝐵𝑛𝑒𝑡 is the net susceptance

𝑌 = 𝑌

where 𝒀 = 𝐺2 + 𝐵𝑛𝑒𝑡 2

= 𝑡𝑎𝑛−1 𝐵𝑛𝑒𝑡

𝐺

6.11 True Power, Reactive Power, and Apparent Power

𝑺 = 𝑃 + 𝑗𝑄𝐿 − 𝑗𝑄𝐶 rectangular form of the apparent power

𝑺 = 𝑃 + 𝑗(𝑄𝐿 − 𝑄𝐶) where 𝑄𝑛𝑒𝑡 = 𝑄𝐿 − 𝑄𝐶 = net reactive power

𝑺 = 𝑆 polar form of the apparent power

where 𝑆 = 𝑃2 + 𝑄𝑛𝑒𝑡2

= 𝑡𝑎𝑛-1 𝑄𝑛𝑒𝑡

𝑃

195

These directions of P, QL and QC are applicable to any type of circuit

Power factor: pf = cos leading (because the total current leads the total voltage by an angle ).

Note: These formulas can be used even if BC < BL only that that in this case the power factor is

lagging.

Example 6.19 A 24-ohm resistor and a 0.0796-henry inductor are connected in parallel across 115-

volt 60-Hz mains. Determine (a) current in resistor; (b) current in inductor; (c) total current; (d)

power factor; (e) power-factor angle. (f) Draw phasor diagram.

Solution

VT = VR = VL = 115 V

XL = 2fL = 2(60Hz)(0.0796 H) = 30

(a) 𝐼𝑅 = 𝑉𝑅

𝑅 =

115 𝑉

24 = 4.79 A

(b) 𝐼𝐿 = 𝑉𝐿

𝑋𝐿 =

115 𝑉

30 = 3.83 A

QL

P

QC

0.0796 H24 ohms115 V

60 Hz

196

(c) 𝐼𝑇 = 𝐼𝑅2 + 𝐼𝐿

2 = (4.79)2 + (3.83)2 = 6.13 𝐴

or 𝑰𝑻 = 𝐼𝑅 – 𝑗𝐼𝐿 = 4.79 – 𝑗3.83 𝐴. Converting to polar form IT = 6.13 A-38.65

The magnitude of the total current is IT = 6.13 A

(d) = tan-1 𝐼𝐿

𝐼𝑅

= tan-1 3.83

4.79 = 38.65

(e) pf = cos = cos 38.65 = 0.78

(f) Phasor diagram

Example 6.20 An 80-ohm resistor and a 4.0-F capacitor are connected in parallel across a 240 volt

400-Hz supply. Determine (a) current in resistor; (b) current in capacitor; (c) total current; (d)

power-factor angle; (e) power-factor. (f) Draw phasor diagram.

VT = VR = VC = 115 V

𝑋𝐶 = 1

2fC =

1

2(400 Hz)(4 F) = 99.47

(c) 𝐼𝑅 = 𝑉𝑅

𝑅 =

240 V

80 = 3 A

(b) 𝐼𝐶 = 𝑉𝐶

𝑋𝐶 =

240 V

99.47 = 2.41 A

80 ohms240 V

400 Hz 4 µF

IR = 4.79 A VT(ref)

= 38.65

IL =

3.83 A IT = 6.13 A

197

𝐼𝑇 = 𝐼𝑅2 + 〰𝐶

2 = (3)2 + (2.41)2 = 3.85 𝐴

Or 𝑰𝑻 = 𝐼𝑅 + 𝑗𝐼𝐶 = 3 + 𝑗2.41𝐴. Converting to polar form IT = 3.85 A38.78

The magnitude of the total current is IT = 3.85 A

(e) = tan-1 𝐼𝐶

𝐼𝑅

= tan-1 2.41

3 = 38.78

pf = cos = cos 38.78 = 0.78

(f) Phasor diagram

Example 6.21 A 25-ohm resistor, a 0.1-henry inductor, and a 160-F capacitor are connected in

parallel across 200-volt 25-Hz mains. Determine (a) current to resistor, inductor, capacitor; (b) total

current; (c) power-factor angle; (d) power factor. (e) Draw vector diagram.

Solution VT = VR = VL = VC

XL = 2fL = 2(25 Hz)(0.1 H) = 15.71

𝑋𝐶 = 1

2fC =

1

2(25 Hz)(160 F) = 39.79

(a) 𝐼𝑅 = 𝑉𝑅

𝑅 =

200 V

25 = 8 𝐴

𝐼𝐿 = 𝑉𝐿

𝑋𝐿 =

200 V

15.71 = 12.73 A

IR = 4.79 A VT(ref)

= 38.65

IL =

3.83 A IT = 6.13 A

IC = 2.41 A IT = 3.85 A

= 38.78

IR = 3 A VT(ref)

25 ohms200 V

25 Hz 160 µF0.1 H

198

𝐼𝐶 = 𝑉𝐶

𝑋𝐶 =

200 V

39.79 = 5.03 A

(b) 𝐼𝑇 = 𝐼𝑅2 + 𝐼𝐶 − 𝐼𝐿

2 = (8)2 + 5.03 − 12.73 2 = 11.10 𝐴

or 𝐼𝑇 = 𝐼𝑅 + 𝑗𝐼𝐶 − 𝑗𝐼𝐿 = 8 + 𝑗5.03 − 𝑗12.73 = 8 – 𝑗7.7 𝐴

Converting to polar form 𝑰𝑻 = 11.10 − 43.91

So , 𝐼𝑇 = 11.10 𝐴

(c) = 43.91

(d) pf = cos 43.91 = 0.72

6.12 Conversion of Parallel or Combination Circuits to Series Circuit

Example 6.22 Transform the circuit below to series circuit:

XL = 48.63 BL = 1

48.63 = 20.56 mS

Vs= 110 V

60 Hz 50 ohms 120 µFL = 129 mH

IL = 5.03 A

IR = 8 A VT (ref)

= 43.79

Inet = 7.7 A

IT = 11.10 A

IL = 12.73 A

199

XC = 22.10 BC = 1

22.10 = 45.25 mS

R = 50 G = 1

50 = 20 mS

Y = G + jBC - jBL = 20 mS + j45.25mS - j20.56 = 20 mS + j24.69 mS

Converting to polar form

Y = 31.774 mS 50.99

where 𝒁 = 𝟏

𝒀 =

1

31.774 mS 50.99 = 31.47 − 50.99

Converting Z to rectangular form

𝒁 = 19.81 − 𝑗 24.45

In this form R = 19.81 and XC = 24.45 (capacitive reactance because the operator is –j)

To find the capacitance;

𝐶 = 1

2𝑓𝑋𝐶 =

1

2(60)(24.45) = 108.49 F

Therefore, the equivalent series of the parallel RLC above is a series components that consists R =

19.81 and capacitance of C = 108.49 F

Example 6.23 Determine the series components of an impedance 60 60 connected across a

60-Hz source.

