Chapter 8 Principles of Electric Circuits, Electron Flow, 9 th ed. Floyd © 2010 Pearson Higher...

Chapter 8Chapter 8

Principles of Electric Circuits, Electron Flow, 9th ed.Floyd

© 2010 Pearson Higher Education,Upper Saddle River, NJ 07458. • All Rights Reserved

Chapter 8Chapter 8



SummarySummary

An ideal voltage source plots a vertical line on the VI characteristic as shown for the ideal 6.0 V source.

Voltage sources

Actual voltage sources include the internal source resistance, which can drop a small voltage under load. The characteristic of a non-ideal source is not vertical.

Cur

rent

(A)

Voltage (V)

0 0

1

2

2

3

4

4 6

5

8 10

Chapter 8Chapter 8



SummarySummary

Voltage sources

A practical voltage source is drawn as an ideal source in series with the source resistance. When the internal resistance is zero, the source reduces to an ideal one.

RS

VS+

Chapter 8Chapter 8



SummarySummary

Voltage sources

If the source resistance of a 5.0 V power supply is 0.5 , what is the voltage across a 68 load?

Use the voltage-divider equation

LL S

L S

68 5 V = 4.96 V

68 0.5

RV V

R R

RS

RL

VS

VOUT

+5.0 V

0.5

68

Chapter 8Chapter 8



SummarySummary

An ideal current source plots a horizontal line on the VI characteristic as shown for the ideal 4.0 mA source.

Current sources

Practical current sources have internal source resistance, which takes some of the current. The characteristic of a practical source is not horizontal.

Cur

rent

(A)

Voltage (V)

0 0

1

2

2

3

4

4 6

5

8 10

Chapter 8Chapter 8



SummarySummary

A practical current source is drawn as an ideal source with a parallel source resistance. When the source resistance is infinite, the current source is ideal.

Current sources

RSIS

Chapter 8Chapter 8



SummarySummary

Current sources

If the source resistance of a 10 mA current source is 4.7 k, what is the voltage across a 100 load?

Use the current-divider equation

SL S

L S

4.7 k10 mA = 9.8 mA

100 4.7 k

RI I

R R

RSIS RL

100 4.7 k10 mA

Chapter 8Chapter 8



SummarySummary

Source conversions

Any voltage source with an internal resistance can be converted to an equivalent current source and vice-versa by applying Ohm’s law to the source. The source resistance, RS, is the same for both.

To convert a voltage source to a current source, SS

S

VI

R

S S SV I RTo convert a current source to a voltage source,

Chapter 8Chapter 8



SummarySummary

Superposition theoremThe superposition theorem is a way to determine currents and voltages in a linear circuit that has multiple sources by taking one source at a time and algebraically summing the results.

What does the ammeter read for I2? (See next slide for the method and the answer).

+-

-

+

-

+

R 1 R 3

R 2

I 2

V S 2V S 1

1 2 V

2 .7 k 6 .8 k

6 .8 k

1 8 V

Chapter 8Chapter 8



SummarySummary

6.10 k

What does the ammeter read for I2?

1.97 mA 0.98 mA

8.73 k 2.06 mA

+-

-

+

-

+

R 1 R 3

R 2

I 2

V S 2V S 1

1 2 V

2 .7 k 6 .8 k

6 .8 k

1 8 V

0.58 mA

1.56 mA

1.56 mA

Source 1: RT(S1)= I1= I2= Source 2: RT(S2)= I3= I2= Both sources I2=

Set up a table of pertinent information and solve for each quantity listed:

The total current is the algebraic sum.

Chapter 8Chapter 8



SummarySummary

Thevenin’s theorem states that any two-terminal, resistive circuit can be replaced with a simple equivalent circuit when viewed from two output terminals. The equivalent circuit is:

Thevenin’s theorem

V T H

R T H

Chapter 8Chapter 8



SummarySummary

V T H

R T H

VTH is defined as


RTH is defined as

the open circuit voltage between the two output terminals of a circuit.

the total resistance appearing between the two output terminals when all sources have been replaced by their internal resistances.

Chapter 8Chapter 8



SummarySummary


R

R

1

R 2R 2 L

V SV S

1 2 V1 0 k

6 8 k 2 7 k

Output terminals

What is the Thevenin voltage for the circuit? 8.76 V

What is the Thevenin resistance for the circuit? 7.30 k

Remember, the load resistor has no affect on the Thevenin parameters.

Chapter 8Chapter 8



SummarySummary


Thevenin’s theorem is useful for solving the Wheatstone bridge. One way to Thevenize the bridge is to create two Thevenin circuits from A to ground and from B to ground. The resistance between point A and ground is R1||R3 and the resistance from B to ground is R2||R4. The voltage on each side of the bridge is found using the voltage divider rule.

R3 R4

R2

RL

R1VS

-

+A B

Chapter 8Chapter 8



SummarySummary


For the bridge shown, R1||R3 = and R2||R4 = . The voltage from A to ground (with no load) is and from B to ground (with no load) is .

