BASIC ENGINEERING – II

Part A ELECTRICAL ENGINEERING

Prepared by Dr.R.Jegatheesan

Professor, EEE Dept.

AC & DC circuits

Circuit parameters, Ohms law, Kirchhoff’s laws. Average and RMS values,

concept of phasor representation. RLC serious circuits and series resonance.

RLC parallel circuits (includes simple problems in DC & AC circuits). Introduction

to three phase system – types of connection, relationship between line and phase

values. (qualitative treatment only).

Electrical machines

Definition of mmf, flux and reluctance, leakage flux, fringing, magnetic materials

and B-H relationship. Problems involving simple magnetic circuits. Faraday’s

laws, induced emfs and inductances, brief idea on Hysteresis and eddy currents.

Working principle, construction and applications of DC machines and AC

machines (1-phase transformers, 3-phase induction motors, single phase

induction motors – split phase, capacitor start and capacitor start & run motors).

Wiring & lighting

Types of wiring, wiring accessories, staircase & corridor wiring, Working and

characteristics of incandescent, fluorescent, SV & MV lamps. Basic principles of

earthing, simple layout of generation , transmission & distribution of power.

TEXT BOOKS

1. Muthusubramanian R, Salaivahanan S, Muraleedharan K A, Basic Electrical, Electronics

and Computer Engineering, Tata McGraw – Hill, 1999

2. Mehta V K, Principles of Electronics, S Chand & Co, 1980

ELECTRIC CIRCUITS

Electric circuits are broadly classified as Direct Current (D.C.) circuits and

Alternating Current (A.C.) circuits. The following are the various elements that

form electric circuits.

D.C. Circuits A.C. Circuits

Elements Representation Elements Representation

Voltage source Voltage source

Current source Current source

Resistor Resistor

Inductor

Capacitor

+ - + - ~

I

V

DC CIRCUITS

The voltage across an element is denoted as E or V. The current through the

element is I.

Conductor is used to carry current. When a voltage is applied across a

conductor, current flows through the conductor. If the applied voltage is

increased, the current also increases. The voltage current relationship is shown

in Fig. 1.

It is seen that I V. Thus we can write

I = G V (1)

where G is called the conductance of the conductor.

Fig. 1 Voltage – current relationship

I = G V (1)

Very often we are more interested on RESISTANCE, R of the conductor, than the

conductance of the conductor. Resistance is the opposing property of the

conductor and it is the reciprocal of the conductance. Thus

R = G

1 or G =

R

1 (2)

Therefore I = R

V (3)

The above relationship is known as OHM’s law. Thus Ohm law can be stated as

the current flows through a conductor is the ratio of the voltage across the

conductor and its resistance. Ohm’s law can also be written as

V = R I (4)

R = I

V (5)

The resistance of a conductor is directly proportional to its length, inversely

proportional to its area of cross section. It also depends on the material of the

conductor. Thus

R = ρ A

(6)

where ρ is called the specific resistance of the material by which the conductor is

made of. The unit of the resistance is Ohm and is represented as Ω. Resistance of

a conductor depends on the temperature also. The power consumed by the

resistor is given by

P = V I (7)

When the voltage is in volt and the current is in ampere, power will be in watt.

Alternate expression for power consumed by the resistors are given below.

P = R I x I = I2 R (8)

P = V x R

V=

R

V 2

(9)

KIRCHHOFF’s LAWS

There are two Kirchhoff’s laws. The first one is called Kirchhoff’s current law,

KCL and the second one is Kirchhoff’s voltage law, KVL.

Kirchhoff’s current law deals with element currents meeting at a junction, which

is a meeting point of two are more elements.

Kirchhoff’s voltage law deals with element voltages in a closed loop also called

as closed circuit.

I4

I3

I2

I1

P

Kirchhoff’s current law

Kirchhoff’s current law states that the algebraic sum of element currents meeting

at a junction is zero.

Consider a junction P wherein four elements, carrying currents I1, I2, I3 and I4, are

meeting as shown in Fig. 2.

Note that currents I1 and I4 are flowing out from the junction while the currents I2

and I3 are flowing into the junction. According to KCL,

I1 – I2 - I3 + I4 = 0 (10)

The above equation can be rearranged as

I1 + I4 = I2 + I3 (11)

From equation (11), KCL can also stated as at a junction, the sum of element

currents that flows out is equal to the sum of element currents that flows in.

Fig. 2 Currents meeting at a junction

V3

V1 + -

V2

V4

+

+

+

-

-

-

Kirchhoff’s voltage law

Kirchhoff’s voltage law states that the algebraic sum of element voltages around

a closed loop is zero.

Consider a closed loop in a circuit wherein four elements with voltages V1, V2, V3

and V4, are present as shown in Fig. 3.

Assigning positive sign for voltage drop and negative sign for voltage rise, when

the loop is traced in clockwise direction, according to KVL

V1 - V2 - V3 + V4 = 0 (12)

The above equation can be rearranged as

V1 + V4 = V2 + V3 (13)

From equation (13), KVL can also stated as, in a closed loop, the sum of voltage

drops is equal to the sum of voltage rises in that loop.

Fig. 3 Voltages in a closed loop

Find the currents I 1 , I 2 , I 3 and I 4 in the circuit shown.

I 4

I 1

2 A 10 A

I 3 I 2

5 A

12 A

Find the voltages V 1 , V 2 and V 3 in the circuit shown.

-

+

V 1

+ 8V -

-

+

V 3

-

+

5V

- V 2 +

+ 2V -

R3 R2 R1

+

+ + +

-

- - - V3 V2 V1

I

E

Resistors connected in series

Two resistors are said to be connected in series when there is only one common

point between them and no other element is connected in that common point.

Resistors connected in series carry same current. Consider three resisters R1, R2

and R3 connected in series as shown in Fig. 4. With the supply voltage of E,

voltages across the three resistors are V1, V2 and V3.

As per Ohm’s law

V1 = R1 I

V2 = R2 I (14)

V3 = R3 I

Fig. 4 Resistors connected in series

+

-

- - V3 V2

E

R2 I R3

+

+ + - V1

R1

+ -

Re q I

E

Applying KVL,

E = V1 + V2 + V3 (15)

= (R1 + R2 + R3) I = Re q I (16)

Thus for the circuit shown in Fig. 4,

E = Re q I (17)

where E is the circuit voltage, I is the circuit current and Re q is the equivalent

resistance. Here

Re q = R1 + R2 + R3 (18)

This is true when two are more resistors are connected in series. When n numbers of

resistors are connected in series, the equivalent resistor is given by

Re q = R1 + R2 + …………. + Rn (19)

Fig. 4 Resistors connected in series

+ + V1 - - V2

R2 R1 I

+ - E

Voltage division rule

Consider two resistors connected in series. Then

V1 = R1 I

V2 = R2 I

E = (R1 + R2) I and hence I = E / (R1 + R2)

Total voltage of E is dropped in two resistors. Voltage across the resistors are

given by

V1 = 21

1

RR

R

E and (20)

V2 = 21

2

RR

R

E (21)

(22)

I2

R1

R2

+ - E

I1

I

A

Resistors connected in parallel

Two resistors are said to be connected in parallel when both are connected

across same pair of nodes. Voltages across resistors connected in parallel will be

equal.

Consider two resistors R1 and R2 connected in parallel as shown in Fig. 5.

As per Ohm’s law,

I1 = 1R

E

I2 = 2R

E

Fig. 5 Resistors connected in parallel

E

I1

I2

R1

R2

+ -

I

A I1 =

1R

E

I2 = 2R

E

I

E

Re q

Applying KCL at node A

I = I1 + I2 = E )R

1

R

1(

21

(23)

From the equivalent circuit shown

I = qeR

E (24)

where E is the circuit voltage, I is the circuit current and Re q is the equivalent

resistance. Comparing eq. (23) and (24)

21qe R

1

R

1

R

1 (25)

Equivalent circuit

21qe R

1

R

1

R

1 (25)

From the above 21

21

qe RR

RR

R

1

Thus 21

21qe

RR

RRR

(26)

When n numbers of resistors are connected in parallel, generalizing eq. (25),

Re q can be obtained from

n21qe R

1................

R

1

R

1

R

1 (27)

(29)

(30)

I2

R1

R2

+ - E

I1

I

A

I1 = 1R

E

I2 = 2R

E

(22)

Current division rule

Referring to Fig. 5, it is noticed the total current gets divided as I1 and I2. The

branch currents are obtained as follows.

From eq. (23)

E = I21

21

RR

RR

Substituting the above in eq. (22)

I1 = I21

2

RR

R

I2 = I21

1

RR

R

Fig. 5 Resistors connected in parallel

30Ω 20Ω 10Ω

+

+ + +

-

- - - V3 V2 V1

I

100V

Example 1

Three resistors 10Ω, 20Ω and 30Ω are connected in series across 100 V supply.

Find the voltage across each resistor.

Solution

Current I = 100 / (10 + 20 + 30) = 1.6667 A

Voltage across 10Ω = 10 x 1.6667 = 16.67 V

Voltage across 20Ω = 20 x 1.6667 = 33.33 V

Voltage across 30Ω = 30 x 1.6667 = 50 V

I2

4Ω

6Ω

I1

30 A

Example 2

Two resistors of 4Ω and 6Ω are connected in parallel. If the supply current is 30 A,

find the current in each resistor.

Solution

Using the current division rule

Current through 4Ω = A1830x64

6

Current through 6Ω = A1230x64

4

Example 3

Four resistors of 2 ohms, 3 ohms, 4 ohms and 5 ohms respectively are connected

in parallel. What voltage must be applied to the group in order that the total power

of 100 W is absorbed?

Solution

Let RT be the total equivalent resistor. Then

120

154

120

24304060

5

1

4

1

3

1

2

1

R

1

T

Resistance RT = Ω0.7792154

120

Let E be the supply voltage. Then total current taken = E / 0.7792 A

Thus 1000.7792x)0.7792

E( 2 and hence E2 = 100 x 0.7792 = 77.92

Required voltage = V8.827277.92

R1

230 V

12 A

16 A

R1

R2

230 V

12 A

4 A

- +

- +

Example 4

When a resistor is placed across a 230 V supply, the current is 12 A. What is the

value of the resistor that must be placed in parallel, to increase the load to 16 A?

Solution

To make the load current 16 A, current through the second resistor = 16 –12 = 4 A

Value of second resistor R2 = 230/4 = 57.5 Ω

R

50 Ω

100 Ω

360 V

7.2 A

+ -

50 Ω

100 Ω

E

7.2 A

+ -

3.6 A

1.3 A

Example 5

A 50 Ω resistor is in parallel with a 100 Ω resistor. The current in 50 Ω resistor is

7.2 A. What is the value of third resistor to be added in parallel to make the line

current as 12.1A?

Solution

Supply voltage E = 50 x 7.2 = 360 V

Current through 100 Ω = 360/100 = 3.6 A

When the line current is 12.1 A, current through third resistor = 12.1 – (7.2 + 3.6)

= 1.3 A

Value of third resistor = 360/1.3 = 276.9230 Ω

3.6 Ω 4.56 Ω

RT = 6 Ω

R1

Example 6

A resistor of 3.6 ohms is connected in series with another of 4.56 ohms. What

resistance must be placed across 3.6 ohms, so that the total resistance of the

circuit shall be 6 ohms?

Solution

3.6 R1 = 6 – 4.56 = 1.44 Ω

Thus 1.5R

3.62.5;

1.44

3.6

R

R 3.6 Therefore1.44;

R3.6

Rx3.6

11

1

1

1

Required resistance R1 = 3.6/1.5 = 2.4 Ω

Power dissipated = 70 W

Example 7

A resistance R is connected in series with a parallel circuit comprising two

resistors 12 Ω and 8 Ω respectively. Total power dissipated in the circuit is 70 W

when the applied voltage is 22 V. Calculate the value of the resistor R.

Solution

Total current taken = 70 / 22 = 3.1818 A

Equivalent of 12 Ω 8 Ω = 96/20 = 4.8 Ω

Voltage across parallel combination = 4.8 x 3.1818 = 15.2726 V

Voltage across resistor R = 22 – 15.2726 = 6.7274 V

Value of resistor R = 6.7274/3.1818 = 2.1143 Ω

8 Ω

+ - 22 V

12 Ω

R

Voltage across 6 Ω = 6 V

12 Ω

6.25 Ω

6 Ω

0.25 Ω E

Example 8

The resistors 12 Ω and 6 Ω are connected in parallel and this combination is

connected in series with a 6.25 Ω resistance and a battery which has an internal

resistance of 0.25 Ω. Determine the emf of the battery if the potential difference

across 6 Ω resistance is 6 V.

Solution

Current in 6 Ω = 6/6 = 1 A

Current in 12 Ω = 6/12 = 0.5 A

Therefore current in 0.25 Ω = 1.0 + 0.5 = 1.5 A

Using KVL E = (0.25 x 1.5) + (6.25 x 1.5) + 6 = 15.75 V

Therefore battery emf E = 15.75 V

4 Ω 4 Ω

6 Ω

0.6 Ω 12 V

3 Ω

Example 9

A circuit consist of three resistors 3 Ω, 4 Ω and 6 Ω in parallel and a fourth

resistor of 4 Ω in series. A battery of 12 V and an internal resistance of 0.6 Ω is

connected across the circuit. Find the total current in the circuit and the terminal

voltage across the battery.

Solution

4 Ω 6 Ω = 24/10 = 2.4 Ω

1.4 Ω 3 Ω = 7.2/5.4 = 1.3333 Ω

Total circuit resistance = 4 + 0.6 + 1.3333 = 5.9333 Ω

Circuit current = 12/5.9333 = 2.0225 A

Terminal voltage across the battery = 12 – (0.6 x 2.0225) = 10.7865 V

F E D

C B A

24 V

13 Ω

1 Ω

22 Ω

18 Ω

9 Ω

14 Ω

11 Ω

5 Ω

F E D

C B A

24 V

13 Ω

1 Ω

22 Ω

18 Ω 14 Ω

11 Ω

14 Ω

F E

C B A

24 V

13 Ω

1 Ω

22 Ω

18 Ω 7 Ω

11 Ω

Example 10

An electrical network is arranged as shown. Find (i) the current in branch AF (ii)

the power absorbed in branch BE and (iii) potential difference across the branch

CD.

Solution

Various stages of reduction are shown.

1

2

F E

24 V

1 Ω

22 Ω

18 Ω 7 Ω

F E

C B A

24 V

13 Ω

1 Ω

22 Ω

18 Ω

18 Ω

11 Ω 13 Ω

2

3

9 Ω

F E

A

24 V

1 Ω

11 Ω

24 V

F E

B A 13 Ω

1 Ω

22 Ω

F E

B A

24 V

22 Ω

1 Ω

22 Ω

F E

C B A

24 V

13 Ω

1 Ω

22 Ω

18 Ω

18 Ω

4 5

6

3

Current in branch AF = 24/12 = 2 A from F to A

Using current division rule current in 13 Ω in Fig. 4= 1 A

Referring Fig. 3, current in branch BE = 0.5 A

Power absorbed in branch BE = 0.52 x 18 = 4.5 W

Voltage across BE = 0.5 x 18 = 9 V

Voltage across CE in Fig. 1 = 9x18

7 = 3.5 V

Referring Fig. given in the problem, using voltage division rule, voltage

across in branch CD = V1.253.5x14

5

6 Ω 3 Ω

25 V 45 V 4 Ω

6 Ω 3 Ω

25 V 45 V 4 Ω I2

A

B

C D

Example 11

Using Kirchhoff’s laws, find the current in various resistors in the circuit shown.