Solution Given Z = 60 60

Convert to rectangular form Z = 30 + j51.96

From this, R = 30 and XL = 51.96 (inductive reactance because the operator is +j)

For 𝐿 = 𝑋𝐿

2𝑓 =

51 .96

260 = 137.83 𝑚𝐻

Therefore the series components are R = 30 and L = 137.83 Mh

Vs= 110 V

60 Hz

19.81 ohms

108.49 µF

200

Problem Set No. 16

PARALLEL CIRCUITS

1. A 40-ohm resistor and an inductor are connected in parallel across 120-volt 50-Hz mains, and the total current is 6.0 amp. Determine the inductance. (Answer: 73.46 mH)

2. A circuit consists of a resistor and a 53.0-F capacitor in parallel across 120-volt 60-Hz mains. The total current is 4 amp. Determine ohms of resistor. (Answer: 37.5 Ω)

3. A 50-ohm resistor, an 80-ohm inductive reactor, and a 60-ohm capacitive reactor are connected in parallel across 240-volt 60-Hz mains. Determine (a) current to resistor, inductor, capacitor; (b) total current; (c) power-factor angle; (d) power factor. (e) Draw phasor diagram. (Answers: a. 4.8 A, 3 A, 4 A, b. 4.9 A, c. 11.77°, d. 0.979)

4. A 30-ohm resistor and a 0.0637-henry inductor are connected in parallel across a 120-V ac supply. Determine (a) frequency at which total current will be 7.22 amp; (b) power factor of the circuit. (c) Draw phasor diagram. (Answers: a. 50 Hz, b. 0.554)

201

Unit 7 Combination Circuits and

Network Theorems

7.1 Combination Circuits

Example 7. 1. A non-inductive resistor and an impedance coil are connected in parallel across 208-volt 60-Hz mains. The resistor takes a current of 2.25 amp; the impedance coil takes a current of 1.5

1. solve the required parameters of a series/parallel combination circuit. 2. analyze a circuit using the network theorems such as Maxwell’s Mesh

Equations, Nodal Analysis, Thevenin’s Theorem, Norton’s Theorem, Superposition Theorem, and Millman’s Theorem

LEARNING OUTCOMES

At the end of the lesson, you

are expected to:

202

amp; the total current is found to be 3.1 amp. Determine (a) power-factor angle and power factor of circuit; (b) power to circuit; (c) power-factor angle and power factor of impedance coil; (d) effective resistance of impedance coil; (e) inductance of impedance coil.

The equivalent components of impedance coil are resistance and inductance.

Let VT = 208 V0

To find circuit , let us consider this triangle

Using cosine law,

(1.5)2 = (2.25)2 + (3.1)2 – 2(2.25)(3.1)coscircuit

R = ?208 V

60 HzLcoil =?

Rcoil = ?

3.1 A

IR = 2.25 A Icoil = 1.5 A

IR = 2.25 A VT (coil)

circuit coil

coil equal to Icoil

Icoil = 1. 5 A IT = 3.1 A

IR = 2.25 A

circuit

IT = 3.1 A Icoil = 1.5 A

203

circuit = 27.06

(a) circuit = 27.06; pf = cos circuit = cos 27.06 = 0.89 lagging

(b) Let us find S: S = VTIT = (208)(3.1) = 644.8 VA

The figure below is the corresponding power triangle of the circuit since the current is lagging.

cos = 𝑃

𝑆 and P = S cos = Spf = 644.8 (0.89) = 573.87 watts

(c) To find coil, consider again this triangle

Using cosine law,

(3.1)2 = (2.25)2 + (1.5)2 - 2(2.25)(1.5)cos

= 109.9

Now coil = 180 - = 180 - 109.9 = 70.1

pfcoil = coscoil = cos 70.1 = 0.34

(d) To find VRcoil and VLcoil , let us consider this triangle

S = 644.8 VA QL = ?

= 27.06

P = ?

IR = 2.25 A

coil

IT = 3.1 A Icoil = 1.5 A

Vcoil = 208 V VL coil

coil = 70.1

VR coil

204

𝑉𝑅𝑐𝑜𝑖𝑙 = 𝑉𝑐𝑜𝑖𝑙 𝑐𝑜𝑠𝑐𝑜𝑖𝑙

= 208 𝑐𝑜𝑠 70.1 = 70.8 𝑉

𝑉𝐿𝑐𝑜𝑖𝑙 = 𝑉𝑐𝑜𝑖𝑙 𝑠𝑖𝑛𝑐𝑜𝑖𝑙

= 208 𝑠𝑖𝑛 70.1 = 195.58 𝑉

𝑅𝑐𝑜𝑖𝑙 = 𝑉𝑅𝑐𝑜𝑖𝑙

𝐼𝑐𝑜𝑖𝑙 =

70.8 V

1.5 A = 47.2

𝑋𝐿 𝑐𝑜𝑖𝑙 = 𝑉𝐿𝑐𝑜𝑖𝑙

𝐼𝑐𝑜𝑖𝑙 =

195.58 V

1.5 A= 130.39

and 𝐿 = 𝑋𝐿

2f =

130.39

2(60) = 0.3459 𝐻

Example 7.2 A unity power-factor load of 1,794 watts is connected in parallel with a load of 1,656

watts operating at a lagging power factor of 0.6. If the line voltage is 115, calculate (a) the current in

each load, (b) the load current, (c) the total power, (d) the overall power factor, (e) the reactive

volt-amperes.

Note: Let

Solution:

Load 1: unity power factor – as defined a load with unity power factor is a purely resistive load.

P1 = IR1 VR1

Lo

ad

1

Vs = 115 V

Lo

ad

2

I1 = 15.6 A VT (ref)

1 =

53.13

I2 = 24 A IT = ?

205

where P1 = 1,794 watts

VR1 = VS = 115 V, since resistance is the only component in load 1

IR = ?

𝐼1 = 𝑃1

𝑉1 =

1,794 W

115 𝑉= 15.6 𝐴

Load 2: lagging power factor ; the components are resistor and inductor.

P2 = 1,656 watts pf = 0.6, with this 2 = cos-1 0.6 = 53.13

The power triangle for load 2

To find S2: 楬2 = 𝑃2

co s 2 =

1,656 W

co s 53.13 = 2,760 VA

S2 = I2V2

𝐼2 = 𝑆2

𝑉2 =

2,760 VA

115 V= 24 𝐴

I2 = 24 A-53.13 (negative because this current is lagging the total voltage)

QL2 = S2 sin 2 = 2760 sin 53.13 = 2,208 VAR

For the total current

IT = I1 + I2

= 15.60 + 24 A-53.13

= (15.6 + j0) + (14.4 – j19.2)

= 30 – j19.2

= 35.62-32.62

QL2 = ?

S2 = ?

= 53.13

P = 1,656 W

206

The equivalent power triangle of the circuit:

PT = 3,450 W

T = tan-1 2,208

3,450

= 32.62

QL2 = QT = 2,208 VARs

Example 7.3 In the circuit shown below, let Z1 = 50 + j75, Z2 = 80 + j25, Z3 = 100 + j70, determine :

a. the total impedance b. the total current c. the total power d. the total reactive power e. apparent power f. the current in each impedance

QL = 2,208 VAR

T = ?

P1 = 1,794 watts P2 = 1,656 watts

PT = 3,450 watts

120 V

60 Hz

50 ohms

75 ohms

80 ohms

25 ohms

100 ohms

70 ohms

I1 I3

I2

207

a. 𝒁𝑻 = 𝒁𝟏 + 𝒁𝟐𝒁𝟑

𝒁𝟐+ 𝒁𝟑 = 50 + 𝑗 75 +

80+𝑗25 100−𝑗70

80+𝑗25 + 100−𝑗70 = 105.03 + 𝑗71.536

= 127.0775 34.2588

b. 𝑰𝑻 = 𝑽𝑻

𝒁𝑻=

1200

127.0775 34.2588= 0.9443 𝐴 − 34.2588

c. 𝑃 = 𝐼𝑇2𝑅𝑇 = 0.9443 2 105.03 = 93.656 𝑊

d. 𝑸 = 𝐼𝑇2𝑋𝐶 = 0.9443 2 71.536 = 63.789 𝑘𝑉𝐴𝑅 (𝑛𝑑𝑢𝑐𝑡𝑖𝑣𝑒)

e. 𝑺 = 𝐼𝑇 𝑉𝑇 = 0.9443 120 = 113.315 VAR

f. 𝑰𝟏 = 0.9443 𝐴 − 34.2588

By current divider

𝑰𝟐 = 0.9443 𝐴 − 34.2588 100−𝑗70

180−𝑗45 = 0.6212 𝐴 − 55.2146

𝑰𝟑 = 0.9443 𝐴 − 34.2588 80+𝑗25

180−𝑗45 = 0.4265 𝐴 − 2.8686

208

Problem Set No. 17

COMBINATION CIRCUITS

1. A 25-ohm resistor and an unknown capacitor are connected in parallel across a 100-volt 50-Hz supply, and the total current is 4.75 amp. When the resistor and capacitor are connected across a 100-volt supply of unknown frequency, the current is 6.41 amp. Determine frequency.

