1 chapter 4 “electrical principles ” bill ryan, kj6igx and glen rikerd, no6w discussion leader

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1 Chapter 4 “Electrical PrinciplesBill Ryan, KJ6IGX and Glen Rikerd, NO6W Discussion Leader

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Page 1: 1 Chapter 4 “Electrical Principles ” Bill Ryan, KJ6IGX and Glen Rikerd, NO6W Discussion Leader

1

Chapter 4

“Electrical Principles” Bill Ryan, KJ6IGX and Glen

Rikerd, NO6W Discussion Leader

Page 2: 1 Chapter 4 “Electrical Principles ” Bill Ryan, KJ6IGX and Glen Rikerd, NO6W Discussion Leader

Chapter 4

Radio Mathematics

2

Page 3: 1 Chapter 4 “Electrical Principles ” Bill Ryan, KJ6IGX and Glen Rikerd, NO6W Discussion Leader

A General Principle

•Each question should be approximately as difficult as the other questions.

Page 4: 1 Chapter 4 “Electrical Principles ” Bill Ryan, KJ6IGX and Glen Rikerd, NO6W Discussion Leader

General Principles

•Question statements that involve more difficult subjects generally can be expected to have easier answers.

Page 5: 1 Chapter 4 “Electrical Principles ” Bill Ryan, KJ6IGX and Glen Rikerd, NO6W Discussion Leader

Look for Simplicity

•Calculations are set up to use simple numbers or combinations of values

•Estimate the answers

Page 6: 1 Chapter 4 “Electrical Principles ” Bill Ryan, KJ6IGX and Glen Rikerd, NO6W Discussion Leader

Central Ideas

•Resonance•Triangles•Ratio•Handling Exponents

Page 7: 1 Chapter 4 “Electrical Principles ” Bill Ryan, KJ6IGX and Glen Rikerd, NO6W Discussion Leader

Resonance

•Capacitive Reactance and Inductive Reactance cancel each other

Page 8: 1 Chapter 4 “Electrical Principles ” Bill Ryan, KJ6IGX and Glen Rikerd, NO6W Discussion Leader

Triangles

•The test uses simple triangles that can be solved by sight

Page 9: 1 Chapter 4 “Electrical Principles ” Bill Ryan, KJ6IGX and Glen Rikerd, NO6W Discussion Leader

Ratio

The ratio of a triangles sides often tells you how to find the answer

Page 10: 1 Chapter 4 “Electrical Principles ” Bill Ryan, KJ6IGX and Glen Rikerd, NO6W Discussion Leader

Handling Exponents

• When multiplying, add the exponents

• When dividing, change the sign and then add the exponents

• For square roots, cut the exponent in half

Page 11: 1 Chapter 4 “Electrical Principles ” Bill Ryan, KJ6IGX and Glen Rikerd, NO6W Discussion Leader

Question E5C11

• Impedance has a real part (resistance) and an imaginary part (reactance)

• The horizontal axis is the real part

• The vertical axis is the imaginary part

Page 12: 1 Chapter 4 “Electrical Principles ” Bill Ryan, KJ6IGX and Glen Rikerd, NO6W Discussion Leader

Polar Coordinates

•Polar Coordinate System has same information as the rectangular axis

•Determined by a distance from a fixed point and an angle with a fixed direction

Page 13: 1 Chapter 4 “Electrical Principles ” Bill Ryan, KJ6IGX and Glen Rikerd, NO6W Discussion Leader

Phase Angle

Page 14: 1 Chapter 4 “Electrical Principles ” Bill Ryan, KJ6IGX and Glen Rikerd, NO6W Discussion Leader

Question E5B06: Time Constant

• Time it takes for a capacitor to charge or discharge by 63.2% is called the time constant

Page 15: 1 Chapter 4 “Electrical Principles ” Bill Ryan, KJ6IGX and Glen Rikerd, NO6W Discussion Leader