Solution

Let the loop current be I1 and I2. We can find element currents in terms of loop

currents.

I1

25 V 45 V 4 Ω I2

A

B D 6 Ω 3 Ω

C

Considering the loop ABCA, KVL yields

6 I1 + 4 (I1 – I2) – 25 = 0

For the loop CDAC, KVL yields

3 I2 – 45 + 4 (I2 – I1) = 0

Thus 10 I1 - 4 I2 = 25

- 4 I1 + 7 I2 = 45

On solving the above I1 = 6.574 A; I2 = 10.1852 A

Current in 4Ω resistor = I1 – I2 = 6.574 – 10.1852 = - 3.6112 A

Thus the current in 4Ω resistor is 3.6112 A from A to C

Current in 6 Ω resistor = 6.574 A; Current in 3 Ω resistor = 10.1852 A

I1

6 Ω 3 Ω

25 V 45 V 4 Ω

5 Ω

A

B

C

D

Example 12

Find the current in 5 Ω resistor in the circuit shown.

25 V 45 V

A

C

Solution

Let the loop current be I1, I2 and I3.

Three loops equations are:

- 25 + 6 (I1 + I3) + 4 (I1 – I2) = 0

- 45 + 4 (I2 – I1) + 3 (I2 + I3) = 0

5 I3 + 6 (I3 + I1) + 3 (I2 + I3) = 0

1436

374

6410

3

2

1

I

I

I

=

0

45

25

: On solving

I3 = -14 A Current in 5 Ω resistor = 14 A from A to C

6 Ω 3 Ω

4 Ω

5 Ω

I1 I2

I3

Example 13

Using mesh analysis find the current i0 and the voltage v ba in the circuit shown.

Answers: 1.7777 A 53.331 V

12 V

1.2 Ω

12 Ω 8 Ω

6 Ω 3 Ω

4 Ω

1 Ω

2 Ω

12 Ω

-

- +

+

Vx

Example 14

In the circuit shown, determine Vx and the power absorbed by 12 Ω resistor.

Answers: 2 V 1.92 W

MAGNETIC CIRCUITS

Introduction

A substance, which when suspended freely, points in the direction of north and

south is called a MAGNET. Magnet attracts iron fillings. It is also called as

permanent magnet. A current passing though a conductor (or a coil) also can

produce magnetic effect and it is called as Electromagnet.

A permanent magnet has one north pole and one south pole. The imaginary lines

which travel from north pole to south pole outside the magnet are called

magnetic lines of force. They are drawn by plotting successive directions pointed

out by a small compass needle in the magnetic field. Magnetic lines of forces are

shown in Fig. 1 and they pass through the magnet.

S N

Fig. 1 Magnetic lines of forces

Flux, Magneto Motive Force and Reluctance

The magnetic lines of force in the magnetic field is called Flux. Its unit is

Weber(Wb). 1 Wb = 108 magnetic lines. Flux is denoted by φ .

Magnetic flux per unit cross sectional area is called Flux density and it is

expressed in Weber / metre2. Flux density is denoted by B.

Magneto Motive Force (mmf) is the source of producing flux in the magnetic

circuit. It can be explained through Electromagnet. When a current of I ampere is

passed through a coil of N turns, results in a mmf of N I . This N I ampere turns is

called the mmf and its unit is ampere turns (AT).

Reluctance is the property of magnetic circuit that opposes the setting of flux.

Reluctance, S = flux

mmf

Its unit is ampere turns / weber.

The following table shows the similarities between magnetic and electric circuits.

Sl.

No. Magnetic circuit Electric circuit

1 Magnetic flux, φ webers Electric current, I ampere

2 Magneto motive force, AT EMF, E volts

3 Reluctance, S AT / Wb Resistance, R ohm

4 φ= reluctance

mmf Current =

resistance

emf

Leakage flux and Fringing effect

Usually we assume that all the flux lines take path of the magnetic medium. But,

practically, some flux lines do not confine to the specified medium. It is because,

to prevent the leakage flux, there is no perfect magnetic insulator. Some flux lines

can pass through air also.

The flux which do not follow the desired path in a magnetic circuit is known as

leakage flux.

All the magnetic flux which complete the desired magnetic circuit are the useful

flux.

To account for the leakage flux, leakage coefficient is introduced. Leakage

coefficient, denoted by λ is defined as follows.

Leakage coefficient, λ = φ

φφ

fluxuseful

fluxleakagefluxuseful

fluxuseful

fluxtotal

Leakage coefficient will be greater than ONE.

An air gap is often introduced in the magnetic circuit out of necessity. When

crossing an air gap, the magnetic lines of force have a tendency to bulge out.

This is because the magnetic lines of force repel each other when they are

passing through non-magnetic material. This phenomenon is known as fringing.

It is shown in Fig. 2

Fringing effect increases the effective area of cross section of the air-gap and as

a result the flux density in the air-gap is reduced.

S

Fig. 2 Fringing effect

Area at

air-gap path

Area at

iron path

N

Problems involving simple magnetic circuits

Before doing problems involving magnetic circuits it is necessary to know some

more terms associated with the magnetic circuit.

Magnetic field intensity, (also called as Magnetizing force) denoted as H, is the

mmf per unit length of magnetic flux path. Thus,

H =

I N

Flux density is proportional to magnetic field intensity. Thus HB . The constant

of proportionality is called permeability, µ. Thus B = µ H or

µ = B / H

Permeability of vacuum or free space is denoted as µ0. Its value is

4 π x 10-7. Permeability of any other medium is given by

µ = µ0 µr

where µr is called the relative permeability of the medium.

An expression for Reluctance, S can be obtained as follows.

r0r0 μμaSgetwe

NIHSince;

aHμμ

IN

aB

IN

φ

INS

Permeance, P is the reciprocal of Reluctance.

An iron core coil with a small air gap is shown in Fig. 3.

Note that

1.Tolal reluctance of magnetic path = reluctance of iron path + reluctance of air gap path.

2. Total source mmf = mmf required to establish flux in iron path + mmf required to

establish flux in air gap.

3. mmf = reluctance x flux OR mmf = H x length of magnetic path

Coil has N turns.

Current through coil = I

Mean radius of magnetic path = Rm

Cross section of core is circular with diameter d

Length of air gap = ℓ g

Fig. 3 Iron core coil

I I

Rm Rm

Example 1

A toroidal air core coil with 2000 turns has a mean radius of 25 cm. The diameter

of each turn is 6 cm. If the current in the coil is 10 A, find (a) MMF (b) flux and (c)

flux density. (Toroidal coil consists of copper wire wrapped around a cylindrical

core)

Solution

Given Air core coil N = 2000 turns;

Rm = 25 cm; d = 6 cm.

(a) MMF = N I = 2000 x 10 = 20000 AT

(b) Flux = MMF / Reluctance

Reluctance, S = r0 μμa

; Since it is air core = µr = 1

ℓ = 2 π x 0.25 = 1.5708 m; a = π r2 = π x 0.032 = 0.002827 m2

Reluctance, S = 8

710x4.4217

10xπ4x0.002827

1.5708

AT / Wb

Flux, φ = 810x4.4217

20000 = 4.5231 x 10-5 Wb

(c) Flux density, B = 0.002827

10x4.5231

a

φ 5

= 0.016 Wb / m2

I

Rm

Rm

I

I

Example 2

The flux produced in the air gap between two magnetic poles is 0.05 Wb. If the

cross sectional area of the air gap is 0.2 m2, find (a) flux density, (b) magnetic

field intensity, (c) reluctance and (d) permeance of the air gap. Find also the mmf

dropped in the air gap, given the length of air gap to be 1.2 cm.

Solution

Given Flux, φ = 0.05 Wb; a = 0.2 m2; ℓ g = 0.012 m

(a) Flux density, B = 2m/Wb0.250.2

0.05

(b) Depending on the data H can be calculated either from H =

NI or

H =μ

B ; Magnetic field intensity, H = m/AT10 x 1.9894

10xπ4

0.25

μ

B 5

7

(c) Reluctance, S = Wb/AT10 x 4.774610xπ4x0.2

0.012

μa

4

70

(d) Permeance, P = AT/Wb10 x 2.094410 x 4.7746

1

S

1 5-

4

MMF = H x ℓ g = 1.9894 x 105 x 0.012 = 2.3873 x 103 AT OR

MMF = Flux x Relectance = 0.05 x 4.7746 x 104 = 2.3873 x 103 AT

Example 3

A ring has mean diameter of 15 cm, a cross section of 1.7 cm2 and has a radial

gap of 0.5 mm cut in it. It is uniformly wound with 1500 turns of insulated wire and

a current of 1 A produces a flux of 0.1 mWb across the gap. Calculate the relative

permeability of iron on the assumption that there is no magnetic leakage.

Solution

Given Dm = 15 cm; a = 1.7 cm2; ℓ g = 0.5 mm; N = 1500 turns; I = 1 A;

φ = 0.1 mWb

Dm = 15 cm; a = 1.7 cm2; ℓ g = 0.5 mm;

N = 1500 turns; I = 1 A; φ = 0.1 mWb

MMF produced = 1500 x 1 = 1500 AT

Total reluctance = MMF / Flux = 1500 / 0.0001 = 1500 x 104 AT / Wb

Total reluctance = Reluctance of iron path + Reluctance of air gap

Reluctance of air gap = Wb/AT10 x 2.340510xπ4 x10x1.7

0.0005

μa

6

740

Reluctance of iron path = 15 x 106 – 2.3405 x 106 = 12.6595 x 106 AT / Wb

Length of iron path = π x 15 x 10-2 – 0.05 x 10-2 = 47.0739 x 10-2 m

Thus 12.6595 x 106 = r

6

r

74

2

r0 μ

10x2203.5423

μx10xπ4x10x1.7

10x47.0739

μμa

Thus µr = 2203.5423 / 12.6595 = 174.0623

I

I

Rm

Rm

Example 4

A series magnetic circuit has an iron path of length 50 cm and an air gap of 1mm.

The cross section of the iron is 6.66 cm2 and the exciting coil has 400 turns.

Determine the current required to produce a flux of 0.9 mWb in the circuit. The

following points are taken from the magnetization curve for the iron.

Flux density (Wb / m2): 1.2 1.35 1.45 1.55

Magnetizing force (AT / m): 500 1000 2000 4000

Solution

Given ℓi = 0.5 m; ℓg= 1 x 10-3 m; a = 6.66 x 10-4 m2; N = 400; Flux = 0.9 mWb

Reluctance of air gap = Wb/AT10 x 1.194910xπ4 x10x6.66

10 x 1

μa

6

74

-3

0

Required air gap mmf = 0.9 x 10-3 x 1.1949 x 106 = 1075.4 AT

Flux density in the iron path = 0.9x 10-3 / (6.66 x 10-4) = 1.3514 Wb / m2

From the given data, for a flux density of 1.3514 Wb / m2

corresponding value of H = 1000 + (1000 x 0.0014 / 0.1) = 1014 AT / m

Required iron path mmf = 1014 x 0.5 = 507 AT

Total mmf required = 1075.4 + 507 = 1582.4 AT

Current required = 1582.4 / 400 = 3.956 A

Example 5

An iron rod of 1 cm radius is bent to a ring of mean diameter 30 cm and wound

with 250 turns of wire. Assume the relative permeability of iron as 800. An air gap

of 0.1 cm is cut across the bent ring. Calculate the current required to produce a

useful flux of 20000 lines if (a) leakage is neglected and (b) leakage factor is 1.1.

Solution

Given r = 1 cm; Dm= 0.3 m; N = 250; µr = 800;

ℓg = 0.001 m; Flux φ= 20000 / (108) = 0.2 mWb

Leakage is neglected Flux in iron path = flux in air gap

Area of cross section, a = π x 10-4 = 0.0003142 m2

Reluctance of air gap = Wb/AT10 x 2.532710xπ4 x0.0003142

0.001

μa

6

70

Required air gap mmf = 0.0002 x 2.5327 x 106 = 506.54 AT

I

I

Rm

Rm

Length of iron path = (π x 0.3) – 0.001 = 0.9415 m

Reluctance of iron path =

Wb/AT10 x 2.9807800x10xπ4 x0.0003142

0.9415

μμa

6

7r0

Required iron path mmf = 0.0002 x 2.9807 x 106 = 596.14 AT

Total mmf required = 506.54 + 596.14 = 1102.68 AT

Current required = 1102.68 / 250 = 4.4107 A

Leakage factor is 1.1

As in previous case, required air gap mmf = 0.0002 x 2.5327 x 106 = 506.54 AT

To maintain useful flux of 0.2 mWb in the air gap,

flux required in the iron path = 1.1 x 0.2 = 0.22 mWb

Required iron path mmf = 0.00022 x 2.9807 x 106 = 655.754 AT

Total mmf required = 506.54 + 655.754 = 1162.294 AT

Current required = 1162.294 / 250 = 4.6492 A

Find the emf of the battery in the circuit shown.

8 A

0.4 Ω

0.6 Ω 0.6 Ω E

0.1 Ω

Example 6

The magnetic circuit shown in Fig. 4 has the following dimensions: ℓ1 = 10 cm,

ℓ2 = ℓ3 = 18 cm, cross sectional area of ℓ1 path = 6.25 x 10 - 4 m2, cross sectional

area of ℓ2 and ℓ3 paths = 3 x 10 - 4 m2, length of air gap = 1 mm. Taking the relative

permeability of the material as 800, find the current in the 600 turn exciting coil to

establish a flux of 100 x 10 - 6 Wb in the air gap, neglecting leakage and fringing.

ℓ3 ℓ2

Fig. 4 - Example 6

ℓ1

Solution

Given ℓ1 = 10 cm; ℓ2 = ℓ3 = 18 cm; a1 = 6.25 x 10- 4 m2; a2 = a3 = 3 x 10=4 m2

ℓg = 1mm; µr = 800; N = 600; 1φ = 100 x 10- 6 Wb

Reluctance of path 1:

Reluctance of air gap = Wb/AT10 x 1.273210xπ4 x10x6.25

10 x 1

μa

6

74

-3

0

Length of iron path = 10 – 0.1 = 9.9 cm

Reluctance of iron path =

Wb/AT10 x 0.1576

800x10xπ4 x10x6.25

0.099

μμa

6

74r0

Thus R1 = (1.2732 + 0.1576) x 106 = 1.4308 x 106 AT / Wb

MMF1 = 100 x 10- 6 x 1.4308 x 106 = 143.08 AT

0.5968x 106 AT/Wb

50 x 10-6

Wb

1.2732 x 106 AT/Wb ((air gap)

0.5968x 106 AT/Wb

172.92 AT

0.1576 x 106 AT/Wb 50 x 10

-6 Wb

100 x 10-6

Wb

Reluctance of path 2:

Reluctance of iron path =

Wb/AT10 x 0.5968

800x10xπ4 x10x3

0.18

μμa

6

74r0

Flux in path 1 will divide equally; Thus flux = 50 x 10-6 Wb

MMF2 = 50 x 10-6 x 0.5968 x 106 = 29.84 AT

Since path 2 and path 3 are in parallel, it is required to consider mmf for only one

of them. Thus, Total MMF = 143.08 + 29.84 = 172.92 AT

Exciting current required = 172.92 / 600 = 0.2882 A

Electrical equivalent of the magnetic circuit considered is shown below.