2. In order to measure the power taken by a small 120-V 60-Hz single-phase induction motor, it is connected in parallel with a non-inductive resistor across 120-V 60-Hz mains. The currents measured are as follows: resistor current, 3.0 amp; motor current, 4.2 amp; total current, 6.7 amp. Determine (a) power factor of circuit; (b) power factor of motor; (c) power to motor; (d) total power to circuit.

3. A parallel circuit consisting of a resistor, an impedance coil, and a capacitor of negligible loss is connected across a 100-volt 25-Hz supply and takes a current of 3.1 amp. The current to the resistor is 2.5 amp, that of the capacitor 2.0 amp and that to the impedance coil, 2.8 amp, Determine (a) power-factor angle and power of entire circuit; (b) power factor and power-factor angle of impedance coil; (c) resistance of impedance coil; (d) reactance and inductance of impedance coil.

4. A parallel circuit consisting of a non-inductive resistor, an impedance coil, and a capacitor of

negligible loss is connected across 120-V 60-Hz mains and takes 6 amp lagging current at a power factor of 0.936. The resistor takes 4 amp., and the capacitor 3 amp. Determine (a) power to the impedance coil; (b) current to impedance coil; (c) power-factor angle and power factor of impedance coil.

5. The following information is given in connection with three loads connected to a 220-V source: load A = 12 A at unity power factor; load B is 40 amp at a power factor of 0.5 lagging; load C is 20 amp at a leading power factor of 0.9. Calculate (a) the total current, (b) the total power, (c) the reactive volt-amperes, (d) the overall power factor.

R motor = ?

L motor

R = ?120 V

60 Hz

IT = 6.7 A

Im = IR =

4.2 A 3 A

209

6. With a secondary load, the primary of a telephone induction coil, having an effective

resistance of 240 ohms and an inductance of 0.01583 henry, is in parallel with a 2-F capacitor of negligible resistance. With 50 volts at a frequency of 1,000 Hz across the primary, determine (a) current to primary; (b) current to capacitor; (c) line, or total, current; (d) power to entire circuit; (e) phase angle between line current and voltage.

7. The currents at the junction point in a circuit have the following values I1 = (16 – j4) A and (8 + j22) A. Calculate (a) I1 + I2, (b) I1 – I2, and (c) I2 – I1.

8. Calculate the equivalent impedance of a circuit in which a coil of wire having a value of 5

53.2 is connected parallel with a capacitive reactance of 6.25 ohms.

9. A coil of wire having a value of (5 + j8) ohms is connected in series with a capacitive reactance XC, and this series combination is then connected in parallel with a resistor R. If the equivalent

impedance of the circuit is 4 0, calculate the values of XC and R.

10. When a small ac motor is operating at rated load from a 115-V 60-Hz source, it takes 287 watts at a lagging power factor of 0.6. (a)

210

Objective Test No. 6

ALTERNATING CURRENT CIRCUITS

1. A sine wave voltage is applied across an inductor. When the frequency of the voltage is increased, the current a. decreases b. increases c. does not change d. momentarily goes to zero

2. An inductor and a resistor are in series with a sine wave voltage source. The frequency is set so

that the inductive reactance is equal to the resistance. If the frequency is increased then a. VR > VL

b. VL < VR c. VL = VR

d. VL > VR

3. A sine wave voltage is applied across a capacitor. When the frequency of the voltage is increased, the current a. increases

b. decreases

c. remains constant

d. ceases

4. A capacitor and a resistor are connected in series to a sine wave generator. The frequency is set so that the capacitive reactance is equal to the resistance and, thus, an equal amount of voltage appears across each component. If the frequency is decreases, a. VR > VL

b. VL < VR

c. VL = VR

d. VL > VR

5. In an ac circuit, the voltage _____________. a. leads the current

b. lags the current

c. is in phase with the current

d. is any of the above, depending on the component

6. The voltage lags behind the current by ¼ cycle in a ___________ circuit. a. pure capacitor

b. pure inductor

c. pure resistor

d. capacitance with inductor

7. The unit of inductive reactance is the _________________. a. henry

211

b. Tesla

c. farad

d. ohm

8. When voltage and current are in phase in an ac circuit, the ____________. a. impedance is zero

b. resistance is zero

c. resistance is zero

d. phase angle is 90

9. The impedance of a circuit does not depend on ____________. a. current

b. frequency

c. resistance

d. capacitance

10. The power dissipated as heated in an ac circuit depends on its ____________. a. resistance

b. capacitive reactance

c. inductive reactance

d. impedance

11. The power factor of a circuit in which XL = XC ____________. a. is zero

b. is one

c. depends on the ratio XC/XL

d. depends on the value of R

12. The inductive reactance of a 1-mH coil in a 5-kHz circuit is ____________.

a. 3.1

b. 6.3

c. 10

d. 31

13. The capacitive reactance of a 5 F capacitor in a 20 kHz circuit is __________.

a. 0.63

b. 1.6

c. 5

d. 16

14. A 2-F capacitor is connected to a 50-V, 400-Hz power supply. The current that flows is a. 0.20 mA

b. 0.25 A

c. 0.30 A

d. 0.40 A

212

15. In a series ac circuit R = 10 , XL = 8 , and XC = 6 when the frequency is f. The impedance at this frequency is __________.

a. 10.2

b. 12

c. 24

d. 104

16. The total voltage in a series RL circuit ________ the current by an angle _____.

a. lags, of 90

b. lags between 0 to 90

c. leads, between 0 to 90

d. leads, between 90 and 180

17. In a series RL circuit, the inductor current _____ the resistor current. a. lags

b. leads

c. is equal

d. is negative

18. The impedance triangle is similar to the __________ triangle with the resistance phasor in place of the ____________.

a. current, resistor current

b. current, resistor voltage

c. voltage, impedance

d. voltage, resistor voltage

19. In the impedance triangle the inductive reactance and impedance phasor are analogous to the ______ and _______ phasor respectively in the voltage triangle. a. inductive voltage, total voltage

b. inductive current, total current

c. inductive voltage, resistive voltage

d. inductive current, resistive current

20. In a series RL circuit phasor diagram, total voltage may be represented by the _____ phasor and the resistor voltage may be represented the ___________ phasor. a. current, voltage

b. current resistance, current

c. impedance, resistance

d. impedance, inductance

21. The phase angle of a series RL circuit is the angle between the ________ phasor and the _________ phasor. a. resistance, inductive reactance

b. resistance, impedance

c. inductive reactance, impedance

d. none of these

213

22. The phase angle of series RL circuit may be computed as ________ or __________ or

___________. a. cos-1R/XL, sin-1XL/R, tan-1R/Z

b. cos-1R/Z, sin-1XL/R, tan-1R/XL

c. cos-1Z/XL, sin-1R/Z, tan-1XL/R

d. cos-1R/Z, sin-1XL/Z, tan-1 XL/R

23. A(n) _______ stores and returns energy to a circuit while a(n) _________ dissipates energy. a. resistor, impedance

b. resistor, inductor

c. inductor, resistor

d. inductor, reactance

24. For an RL circuit, the power factor cannot be less than ______ or greater than ________. a. 0, 1

b. 1,0

c. 0, -1

d. -1, 0

25. The voltage across the capacitor _________ the current through it by ______.

a. lags, 45

b. lags, 90

c. leads, 0

d. leads, 90

26. If the resistance in a series RC circuit is increased the magnitude of the phase angle ______. a. increases

b. remains the same

c. decreases

d. changes in an indeterminate manner

27. In a series RC circuit, the current _______ the total voltage by an angle _______.

a. lags, of 45

b. lags, of 0

c. leads, between 0 to 90

d. leads, of 90

28. The resistance phasor of a series circuit points to the right. The capacitor reactance phasor points ______ while the diagonal of the rectangle having there two phasors as sided represents their ________. a. up, impedance