Question E5B06: Time Constant

• Tau = Resistance x Capacitance

Page 16: 1 Chapter 4 “Electrical Principles ” Bill Ryan, KJ6IGX and Glen Rikerd, NO6W Discussion Leader

Handling Exponents

•1 megohm x 1 microfarad =

•1 x 106 x 1 x 10-6 =

•106 + (-6) = 100 = 1 second

Page 17: 1 Chapter 4 “Electrical Principles ” Bill Ryan, KJ6IGX and Glen Rikerd, NO6W Discussion Leader

E5B06: Time Constant Calculation

• Given 800 V DC charge decreases to 294 V DC. (Looks like – 63.2%)

• R x C = 1 megohm x 450 microfarads

• 1 x 106 x 450 x 10-6 = 450 seconds

Page 18: 1 Chapter 4 “Electrical Principles ” Bill Ryan, KJ6IGX and Glen Rikerd, NO6W Discussion Leader

Phase

• Relationship of current and voltage is the phase of the two waves

• Phase means time

• The effect that occurs first leads the second, the trailing effect lags the first

Page 19: 1 Chapter 4 “Electrical Principles ” Bill Ryan, KJ6IGX and Glen Rikerd, NO6W Discussion Leader

Voltage jumps from coil to coil

Since voltage does not travel along the whole wire, it moves faster than the current in an inductor

Voltage leads current in an inductor

Page 20: 1 Chapter 4 “Electrical Principles ” Bill Ryan, KJ6IGX and Glen Rikerd, NO6W Discussion Leader

• Voltage must build up in a capacitor. This causes a delay. Current seems to go right through the capicator

Current leads voltage in a capacitor

Page 21: 1 Chapter 4 “Electrical Principles ” Bill Ryan, KJ6IGX and Glen Rikerd, NO6W Discussion Leader

• When reactance is inductive, voltage leads current

• When reactance is capacitive, current leads voltage

Page 22: 1 Chapter 4 “Electrical Principles ” Bill Ryan, KJ6IGX and Glen Rikerd, NO6W Discussion Leader

Question E5B10: Phase Relationship

What is the Relationship between current through and voltage across an inductor

• Voltage moves from loop to loop, current must move through the wire

• In this question, voltage leads current by 90 degrees

Page 23: 1 Chapter 4 “Electrical Principles ” Bill Ryan, KJ6IGX and Glen Rikerd, NO6W Discussion Leader

Understanding Right Triangles:The central angle is the phase angle

• Two sides are equal: opposite angles = 45 degrees

• Opposite side longer than adjacent: central angle is more than 45 degrees

• Opposite side shorter: central angle is less than 45 degrees

Page 24: 1 Chapter 4 “Electrical Principles ” Bill Ryan, KJ6IGX and Glen Rikerd, NO6W Discussion Leader

Equal sides of triangle, phase angle = 45

• When reactance and resistance are the same number of ohms, then the triangle has equal sides

• 100 ohms• 45• 100 ohms

Page 25: 1 Chapter 4 “Electrical Principles ” Bill Ryan, KJ6IGX and Glen Rikerd, NO6W Discussion Leader

Reactance smaller, phase angle = <45

• When reactance is smaller than the resistance, the phase is less than 45

• 250 ohms• • < -45 100 ohms

• Net Capacitative Reactance is a negative phase angle

Page 26: 1 Chapter 4 “Electrical Principles ” Bill Ryan, KJ6IGX and Glen Rikerd, NO6W Discussion Leader

Reactance larger than resistance

• When reactance is larger than the resistance, the phase is more than 45

• 100 ohms - > -45

• 250 ohms XC

• •

Page 27: 1 Chapter 4 “Electrical Principles ” Bill Ryan, KJ6IGX and Glen Rikerd, NO6W Discussion Leader

Question E5B13

• Resistance = 1000 ohm• Reactance is inductive 250 ohms• Phase angle is less than 45 degrees

and voltage leads the current

• Inductive Reactance• 250 ohms• <45• Resistance 1000 ohms

Page 28: 1 Chapter 4 “Electrical Principles ” Bill Ryan, KJ6IGX and Glen Rikerd, NO6W Discussion Leader