Hmax, Bmax

H

B

N

O

R

P

S

M

Q

- Hmax, - Bmax

Hysteresis loss

Consider an iron bar which can be magnetized as shown in Fig. 6. Magnetizing

force, H can be varied by controlling the current through the coil. Corresponding

values of flux density B can be noted. First the B-H curve will follow OM shown if

Fig. 7. Now if H is decreased gradually, B will not decrease along MO. Instead it

will decrease along MN. Even when H is zero, B has a definite value ON. This

implies that even on removing the magnetizing force, H, the iron bar is not

getting demagnetized completely. The value of ON measures the retentivity of the

material.

Fig. 6 – Circuit for B-H curve Fig. 7 – Hysteresis loop

- Hmax O P

M

Q

R

S

N

Hmax

To demagnetize the iron bar, the magnetizing force has to be applied in the

reverse direction. Flux density, B becomes zero at P. The value of H as measured

by OP is known as coercive force. If H is further increased, the curve will follow

the path PQ. By taking H back from – Hmax, a similar curve QRSM is obtained. It is

seen that B always lags behind H. This lagging character of B with respect to H is

called hysteresis and the complete loop is called hysteresis loop. Different

magnetic material will have different hysteresis loop. Fig. 8 shows the hysteresis

loop of cast steel and alloyed steel.

Area of hysteresis loop gives the hysteresis loss per unit volume of the material.

About 4% addition of Slican to steel give rise to reduction in hysteresis loop area

and hence hysteresis loss.

Fig. 8 – Hysteresis loop of cast steel and alloyed steel

Eddy current loss

Whenever a conducting material cuts the magnetic flux (armature core in the

case of rotating machines) an emf is induced in the core. This emf sets up

current through the solid mass. Such current is known as eddy current. Flow of

eddy current results in eddy current loss.

The eddy current loss is proportional to square of the thickness of the material.

This loss can be minimized by using a laminated core, which offers high

resistance for the flow of eddy current.

A B

K

Faraday’s Laws of Electromagnetic Induction

When a current flows in a conductor, magnetic field is produced. The reverse

phenomenon, whereby an Electro Motive Force (EMF) and hence current is

produced in an electric circuit by some action on magnetic field. This is called

electromagnetic induction. Consider the setup shown in Fig. 9.

When the switch, K is closed from the open position, there will be induced

voltage and hence current in coil B as indicated by the galvanometer G. When the

key is opened from the closed position, the current flow will be in the opposite

direction. This illustrates the production of static induced emf.

A

G

Fig. 9 – Static induced emf

Consider the setup shown in Fig. 10.

When the conductor AB is moved from the top position in the downward

direction, it cuts the magnetic field at right angle. An emf is induced in the

conductor resulting current flow as indicated by the galvanometer. When the

conductor is moved from the bottom position in the upward direction, there will

be current flow in the opposite direction. This illustrates the production of

dynamic induced emf.

S

A

B

N

G

Fig. 10 – Dynamic induced emf

The results of the above two experiments can be summed up into two laws,

known as Faraday’s Laws of Electromagnetic Induction.

First Law: Whenever the flux linking with a coil changes, a static emf is induced

in it and as such the emf lasts only for the time the change is taking place.

OR

When a moving conductor cuts the magnetic field, an emf induced in it which is

called as dynamic emf.

Second Law: The magnitude of the induced emf is equal to the rate of change of

flux linkage. Flux linkage = Flux x Number ot turns having units as Wb. turns.

Induced emf

An emf is induced in a coil or conductor whenever there is a change in flux

linkages. The change in flux linkages can occur in two ways.

(i) The coil is stationary and the magnetic field is changing. Resulting

induced emf is known as static induced emf. Transformer works on this

principle.

(ii) The conductor is moved in a stationary magnetic field in such a way

that there is change in flux linkage. Resulting induced emf is known as

dynamic induced emf. Generator works on this principle.

Static induced emf

In this case, the coil is held stationary and the magnetic field is varied. The

induced emf may be self induced or mutually induced.

A B

K

Two coils are wound over a magnetic specimen. Coil A is energized using a

battery. If switch K is initially closed, then a steady current of I ampere will flow

through the coil A. It produces a flux of φ Wb. Let us assume that the entire flux

links coils A and B. When the switch is suddenly opened, the current reduces to

zero and the flux linking both the coils becomes zero. As per Faraday’s law, emf

is induced in both the coils A and B. Such emfs are known as static induced

emfs. Static induced emf can be classified into two categories, namely self

induced emf and mutually induced emf.

Fig. 9 – Static induced emf

A

G

Self induced emf

If a single coil carries a current, flux will be set up in it. If the current changes,

the flux will change. This change in flux will induce an emf in the coil. This kind of

emf is known as self induced emf. In other words, self induced emf is the emf

induced in a circuit when the magnetic flux linking it changes because of the

current changes in the same circuit.

The magnitude of this self induced emf e = N dt

dφ

td

dψ

Mutually induced emf

Mutually induced emf is the emf induced in one circuit due to change of flux

linking it, the flux being produced by the current in another circuit.

Referring to Fig. 9, when a change in current though coil A occurs, we find the

flux linking coil B changes. Hence, an emf is induced in coil B and it is called as

mutually induced emf.

S N

A

B

Dynamic induced emf

Consider the experimental setup shown in Fig. 10.The magnetic poles, produce a

stationary flux density of B Wb. / m2. Let the conductor length be ℓ meters. The

conductor is moved at right angle to the field. Let the distance moved in dt

second be dx meters.

Area swept by the conductor in dt sec. = ℓ dx m2

Magnetic flux cut by the conductor = B ℓ dx Wb.

G

Fig. 10 – Dynamic induced emf

Taking the conductor has one turn, corresponding

flux linkage, ψ = B ℓ dx Wb Turn

Rate of change of flux linkage = B ℓ dt

dx

According to Faraday’s Law, this is the induced emf, e in the conductor.

Thus induced emf, e = B ℓ v volts

where v = linear velocity = dt

dx

Let the conductor be moved with velocity v m / sec. in an inclined direction,

making an angle ө to the direction of field. Then

Induced emf, e = B ℓ v sin ө volts

This is the basic principle of working of a generator.

S N

A

B F I

Force on current carrying conductor

Consider the setup shown in Fig. 11. When a current of I ampere flows in the

conductor from A to B, it will experience a force, F given by

F = B ℓ I Newton

This relation is true if the conductor is at right angle to the magnetic field. In case

if the conductor is an inclined direction, making an angle ө to the direction of

field, then

F = B ℓ I sin ө Newton

This is the basic principle of working of a motor.

Fig. 11 – Force on current carrying conductor

Self inductance, L

Self inductance of a coil, L is the rate of change of flux linkages with respect to

the current in it. Its unit is Henry. Thus

L = Id

dψ = N

Id

dφ Henry

Equation for self inductance

Consider a magnetic circuit shown in Fig. 12.

Fig. 12 Self inductance

I

With usual notations

Magnetizing force, H =

IN AT / m

Flux density, B = µ0 µr H = µ0 µr (

IN) Wb. / m2

Magnetic flux, φ = µ0 µr (

IN) a Wb.

Flux linkage = N φ = µ0 µr (

I2N

) a Wb. Turns

Self inductance, L = N Id

dφ = N

I

φ =

aNμμ 2

r0 =

)μμa/(

N

r0

2

= Reluctance

N2

Thus self inductance L =

Id

dψ = N

Id

dφ =

Reluctance

N2

I

Expression for self induced emf in terms of self inductance

The magnitude of self induced emf, e = N dt

dφ

Thus self induced emf, e = N dt

dx

d

dφ I

I

= L dt

dI

The self induced emf in a circuit is directly proportional to the rate of change of

current in the same circuit.

I1

Mutual inductance

Mutual inductance between two circuits is defined as the flux linkages of one

circuit per unit current in the other circuit.

Reluctance of the magnetic circuit = S

Flux in coil 1 φ1 = S

N

Reluctance

N 11 11 II

Assuming that all the flux φ1 links the entire coil 2,

flux linkage of circuit 2 due to current in circuit 1 ψ21 = S

NN 21 1I

Mutual inductance M = S

NNψ 21

1

21 I

Coil 2 Coil 1

21φ

21 φφ

I1 I2

Consider two air core coils having self inductances L1 and L2 that are closer to

each other as shown in Fig. 12. When current passes through coil 1, flux φ11

is

produced in coil 1. Only a part of this flux links with coil 1 and the remaining flux

links both the coils 1 and 2. Generally, the flux linking both the coils is useful and

it is called mutual flux and represented by φ21

. The other part of the flux is called

leakage flux represented by φℓ1

. When the coil 2 carries current, flux produced in

it is φ22

and leakage flux is φℓ2

and the mutual flux is φ12

. Fluxes φℓ1

, φ21

,φℓ2

and φ12

are shown in Fig. 13.

12φ

Fig. 13 Two coils in proximity

The operation of many useful devises which utilizes mutual inductance

phenomenon depends upon how close the coils are coupled to each other. A

fraction of total flux produced by a coil links both the coils and this coefficient

represented by k. The coefficient of coupling depends on the relative position of

coils 1 and 2. Thus, coefficient of coupling, k = 22

21

11

12

φ

φ

φ

φ . It is to be noted that

coefficient of coupling is always ≤ 1. If both the coils are far apart, then k = 0. On

the other hand if both the coils are wound over the same core, then k = 1. Similar

to the definition of self inductance, mutual inductances can be written as

M12 = N1 2

21

d

φd

I and M21 = N2

1

12

d

φd

I

Using energy criteria, it can be proved that M12 = M21 = M

Then M2 = N1 N2 2

21

d

φd

I

1

12

d

φd

I = N1 N2 k

2

22

d

φd

I k

1

11

d

φd

I

= k2 N1 1

11

d

φd

IN2

2

22

d

φd

I = k2 L1 L2

Thus M = k 21 LL

Certain formulae

Static induced emf:

e = N dt

dφ volts e = L

dt

dI volts

Dynamic induced emf:

e = B ℓ v sin ө volts

Force on a current carrying conductor:

F = B ℓ I sin ө Newton

Self inductance:

L = N Id

dφ Henry L =

Reluctance

N2

Henry

Mutual Inductance:

M = S

NN 21

M = k 21 LL

Example 7

A coil of resistance 150 Ω is placed in a magnetic flux of 0.1 m Wb. It has 500

turns and a galvanometer of 450 Ω resistance is connected in series with it. The

coil is moved from the given field to another field of 0.3 m Wb. In 0.1 sec. Find the

average induced emf and the average current through the coil.

Solution

Given Rc = 150 Ω; 1φ = 0.1x 10-3 Wb.; N = 500 turns; Rg = 450 Ω; 2φ = 0.3x 10-3 Wb.;

t = 0.1 sec.

Induced emf, e = N dt

dφ = 500 x

0.1

)10x0.110x(0.3 33 = 500 x 2 x 10-3 = 1.0 Volt

Current, I = induced emf / total resistance = 1.0 / (150 + 450) = 0.001667 A

Example 8

A conductor of length 100 cm moves at right angle to a uniform magnetic field of

flux density 1.5 Wb. / m2 with a velocity of 30 m / sec. Calculate the emf induced

in it.

Find also the value of induced emf when the conductor moves at an angle of 600

to the direction of the magnetic field.

Solution

Given ℓ = 1.0 m; ө = 900 ; B = 1.5 Wb. / m2; v = 30 m / sec.; ө = 600

Induced emf, e = B ℓ v = 1.5 x 1.0 x 30 = 45 V

With ө = 600. Induced emf, e = B ℓ v sin ө = 45 x sin 600 = 38.9711 V

Example 9

A conductor of 10 cm long lies perpendicular to a magnetic field of strength 1000

AT / m., Find the force acting on it when it carries a current of 60 A.

Solution

Given ℓ = 0.1 m; ө = 900 ; H = 1000 AT / m; I = 60 A

Flux density, B = µ0 H = 4 π x 10-7 x 1000 = 0.001257 Wb. / m2

Force, F = B ℓ I = 0.001257 x 0.1 x 60 = 0.00754 Newton

Example 10

An air cored toroidal coil has 480 turns, a mean length of 30 cm and a cross-

sectional area of 5 cm2. Calculate (a) the inductance of the coil and (b) the

average induced emf, if a current of 4 A is reversed in 60 m sec.

Solution

Given N = 480 turns; ℓ = 0.3 m; a = 5 x 10-4 m2; dI = 8 A; dt = 60 x 10-3 sec.

Inductance, L = N2 / Reluctance

Reluctance, S = ℓ / ( a µ0) = AT/Wb10x0.477510xπ4x10x5

0.3 9

74

Inductance, L = Hm0.482510x0.482510x0.4775

480 3

9

2

Induced emf, e = L V0.06433x1060

8x10x0.4825

dt

d3

3

I

Example 11

A current of 5 A when flowing through a coil of 1000 turns establishes a flux of

0.3 m Wb. Determine the self inductance if the coil.

Solution

Given I = 5 A; N = 1000 turns; φ = 0.3x 10-3 Wb.;

Self inductance, L = N dI

dφH0.06

5

10x0.31000

3

x

Example 12

A coil has a self inductance of 30 mH. Calculate the emf in the coil when the

current in the coil (a) increases at the rate of 300 A / sec. (b) raises from 0 to 10 A

in 0.06 sec.

Solution

Given L = 30 x 10-3 H;

(a) Induced emf, e = L V9300x10x30dt

d 3 I

(b) Induced emf, e = L V50.06

10x10x30

dt

d 3 I

Example 13

The number of turns in a coil is 250. When a current of 2 A flows in this coil, the

flux in the coil is 0.3 m Wb. When this current is reduced to zero in 2 m sec., the

voltage induced in another coil is 63.75 V. If the coefficient of coupling between

the two coils is 0.75, find the self inductances of the two coils, mutual inductance

and the number of turns in the second coil.

Solution

Given N1 = 250; I1 = 2 A; 1φ = 0.3x 10-3 Wb.; dI1 = 2 A ; dt1 = 2 m sec; e2 = 63.75 V;

k = 0.75

N1 = 250; I1 = 2 A; 1φ = 0.3x 10-3 Wb.; I1’ = 0 ; dt = 2 m sec; e2 = 63.75 V; k = 0.75

Self inductance, L1 = H0.03752

10x0.3x250

d

dφN

3

1

11

I

Induced emf in coil 2, e2 = M 63.750.002

2xM

dt

d1I

Thus mutual inductance, M = 63.75 mH

Since M = k 21 LL

0.063752 = 0.752 x 0.0375 x L2

Thus self inductance of coil 2, L2 = 0.1927 H

Flux 1φkφ 2 = 0.75 x 0.3x 10-3 Wb = 0.225 x 10-3 Wb; 0φkφ 1 ''

2

Also, e2 = N2 x 63.7510x2

10x0.225xN

dt

dφ3

3

22

Thus N2 = 567

I1 = 2 A; I1’ = 0

φ1 = 0.3 m Wb φ1

’ = 0;

dt = 2 m sec.