b. down, impedance

c. left, current

d. up, total voltage

29. The phase angle for a series RC circuit is defined as the angle between the _______ and the _______ phasors.

214

a. current, resistance voltage

b. current, total voltage

c. resistance voltage, capacitor voltage

d. R, XC

30. The phase angle for a series RC circuit may be computed as the angle between the _______ and _________ phasors.

a. resistance, impedance

b. reactance, impedance

c. reactance, impedance

d. none of these

31. The power dissipated in a series RC circuit with R = 10 ohms and XC = 10 ohms carrying an effective current of 3 amperes is ________ watts. a. 30

b. 30/2

c. 90

d. 90/2

32. The magnitude of the power factor of an RC circuit with R = 10 ohms and XC = 10 ohms, 1.2 amp effective is a. 1

b. 0.5

c. 0.707

d. 0

33. The magnitude of the power factor of an RC circuit with R = 30 ohms, XC = 40 ohms and E = 100 V effective is a. 60

b. 80

c. 100

d. 120

34. The net reactance in an RLC circuit is a. XL

b. XC

c. XC + XL

d. XL – XC

35. The impedance of a series RLC circuit is

a. 𝑅2 + 𝑋𝐿2 + 𝑋𝑐

2

b. 𝑅2 + 𝑋𝐿2 + 𝑋𝑐

2

c. 𝑅2 + 𝑋𝐿2 − 𝑋𝑐

2 2

d. 𝑅2 + 𝑋𝐿 − 𝑋𝑐 2

215

7.2 Network Theorems

To recall Kirchhoff’s laws solve for the current in each impedance in the circuit below so that you

will have a basis for the answers in the network theorems. Remember that whatever network

theorem you use will arrive at the same result.

Let Vs = 120 V0

By KVL at loop I

120 0 − (50 + 𝑗75) 𝐼1 − (80 + 𝑗25) 𝐼2 = 0

50 + 𝑗75 𝐼1 + (80 + 𝑗25) 𝐼2 = 120 equation 1

By KVL at loop 2

(80 + 𝑗25) 𝐼2 − (100 − 𝑗70) 𝐼3 = 0 equation 2

By KCL at node a

𝐼1 − 𝐼2 − 𝐼3 = 0 equation 3

Equating (1) , (2) and (3)

50 + 𝑗75 𝐼1 + (80 + 𝑗25) 𝐼2 = 120

(80 + 𝑗25) 𝐼2 − (100 − 𝑗70) 𝐼3 = 0

𝐼1 − 𝐼2 − 𝐼3 = 0

120 V

60 Hz

50 ohms

75 ohms

80 ohms

25 ohms

100 ohms

70 ohms

I II

IB

IA

I1

a

I3

I2

216

Then using determinant method

𝐷 = 50 + 𝑗75 (80 + 𝑗25) 0

0 (80 + 𝑗25) − (100 − 𝑗70)1 −1 −1

= −22,125 − 𝑗8,150

𝐼1 =

120 (80 + 𝑗25) 0

0 (80 + 𝑗25) − (100 − 𝑗70)0 −1 −1

−22,125 − 𝑗8,150 =

−21,600 + 𝑗5,400

−22,125 − 𝑗8,150=

= 0.7805 − 𝑗0.5316 = 0.9443 − 34.2589

𝐼2 =

50 + 𝑗75 120 0

0 0 − (100 − 𝑗70)1 0 −1

−22,125 − 𝑗8,150 =

−12,000 + 𝑗8,400

−22,125 − 𝑗8,150=

= 0.3544 − 𝑗0.5102 = 0.6214 − 55.2127

𝐼3 =

50 + 𝑗75 (80 + 𝑗25) 120

0 (80 + 𝑗25) 01 −1 0

−22,125 − 𝑗8,150 =

−21,600 + 𝑗5,400

−22,125 − 𝑗8,150=

= 0.4265 𝐴 − 2.862

7.2.1 Maxwell’s Mesh Equations

Example 7.3 In the circuit below, determine the current I1, I2, and I3. Solution:

120 V

60 Hz

50 ohms

75 ohms

80 ohms

25 ohms

100 ohms

70 ohms

I II

IB

IA

217

Let Vs = 120 V0

At mesh I

1200 − 𝐼𝐴 50 + 𝑗75 – 𝐼𝐴 80 + 𝑗25 + 𝐼𝐵 80 + 𝑗25 = 0

120 + 𝑗0 − 𝐼𝐴[(50 + 𝑗75) + (80 + 𝑗25)] + 𝐼𝐵(80 + 𝑗25) = 0

120 – 𝐼𝐴 (130 + 𝑗100) + 𝐼𝐵(80 + 𝑗25) = 0

(130 + 𝑗100)𝐼𝐴 − (80 + 𝑗25)𝐼𝐵 = 120 equation 1

At mesh II

− 𝐼𝐵 80 + 𝑗25 + 𝐼𝐴 80 + 𝑗25 − 𝐼𝐵 100 – 𝑗70 = 0

− 𝐼𝐵[(80 + 𝑗25) + (100 – 𝑗70)] + 𝐼𝐴 (80 + 𝑗25) = 0

− 𝐼𝐵(180 – 𝑗45) + 𝐼𝐴 (80 + 𝑗25) = 0

(80 + 𝑗25)𝐼𝐴 − (180 – 𝑗45)𝐼𝐵 = 0 equation 2

Equate equations 1 and 2

130 + 𝑗100 𝐼𝐴 − 80 + 𝑗25 𝐼𝐵 = 120

(80 + 𝑗25)𝐼𝐴 − (180 – 𝑗45)𝐼𝐵 = 0

Using determinant method

𝐷 = 130 + 𝑗100 − 80 + 𝑗25

(80 + 𝑗25) − (180 – 𝑗45) = −22,125 − 𝑗8150

𝐼𝐴 =

120 − 80 + 𝑗25

0 − (180 – 𝑗45)

−22,125 − 𝑗8150=

−21,600 + 𝑗5,400

−22,125 − 𝑗8150= 0.7805 − 𝑗0.5316 𝐴

= 0.9443 𝐴 − 34.2589

𝐼𝐵 =

130 + 𝑗100 120

(80 + 𝑗25) 0

−22,125 − 𝑗8150=

−9,600 − 𝑗3,000

−22,125 − 𝑗8,150= 0.426 − 𝑗0.0213 𝐴

= 0.4265 𝐴 − 2.862

218

To find 𝐼1 , 𝐼2 , and 𝐼3

𝐼1 = 𝐼𝐴 = 0.9443 𝐴 − 34.2589

𝐼2 = 𝐼𝐴 − 𝐼𝐵 = 0.7805 − 𝑗0.5316 − 0.426 − 𝑗0.0213 𝐴 = 0.3545 − 𝑗0.5103

= 0.6214 − 55.2127

𝐼3 = 0.4265 𝐴 − 2.862

For problem set No. 20 solve the load current in each network using the network theorem being required to apply.

(a )

(b)

E

120 Vrms

60 Hz

0°

R1

10Ω

L1

150mH

L

250mH

C130µF

R50Ω

R325Ω

E

120 Vrms

60 Hz

30°

R1

10Ω

L1

150mH

L

250mH

C1

30µF

R50Ω

R3

25Ω

E1

150 Vrms

60 Hz

-45°

L

O

A

D

L

O

A

D

219

Problem No. 18

NETWORK THEOREMS

7.2.2 Mesh Analysis

7.2.3 Nodal Analysis

7.2.4 Thevenin’s Theorem

7.2.5 Norton’s Theorem

7.2.6 Superposition Theorem

7.2.7 Millman’s Theorem

220

Unit 8

Power Factor Correction or Improvement

1. discuss power factor correction or improvement. 2. determine sizes of capacitors to improve or correct the power factor

of a system.