Question E5B07

Resistance = 1000 ohm• Reactance is capacitive 250 ohms• Phase angle is less than 45 degrees

and current leads the voltage 1000• <-45 250

• Capacitive reactance is always drawn downwards to show a negative angle

Page 29: 1 Chapter 4 “Electrical Principles ” Bill Ryan, KJ6IGX and Glen Rikerd, NO6W Discussion Leader

Question E5B13 Complex Impedance

• Phase angle between voltage across and current through series RLC circuit

• XL =500 ohms, XC = 250 ohms R = 1000 ohms

• Net reactance: XL – XC = 250 ohms XL

• Phase angle is less than 45 degrees and voltage leads current

Page 30: 1 Chapter 4 “Electrical Principles ” Bill Ryan, KJ6IGX and Glen Rikerd, NO6W Discussion Leader

Tank Circuits

• Parallel RLC circuits at resonance have a high impedance and appear as an open circuit.

• It also has a circulating current and builds a maximum voltage at resonance.

Page 31: 1 Chapter 4 “Electrical Principles ” Bill Ryan, KJ6IGX and Glen Rikerd, NO6W Discussion Leader

3 4 5 Right Triangles

• Pythagorean Theorem for series RLC

• If shortest sides are 3 and 4 units long, then the longest side is 5 units

• Resistance = 400 ohms and • Reactance = 300 ohms

• Impedance = 500 ohms

Page 32: 1 Chapter 4 “Electrical Principles ” Bill Ryan, KJ6IGX and Glen Rikerd, NO6W Discussion Leader

Pythagorean Triangles

• Any combination of 300 and 400 for the sides will produce an impedance of 500

• 500• 300•

• 400•

Page 33: 1 Chapter 4 “Electrical Principles ” Bill Ryan, KJ6IGX and Glen Rikerd, NO6W Discussion Leader

30 60 90 Triangles

• When the phase angle is 30 degrees, the reactance is always half of the impedance

• 200 • 30 100• 186

Page 34: 1 Chapter 4 “Electrical Principles ” Bill Ryan, KJ6IGX and Glen Rikerd, NO6W Discussion Leader

Equilateral Triangles

When the phase angle is 45 degrees, the impedance is 1.41 times either the reactance or resistance

141 100 45 100

Page 35: 1 Chapter 4 “Electrical Principles ” Bill Ryan, KJ6IGX and Glen Rikerd, NO6W Discussion Leader

Question: E5D19 Power Factor

• Real Power = PF x Apparent Power

• Power Factor (PF) = 0.71

• Apparent Power = 500 watts

• Watts consumed = 0.71 x 500 watts = 355 watts

Page 36: 1 Chapter 4 “Electrical Principles ” Bill Ryan, KJ6IGX and Glen Rikerd, NO6W Discussion Leader

Resonant Frequency

• Same for Series and Parallel Circuits•

Fr = 1

2 pi L x C

Page 37: 1 Chapter 4 “Electrical Principles ” Bill Ryan, KJ6IGX and Glen Rikerd, NO6W Discussion Leader

Resonant Frequency Strategy

•Calculate L x C alone first•Take square root•Plug values into formula•Estimate the result

Page 38: 1 Chapter 4 “Electrical Principles ” Bill Ryan, KJ6IGX and Glen Rikerd, NO6W Discussion Leader

Question E5A14 Resonant Freq

• L x C = • 40 microhenrys x 200 picofarads =

• 40 x 10-6 x 200 x 10-12 = 8,000 x 10-12

• Take square root of 8,000 x 10-18 =

• Square root of 80 x 10-16 = 9 x 10-8

Page 39: 1 Chapter 4 “Electrical Principles ” Bill Ryan, KJ6IGX and Glen Rikerd, NO6W Discussion Leader