N1 = 250

φ2 = 0.225 m Wb φ2

’ = 0;

k = 0.75

FUNDAMENTALS OF AC

Fig. 1

(a) Sinusoidal waveform (c) Sawtooth waveform (b) Rectangular waveform

Electrical appliances such as lights, fans, air conditioners, TV, refrigerators, mixy,

washing machines and industrial motors are more efficient when they operate

with AC supply. The required AC voltage is generated by AC generator also called

as alternator.

A waveform is a graph in which the instantaneous values of any quantity are

plotted against time. A periodic waveform is the one which repeats itself at

regular intervals. A waveform may be sinusoidal or non sinusoidal. Examples of a

few periodic waveforms are shown in Fig.1.

xm

x(t)

ωt

Fig. 2

Alternating waveform is a waveform which reverses its direction at regular

intervals. Sinusoidal and rectangular waveforms shown above are alternating

waveforms. Let us see more details about sinusoidal waveform.

Fig. 2 shows a sinusoidal waveform, which can be called as a sinusoid. It can

represent a voltage or current. Its equation can be written as

x(t) = xm sin (ωt + φ) (1)

Thus a sinusoid is described in terms of

i) its maximum value

ii) its angular frequency, ω and

iii) its phase angle φ

It is evident that sinusoid repeats in a cyclic manner. The number of cycles it

makes in one second is called the frequency (f). Thus the unit for frequency is

cycles per second which is also commonly known as hertz (Hz). Electric supply

has a frequency of 50 or 60 Hz. In communication circuit, the frequency will be in

the order of Mega Hz.

The time taken by the sinusoid to complete one cycle is called the period (T) of

the sinusoid. When the supply frequency is 50 Hz, the sinusoid makes 50 cycles

in one second. Thus the period is 1/50 = 0.02 second. The frequency and the

period are related as

T = f

1 or f =

T

1 (2)

The angular frequency of sinusoid is represented by ω and its unit is radians per

second. In one cycle the angle covered is 2π radians. When the frequency is f

cycles per second, the angle covered in one second will be 2πf radians. Thus

ω = 2 π f (3)

While drawing a sinusoid, instead of ωt, time t can be taken in the x-axis.

Example 1

Consider the voltage sinusoid

v(t) = 70 sin ( 60 t + 20 0 ) V

Find the amplitude, phase, angular frequency, frequency, period and the value of

voltage at time t = 0.25 s.

Solution

Amplitude v m = 70 V

Phase φ = 20 0

Angular frequency ω = 60 rad / s

Frequency f = 2π

60

2π

ω = 9.5511 Hz

Period T = f

1 =

9.5511

1 = 0.1047 s

Voltage value at t = 0.25 s is

v (0.25) = 70 sin ( 60 x 0.25 xπ

180 + 20 0 ) = 24.59 V

ωt

x(t)

Fig. 3

x(t) = xm sin(ωt)

x(t) = xm sin(ωt + φ)

The two sinusoids shown in Fig. 3 are x(t) = xm sin ωt and x(t) = xm sin(ωt + φ)

The sinusoid x(t) = xm sin(ωt + φ) leads the sinusoid x(t) = xm sin ωt by an angle

of φ. The sinusoids can also be written as

x(t) = xm sin(θ + φ) (4)

The average value of the periodic waveform can be obtained as:

Average value = Period

cyclecompleteoneunderArea (5)

Average value is also called as mean value.

The Root Mean Square (RMS) value of periodic waveform is:

RMS value = Period

cycleoneforcurvesquaredunderArea (6)

Form Factor is defined as

Form Factor = value Average

ValueRMS (7)

Peak Factor is defined as

Peak Factor = value SMR

ValuePeak (8)

Consider a current waveform described by

i(t) = Im sin θ (9)

Its positive half cycle and negative half cycle of such sinusoids are negative of

each other. Hence the area in one cycle is zero. For such sinusoidal wave form

the average value is the average value over half cycle.

Thus

Area of the curve = sinI0

m

θ dθ = cos(Im θ)

0

= 1)1(Im = 2 Im

Iav = π

I2 m = 0.6366 Im (10)

When we square the waveform i(t) = Im sin θ, the first and the second half of the

cycle will be same. Therefore while computing the R M S value of i(t) = Im sin θ it

is enough to consider only one half cycle.

Area of square curve = 2

0

2

m sinI

θ dθ

= 2cos(12

I2

m

0

θ) dθ = 2

I2

m ( θ - )2

2sin

0

= 2

I2

m [π – 0] = 2

mI2

π

Mean square value = 2

I2

m

RMS value =2

Im (11)

= 0.7071 Im (12)

Form factor = value Average

ValueRMS = 1.11

I0.6366

I0.7071

m

m (13)

Peak factor = value SMR

ValuePeak = 1.414

I0.7071

I

m

m (14)

v

x

Vm

t

Fig. 6

v2

x

V2m

t

Fig. 5

A1

We may be calculating average and RMS values of waveforms in which inclined

straight line variations are present. Consider the waveform shown in Fig. 4. Its

square curve is shown in Fig. 5. Area A1 of the square curve can be calculated as

follows.

Equation of the straight line is: v = tx

vm ; Then v2 = 2

2

2

m tx

v

Area A1 = x

0

dttx

V 2

2

2

m = 3

t

x

v 3

2

2

m x

0

= xV3

1 2

m

It can be verified that the above result is true for the waveform shown in Fig. 6

also.

v

x

Vm

t

Fig. 4

2

25

i2

Fig. 8

Find the average and RMS

values of the waveform

shown in Fig. 7 t

5 A

2 4

Fig. 7

Example 2

Solution

Iav = A2.52x5x2

1x

2

1triangletheofareax

2

1

The square curve is shown in Fig. 8.

Area of square curve = 16.66632x25x3

1

Mean square value = 8.33322

16.6663

RMS value = A2.88678.3332

π/3 2π/3

π t

Vm

v

Fig. 9

Vm2

3

π

3

π 2

π

A3

A2

A1

Fig. 10

Example 3

Find the average and RMS value of the waveform shown in Fig. 9.

Solution

Area of positive half cycle = mmm V3

π

2

1V

3

πV

3

π

2

1 = mV

3

π2

Average value = mm V0.6667V3

2

The square curve is shown in Fig. 10.

Area of square curve

= 2

m

2

m

2

m

2

m Vπ9

5V

9

πV

3

πV

9

π

Mean Square value = 2

mV9

5; Thus RMS value = 0.7454 Vm

Im

i

θ

Fig. 11

2π 0

π

Example 4

Find the average and RMS values of the half wave rectified sine wave shown in

Fig. 11.

Solution

As seen earlier, area of half sine wave = 2 Im

Total area = 2 Im + 0 = 2 Im

Average value Iav = π2

I2 m = 0.3183 Im

As seen earlier, area of square of half sine wave = 2

mI2

π

Total area of square curve = 2

mI2

π

Mean of square curve = π2

1 2

mI2

π = 2

mI4

1= 0.25 2

mI

RMS value IRMS = 0.5 Im

Im

i

θ

Fig. 12

2π 0

π

Example 5

Find the average and RMS values of the full wave rectified sine wave shown in

Fig. 12.

Solution

As seen earlier, area of half sine wave = 2 Im

Total area = 2 Im + 2 Im = 4 Im

Average value Iav = π2

I4 m = mIπ

2 = 0.6366 Im

As seen earlier, area of square of half sine wave = 2

mI2

π

Total area of square curve = 2

mIπ

Mean of square curve = π2

1 2

mIπ = 2

mI2

1

RMS value IRMS = 2

Im = 0.7071 Im

If the waveform is the sum of several waveforms, its AVERAGE value is the sum

of average values of its components and its RMS values can be obtained as

follows. Let

W = W1 + W2 + W3 and their RMS values be W1 RMS, W2 RMS and W3 RMS respectively.

Then

WRMS = 2

RMS3

2

RMS2

2

RMS1 WWW

Example 6

A conductor carries simultaneously a direct current of 10 A and a sinusoidal

alternating current with a peak value of 10 A. Find the RMS value of the conductor

current.

Solution

Conductor current i(t) = (10 + 10 sin ωt) A

Here W1 = 10 A and W2 = 10 sin ωt A

Therefore W1 RMS = 10 A; W2 RMS = 7.071 A

RMS value of conductor current = 12.2474A7.07110 22

E2

1 A

E1

8 Ω

2 V 5 V

+ -

+

-

-

-

+

+

3 A

Cycle Test 1 28 – 02 – 2011

1. Briefly explain the fringing effect.

2. Define Magnetic field intensity.

3. State and explain Kirchhoff’s current law and Kirchhoff’s voltage law

with suitable examples.

4. Using Kirchhoff’s laws, find the voltages E1 and E2.

3 k Ω

6 k Ω

18 k Ω 8 k Ω

20 k Ω

- +

12 V

5. Find the power delivered by the voltage source.

6. Find the ampere turns required to produce a flux of 0.4 mWb. in the

air gap of 0.5 mm in a circular magnetic circuit. The iron ring has a

cross section of 4 mm2 and 63 cm mean length. Assume the relative

permeability of iron as 800. Neglect flux leakage.

Reluctance of air gap = Wb/AT10 x 9.947210xπ4 x10x4

10 x 0.5

μa

7

7

-2

0

6

Reluctance of iron path = r0 μμa

Wb/AT10 x 1.5667800x10xπ4 x10x4

10 x 63 8

76

-2

Total mmf required = 0.4 x 10-3 x 25.6142 x 107 = 1.0246 x 105 AT

SINGLE PHASE AC CIRCUITS

PHASORS

Consider a linear ac circuit having one or more sinusoidal inputs having same

frequency as shown in Fig. 1. The amplitudes and phase angles of the inputs may

be different while their frequency should be same.

The output what we may be interested may be voltage across an element or

current through an element. The output waveform will be sinusoidal with the same

frequency as the input signals. This could be easily verified experimentally.

The steady-state analysis of such circuits can be carried out easily using phasors.

Linear ac circuit

Sinusoidal

inputs of

same

frequency

Sinusoidal

outputs of

same

frequency

Fig. 1

A sinusoid is fully described when its maximum value, angular frequency and

phase are specified.

A question may arise whether we should always deal with such sinusoidal time

function to represent voltage and current in ac circuits. When all the inputs are

sinusoidal time function with the same angular frequency ω, the voltage or the

current in any part of the circuit will also be of sinusoidal time function with the

SAME ANGULAR FREQUENCY ω. Hence it is redundant to carry information of

ω, while representing voltages and currents in ac circuits. This idea gives birth to

the concept of PHASORS.

The phasor corresponding to sinusoid x(t) = x m cos (ωt + φ) is X = 2

xm φ (1)

In case x(t) is expressed as x(t) = x m sin (ωt + φ), it can be written as

x(t) = x m sin (ωt + φ) = x m cos (ωt + φ – 2

π) and the corresponding phasor is

X = 2

xm 2

π (2)

In a similar way we can state:

If x(t) = - x m cos (ωt + φ) its phasor is X = 2

xm φ- π (3)

If x(t) = - x m sin (ωt + φ) its phasor is X = 2

xm 2

π (4)

Eqs. (1) to (4) are useful to find the phasor for a given sinusoid.

Fig. 1 Quadrants for Phasor sin

Fig. 1 is useful to locate

the quadrant in which

the phasor lies

- cos cos

- sin

A few sinusoids and the corresponding phasors are;

x1 (t) = 2 150 cos ( ωt + 15 0 ) X

1 = 150 15 0

x2 (t) = 2 150 cos ( ωt - 75 0 ) X

2 = 150 - 75 0

x3 (t) = 2 100 sin ωt X

3 = 100 - 90 0

x4 (t) = 2 100 sin (ωt + 30 0 ) X

4 = 100 - 60 0

x5 (t) = 2 100 sin ( ωt - 150 0 ) X

5 = 100 - 240 0

x6 (t) = - 2 80 cos ( ωt + 30 0 ) X

6 = 80 210 0

x7 (t) = - 2 80 sin (ωt - 30 0 ) X

7= 80 60 0

.

Fig. 2 Phasors of given sinusoids

The above Phasors are

shown in Fig. 2

X7

X6

X5

X4 X

3

X2

X1

The important motivation for the use of phasors is the ease with which two or

more sinusoids at the same frequency can be added or subtracted. In the

sinusoidal steady state, all the currents and voltages are of same frequency.

Hence phasors can be used to combine currents or voltages. KCL and KVL can

be easily interpreted in terms of phasor quantities.

A phasor is a transformed version of a sinusoidal voltage or current waveform

and it contains the amplitude and phase angle information of the sinusoid.

Phasors are complex numbers and can be depicted in a complex plane. The

relationship of phasors on a complex plane is called a phasor diagram.

Example 1

Using phasor concept, find the sum of 4 voltages given by :

v 1 = 2 50 sin ωt v 2 = 2 40 sin (ωt + π / 3)

v 3 = 2 20 sin (ωt – π / 6) v 4 = 2 30 sin (ωt + 3π / 4)

Solution

In phasors corresponding to the sinusoids are:

V 1 = 50 - 90° = 0.0 - j 50.0

V 2 = 40 - 30° = 34.6410 - j 20.0

V 3 = 20 - 120° = -10.0 - j 17.3205

V 4 = 30 45° = 21.2132 + j21.2132

V 1 + V 2 + V 3 + V 4 = 45.8542 - j66.1073

= 80.4536 - 55.250

Corresponding sinusoid is obtained as

v T = 2 80.4536 cos (ω t - 55.250) = 113.7786 cos (ω t - 55.25

0)

= 113.7786 sin (ω t + 34.750)

3

j3

3

j4 3 +j 4

OPERATOR j

Operator j is useful in dealing with COMPLEX NUMBERS.

11801901x901j901j 00020

1j901jj 403

036.8753j4

- j 4.4

4

j 3

2.4

- 3.3

j 2.2

- 3

- j 4

COMPLEX NUMBERS

4 + j 3 - 3.3 + j 2.2 - 3 – j 4 2.4 – j 4.4 are a few complex numbers.

They can be represented either in RECTANGULAR FORM or POLAR

FORM. The numbers shown above are in rectangular form. There is one

phasor corresponding to each complex number as shown below.

0146.342.2j3.3-

0126.9-54j3- 061.4-54.4j2.4

SINGLE ELEMENT IN STEADY STATE Voltage-current relationship of resistor, inductor and capacitor can be obtained in

phasor form. Such phasor representations are useful in solving ac circuits.