LEARNING OUTCOMES

At the end of the lesson,

you are expected to:

221

Introduction

Most industries use a large number of electric motors; therefore, industrial plants represent highly

inductive loads. This means that industrial power systems operate at a power factor of less than

unity (1.0). However, it is undesirable for an industry to operate at a low-power factor, since the

electrical power system will have to supply more power to the industry than is actually used.

A given value of volt-amperes (voltage x current) is supplied to an industry by the electrical

power system. If the power factor (pf) of the industry is low, the current must be higher, since the

power converted by the total industrial load equals VA x pf. The value of the power factor decreases

as the reactive power (unused power) drawn by the industry increases. This is shown in Figure 8-1.

We will assume a constant value of true power, in order to see the effect of increases in reactive

power drawn by a load. . The smallest reactive power shown (VAR1) results in the volt-ampere value

of VA1. As reactive power is increased, as shown by the VAR2 and VAR3 values, more volt-amperes

(VA2 and VA3) must be drawn from the source. This is true sine the voltage component of the

supplied volt-amperes remains constant. This example represents the same effect as a decrease in

the power factor, since pf = W/VA, and, as VA increases, the pf will decrease if W remains constant.

Utility companies usually charge industries for operating at power factors below a specified

level. It is desirable for industries to “correct” their power factor to avoid such charges to make

more economical use of electrical energy. Two methods may be used to cause the power factor to

increase: (1) power-factor corrective capacitors, and (2) three-phase synchronous motors. Since the

effect of capacitive reactance is opposite to that of inductive reactance, their reactive effects will

counteract one another. Either power-factor-corrective capacitors, or three-phase synchronous

motors, may be used to add the effect of capacitance to an AC power line.

Figure 8-1. Effect of increases in reactive power (VAR) on apparent power (VA).

Q3 (vars)

S3

Q2 (vars)

S2 3

2 Q1 (vars)

S1 1

PT (watts)

222

An example of power factor of power factor correction is

shown in Figure 8-2. We will assume from the example that

both true power and inductive reactive power remain constant at values of 10 kW and 10 kVAR. In

Figure 8-2(a), the formulas show that the power factor equals 70 per cent. However, if 5-kVAR

capacitive reactive power is introduced into the electrical power system, the net reactive power

becomes 5 kVAR (10 kVAR inductive minus 5 kVAR capacitive), a shown in Figure 8-2(b). With the

addition of 5 kVAR capacitive to the system, the power factor is increased to 89 per cent. Now, in

Figure 8-2(c), if 10-kVAR capacitive is added to the power system, the total reactive power (kVAR)

becomes zero. The true power is now equal to the apparent power; therefore, the power factor is

1.0, or 100 per cent, which is characteristic of a purely resistive circuit. The effect of the increased

capacitive reactive power in the system is to increase or “correct” the power factor and, thus, to

reduce the current drawn from the power distribution lines that supply the loads. In many cases, it is

beneficial for industries to invest in either power-factor-corrective capacitors, or three-phase

synchronous motors to correct their power factor.

Capacitors for Power Factor Correction

Static capacitors are used for power factor correction in the system. They are constructed similarly

to the smaller capacitors used in electrical equipment, which have metal-foil plates separated by

paper insulation. Ordinarily, static capacitors are housed in metal tanks, so that the plates can be

immersed in an insulating oil to improve high-voltage operation. The usual operating voltages of

static capacitors are from 230 volts to 13.8 kilovolts. These units are connected in parallel with

power lines, usually at the industrial plants, to increase the system power factor. Their primary

disadvantage is that their capacitance cannot be adjusted to compensate or changing power factors.

Power factor correction can also be accomplished by using synchronous capacitors

connected across the power lines. (Three-phase synchronous motors are also called synchronous

capacitors). The advantage of synchronous capacitors over static capacitors is that their capacitive

effect can be adjusted as the system power factor increases or decreases. The capacitive effect of a

synchronous capacitor is easily changed by varying the DC excitation voltage applied to the rotor of

the machine. Industries considering the installation of either static or synchronous capacitors should

first compare he initial equipment cost and the operating cost against the savings brought about by

an increased system power factor.

223

Figure 8-2. Illustration of the effect of capacitive reactance on an inductive circuit:

(a) Reactive power = 10 kVAR inductive, (b) Reactive power = 10 kVAR inductive, and 5 kVAR

capacitive, (c) Reactive power = 10 kVAR inductive, and 10 kVAR capacitive.

𝑝𝑓 = 10 kW

10 kVA = 1 or 100%

𝑆 = 𝑃2 + 𝑄𝑛𝑒𝑡2 = 102 + 02 = 10 𝑘𝑉𝐴

where 𝑄𝑛𝑒𝑡 = 𝑄𝐿 − 𝑄𝐶 = 10 − 10 = 5 𝑘𝑉𝐴𝑅

𝑄𝐿

𝑆 = 𝑃

𝑄𝐶

𝑝𝑓 = 10 kW

11.18 kVA = 0.8945 or 89.45 %

攠 = 𝑃2 + 𝑄𝑛𝑒𝑡2 = 102 + 52 = 11.18 𝑘𝑉𝐴

where 𝑄𝑛𝑒𝑡 = 𝑄𝐿 − 𝑄𝐶 = 10 − 5 = 5 𝑘𝑉𝐴𝑅

𝑄𝐿

𝑆

𝑃

𝑄𝐿

𝑆

𝑃

𝑄𝐶

𝑝𝑓 = true power

apparent power =

𝑃

𝑆

𝑝𝑓 = 10 kW

14.14 kVA = 0.707 or 70.7 %

𝑆 = 𝑃2 + 𝑄𝐿2 = 14.14 kVA

224

What is Power Factor Correction or Improvement?

Power factor correction or improvement consists of adding a capacitive reactive power to an ac

circuit in such a manner that the apparent power drawn from the source is reduced without altering

the current through or the voltage across the load itself.

In commercial use, the power factor should be close to unity for efficient distribution of

electric power. However, the inductive load of motors may result in a power factor of 0.70, as an

example, for the phase angle of 45. To correct for this lagging inductive component of the current

in the main line, a capacitor can be connected across the line to draw leading current from the

source. To bring the power factor up to 1.0, that is unity pf, the value of capacitance is calculated to

take the same amount of volt-amperes as the VARS of the load.

What is the purpose of correcting power factor?

1. To make use of electrical energy more economical. This can be proven by analyzing Figure 8-3 and Figure 8-4(b). In Figure 8-3 without the capacitor for power factor correction the current is 4.55 A. If a capacitor is connected across the load the current becomes 3.13 A. Lowering the current means lowering the energy used by the load.

2. To reduce the cost of materials since cost of wires and cables depend on their sizes. In a high current application large size of wire should be used. Small current uses small sizes of wires.

3. It is desirable for industries to “correct” their power factor to avoid such charges. Electrical utility companies impose penalties to the users in which electrical loads have very low power factor.

What is the proper connection of the power factor devices? Series or Parallel

To find out, let us consider this problem: In the circuit below, an induction motor is connected across

a 220-V 60-Hz source, find the current through the motor or the current in the line.

Figure 8-3. An induction motor in which the power factor is to be corrected.

220 V

60 Hz

Induction Motor

700 watts, 70 % lagging

power factor

225

The power triangle is shown below

To find

𝑝𝑓 = 𝑐𝑜𝑠 = 0.70 ; = 𝑐𝑜𝑠−1 0.70 = 45.57

To find the current,

𝑝𝑓 = 𝑐𝑜𝑠 = 𝑃

𝑆 =

𝑃

𝐼𝑇𝑉𝑇 𝑠𝑜 𝑡𝑕𝑎𝑡 𝐼𝑇 =

𝑃

𝑉𝑇𝑝𝑓=

700 W

(220 V)(0.70) = 4.55 𝐴

To find the resistance and inductance of the motor,

Note: The motor is composed of resistance and inductance since the current is lagging by 70%.