Plug in values and estimate for Fr

Fr = 1 = 10+8

6 x 9 x 10-8 50

Fr = .02 x 10+8 = 2 x 10+6 = 2 MHz estimate

Page 40: 1 Chapter 4 “Electrical Principles ” Bill Ryan, KJ6IGX and Glen Rikerd, NO6W Discussion Leader

Answer to Question E5A14

•Estimate of Fr = 2 MHZ

•Actual value of Fr = 1.78 MHz

•Just estimate freely and pick the closest answer

Page 41: 1 Chapter 4 “Electrical Principles ” Bill Ryan, KJ6IGX and Glen Rikerd, NO6W Discussion Leader

Q

• Q is a quality factor for a circuit for low resistive losses

• Q is the ratio of reactance to resistance

• Q = X • R

Page 42: 1 Chapter 4 “Electrical Principles ” Bill Ryan, KJ6IGX and Glen Rikerd, NO6W Discussion Leader

Resonant Circuit Bandwidth

•Bandwidth is the freqency range within 3 dB below peak response

•Delta f = bandwidth = fr

• Q

Page 43: 1 Chapter 4 “Electrical Principles ” Bill Ryan, KJ6IGX and Glen Rikerd, NO6W Discussion Leader

Question E5A13 Half Power BW

• Half Power BW of a parallel resonant circuit:

• Resonant Frequency = 14.25 MHz• Q = 187

• BW = 14.25 MHz = 15 MHz = 100 kHz• 187 150

• Actual answer: 76.2 kHz

Page 44: 1 Chapter 4 “Electrical Principles ” Bill Ryan, KJ6IGX and Glen Rikerd, NO6W Discussion Leader

Encouraging Thoughts

• We want you to pass the Extra Exam

• Try to focus on simplifying methods

• True mastery of this material will take a long time

• Commit to taking the test at end of class despite whether feeling ready

Page 45: 1 Chapter 4 “Electrical Principles ” Bill Ryan, KJ6IGX and Glen Rikerd, NO6W Discussion Leader

In Class Practice PRoblems

• Here is a list of possible in class seminar questions

Page 46: 1 Chapter 4 “Electrical Principles ” Bill Ryan, KJ6IGX and Glen Rikerd, NO6W Discussion Leader

Chapter 4

E5C11 What do the two numbers represent that are used to define a point on a graph using rectangular coordinates?•The sine and cosine values•The tangent and cotangent values•The coordinate values along the horizontal and vertical axes•The magnitude and phase of the point

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Page 47: 1 Chapter 4 “Electrical Principles ” Bill Ryan, KJ6IGX and Glen Rikerd, NO6W Discussion Leader

Chapter 4

E5D05 What is a magnetic field?•The space between the plates of a charged capacitor, through which a magnetic force acts•The force that drives current through a resistor•The region surrounding a magnet through which a magnetic force acts•Electric current through the space around a permanent magnet 47

Page 48: 1 Chapter 4 “Electrical Principles ” Bill Ryan, KJ6IGX and Glen Rikerd, NO6W Discussion Leader

Chapter 4

E5B05 (A) How long does it take for an initial charge of 20 V DC to decrease to 7.36 V DC in a 0.01-microfarad capacitor when a 2-megohm resistor is connected across it?•0.02 seconds•0.04 seconds•20 seconds•40 seconds 48

Page 49: 1 Chapter 4 “Electrical Principles ” Bill Ryan, KJ6IGX and Glen Rikerd, NO6W Discussion Leader

Chapter 4

E5D13 (B) How many watts are consumed in a circuit having a power factor of 0.2 if the input is 100-V AC at 4 amperes?•2000 watts• 400 watts• 80 watts• 50 watts

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Page 50: 1 Chapter 4 “Electrical Principles ” Bill Ryan, KJ6IGX and Glen Rikerd, NO6W Discussion Leader

Chapter 4

E5C19 (B) Which point on Figure E5-2 best represents that impedance of a series circuit consisting of a 400 ohm resistor and a 38 picofarad capacitor at 14 MHz?•Point 6•Point 5•Point 4•Point 2 50

Page 51: 1 Chapter 4 “Electrical Principles ” Bill Ryan, KJ6IGX and Glen Rikerd, NO6W Discussion Leader

Chapter 4

Figure E5-2 refers to question E5C19

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Page 52: 1 Chapter 4 “Electrical Principles ” Bill Ryan, KJ6IGX and Glen Rikerd, NO6W Discussion Leader

Chapter 4

• E5A12 What is the half-power bandwidth of a parallel resonant circuit that has a resonant frequency of 3.7 MHz and a Q of 118?