RESISTOR

Let the voltage v(t) across the resistor terminals be

v (t) = V m cos ωt (5)

The current through it is given by

tωcosR

V

R

v(t)i(t) m (6)

Expressing the equations (5) and (6) in phasor form we get

0m 02

VV (7)

0m 0R2

VI (8)

The impedance of an element is defined as the ratio of the phasor voltage across

it to the phasor current through it. Thus

I

VZ (9)

For a resistor I

VZ 00R (10)

Thus in the case of a resistor, voltage-current relationship is

V = R I (11)

Representation of resistor in time frame and its phasor form are shown in Fig. 3.

It is to be noted that as seen by the Eqns. (7) and (8),both the voltage V and the

current I have the same phase angle of 0 0 . The phasor diagram showing the

voltage and current in a resistor is shown in Fig. 4.

R

+ v(t) -

I

+ V -

i(t)

+ v(t) -

R

Fig. 3 Representation of a resistor

V

I

Fig. 4 Phasor diagram - resistor

In the phasor diagram shown in Fig. 4, importance must be given to the phase

angles of the voltage V and the current I. The lengths of the phasors depend on

their magnitude and the scale chosen. In no occasion length of a voltage phasor

and the length of a current phasor can be compared since they have different

units. The scale for current phasors will be like

1 cm = x Volts while the scale for the voltage phasors will be like

1 cm = y Ampere.

The impedance of the resistor is R 0°. In a general network where R is

embedded, the phasor corresponding to the voltage across R and the phasor

corresponding to the current through R are always in phase.

V

I

Fig. 4 Phasor diagram - resistor

INDUCTOR

For the inductor, the voltage-current relationship is

dt

di(t)Lv(t) (12)

In steady state, let the current through it be

i (t) = Im cos ωt (13)

Then dt

di(t)Lv(t) = - ωL Im sin ωt (14)

Expressing the above two equations in phasor form we have

0m 02

II (15)

and 0m 902

ILωV (16)

The impedance of the inductor is given by

I

VZ = ωL 90 0 = j ωL = j X L (17)

where X L = ωL (18)

Thus, the terminal relationship of an inductor in phasor form is

V = j X L I (19)

Representation of inductor in time frame and its phasor form are shown in Fig. 5.

It is to be noted that as seen by the Eqns. (15) and (16), the voltage V leads the

current I by a phase angle of 90 0 . The phasor diagram showing the voltage and

current in an inductor is shown in Fig. 6.

L I

+ V -

i(t)

+ v(t) -

j LX

Fig. 5 Representation of a inductor

V

I

Fig. 6 Phasor diagram - inductor

It is to be noted that the voltage V leads the current I by 90° or we can also state

that the current I lags the voltage V by 90°. The steady state impedance

corresponding to the inductance L is jX L where X L = ωL. The quantity X L is

known as the INDUCTIVE REACTANCE.

V

I

Fig. 6 Phasor diagram - inductor

CAPACITOR

For the capacitor, the voltage-current relationship is

dt

dv(t)Ci(t) (20)

In steady state, let the voltage across it be

v (t) = Vm cos ωt (21)

Then dt

dv(t)Ci(t) = - ωC Vm sin ωt (22)

Expressing the above two equations in phasor form we have

0m 02

VV (23)

and 0m 902

VCωI (24)

The impedance of the capacitor is given by

I

VZ

= Cω

1 - 90 0 = -

Cω

j = - j XC (25)

where XC = Cω

1 (26)

Thus, the terminal relationship of a capacitor in phasor form is

V = - j XC I (27)

Representation of capacitor in time frame and its phasor form are shown in Fig. 7.

C I

+ V -

i(t)

+ v(t) -

- j CX

Fig. 7 Representation of a capacitor

It is to be noted that as seen by the Eqns. (23) and (24), the current I leads the

voltage V by a phase angle of 90 0 . The phasor diagram showing the voltage and

current in a capacitor is shown in Fig. 8.

It is to be noted that the current I leads the voltage V by 90° or we can also state

that the voltage V lags the current I by 90°. The steady state impedance

corresponding to the capacitance C is - jXC where XC = Cω

1. The quantity XC is

known as the CAPACITIVE REACTANCE.

It is conventional to say how the current phasor is relative to voltage phasor.

Thus for the resistor, the current phasor is in phase with the voltage phasor. In

an inductor, the current phasor lags the voltage phasor by 900. In the case of a

capacitor, the current phasor leads the voltage phasor by 90 0 .

V

I

Fig. 8 Phasor diagram of - capacitor

Example 2

The voltage of v = 2 80 cos (100 t - 55 0 ) V is applied across a resistor of 25Ω.

Find the steady state current through the resistor.

Solution

Here V = 80 055 and R = 25 Ω

Thus, current I = A553.225

5580

R

V 00

Current i(t) = 4.5255 cos (100 t - 550) A

Example 3

The voltage of v = 2 20 sin (50 t - 25 0 ) V is applied across an inductor of

0.1 H. Find the steady state current through the inductor.

Solution

Phasor voltage V = 20 - 25 0 - 90 0 = 20 - 115 0 V

Impedance Z = j ω L = j 50 x 0.1 = 5 90 0 Ω

Current I = A2054905

11520

Z

V 0

0

0

Converting this to the time domain

Current i(t) = 5.6569 cos (50 t - 2050) A

= - 5.6569 cos (50 t - 250) A

Example 4

The voltage of v = 2 12 cos (100 t - 25 0 ) V is applied across a capacitor of

50 F. Find the steady state current through the capacitor.

Solution

Phasor voltage V = 12 - 25 0 V

Impedance Z = - j Ω200j10x50x100

1j

Cω

16

Current I = mA6560A650.06A90200

2512

Z

V 00

0

0

Converting this to the time domain

Current i(t) = 84.8528 cos (100 t + 65 0 ) mA

ANALYSIS OF RLC CIRCUITS

An ac circuit generally consists of resistors, inductors and capacitors connected

in series, parallel and series-parallel combinations. Often we need to simplify the

circuit by finding the equivalents. Further to this, we have to make use of KVL,

KCL, source transformation, voltage division and current division what we

discussed in previous chapter, by replacing resistors by impedances and dc

voltages and currents by voltage phasors and current phasors.

A coil used in ac circuit will have its own resistance in addition to the inductive

reactance due to its inductance. One such coil is shown in Fig. 9.

It is clear that the resistance R and inductive reactance j LX are connected in

series. The impedance of this coil is

Z = R + j LX (28)

j LX

R

Fig. 9 A coil in an ac circuit

Now consider a case where a resistance R and a capacitance having a capacitive

reactance - j CX are connected in series as shown in Fig. 10.

The impedance of the circuit is

Z = R - j CX (29)

IMPEDANCE AND ADMITTANCE

The steady state impedance ( a complex quantity ) can be written in two forms,

namely Rectangular form and Polar form as

Rectangular form: Z = R + j X

Polar form: Z = Z

- j CX R

Fig. 10 A resistance and a capacitance in series

If two impedances Z 1 and Z 2 are connected in series as shown in Fig. 11, then

the equivalent impedance is

Z qe = Z 1 + Z 2 (30)

This could be generalized to n number of impedances connected in series as

Z qe = Z 1 + Z 2 + …………….+ Z n (31)

If n number of impedances Z 1 , Z 2 , …., Z n are connected in parallel as shown in

Fig. 12, the equivalent impedance is obtained from

n21qe Z

1...........

Z

1

Z

1

Z

1 (32)

Z1

Z qe

Z 2

Fig. 11 Two impedances connected in series

If two impedances Z 1 and Z 2 are connected in parallel, then the equivalent

resistance is obtained from

21qe Z

1

Z

1

Z

1 (33)

21

21

ZZ

ZZ

Therefore

Z qe = 21

21

ZZ

ZZ

(34)

…… nZ 1Z 2Z Z qe

Fig. 12 n impedances connected in parallel

While dealing with the parallel circuit, it is also useful to define another quantity

called ‘admittance’. ADMITTANCE is defined as the reciprocal of the impedance

and it is denoted by Y. Thus

Y = Z

1 (35)

When the admittance Y is written in rectangular form as

Y = G + j B (36)

G is called as ‘conductance’ and B is called as ‘susceptance’. The unit of G, B

and Y are mho or siemens and is denoted by .

When two impedances Z 1 and Z 2 are connected in parallel, referring to Eqn. (33)

the equivalent admittance Y qe is given by

Y qe = Y 1 + Y 2 (37)

where Y 1 and Y 2 are the admittances corresponding to the impedances Z 1 and

Z 2 respectively.

When n number of admittances Y 1 ,Y 2 ,…….., Y n are connected in parallel, Eqn.

(3.44) can by generalized as

Y qe = Y 1 + Y 2 +…………..+ Y n (38)

If there are n equal impedances Z are connected in series, then the equivalent

impedance is

Z qe = n Z (39)

Similarly if there are n equal admittances Y are connected in parallel, then the

equivalent admittance is

Y qe = n Y (40)

RL CIRCUIT

Having studied how to combine the series and parallel impedances we shall now

see how the RL, RC and RLC circuits can be analyzed.

Let us consider a simple circuit in which a resistor and an inductor are connected

in series as shown in Fig. 13.

Taking the supply voltage as reference

00EE (41)

j X L R

E

I

+ V L - + V R -

Fig. 13 RL circuit

Circuit impedance θZXjRZ L (42)

Circuit current θZ

E

θZ

0E

Z

EI

0

(43)

= θI (44)

where Z

E I (45)

Further θIRIRVR (46)

0

LLL 90θIXIXjV (47)

Using KVL, we get V R + V L = E (48)

The phasor diagram for this RL circuit can be got by drawing the phasors E, I, V R

and V L as shown in Fig. 14.

Consider the triangle formed by the phasors V R , V L and E. Recognizing that

IZEandIXV,IRV LLR if each side of the triangle is divided

by I then R, X L and Z will form a triangle as shown in Fig, 15. This triangle

is known as the IMPEDANCE TRIANGLE.

V L

900

V L

E

V R I

θ

Fig.14 Phasor diagram of RL circuit

E = E 00 V

I =

VL = XL 090θI

I =

I = I θ-

I = VR = R θI

I =

Z

X L

R

θ

Fig. 15 Impedance diagram of RL circuit

By knowing any two of

4 quantities, other two can

be calculated

RC CIRCUIT

Let us now consider the circuit in which a resistor and a capacitor are connected

in series as shown in Fig. 16.

Taking the supply voltage as reference

00EE (49)

Circuit impedance θZXjRZ C (50)

Circuit current θZ

E

θZ

0E

Z

EI

0

(51)

= θI (52)

where Z

E I (53)

- j X C R

E

I

+ V C - + V R -

Fig. 16 RC circuit

Further θIRIRVR (54)

0

CCC 90θIXIXjV (55)

Using KVL, we get V R + V C = E (56)

The phasor diagram for this RC circuit can be got by drawing the phasors V R , V C ,

E and I as shown in Fig. 17.

Consider the triangle formed by the phasors V R , V C and E. Recognizing that

IZEandIXV,IRV CCR if each side of the triangle is divided

by I then R, X C and Z will form a triangle as shown in Fig, 18. This triangle

is known as the IMPEDANCE TRIANGLE.

R

θ

Z X C

Fig. 18 Impedance triangle of RC circuit

900

V C

V R

V C

E

I

θ

Fig. 17 Phasor diagram of RC circuit

GE 0106 BASIC ENGINEERIN II SURPRISE TEST 1 March 2010

1. When the current through a coil of 1000 turns is increased by 5 A, the flux in it

increases by 0.3 m Wb. Determine the self inductance of the coil.

2. The coefficient of coupling between coil A of 1200 turns and coil B of 1000 turns is

0.6. A current of 12 A in coil A establishes in it a flux of 0.12 m Wb. If the current in

coil A changes from 12 A to -12 A in 0.02 sec., what would be the emf induced in

coil B?

3. For the wave form shown, find the average value, RMS value, Form factor and

Peak factor.

v

5 sec. 2 sec. 0

12 V

t (sec.)

When the current through a coil of 1000 turns is increased by 5 A, the flux in it

increases by 0.3 m Wb. Determine the self inductance of the coil.

L = N H0.065

10x0.3x1000

dI

dφ 3

The coefficient of coupling between coil A of 1200 turns and coil B of 1000 turns is

0.6. A current of 12 A in coil A establishes in it a flux of 0.12 m Wb. If the current in

coil A changes from 12 A to -12 A in 0.02 sec., what would be the emf induced in

coil B?

NA = 1200; NB = 1000; k = 0.6; When dI = 12 A, d Aφ = 0.12 m Wb.

When dI = 24 A, d Aφ = 0.24 m Wb.; Corresponding dt = 0.02 sec.

d Bφ = 0.6 x 0.24 = 0.144 m Wb

eB = NB V7.20.02

10x0.144x1000

dt

dφ 3

B

For the wave form shown, find the average value, RMS value, Form factor and

Peak factor.

Average value = 5

1 [ ]12x5x

2

1 = 6 V

Area of square curve = ( )144x2x3

12405x

3

144)144x3x

3

1(

Mean square = 485

240 ; RMS value = V6.928248

Form Factor = 1.15476

6.9282

value Average

value RMS

Peak Factor = 1.73216.9282

12

value RMS

value Peak

v

5 sec. 2 sec. 0

12 V

t (sec.)

RLC CIRCUITS

Analysis of RLC circuits is the series, parallel and series-parallel combination of

RL and RC circuits. Equivalent of RLC circuit will be R, or RL or RC circuit as

illustrated in the examples to be discussed.

POWER AND POWER FACTOR

Let E 00 be the supply voltage in an AC circuit. The supply current may lag or

lead the supply voltage. Let the supply current be I - θ. The supply current can

be resolved into two components (i) A component Ip in phase with the voltage

and (ii) A component Iq at right angle to the voltage as shown in Fig. 19.

θ

Iq

Ip

I - θ0

E 00

Fig. 19 Power and Power factor

Current Ip is called the active or in-phase component while Iq is known as reactive

or quadrature component. As seen from Fig. 19

Ip = I cos θ and (57)

Iq = I sin θ (58)

It is to be noted that

20.Fig.inastriangleanglerightaofsidesthreeformIandθsinIθ,cosI

Iq

Ip

Fig. 20 Components of current

I

θ

Active Power (P)

Active power is the real power consumed by the circuit. This is due to the in-

phase component.

Active or real power P = E Ip

= E I cos θ Watts (59)

Reactive Power (Q)

The power associated with the reactive component of current Iq is known as

reactive power. Its unit is Volt Ampere Reactive (VAR).

Reactive power Q = E Iq

= E I sin θ VAR (60)

Apparent Power and Power Factor

The product of voltage and current, E I is called as Apparent Power, denoted

by S . Its unit is Volt Ampere (VA).

Apparent power S = E I VA (61)

Similar to Fig. 20, real power P, reactive power Q and apparent power S form

three sides of a right angle triangle as shown in Fig. 21.

Power Factor (pf) is the ratio of real power to apparent power.

Thus power factor = IE

θcosIE

= cos θ (62)

Q

P

Fig. 21 Components of power

S

P = θcosIE

Q = θsinIE

S = IE

θ

By the above definition, it is not possible to distinguish whether the load is

inductive or capacitive. If the load is inductive, the current is lagging the voltage

and the nature of the power factor is LAGGING. On the other hand if the load is

capacitive, the current is leading the voltage and hence the nature of the power

factor is LEADING.