To find R,

Using the formula 𝑃 = 𝐼2𝑅 𝑎𝑛𝑑 𝑅 =𝑃

𝐼2

𝑅 =㥤

𝐼2 = 700 W

(4.55 A)2 = 33.81 𝐴

For the inductance,

𝑋𝐿 = 𝑅 𝑡𝑎𝑛 = 33.81 𝑡𝑎𝑛 45.57 = 34.49

𝐿 = 𝑋𝐿

2f =

34.49

2(60) = 91.49 𝑚𝐻

QL

S

= 45.57

P = 700 W

XL XL

Z

= 45.57

R

226

The equivalent circuit for the motor would be a series RL of resistance 33.81 and inductance 91.49

mH.

Now connect a 32-F in series with the motor across the same source (see Figure 8-4a). To find the

current.

𝑍 = 𝑅2 + 𝑋𝐿 − 𝑋2 2

where 𝑅 = 33.81

𝑋𝐿 = 34.49 because the frequency has not changed

𝑋𝐶 = 1

2(60)(32x10−6) = 82.89

𝑍 = (33.81)2 + 34.49 – 82.89 2

= 59.04

Then 𝐼 = 𝑉𝑇

𝑍=

220 V

59.04 = 3.73 𝐴

As you can see, the current is lowered but this current is not the rated current in the motor. The

load current is altered in this case, thus, this is not the correct connection of the capacitor.

(a) (b)

Figure 8-4. (a) A capacitor connected in series with the motor, (b) A capacitor connected in parallel

with the motor. This is the correct connection of the power factor correction device because in

this circuit the current in the load is not altered and the line current is lowered.

R = 33.81 ohms

91.49 mH

R = 33.81 ohms

91.49 mH

C = 88 µF

220 V

60 Hz

R = 33.81 ohms

91.49 mH

C = 88 µF220 V

60 Hz

3.73 A 3.13 A

227

Now let us connect the capacitor in parallel with the motor (Figure 8-4b) The corresponding power

triangle is shown below.

𝑄𝐿 = 𝑃 𝑡𝑎𝑛 = 700 𝑡𝑎𝑛 45.57 = 714.07 VARS

The capacitive reactive power is calculated by using the formula QC = VC2/XC

𝑄𝐶 = (220)2

82.89 = 583.91 VARS

𝑄𝑛𝑒𝑡 = 𝑄𝐿 – 𝑄𝐶 = 714.07 − 583.91 = 130.16 VARS

= 𝑡𝑎𝑛−1 130.16

700 = 10.53

QL = 714.07 VARS

S

= 45.57

P = 700 W

QC = 583.91 VARS

QL QC

Snew Qnet = 130.16 VARS

= 45.57

P = 700 W

228

pf = cos = cos10.53 = 0.983 or 98.3%, the power factor is increased from 70% to 98.3%.

For the total current

𝑆 = 𝑃 𝑐𝑜𝑠 = 700 𝑐𝑜𝑠 10.53 = 688.21 VA

𝐼𝑇 = 𝑆

ㇲ𝑇=

688.21 VA

220 V= 3.13 𝐴

In this circuit the current is lowered without altering the current in the motor and the power factor is

increased. Therefore the correction of capacitor and other power factor correction devices is parallel

with the load or across the power lines.

How to Obtain a Unity Power Factor

In order to obtain a unity power factor QL = QC so that the P = S

Example 8.1 An electric motor has an inductance of 0.045 H and a resistance of 30 ohms.

a) Calculate the current though the motor when connected to a 24-V 60-Hz power source; b) Calculate the power used by the motor; c) Calculate the power factor of the load; d) A capacitor is to be connected in across the load so that its power factor becomes 96%

lagging. Calculate the capacitance of the capacitor. e) Repeat (c) for 96% leading. f) Repeat (c) for unity power factor.

Solution:

(a) 𝑋𝐿 = 2𝑓𝐿 = 2(60)(0.045) = 16.96

QL

S QC - QL = 0

P

229

𝒁 = 𝑅2 + 𝑋𝐿2 = (30)2 + (16.96)2 = 34.36

𝑰 = 𝑽𝑻

𝒁 =

24 𝑉

34.36 = 0.6985 A

(b) 𝑃 = 𝐼2 𝑅 = (0.6985)2(30) = 14.64 W

(c) 𝑝𝑓 = 𝑃

𝑆 =

𝑃

IT VT =

14.64

(0.6985)(24)= 0.8732 or 87.32%

or

= 𝑡𝑎𝑛−1 𝑋𝐿

R = 𝑡𝑎𝑛−1

16.96

30= 29.48

then 𝑝𝑓 = 𝑐𝑜𝑠 = 𝑐𝑜𝑠 29.48 = 0.87

(d)

The power triangle of the load,

By connecting a capacitor so that its power factor becomes 0.96 lagging,

For a 0.96 pf: = 𝑐𝑜𝑠 −10.96 = 16.26

24 V

60 Hz

R = 30 ohms

L = 0.045 H

= 14.64 tan 29.48

𝑄𝐿 = 𝑃 tan 𝑜𝑙𝑑

𝑆𝑜𝑙𝑑 = 8.276 𝑉𝐴𝑅

old = 29.48

𝑃 = 14.64 𝑊

P = 14.64 W

230

𝑄𝑛𝑒𝑡 = 𝑄𝐿 − ち𝐶 = 8.276 − 𝑄𝐶

𝑡𝑎𝑛 𝑛𝑒𝑤

=𝑄𝐿 − 𝑄𝐶

𝑃

𝑡𝑎𝑛 16.26 =8.276 − 𝑄𝐶

14.64

𝑄𝐶 = 8.276 – 14.64 𝑡𝑎𝑛 16.26 = 4.006 VARS

𝑋𝐶 = 𝑉𝐶

2

𝑄𝐶 =

(24)2

4.006 = 143.78

𝐶 = 1

2(60)(143.78) = 18.45 F

(e) new = 16.26 (still 16.26 because the power factor is 0.96)

𝑡𝑎𝑛 𝑛𝑒𝑤

= 𝑄𝑛𝑒𝑡

𝑃

𝑡𝑎𝑛 𝑛𝑒𝑤

= 𝑄𝐶 − 𝑄𝐿

𝑃

QL

Sold

old Qnet = 8.276 - QC

new= 16.26

QC = ?

QC = ?

QL

Sold

old

new= 16.26 Qnet = QC - 8.276

Snew

QC

231

𝑡𝑎𝑛 16.26 = 𝑄𝐶 − 8.276

14.64

𝑄𝐶 = 12.55 VARS

𝑋𝐶 = 𝑉𝐶

2

𝑄𝐶 =

𝑉𝐶2

12.55 = 45.9

𝐶 = 1

2(60)( 45.9) = 57.79 F

(f) To make the power factor unity, QL = QC. Since QL = 8.276 VARS, QC is also equal to 8.276 VARS.

QC = 8.276 VARS

𝑋𝐶 = 𝑉𝐶

2

𝑄愲 =

(24)2

8.276 = 69.6

𝐶 = 1

2(60)( 69.6) = 38.11 F

.

232

Problem Set No. 19

POWER CORRECTION OR IMPROVEMENT

1. A 5-kW AC motor has a power factor of 65% lagging when it is connected to a 200-V 50-Hz mains. What effective current does the motor draw? What is the current through the motor

when a capacitor of 80 F is connected in parallel to the motor? What is the power factor in this case?

2. In ac circuit takes a load of 160 kVA at a lagging power factor of 0.75 when connected to a 460-V 60-Hz source. What value of capacitance that must be connected across the load to make the power factor 0.89 lagging?

3. A small ac motor used in a washing is, in effect, an RL circuit. If the machine takes 311 watts and 4.5 amp from a 115-V source when operating normally, calculate its power factor.