• 15.7 kHz• 31.4 kHz• 218.3 kHz• 436.6 kHz

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Page 53: 1 Chapter 4 “Electrical Principles ” Bill Ryan, KJ6IGX and Glen Rikerd, NO6W Discussion Leader

Next Class Session

• Study Chapter 5 “Components and Building Blocks”

• Study the Chapter 5 Question Pool questions found in the Blue Boxes

• Prepare your Chapter 5 Study Guide materials

53

Page 54: 1 Chapter 4 “Electrical Principles ” Bill Ryan, KJ6IGX and Glen Rikerd, NO6W Discussion Leader

Chapter 4

E5C11 What do the two numbers represent that are used to define a point on a graph using rectangular coordinates?•The sine and cosine values•The tangent and cotangent values•The coordinate values along the horizontal and vertical axes•The magnitude and phase of the point

54

Page 55: 1 Chapter 4 “Electrical Principles ” Bill Ryan, KJ6IGX and Glen Rikerd, NO6W Discussion Leader

Chapter 4

E5D05 What is a magnetic field?•The space between the plates of a charged capacitor, through which a magnetic force acts•The force that drives current through a resistor•The region surrounding a magnet through which a magnetic force acts•Electric current through the space around a permanent magnet

55

Page 56: 1 Chapter 4 “Electrical Principles ” Bill Ryan, KJ6IGX and Glen Rikerd, NO6W Discussion Leader

Chapter 4

E5B05 (A) How long does it take for an initial charge of 20 V DC to decrease to 7.36 V DC in a 0.01-microfarad capacitor when a 2-megohm resistor is connected across it?•0.02 seconds•0.04 seconds•20 seconds•40 seconds 56

Page 57: 1 Chapter 4 “Electrical Principles ” Bill Ryan, KJ6IGX and Glen Rikerd, NO6W Discussion Leader

Chapter 4

E5D13 (B) How many watts are consumed in a circuit having a power factor of 0.2 if the input is 100-V AC at 4 amperes?•2000 watts• 400 watts• 80 watts• 50 watts

57

Page 58: 1 Chapter 4 “Electrical Principles ” Bill Ryan, KJ6IGX and Glen Rikerd, NO6W Discussion Leader

Chapter 4

E5C19 (B) Which point on Figure E5-2 best represents that impedance of a series circuit consisting of a 400 ohm resistor and a 38 picofarad capacitor at 14 MHz?•Point 6•Point 5•Point 4•Point 2 58

Page 59: 1 Chapter 4 “Electrical Principles ” Bill Ryan, KJ6IGX and Glen Rikerd, NO6W Discussion Leader

Chapter 4

Figure E5-2 refers to question E5C19

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Page 60: 1 Chapter 4 “Electrical Principles ” Bill Ryan, KJ6IGX and Glen Rikerd, NO6W Discussion Leader

Chapter 4

• E5A12 (B) What is the half-power bandwidth of a parallel resonant circuit that has a resonant frequency of 3.7 MHz and a Q of 118?

• 15.7 kHz• 31.4 kHz• 218.3 kHz• 436.6 kHz

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Page 61: 1 Chapter 4 “Electrical Principles ” Bill Ryan, KJ6IGX and Glen Rikerd, NO6W Discussion Leader

Next Class Session

• Study Chapter 5 “Components and Building Blocks”

• Study the Chapter 5 Question Pool questions found in the Blue Boxes

• Prepare your Chapter 5 Study Guide materials

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