Whenever power factor is furnished, it must be clearly stated whether it is lagging

or leading. For inductive load, the power factor is cos lagging; for capacitive

load, the power factor is cos leading; for resistive load since the voltage and

current are in-phase, power factor angle is zero and the power factor is said to

be UNITY.

Power associated with R, L and C can be obtained as follows.

In the case of resistor, p.f. angle is zero and hence

P = E I cos θ = E RIIIZI2

(63)

EQ I sin θ = 0 (64)

In the case of pure inductor and pure capacitor, p.f. angle = 900 and hence

P = E I cos θ = 0 (65)

Q = E I sin θ = E XIIIZI2

(66)

Example 5

In a series circuit containing pure resistance and pure inductance, the current

and voltage are: i(t) = 5 sin (314t + 3

π2) and v(t) = 20 sin (314t +

6

π5). (i) What is

the impedance of the circuit? (ii) What are the values of resistance, inductance

and power factor? (iii) What is the power drawn by the circuit?

Solution Phasor current and phasor voltages are

Current I = 2

5120-90 =

2

5300; Voltage V =

2

20150-90 =

2

20600

Impedance Z = Ω2)j3.4641(Ω304305

6020

I

V 0

Resistance R = 3.4641 Ω

XL = 2 Ω; 314 L = 2; L = H314

2; Inductance L = 6.3694 mH

Angle between voltage and current = 300

p.f. = cos 300 = 0.866 lagging

Power P = θcosIV = 2

20 x

2

5 x 0.866 = 43.3 W

V I

10 A 1000 W

R

X

|Z|

θ

Example 6

An inductive coil takes 10 A and dissipates 1000 W when connected to a supply

of 250 V, 25 Hz. Calculate the impedance, resistance, reactance, inductance and

the power factor.

Solution

P = RI2

; Resistance R = 100

1000 = 10 Ω

Ω2510

250Z ; From impedance triangle X = Ω22.91281025 22

Thus impedance Z = (10 + j 22.9128) Ω = 25 66.420 Ω

Resistance R = 10 Ω Reactance (Inductive) XL = 22.9128 Ω

Inductance L = H0.145925xπ2

22.9128

fπ2

XL

From impedance triangle, power factor = lagging0.425

10

Z

R

250 V, 25 Hz

Example 7

A resistance is connected in series with a coil. With a supply of 250 V, 50 Hz, the

circuit takes a current of 5 A. If the voltages across the resistance and the coil are

125 V and 200 V respectively, calculate (i) impedance, resistance and reactance

of the coil (ii) power absorbed by the coil and the total power. Draw the phasor

diagram.

Resistance R = Ω255

125

O CZ = Ω405

200 ; TZ = Ω50

5

250

Since O CZ = 40 Ω, 40XjR O CO C ; 1600XR 2

O C

2

O C

Since TZ = 50 Ω; 50Xj)R(25 O CO C

625 + 50 RC O + 2500XR 2

O C

2

O C i.e. 50 RC O = 2500 – 625 -1600 = 275

Resistance of the coil RC O = 5.5 Ω Also 1569.755.5-1600X 22

O C

XC O RC O

200 V 125 V

250 V 50 Hz

5 A R

Fig. 21 Example 7

Reactance of the coil XC = 39.62 Ω

Impedance of the coil ZC O = (5.5 + j 39.62) = 40 82.10 Ω

Power absorbed by the coil PC O = 52 x 5.5 = 137.5 W

Total power PT = (52 x 25) + 137.5 = 762.5 W

Total impedance ZT = (30.5 + j 39.62) = 50 52.410Ω

V27.55.5x5RI O C ; V198.139.62x5XI O C

Phasor diagram is shown in Fig. 22.

198.1 V

27.5 V

52.410

250 V

200 V

125 V

I

Fig. 22 Phasor diagram-Example 7

RCO XCO

Example 8

When a resistor and a seriesly connected inductor coil, are supplied with 240 V, a

current of 3 A flows lagging behind the supply voltage by 370. The voltage across

the coil is 171 V. Find the value of the resistor, resistance and reactance of the

inductor coil.

Solution

Supply voltage E = 240 00 V; Supply current I = 3 - 370 A

Circuit impedance Z = I

E = 80 370 Ω = (63.8908 + j 48.1452) Ω

Thus R + RCO = 63.8908 Ω and XCO = 48.1452 Ω

For the coil, Ω573

171ZCO ; From impedance triangle of the coil

RCO = Ω30.512948.145257 22

Value of resistor R = 63.8908 – 30.5129 = 33.3779 Ω

Resistance of coil RCO = 30.5129 Ω; Reactance of coil XCO = 48.1452 Ω

3 - 370 A 171 V

240 V

R

Fig. 23 Example 8

Example 9

When a voltage of 100 V at 50 Hz is applied to choking coil 1, the current taken is

8 A and the power is 120 W. When the same supply is applied to choking coil 2,

the current is 10 A and the power is 500 W. Find the current and power when the

supply is applied to two coils connected in series.

Solution

Resistance R1 = Ω1.8758

1202

Impedance Ω12.58

100Z1 ; Therefore X1 = Ω12.35861.87512.5 22

Resistance R2 = Ω510

5002

Impedance Ω1010

100Z2 ; Therefore X2 = Ω8.6603510 22

Total resistance RT = 6.875 Ω; Total reactance XT = 21.0189 Ω

Total impedance ZT = (6.875 + j 21.0189) = 22.1147 71.890 Ω

Total current A4.521922.1147

100IT Power PT = 4.52192 x 6.875 = 140.5771 W

Example 10

A resistance of 100 ohm is connected in series with a 50 µF capacitor. When the

supply voltage is 200 V, 50 Hz, find the (i) impedance, current and power factor (ii)

the voltage across resistor and across capacitor. Draw the phasor diagram.

Solution

Resistor R = 100 Ω; Reactance of the capacitor XC = 50x50x2π

106

= 63.662 Ω

Impedance Z = (100 – j 63.662) = 118.5447 - 32.480

Taking the supply voltage as reference, E = 200 00 V

Current I = A32.481.687132.48118.5447

0200

Z

E 0

0

0

Power factor = cos 32.480 = 0.8436 leading

Voltage across resistor VR = 100 x 1.6871 32.480 = 168.71 32.480

V

Voltage across capacitor VC = - j 63.662 x 1.6871 32.480 = 107.4042 - 57.520

V

Phasor diagram is shown in Fig. 24.

900

VC = 107.4042 - 57.520 V

VR = 168.71 32.480 V I =1.6871 32.480 A

E = 200 00 V

32.480

Fig. 24 Phasor diagram - Example 10

Example 11

In a circuit, the applied voltage of 150 V lags the current of 8 A by 400. (i) Find the

power factor (ii) Is the circuit inductive or capacitive? (iii) Find the active and

reactive power.

Solution

Power Factor = 0.766 leading

Circuit is capacitive.

Active Power P = 150 x 8 x 0.766 = 919.2 W

Reactive Power Q = 150 x 8 x 0.6428 = 771.36 VAR

Example 12

Find the circuit constants of a two elements series circuit which consumes 700 W

with 0.707 leading power factor. The applied voltage is V = 141.4 sin 314 t volts.

Solution

V99.98492

141.4VW700P ; ; cos θ = 0.707; Since Power P = θcosIV

A9.90250.707x99.9849

700I and Ω10.0969

9.9025

99.9849Z

From the impedance triangle

Resistance R = Ω7.13850.707x10.0969θcosZ

Reactance XC = Ω7.13850.707x10.0969θsinZ

Capacitance C = Fμ446.1327.1385x314

1

Example 13

A series R-C circuit consumes a power of 7000 W when connected to 200 V, 50 Hz

supply. The voltage across the resistor is 130 V. Calculate (i) the resistance,

impedance, capacitance, current and p.f. (ii) Write the equation for the voltage

and current.

Solution

7000R)R

130 2 (

Resistance R = Ω2.41437000

1302

; Current A53.84582.4143

130I

Since 200 x 53.8458 x cos θ = 7000, p.f. = 0.65 leading; θ = 49.460

From impedance triangle, reactance XC = R tan θ = 2.4143 x 1.1691 = 2.8226 Ω

Impedance Z = (2.4143 – j 2.8226) Ω = 3.7143 - 49.460Ω

Capacitance C = Fμ1127.722.8226x50xπ 2

1

Current I = 53.8458 49.460 A; Power Factor = 0.65 leading

Taking supply voltage as reference

v(t) = t314.16cos282.8427t)50x(2πcos200x2

i(t) = A)49.46t314.16cos76.15)49.46t50x(2πcos53.8458x2 00 (

θ

CX

Power P = 7000 W

- j CX

R

Fig. 25 Circuit – Example 13

130 V

200 V 50 Hz

R

Example 14

A coil of resistance 10 Ω and inductance 0.1 H is connected in series with a 150

μF capacitor across 200 V, 50 Hz supply. Calculate (i) inductive reactance,

capacitive reactance, impedance, current and power factor and (ii) the voltage

across the coil and capacitor.

Solution

Data are marked in Fig. 26

Inductive reactance XL = 2 π x 50 x 0.1 = 31.4159 Ω

Capacitive reactance XC = Ω21.2207150x50xπ2

106

Impedance Z = (10 + j 31.4159 – j 21.2207) = (10 + j 10.1952) Ω = 14.2808 45.550 Ω

Taking supply voltage as reference

Current I = A45.5514.004845.5514.2808

200 0

Power factor = 0.7003 lagging

Impedance ZCoil = (10 + j 31.4159) = 32.9691 72.340 Ω

Voltage VCoil = ZCoil x I = 461.7257 26.790 V

Voltage VCap = ZCap x I = - j 21.2207 x 14.0048 - 45.550 = 297.1916 - 135.550 V

150 μF

10 Ω

Fig. 26 Circuit – Example 14

200 V 50 Hz

0.1 H

Example 15

In the circuit shown in Fig. 27, the current leads the voltage by 500. Find value of

R and the voltages across each circuit element. Draw the phasor diagram.

Solution

Circuit is capacitive

Power factor angle θ = 500

Taking the supply voltage as reference V = 200 00 volts

XL = 31.4159 Ω; XC = 63.662 Ω

Impedance Z = R + j 31.4159 – j 63.662 = R – j 32.2461

From the impedance triangle

tan θ = 1.19175R

32.2461 ; Thus Resistance R = 27.0578 Ω

Impedance Z = (27.0578 – j 32.2461) = 42.0944 - 500 Ω

Current I = Z

V = A504.7512

5042.0944

200 0

5 μF R

Fig. 27

200 V 500 Hz

10 m H

500

32.2461 Ω

R

Voltage across R VR = R x I = 128.557 500 V

Voltage across L VL = j 31.4159 x 4.7512 500 = 149.2632 1400 V

Voltage across C VC = - j 63.662 x 4.7512 500 = 302.4708 - 400 V

Phasor diagram is shown in Fig. 28.

VC

VL VR

V

I

V = 200 00

I = I = 4.7512 050 A

I = VR = 128.557 050 V

I = VL = 149.2632 0140 V

I = VC = 302.4708 040- V

I = Fig. 28 Phasor diagram – Example 15

Example 16

A 230 V, 50 Hz voltage is applied to a coil of L = 5 H and R = 2 Ω in series with a

capacitance C. What value must C have in order that the p.d. across the coil shall

be 250 V?

Solution

Coil impedance ZCO = (2 + j 1570.7963) = 1570.7975 89.930 Ω

A0.15921570.7975

250I ; Ω 1444.7236

0.1592

230ZT ;

Therefore 22 + XT2 = 1444.72362; Thus XT = ± 1444.7 Ω

i.e. 1570.8 – XC = ± 1444.7; Thus XC = 126.1 Ω or XC = 3015.5 Ω

If XC = 126.1 Ω circuit is INDUCTIVE; C = Fμ25.243126.1x50x2π

1

If XC = 3015.5 Ω circuit is CAPACITIVE; C = Fμ1.0563015.5x50x2π

1

Refer Fig. 29

XL = 2 π x 50 x 5 = 1570.7963 Ω

- j XC

2 Ω

Fig. 29 Circuit – Example 16

230 V 500 Hz

j XL

250 V

Example 17

A resistance R, an inductance L = 0.5 H and a capacitance C are connected in

series. When a voltage v = 350 cos(3000 t – 200) volt is applied to this series

combination, the current flowing is 15 cos(3000 t – 600) amperes. Find R and C.

Solution

Supply voltage V = 247.4873 - 200 Volts

Current I = 10.6066 -600 A

Inductive reactance XL = 3000 x 0.5 = 1500 Ω

Impedance Z = R + j (XL – XC) = Ω4023.33336010.6066

20247.4873V 0

I

= (17.8743 + j 14.9984) Ω

Thus R = 17.8743 Ω

XL – XC = 14.9984; Thus XC = 1500 – 14.9984 = 1485.0016 Ω

Therefore C = Fμ0.22451485.0016x3000

1

Example 18

Calculate the admittance Y, the conductance G and the susceptance B of a circuit

consisting of 10 Ω in series with an inductor of 0.1 H when the frequency is 50 Hz.

Solution

Inductive reactance XL = 2 π x 50 x 0.1 = 31.4159 Ω

Circuit impedance Z = (10 + j 31.4159) Ω

Circuit admittance Y = Z

1 = (0.0092 – j 0.0289) = 0.03033 -72.34 mho

Thus Conductance G = 0.0092 mho; Susceptance B = 0.0289 mho Inductive

Example 19

An impedance of (7 + j 5) Ω is connected in parallel with another circuit having

impedance of (10 – j 8) Ω. The supply voltage is 230 V, 50 Hz. Calculate (i) the

admittance, conductance and susceptance of the combined circuit and (ii) the

total current taken from the mains and the p.f.

Solution

Impedance Z1 = (7 + j 5) Ω; Admittance Y1 = (0.0946 – j 0.0676) mho

Impedance Z2 = (10 - j 8) Ω; Admittance Y2 = (0.061 – j 0.0488) mho

Total admittance YT = Y1 + Y2 = (0.1556 – j 0.0188) mho = 0.1567 - 6.890 mho

Conductance G = 0.1556 mho; Susceptance B = 0.0188 mho Inductive

Taking the supply voltage as reference, V = 230 00 volts

Total current taken I = Y V = 36.041 - 6.890 A

p.f. = 0.9928 lagging

Example 25

Consider RLC series circuit with R = 100 Ω, L = 1.0 H and C = 1.0 µF. It is

connected to 500 V variable frequency supply. For a range of ω = 800 to 1200 rad.

per sec. in steps of 100 rad. per sec., compute the values of XL, XC, IandZ and

plot them. Mark the region of leading and lagging power factor. For ω = 1000 rad.

per sec., find the values of voltages across the inductance and capacitance.

Solution

The calculated values are shown in the table.