4. An induction motor draws 6.0 A at 0.8 lagging power factor from 208-V 60-Hz source. (a) What value of capacitance must be place in parallel with the motor to raise the overall

power factor to unity? (b) What are the magnitudes of the final motor current, capacitor current, and line

current?

5. What value of capacitance is required to produce an overall power factor of 0.96 lagging with the motor of Problem 4?

6. A synchronous motor capable of operating with a leading power factor draws 15 kW from a distribution transformer while driving an air compressor. The remainder of the load on the transformer is 80 kW at 0.85 lagging power factor. (a) How many kilovars of capacitive reactive power must the synchronous motor produce to

raise the overall power factor to 0.96 lagging? (b) What is the reactive factor of the synchronous motor operating in this manner?

7. The power factor of a load on a 120-V 60-Hz source is raised from 0.707 lagging to 0.866

lagging by connecting a 53-F capacitor across the load. What is the active power of the load?

8. The power factor of a load on a 120-V 60-Hz is raised from 0.866 lagging to 0.966 leading by

connecting a 110 ½ - F capacitor in parallel with the load. What is the rms load current?

233

Objective Test No. 7

POWER FACTOR CORRECTION OR IMPROVEMENT

1. A power factor of 1 indicates that the circuit phase angle is

a. 90

b. 45

c. 180

d. 0

2. Energy sources are normally rated in a. watts b. volt-amperes c. volt-amperes reactive

3. The magnitude of the power factor of an RC circuit with R = 10 , XC = 10 and 12 A effective is a. 1

b. 0.5

c. 0.707

d. 0

4. The current in a series inductive circuit is 7.5 A at 25 Hz. The circuit takes 425 W, and the power factor is 0.47. The resistance of the circuit is

a. 7.32

b. 7.56

c. 7.86

d. 7.98 g

5. A circuit with a resistor and capacitor in series takes 200 W at a power factor of 0.40 from 200-V 50-Hz supply. The capacitance of the circuit is

a. 32 F

b. 43.4 F

c. 66.4 F

d. 80 F

6. In a pure reactive circuit, the power factor is a. lagging

b. zero

c. leading

d. unity

7. Power factor is defined as the ration of a. volt-amperes to watts

b. watts to volt-amperes

c. volt-ampere reactive to watts

234

d. watts to volt-ampere reactive

8. In a series circuit consisting of resistance and reactance, power factor is defined as the ratio of a. resistance to impedance b. resistance to reactance c. reactance to impedance

9. For a parallel circuit consisting of resistance and reactance the value of power factor is the ratio of a. impedance to reactance

b. reactance to impedance

c. resistance to impedance

d. impedance to resistance

10. It is not easy to find the value of impedance in parallel circuit but power factor can be easily obtained as the ratio of a. active current to line current

b. reactive current to line current

c. line current to active current

11. The power factor of ac circuit containing both a resistor and a capacitor is a. more than unity

b. leading by 90 degrees

c. leading power factor

12. In an ac circuit, a low value of reactive volt-ampere compared and with watts indicates a. high power factor b. unity power factor c. leading power factor

13. In a given ac circuit when power factor is unity the reactive power is

a. maximum

b. equal to I2R

c. zero

14. The capacitor for power factor correction are rated in terms of a. voltage

b. VA

c. kW

d. kVAR

235

Unit 9 RESONANCE

1. define resonance. 2. discuss the characteristics of a series resonance. 3. discuss the characteristics of a parallel resonance. 4. find the frequency for resonance. 5. solve the parameters of a circuit for resonance.

LEARNING OUTCOMES

At the end of the lesson,

you are expected to:

236

9.1 Resonant Effect

Inductive reactance increases as the frequency is increased, but capacitive reactance decreases

with higher frequencies. Because of these opposite characteristics, for any LC combination there

must be a frequency at which the XL equals the XC, as one increases while the other decreases. This

case of equal and opposite reactances is called resonance, and the ac circuit is then a resonant

circuit.

Any LC circuit can be resonant. It all depends on the frequency. At the resonant frequency,

an LC combination provides the resonance effect. Off the resonant frequency, either below or

above, the LC combination is just another ac circuit.

The frequency at which the opposite reactances are equal is the resonant frequency. This

frequency can be calculated as fr = 1/(2LC) where L is the inductance in henrys, C is the

capacitance in farads, and fr is the resonant frequency in hertz that makes XL = XC.

In general, we can say that large values of L and C provide a relatively low resonant

frequency. Smaller values of L and C allow higher values for fr. The resonance effect is most useful

for radio frequencies, where the required values of microhenrys for L and picofarads for C are easily

obtained.

The most common applications of resonance in RF circuits is called tuning. In this use, the LC

circuit provides maximum voltage output at the resonant frequency, compared with the amount of

output at any other frequency either below or above resonance. This idea is illustrated in Figure 9-

1, where the LC circuit resonant at 1000 kHz magnifies the effect of this particular frequency. The

result is maximum output at 1000 kHz, compared with lower or higher frequencies.

(a) (b)

Figure 9-1. Applications of resonance. (a) LC circuit resonant at fr of 1000 kHz to provide

maximum output at this frequency. (b) Wavemeter as an example of tuning an LC circuit for

resonance at different frequencies.

Resonant

LC circuit

fr = . 1 .

2LC

= 1000 kHz

237

For the wavemeter in Figure 9-1(b), note that the capacitance C can be varies to provide resonance

at different frequencies. The wavemeter can be tuned to any one frequency in any range

depending on the LC combination.

Tuning in radio and television are applications of resonance. When you tune a radio to one

station, the LC circuits are tuned to resonance for that particular carrier frequency. Also, when you

tune a television receiver to a particular channel, the LC circuits are tuned to resonance for that

station. There are almost unlimited uses for resonance in ac circuits.

From Wikipedia

Resonance of a circuit involving capacitors and inductors occurs because the collapsing magnetic field of the inductor generates an electric current in its windings that charges the capacitor, and then the discharging capacitor provides an electric current that builds the magnetic field in the inductor. This process is repeated continually. An analogy is a mechanical pendulum.

At resonance, the series impedance of the two elements is at a minimum and the parallel impedance is at maximum. Resonance is used for tuning and filtering, because it occurs at a particular frequency for given values of inductance and capacitance. It can be detrimental to the operation of communications circuits by causing unwanted sustained and transient oscillations that may cause noise, signal distortion, and damage to circuit elements.

Parallel resonance or near-to-resonance circuits can be used to prevent the waste of electrical energy, which would otherwise occur while the inductor built its field or the capacitor charged and discharged. As an example, asynchronous motors waste inductive current while synchronous ones waste capacitive current. The use of the two types in parallel makes the inductor feed the capacitor, and vice versa, maintaining the same resonant current in the circuit, and converting all the current into useful work.

Since the inductive reactance and the capacitive reactance are of equal magnitude, ωL = 1/ωC, so:

where ω = 2πf, in which f is the resonance frequency in hertz, L is the inductance in henries, and C is the capacitance in farads when standard SI units are used.

The quality of the resonance (how long it will ring when excited) is determined by its Q factor, which is a function of resistance. A true LC circuit would have infinite Q, but all real circuits have some resistance and smaller Q and are usually approximated more accurately by an RLC circuit.

238

9.2 Series Resonance

(a)

(b)

(

b)

Figure 9-2. Series Resonance. (a) Schematic diagram of series RLC, (b) Graph to show reactances

XC and XL are equal and opposite at the resonant frequency fr.

XL, XL = 2fL

XL = XC at fr

fr frequency

XC = 1/(2fC)

XC,

C = 106 pF

XC = 1500 ohms

L = 239 µH

XL = 1500 ohms

300 µV

10

239

Figure 9-3. Graphs showing maximum current at resonance for the series circuit

shown in Figure 9-2. (a) Amplitudes of individual cycles. (b) response curve to show

amount of I below and above resonance. Values of I are in Table 9-1.