ω rad. / sec. XL Ω XC Ω Z Ω Z Ω I A

800 800 1250 100 – j 450 461 1.08

900 900 1111 100 – j 211 233 2.15

1000 1000 1000 100 100 5.0

1100 1100 909 100 + j191 216 2.31

1200 1200 833 100 + 367 380 1.32

R = 100 Ω

L = 1.0 H

C = 1.0 μF

o

*

Ι Z Ι Ι I Ι

Ι Z Ι

ω rad. / sec. XL Ω XC Ω Z Ω Z Ω I A

800 800 1250 100 – j 450 461 1.08

900 900 1111 100 – j 211 233 2.15

1000 1000 1000 100 100 5.0

1100 1100 909 100 + j191 216 2.31

1200 1200 833 100 + 367 380 1.32

1100 XC

XL

I

XC XL

lagging

p.f.

leading

p.f.

o

o

X

* X

X

X

X

200 2.0

300 3.0

400 4.0

500 5.0

600 6.0

100 1.0

1300

*

*

*

*

1200

1000

900

800

ω

1200 1000 900 800 1100

Fig. 35 Resonance charecteristics

R = 100 Ω

L = 1.0 H

C = 1.0 μF

When ω = 1000 rad. per sec., XL = 1000 Ω and XC = 1000 Ω. Taking the supply

voltage as reference, current I = 5 A00 .

Therefore

VL = j 1000 x 5 = j 5000 V

VC = - j 1000 x 5 = - j 5000 V

Even though the supply voltage is 500 V, voltages across the inductance and

capacitance are 5000 V.

It is to be noted that VR = 100 x 5 = 500 V and VR + VL + VC = 500 V The phasor

diagram is shown in Fig. 36.

I

VC

VL

VR

Fig. 36 Phasor diagram

SERIES RESONANCE

An R – L – C series circuit is said to be at resonance when the applied voltage

and the resulting current are in phase. Thus at resonance condition, the

equivalent impedance of the circuit will be a resistance only. Since the supply

voltage and the circuit current are in phase, the power factor of the resonant

circuit is unity.

Consider an R – L – C series circuit connected to a variable frequency supply as

shown in Fig. 37.

Impedance of the circuit Z = R + j (XL – XC) (67)

C

R

Fig. 37 Resonance Circuit

Variable frequency

AC supply

L - j XC

R j XL

Variable frequency

AC supply

While the frequency is increased from a low value, XL increases and XC

decreases. At one particular frequency, called resonance frequence, XL and XC

are equal and the total impedance will be equal to R. This angular frequency is

designated as ω0. For other frequencies

2

CL

2 )X(XRZ (68)

Thus at resonance frequency, the impedance is minimum and is equal to R and

the current is at the maximum value given by R

E. When ω < ω0 , XC > XL and

hence p.f. is of leading nature. When ω > ω0 , XL > XC and hence the p.f. is lagging

in nature. At resonance frequency since the circuit impedance is of resistance

only, the p.f is unity. Expression for angular resonance frequency ω0 can be

obtained as follows. At resonance condition, XL = XC. Therefore

ω0 L = Cω

1

0

; Thus ω0 = CL

1 rad. / sec. (69)

Resonance frequency f0 = π2

1

CL

1 Hz (70)

Resonance condition can be achieved by varying the values of either L or C also.

QUALITY FACTOR

In Example 26, it was found that, at resonance condition, voltage across L or C is

much larger than the supply voltage. The property of developing high voltage

during resonance condition is defined by QUALITY FACTOR, also referred as Q

Factor.

Q Factor, Q = E

V

E

V CL at resonance condition. (71)

Thus Q = I

I

R

XL

R

Lω

R

X 0L (72)

Also Q = I

I

R

XC

RCω

1

R

X

0

C (73)

BANDWIDTH

At resonance condition, the circuit current is maximum and is given by I0 = R

E.

The entire power input is absorbed by R and this power is given by P0 = I02 R. For

all other frequencies around ω0, circuit current is less than I0 and hence the

power absorbed by the circuit will be less than P0. The power absorbed will be 2

P0

when the circuit current is 02

1I

BANDWIDTH is defined as that range of frequencies around the resonance

frequency ω0, within which the power absorbed by the circuit is greater than or

equal to 2

P0 , where P0 is the power absorbed by the circuit at resonance condition

OR within which the circuit current is greater than or equal to 2

0I, where I0 is the

circuit current at resonance condition.

Referring to Fig. 38,

Bandwidth ωB W = ω2 - ω1 (74)

It can be shown that ωB W = L

R (75)

Also from eq. (72), it can be seen that

Q = R

Lω0 = WB

0

ω

ω (76)

I

2

0I

I0

ω2 ω1

ω

Fig. 38 Bandwidth

Example 26

A series R-L-C circuit with R = 10 Ω, L = 10 mH and C = 1 μF is connected to 200 V

variable frequency supply. Calculate the resonance frequency. Also find the

circuit current and the voltages across the elements. Determine the Quality factor

and Bandwidth.

Solution

ω02 = 8

6

101x0.01

10

CL

1 ; Therefore ω0 = 104 rad. / sec.

Resonant frequency f0 = π2

ω0 = 1591.55 Hz.; At resonance XL = XC = ω0 L = 100 Ω

Circuit current at resonance I0 = 10

200 = 20 A

Voltage across resistor VR = 10 x 20 = 200 V

Voltage across L and C: V200020x100XVV 0LCL I

Quality factor Q = 1010

100

R

XL

Bandwidth ωB W = 10000.01

10

L

R rad. / sec. = 159.155 Hz

Example 27

A series R-L-C circuit is connected to a 200 V, 50 Hz supply. When L is varied, the

maximum current obtained is 0.4 A. At that condition, the voltage across the

capacitor is 330 V. Find the circuit constants.

Solution

Resistor R = 200 / 0.4 = 500 Ω

330XVV 0LCL I ; Therefore XL = XC = 330 / 0.4 = 825 Ω

ω0 = 2 π x 50 = 314.1593

Inductance L = H 2.626314.1593

825

ω

X

0

L

Capacitance C = 3.8583 F825 x 314.1593

1

Xω

1

C0

μF

Ι IL Ι

Ι IL Ι

Example 20

When a 240 V, 50 Hz supply is applied to a resistor of 15 Ω which is in parallel

with an inductor, total current is 22.1 A. What value must the frequency has for

the total current to be 34 A?

Solution

A1615

240R I ; 162 +

2

LI = 22.12; Thus LI = 15.245 A

Therefore XL = Ω15.742915.245

240 ; Thus L = H0.05011

50 x π 2

15.245

With new frequency

162 + 2

LI = 342; Thus LI = 30 A and XL = Ω830

240

New frequency f = Hz25.40890.05011 x π 2

8

Current in resistor and

current in inductor will

have a phase difference

of 900

15 Ω XL 240 V Ω, 50 Hz

Fig. 30 – Example 20

16 A

22.1 A

16 A

34 A

Example 21

A coil of resistance 15 Ω and inductance 0.05 H is connected in parallel with non-

inductive resistor of 20 Ω. Find (i) the current in each branch and the total current

supplied and (ii) the phase angle of the combination when a voltage of 200 V at 50

Hz is applied. Draw the phasor diagram.

Solution

Current in the coil I1 = A46.32-9.2083 6.6596 j - 6.359415.708j 15

200 0

Current in non-inductive resistor I2 = 10 A

Current lags the voltage by 22.150

20 Ω

I2

X L = 2 π x 50 x 0.05

= 15.708 Ω

Taking supply voltage as

reference

A22.15-17.663 6.6596 j - 6.35941currentTotal 0

T I

Fig. 32 – Phasor diagram - Example 22

22.150 46.320

V

I1

IT

15 Ω

Fig. 31 – Example 21

X L

200 V Ω, 50 Hz

6 mH

Example 22

A coil of inductance 6 mH and resistance 40Ω is connected across a supply of

100 V, 800 Hz. Also across the supply is a circuit consisting of 4 μF in series with

50 Ω resistor. Find (i) the total current taken from the supply and (ii) the phase

angle between the currents in the coil and the capacitance. Draw the phasor

diagram.

Solution

X L = 2 π x 800 x 0.006

= 30.1593 Ω

XC = Ω49.73594x800x2π

106

40 Ω

Fig. 33 – Example 22

100 V

800 Hz

50 Ω

4 μF

Ω37.0250.0957 30.1593 j 40Z 0

1

Ω44.8570.5242 49.7359 j 50 Z 0

2

Current I1 = A37.02-1.9962 1.2019 j - 1.593830.1593j 40

100 0

Current I2 = A44.851.418 1.0 j - 1.005349.7359j 50

100 0

Total current IT = (2.5991 – j 0.2019) A = 2.6069 - 4.440 A

Angle between currents I1 and I2 = 81.870

4.440 37.020

44.850 V (Ref)

I2

I1

I T

Fig. 34 – Phasor diagram - Example 22

Example 23

Two circuits, with impedances Z1 = (10 + 15) Ω and Z2 = (6 – j 8) Ω are connected

in parallel. If the total current supplied is 15 A, find the power consumed by each

impedance.

Solution

Taking the supply current as reference, IT = 15 A00

Current I1 = A76.768.589)8.3607j(1.967215x7j16

8j6 0

Current I2 = A32.6815.484)8.3607j(13.032815x7j16

15j10 0

Power P1 = |I1|2 R1 = 8.5892 x 10 = 737.7092 W

Power P2 = |I2|2 R2 = 15.4842 x 6 = 1438.5255 W

XL1

Example 24

Two coils are connected in parallel across a voltage of 200 V, 50 Hz. The coils

have resistances of 10 Ω and 5 Ω and inductances of 0.023 H and 0.035 H

respectively. Find (i) current in each coil and total current and (ii) p.f. of the

combination.

Solution

XL1 = 2 π x 50 x 0.023 = 7.2257 Ω; XL2 = 2 π x 50 x 0.036 = 10.9956 Ω

Z1 = (10 + j 7.2257) Ω = 12.3374 Ω35.850

Z2 = (5 + j 10.9956) Ω = 12.079 Ω65.550

Current I1 = V / Z1 = 16.2109 A35.850

Current I2 = V / Z2 = 16.5577 A65.850

Total current IT = 31.6743 A50.860

p.f. of the combination = 0.6312 lagging

10 Ω 200 V

50 Hz

5 Ω

XL2

E 'CC

E AA'

E BB'

THREE PHASE SYSTEM

In general, generation, transmission and utilization of electric power is more

economical in three phase system compared to single phase system.

The windings of three phase alternators are designated as AA’, BB’ and CC’. The

voltages generated in these windings are

)240tωcos(Ee

)120tωcos(Ee

ωtcosEe

0

mCC'

0

mBB'

mAA'

(77)

The phasor descriptions of three voltages are shown in Fig. 39. Here E AA' is

taken as reference. Each voltage phasor is lagging the previous one by 120 0 .

2

EEEEE m

CC'BB'AA'

Fig. 39 Phasor representation of 3 phase voltages

0

AA0EE '

0

CC

0

BB

240EE

120EE

'

'

C

A

B

N

CI

AI

BI

AE

ABE

Generally E AA' is written as E A . Other phasors are represented likewise. Thus

0

C

0

B

0

A

240EE

120EE

0EE

(78)

The three generator windings are connected either in STAR (wye) or in DELTA.

STAR CONNECTED GENERATOR

Fig. 40 shows the winding connections of star connected generator. The

generator is connected to a 3 phase load.

E A , E B and E C are called the PHASE VOLTAGES. E AB , E BC and E CA are called

the LINE VOLTAGES or line to line voltages. The current flowing in each phase is

called PHASE CURRENT and the current flowing in each line called LINE

CURRENT.

L

O

A

D

Fig. 40 Star connected generator

E C - BE

E AB

E A

E B

Let lI and hpI be the magnitude of line current and phase current and lE and

hpE be the magnitude of line voltage and phase voltage. In case of star connected

system

Line current = Phase current i.e. lI = hpI (79)

Taking E A as the reference, the voltage phasors are shown in Fig. 41.

0

phC

0

phB

0

phA

240EE

120EE

0EE

The relationship between line voltage and phase voltage can be obtained as

follows.

0

ph

ph

phphBAAB

30E3

j0.866)(1.5E

j0.866)0.5(EEEEE

The above result can be seen from the Fig. 41.

Fig. 41 Voltage phasors - Star system

Similar expression can be obtained for E BC and E CA also. Collectively, we have

0

phCA

0

phBC

0

phAB

150E3E

90E3E

30E3E

(80)

Thus lE = phCABCAB E3EEE

Therefore for star connected system

phl

phl

II

E3E

(81)

Power supplied by the

three phase alternator (82)

(83)

Above results are true for star connected load also, except that the power is

consumed by the load.

θcosIE3

θcosIE3

powerphasex3

ll

phph

'C

C

CI

AI

BI

A

B

'A

'B BCI

ABI

CAI

DELTA CONNECTED GENERATOR

Delta connected generator is shown in Fig. 42.

I A , I B and I C are called LINE CURRENTS. I AB , I BC and I CA are called PHASE

CURRENTS. The voltage across each phase is called PHASE VOLTAGE and

voltage across two lines is called LINE VOLTAGE or line to line to line voltage.

In case of delta connected system, line voltage is equal to phase voltage. i.e.

E l = E hp (84)

L

O

A

D

Fig. 42 Delta connected generator

I CA

I AB

I BC

I A

Taking I BC as reference, current phasors are shown in Fig. 43.

0

phAB

0

phCA

0

phBC

240II

120II

0II

Considering the junction point formed by A and C'

I A = I AC - I BA = I hp ( - 0.5 - j 0.866 ) - I hp ( - 0.5 + j 0.866 ) = - j 3 I hp

The above result can be seen from Fig. 43. Similar expression can be obtained for

I B and I C . Collectively, we have

Fig. 43 Current phasors – Delta connected system

0

phC

0

phB

0

phA

30I3I

210I3I

90I3I

(85)

Therefore lI = phCBA I3III

Thus for delta connected system

E l = E hp

I l = 3 I hp

Power supplied by the

three phase alternator (87)

(88)

Above results are true for delta connected load also except that the power is

consumed by the load.

(86)

θcosIE3

θcosIE3

powerphasex3

ll

phph

Working principle, construction and applications of DC Generator

The dc generator is rotating electrical machine which converts mechanical

energy into electrical energy. The generator is usually driven by a steam turbine

or water turbine which is called as prime mover.

The dc generator operates on the principle based on the Faraday’s Law of

electromagnetic induction. The generator should have (i) magnetic field (ii)

conductors capable of carrying current (iii) movement of conductors in the

magnetic field.

Necessary magnetic field is produced by field coil. The set of conductors in

which the voltage is induced, is called the armature.

The voltage induced in the coil will be as shown in Fig. 15.

Fig. 15 EMF induced in an armature coil

ZZ

YY Y

AA

A A

AA

YY Y

AA

A Z

Depending on how the Armature and Field windings are connected, we have

different types of dc generators. They are shown in Fig. 18.

ZZ

ZZ

Z

Fig. 18 (b) Shunt generator Fig. 18 (a) Series generator

L

O

A

D

G

L

O

A

D

G

Fig. 18 (c) Short shunt compounded generator

G

L

O

A

D

Fig. 18 (d) Long shunt compounded generator

Z L

O

A

D

AA

G

A

Y

YY

Application of dc generators

Shunt generators are used in supplying nearly constant loads. They are used for

charging batteries and supplying the fields of synchronous machines.