I, A 30 Current maximum at fr

20

10

600 800 1000 1200 1400

Frequency, kHz

Highest current

value at fr

Small current values Small current values

below fr above fr

(a)

(b)

240

.

Table 9-1

Series-Resonance calculations for the Circuit in Figure 9-2.

f

kHz

XL =

2fL

XC =

1/(2fL)

Net Reactance, Z, I =

VT/Z

A

VL =

IXL

V

VC =

IXC

V

XC - XL XL – XC

600 900 2500 1600 1600 0.19 171 475

800 1200 1875 675 675 0.44 528 825

1000 1500 1500 0 0 10 30 45,000 45,000

1200 1800 1250 550 550 0.55 990 688

1400 2100 1070 1030 1030 0.29 609 310

9.3 Characteristics of a Series Resonance:

a. The inductive reactance and the capacitive reactance of the circuit are equal. b. Therefore, the impedance is minimum and is equal to the resistance of the circuit. c. Hence, the current in the circuit is maximum and is equal to E/R. d. The current through the circuit is in phase with the applied voltage.

This condition is known as resonance. The series RLC circuit if Figure 9-2(a) is called a

resonant circuit, and the frequency at which resonance occurs is called the resonant frequency of

the circuit.

The letter symbol for resonant frequency is fr.

Since at resonance XL = XC, then 2fL = 1/2fC, for which

𝑓𝑟 = 1

2 LC

where fr is in hertz, L is in henrys, and C is in farads.

241

9.4 Q-factor of a Series Circuit

In the case of an RLC circuit it is defined as equal to the voltage magnification in the circuit at

resonance.

𝑄 − 𝑓𝑎𝑐𝑡𝑜𝑟 = 1

𝑅

𝐿

𝐶

where R = resistance; L = inductance; C = capacitance

In the case of series resonance, the higher quality factor, i.e., Q factor means not only higher

voltage magnification but also higher selectivity of the tuning coil.

Example 9.1 A coil in a tined circuit in a radio receiver has an inductance of 300 H and a

resistance of 15 . What value of capacitance must be connected in series with the coil for the

circuit to be series resonant at 840 kHz?

Solution:

𝑋𝐿 = 2𝑓𝐿 = 2 𝑥 (840 𝑘𝐻𝑧) 𝑥 (300 𝐻) = 1583

At resonance,

𝑋𝐶 = 𝑋𝐿 = 1583

C = 1

2 x (840 kHz ) x (1583 ) = 120 𝑝𝐹

9.5 Parallel Resonance

At 65 IL = IC and IT = IR. This condition is called parallel resonance. Other characteristics of parallel

resonance are:

1. The admittance of the circuit is minimum and is equal to the conductance of the circuit.

2. The current drawn is minimum.

E120 Vrms

65 Hz

0°

R100Ω

L200mH

C30µF

242

3. The phase angle between the current and voltage is zero, the power factor is unity.

4. The resonant frequency is given by 𝑓𝑟 = 1

2 LC if the resistance in the inductance and

capacitance branches is negligible.

9.6 Half-Power Frequencies

The half-power frequencies are those frequencies at which the power dissipation in the circuit is half

of the power dissipation at resonant frequency fr. They are the corresponding frequencies f1 and f2 at

the value of current 𝐼 = 𝐼𝑜

2 ; where Io is the current at resonance in RLC series circuit.

Hence power Po drawn by the circuit at resonance is

𝑃0 = 𝐼02𝑅

Power in the circuit at f1 = 𝐼𝑜

2

2

𝑅 = 1

2 𝐼𝑜

2𝑅 (=half the power at resonance)

Power in the circuit at f2 = 𝐼𝑜

2

2

𝑅 = 1

2 𝐼𝑜

2𝑅 (=half the power at resonance)

9.7 Bandwidth and Selectivity

frequency

I0

current

0.707 I0

f1 fr f2

10

600 800 1000 1200 1400

Frequency, kHz

243

The difference (f2 –f1) is called the bandwidth of the resonant network. The ratio of the bandwidth to

the resonant frequency is defined as the selectivity of the circuit.

When frequency is varied in RLC circuit, the selectivity becomes

𝑓2 − 𝑓1

𝑓𝑟 =

1

𝑄0

Example 9.2 A parallel resonant circuit having L = 100 H and C = 10 pF. If Q of the circuit is 50,

what is the bandwidth of the resonant circuit?

Solution:

𝑓− = 1

2 LC

= 1

2 100 x 10−6 10 x 10−12

= 503 kHz

244

Problem Set No. 20

RESONANCE

1. A series circuit with a 12-ohm resistor and a 32-ohm inductive reactance is connected across 240-volt 60-Hz mains. Determine (a) capacitive reactance that will make circuit resonant; (b) current; (c) circuit power; (d) value of capacitance and inductance.

2. Prob. 1 determine for resonance (a) value of capacitance C with resistance R, inductance L, and frequency as given; (b) value of inductance L with C and f as given; (c) value of f with L and C as given; (d) voltage across inductor and capacitor in (a) , (b) , (c); (e) current and power in (a), (b), (c).

3. In a series circuit the resistance is 1,000 ohms and the inductance 0.008 henry and the

capacitance is adjustable. Determine (a) value of capacitance to give resonance at 1,000 Hz; (b)

current if emf is 40 volts; (c) voltage across inductance and across capacitance; (d) power.

245

Objective Test No. 8

RESONANCE

1. There will ________ be a frequency called the ______ frequency, at which _______.

a. sometimes, natural, XL = XC

b. always, natural, R = 0

c. always, resonate, XL = XC

d. sometimes, resonant, R = 0

2. For the series RLC circuit at resonance the current amplitude is _____ for a fixed voltage

amplitude and power factor.

a. minimum, zero

b. minimum, unity

c. maximum, zero

d. maximum, unity

3. In an RLC circuit, the current at resonance is

a. maximum

b. minimum

c. infinity

d. zero

4. In RLC circuits, the current at resonance is

a. minimum in parallel circuit and minimum in series circuit. b. maximum in parallel circuit and minimum in series circuit. c. maximum in either parallel or series circuit. d. minimum in either parallel or series circuit.

5. A series resonance is capacitive at f = 100 Hz. The circuit will be inductive somewhere at a. f < 100 Hz

b. f > 100 Hz

c. f = 100 Hz by increasing the value of the resistance

d. none of these

6. At frequency less than the resonant frequency a. series circuit is capacitive and parallel circuit is inductive b. series circuit is inductive and parallel circuit is capacitive c. both circuits are inductive d. both circuits are capacitive.

7. The value of current at resonance in a series RLC circuit affected by the value of

246

a. R

b. C

c. L

d. all of these

8. In resonant circuit, the power factor at resonance is a. zero

b. 1

c. 0.5

d. 0.707

9. Which of the following statement is true for a series RLC circuit tuned at resonant frequency a. the voltage across C > applied b. the voltage across L > applied voltage c. the voltage across L and C > applied voltage d. the voltage across both L and C < applied voltage

10. The power factor at resonance in RLC parallel circuit is

a. zero b. 0.08 lagging c. 0.8 lagging d. unity

11. The quality factor of RLC circuit will increase if

a. R increases b. R decreases c. impedance increases d. voltage increases

12. Higher the Q of a series circuit,

a. broader it resonance curve b. b. narrower its pass band c. greater its bandwidth d. sharper its resonance

13. Selectivities of different resonant circuits are compared in terms of their

a. impedances b. reactances c. frequencies d. bandwidths

247

References

“Electrical Power Systems Technology”. Stephen W. Fardo, Dale R. Patrick

“Electrical Engineering”, Chester Dawes

“Electric Circuits”, Charles Siskind

“Understanding AC Circuits”, Dale R. Patrick

”Electric Circuit Fundamentals”, Thomas L. Floyd

“1001 Solved Problems in Electrical Engineering”, Romeo Rojas

“Textbook-Reviewer in Electrical Engineering”, Marcialito M. Valenzona

“Electricity: Principles and Application”, Richard J. Fowier