Series generators are used as boosters for adding voltage to transmission lines

to compensate for the line drop.

Cumulative compound generators are used for drives which require constant dc

voltage supply.

Differential compound generators are used in arc welding.

Working principle, construction and applications of DC Motor

Whenever a current carrying conductor is kept in a stationary magnetic field, an

electromotive force is produced. This force is exerted on the conductor and

hence is moved away from the field. This is the principle used in dc motors.

Construction of dc motor is exactly similar to dc generator.

In a dc motor, both the armature and the field windings are connected to a dc

supply. Thus, we have current carrying armature conductors placed in a

stationary magnetic field. Due to electromagnetic torque exerted on the armature

conductors, the armature starts revolving. Thus, electrical energy is converted

into mechanical energy in the armature.

When the armature is in motion, we have revolving conductors in a stationary

magnetic field. As per Faraday’s Law of electromagnetic induction, an emf is

induced in the armature conductors. As per Lenz’s law, this induced emf opposes

the voltage applied to the armature. Hence it is called back emf. There will be

small voltage drop due to armature resistance. Thus, the applied voltage has to

overcome the back emf in addition to supplying the armature voltage drop. The

input power is used to produce necessary torque for the continuous rotation of

the armature.

Depending on how the Armature and Field windings are connected, we have

different types of dc motors. They are shown in Fig. 19.

ZZ

Z

Z

YY

DC

supply

voltage

AA

A Y

-

+

A

AA

+

DC

supply

voltage

+

YY Y

AA

A

ZZ -

DC

supply

voltage

Z

ZZ

Y

YY

+

-

DC

supply

voltage

YY Y

-

Depending on how the Armature and Field windings are connected, we have

different types of dc motors. They are shown in Fig. 19.

M

M

M

M

Fig. 19 (a) Series motor Fig. 19 (a) Shunt motor

G

Fig. 19 (c) Short shunt compounded motor

Fig. 19 (d) Long shunt compounded motor

AA

G

A

Application of dc motors

DC series motors are used in electric trains, cranes, hoists, conveyors etc. where

high starting torque is required.

Shunt motors are used where the speed has to remain constant under loaded

condition.

Compound motors are used for driving heavy tools for intermittent heavy loads

such as rolling mills, printing machines etc.

Working principle, construction and applications of 1- phase transformer

The transformer works on the principle of electromagnetic induction. The

induced emf in a transformer comes under the classification of statically induced

emf.

The transformer is a static apparatus used to transfer electrical energy from one

circuit to another. The two circuits are magnetically coupled. One of the circuits,

namely Primary, is energized by connecting it to an ac supply at specific voltage

magnitude, frequency and waveform. Then we have a mutually induced voltage

available across the second circuit, namely Secondary, at the same frequency

and waveform but with a desired voltage magnitude. These aspects are indicated

in Fig. 20.

EMF induced in primary side E1 = N1 dt

dφ

Since same flux is linking both the primary and secondary coils

EMF induced in primary side E2 = N2 dt

dφ

Voltage ratio 2

1

2

1

N

N

E

E

Since losses in the transformer are very less, Voltampere in both the sides are

equal. i.e.

E1 I1 = E2 I2

Then the current ratio 1

2

1

2

2

1

N

N

E

E

I

I

Apart from primary and secondary windings, transformer has a good magnetic

core.

The transformer core is generally laminated and is made out of a good magnetic

material such as transformer steel or silicon steel. Such a material has high

relative permeability and low hysteresis loss. There are two types of transformer

cores. They are known as Core Type and Shell type. In core type transformer, L –

shaped stampings as shown in Fig. 21 are used. One core type transformer is

shown in Fig. 22.

Fig. 21 L – type stampings

Laminated core of a shell type transformer is shown in Fig. 23. In this E – type

and I type laminations are used. Fig. 24 shows a shell type transformer.

Fig. 23 Laminated core of shell type transformer

Application of transformers

The transformers are classified as Step-up transformers and Step-down

transformers. When the secondary voltage is more than the primary voltage,

transformer is called a step-up transformer. In step-down transformer, the

secondary voltage is less than the primary voltage.

Transformers are used in the following applications:

(i) Power transformers located in Power Plants are used to step-up the

generated voltage to a high transmission voltage.

(ii) Transformers are used in distribution circuit to step-down voltages to

the desired level.

(iii) Almost all electronic circuits use transformers.

(iv) Potential transformers are used to measure high voltages and current

transformers are used to measure high currents.

(v) Furnace transformers and welding transformers are some special

applications of transformers.

Working principle, construction and applications of 3- phase

induction motor

When a three phase balanced voltage is applied to a three phase

balanced winding, a rotating magnetic field is produced. This field

has a constant magnitude and rotates in space with a constant speed.

If a stationary conductor is placed in this field, an emf will be induced

in it. By creating a closed path for the current to flow, an

electromagnetic torque can be exerted on the conductor. Thus the

conductor is put in rotation.

A three phase balanced voltage is applied across the three phase balanced stator

winding. A rotating magnetic field is produced. This magnetic field completes its

path through the stator, the air gap and the rotor. The rotor conductors, which

are stationary at the time of starting, are linked by time varying magnetic field.

Therefore emf is induced in the rotor conductors. Since the rotor circuit forms a

closed path, rotor current is circulated. Thus the current carrying conductors are

placed in a rotating magnetic field. Hence an electromotive force is exerted on

the rotor conductors and the rotor starts rotating.

According to Lenz’s law, the nature of the induced current is to oppose the cause

producing it. Here the cause is the relative motion between the rotor conductors

and the rotating magnetic field. Hence the rotor rotates in the same direction as

that of the rotating magnetic field.

In practice, the rotor speed never equals to the speed of the rotating magnetic

field. The difference in the two speeds is called the slip. The current drawn by the

stator gets adjusted according to the load on the motor.

Three phase induction motors are used in industry for very many purposes. They

are used in lathes, drilling machines, agricultural and industrial pumps,

compressors and industrial drives.

The important parts of a three phase induction motor are schematically

represented in Fig. 25. Broadly classified, they are stator and rotor which are

described below.

Stator is the stationary part of the motor. The stator core consist of high grade,

low loss electrical sheet-steel stampings assembled in the frame. Slots are

provided on the inner periphery of the stator to accommodate the stator

conductors. Required numbers of stator conductors are housed in the slots.

These conductors are arranged to form a balanced three phase winding. The

stator winding may be connected in star or delta.

Rotor is the rotating part of the induction motor. The air gap between the stator

and rotor is as minimum as possible. The rotor is also in the form of slotted

cylindrical structure. There are to types of rotors, namely Squirrel Cage rotor and

Slip-ring or Wound rotor.

Fig. 26 shows the construction of a squirrel cage rotor.

In this type, each rotor slot accommodates a rod or bar made of good conducting

material. These rotor bars are short circuited at both ends by means of end rings

made of the same metal as that of rotor conductors. Thus the rotor circuit forms a

closed path for any current to flow through.

Fig. 26 Squirrel cage rotor of three phase induction motor

Fig. 27 shows the rotor of slip-ring induction motor. In this case conductors are

housed in rotor slots. These conductors are connected to form a star connected

balanced three phase winding. The rotor is wound to give same number of poles

as the stator. The three ends of the rotor winding are connected to the three slip-

rings. The brushes are riding over the slip-rings. Slip-rings are short circuited at

the time of starting. External resistances can be connected to control the speed

of the motor. Although the wound rotor motor costs more than a squirrel cage

motor, it has the features of controlling the torque and the speed.

Fig. 27 Rotor of slip-ring induction motor

Starting resistance and speed controller

Working principle, construction and applications of single phase induction motor

Single phase induction motors are used in variety of applications at home,

factory, office and business establishments. Single phase induction motor is not

self starting. Additional arrangement has to be made to make it self-starting. This

could be achieved by using two windings, main winding and starting winding,

with large phase difference between the currents carried by them. This kind of

split-phase motor produces a revolving flux and hence makes the motor self

starting. Depending on the circuit element connected in series with the starting

winding, the split-phase motors are classified into

(i) Resistance-start induction motor

(ii) Capacitance-start induction motor

(iii) Capacitance-start-and-run motor

Resistance-start induction motor

Main winding

Rotor

Starting winding

Single phase a.c. supply

Is

S

Im

Fig. 28 Resistance start induction motor

Im

Is

Resistance start induction motor is shown in Fig. 28. The starting winding has a

high resistance connected in series with it. The current flowing through it is given

by Is. The main winding has low resistance and high reactance and it carries

current Im. Current in starting winding is Is. The torque developed by the motor is

proportional to sinα whereα is the angle between Im and Is as shown in Fig. 29.

For obtaining high torque, angle α should be as high as possible. Here θ is the

power factor angle. The centrifugal switch S disconnects the starting winding

when the motor speed reaches 80% of full load speed.

θ

α

V

I

Fig. 29 Phasor diagram of Resistance start induction motor

Capacitor-start induction motor

In the capacitor-start induction motor, a capacitor is connected in series with the

starting winding as shown in Fig. 30.

Im

Main winding

Starting winding

C

Rotor

Single phase a.c. supply

Is

S

Fig. 30 Capacitor start-induction motor

Is

θ

α

I

Fig. 31 Phasor diagram of capacitor-start induction motor

V

Im

The phasor diagram of capacitor-start induction motor is shown in Fig. 31.

The following are the advantages of capacitor-start induction motor:

(i) Increase in starting torque

(ii) Better starting power factor

Capacitor-start-and- run motor

Capacitor-start-and-run motor is similar to that of the capacitor-start motor

except that the capacitor in the starting winding circuit remains there through out

the operation of the motor. The advantages of this type of motor are:

(i) Low noise in the motor while running

(ii) Higher power factor

(iii) Higher efficiency

(iv) Improved over-load capacity

HOUSE WIRING

House wiring deals with the distribution system arranged within the domestic

premises. Wiring requirement varies with customer to customer. House wiring

generally done on either 230 V single phase or 400 V three phase supply. In the

latter case, total load is divided among the three phases. An earth wire is also run

connecting all the power plugs from where large quantity of electrical energy is

tapped by using electrical appliances like heater, electric iron, hot plate, air

conditioner etc.

Wiring materials and accessories

The following are the wiring material used for house wiring:

Switches Lamp holders Ceiling roses

Socket out-lets Switch boards Wires Conduits

Miniature circuit breaker Fuse unit

The accessories used for house wiring are:

Screw driver Cutting pliers Nose pliers Wire stripper Knife

Hammer Drilling machine Test lamp Wood saw Hack saw

Types of wiring

The type of wiring depends on environment, durability, safety, appearance and

cost.

Cleat Wiring: In this system, V I R (Vulcanised India Rubber) conductor are

supported in porcelain cleats. It is much cheaper; but will not provide good

appearance.

Wooden Casing Capping: This system is more commonly used. It consists of

rectangular wooden blocks, called casing. It has two grooves into which the

wires are laid. Two or three wires of same polarity may be run in one groove.

Wires of opposite polarity are not run in the same groove. The wooden casing at

the top is covered by means of capping and is screwed on it. Nowadays the

wooden casing and cappings are replaced by plastic to give good appearance

and long life.

Conduit Wiring: In this system of wiring, V I R conductors are run inside metallic

pipes called conduit. The conduits are buried into the walls. This system of wiring

provides mechanical protection and good appearance. Nowadays instead of

metal, PVC pipes are used.

Staircase Wiring

In staircase wiring a single lamp, placed at the middle of the staircase, is

controlled by switches at two places, one at the beginning of the staircase and

the other at the end of the staircase. For this purpose two-way switches are

required. The wiring circuit is shown below.

Position of switch S1 Position of switch S2 Condition of lamp

1 1 ON

1 2 OFF

2 1 OFF

2 2 ON

L

2

2

1

1

N

P

S1

S2

Corridor Wiring

The diagram for corridor wiring is shown below.

Moving from left to right:

Enters Closes S1 L1 ON

Reaches S2 Put S2 to 2 L1 OFF and L2 ON

Reaches S3 Put S3 to 2 L2 OFF and L3 ON

Reaches S4 Opens S4 L3 OFF

Moving from right to left:

Enters Closes S4 L3 ON

Reaches S3 Put S3 to 1 L2 ON and L3 OFF

Reaches S2 Put S2 to 1 L1 ON and L2 OFF

Leaves Opens S1 L1 OFF

2

2

1

1

L3

L1

L2

S1

S2

S3

S4

P

N

Incandescent lamp

This works on the principle that any hot body radiates energy. An electric current

passes through a thin filament, heating it to a temperature that produces light.

The enclosing glass bulb contains either a vacuum or an inert gas to prevent

oxidation of the hot filament.

Incandescent bulbs are made in a wide range of sizes and voltages, from 1.5 volts to

about 300 volts. They require no external regulating equipment and have a low

manufacturing cost, and work well on either alternating current or direct current. As a

result the incandescent lamp is widely used in household and commercial lighting, for

portable lighting such as table lamps, car headlamps, and flashlights, and for decorative

and advertising lighting.

Fluorescent lamp

A fluorescent lamp or fluorescent tube is a gas-discharge lamp that uses

electricity to excite mercury vapor. The excited mercury atoms produce short-

wave ultraviolet light that then causes a phosphor to fluoresce, producing visible

light. A fluorescent lamp converts electrical power into useful light more

efficiently than an incandescent lamp. Lower energy cost typically offsets the

higher initial cost of the lamp. The lamp is more costly because it requires a

ballast to regulate the flow of current through the lamp.

While larger fluorescent lamps have been mostly used in commercial or institutional

buildings, the compact fluorescent lamp is now available in the same popular sizes and

is used as an energy-saving alternative in homes.

Sodium vapor lamp

A Sodium vapor lamp is a gas discharge lamp which uses sodium in an excited

state to produce light. There are two varieties of such lamps: low pressure and

high pressure. Because sodium vapor lamps cause less light pollution than

mercury-vapor lamps, many cities that have large astronomical observatories

employ them.

Mercury vapor lamp

A mercury-vapor lamp is a gas discharge lamp that uses mercury in an excited

state to produce light. The arc discharge is generally confined to a small fused

quartz arc tube mounted within a larger borosilicate glass bulb. The outer bulb

may be clear or coated with a phosphor; in either case, the outer bulb provides

thermal insulation, protection from ultraviolet radiation, and a convenient

mounting for the fused quartz arc tube.

Mercury vapor lamps (and their relatives) are often used because they are

relatively efficient. Phosphor coated bulbs offer better color rendition than either

high- or low-pressure sodium vapor lamps. Mercury vapor lamps also offer a very

long lifetime, as well as intense lighting for several special purpose applications.

Earthing

Earthing provides safe discharge of electric current due to leakages and faults to

ground.

All metallic parts of electrical appliances shall be connected by earth wire made

of very good conductor and finally the earth wire is connected to ground.

Earthing can be done through G.I. pipe or G.I. plate buried in the ground and

surrounded by charcoal and common salt to provide good conductivity. To

ensure safety earth resistance should be checked now and then and it is kept at a

very low value.