The Thevenin circuits for each of the bridge are shown on the following slide.

165 179

7.5 V 6.87 V

R3 R4

R2

RL

R1VS

-

+A B

330 390

330 330

+15 V150

Chapter 8Chapter 8



SummarySummary


Putting the load on the Thevenin circuits and applying the superposition theorem allows you to calculate the load current.

RLA B

150 VTH VTH

RTH RTH'

'165 179

7.5 V 6.87 V

A B

VTH VTH

RTH RTH'

'165 179

7.5 V 6.87 V

1.27 mAThe load current is:

Chapter 8Chapter 8



SummarySummary

Norton’s theorem states that any two-terminal, resistive circuit can be replaced with a simple equivalent circuit when viewed from two output terminals. The equivalent circuit is:

Norton’s theorem

RNIN

Chapter 8Chapter 8



SummarySummary

Norton’s theorem

the output current when the output terminals are shorted.

the total resistance appearing between the two output terminals when all sources have been replaced by their internal resistances.

IN is defined as

RN is defined as

RNIN

Chapter 8Chapter 8



SummarySummary

Norton’s theorem

Output terminals

What is the Norton current for the circuit? 17.9 mA

What is the Norton resistance for the circuit? 359

R2

R1

RL

VS +10 V

560

820 1.0 k

The Norton circuit is shown on the following slide.

Chapter 8Chapter 8



SummarySummary

Norton’s theorem

RNIN

17.9 mA 359

The Norton circuit (without the load) is:

Chapter 8Chapter 8



SummarySummary

Maximum power transfer

The maximum power is transferred from a source to a load when the load resistance is equal to the internal source resistance.

The maximum power transfer theorem assumes the source voltage and resistance are fixed.

RS

RL

VS +

Chapter 8Chapter 8



SummarySummary


What is the power delivered to the matching load?

The voltage to the load is 5.0 V. The power delivered is

RS

RL

VS + 50

50 10 V

22

LL

5.0 V= 0.5 W

50

VP

R

Chapter 8Chapter 8



SummarySummary

-to-Y and Y-to- conversion

The -to-Y and Y-to- conversion formulas allow a three terminal resistive network to be replaced with an equivalent network.

For the -to-Y conversion, each resistor in the Y is equal to the product of the resistors in the two adjacent branches divided by the sum of all three resistors.

RC

RA RB

R1 R2

R3

Chapter 8Chapter 8



SummarySummary

-to-Y and Y–to- conversion

The -to-Y and Y-to- conversion formulas allow a three terminal resistive network to be replaced with an equivalent network.

For the Y-to- conversion, each resistor in the is equal to the sum of all products of Y resistors, taken two at a time divided by the opposite Y resistor.

RC

RA RB

R1 R2

R3

Chapter 8Chapter 8



Current source


Norton’s theorem

Superposition theorem

Transfer of maximum power from a source to a load occurs when the load resistance equals the internal source resistance.

A method for simplifying a two-terminal linear circuit to an equivalent circuit with only a current source in parallel with a resistance.

A device that ideally provides a constant value of current regardless of the load.

Key Terms

A method for analysis of circuits with more than one source.

Chapter 8Chapter 8



Terminal equivalency


Voltage source

A method for simplifying a two-terminal linear circuit to an equivalent circuit with only a voltage source in series with a resistance.

The concept that when any given load is connected to two sources, the same load voltage and current are produced by both sources.

Key Terms

A device that ideally provides a constant value of voltage regardless of the load.

Chapter 8Chapter 8



Quiz

1. The source resistance from a 1.50 V D-cell is 1.5 . The voltage that appears across a 75 load will be

a. 1.47 V

b. 1.50 V

c. 1.53 V

d. 1.60 V

RS

RL

VS

VOUT

+1.5 V

1.5

75

Chapter 8Chapter 8



Quiz

2. The internal resistance of an ideal current source

a. is 0

b. is 1

c. is infinite

d. depends on the source

Chapter 8Chapter 8



Quiz

3. The superposition theorem cannot be applied to

a. circuits with more than two sources

b. nonlinear circuits

c. circuits with current sources

d. ideal sources

Chapter 8Chapter 8



Quiz

4. A Thevenin circuit is a

a. resistor in series with a voltage source

b. resistor in parallel with a voltage source

c. resistor in series with a current source

d. resistor in parallel with a current source

Chapter 8Chapter 8



Quiz

4. A Norton circuit is a

a. resistor in series with a voltage source

b. resistor in parallel with a voltage source

c. resistor in series with a current source

d. resistor in parallel with a current source

Chapter 8Chapter 8



Quiz

5. A signal generator has an output voltage of 2.0 V with no load. When a 600 load is connected to it, the output drops to 1.0 V. The Thevenin resistance of the generator is

a. 300

b. 600

c. 900

d. 1200 .

Chapter 8Chapter 8



Quiz

6. A signal generator has an output voltage of 2.0 V with no load. When a 600 load is connected to it, the output drops to 1.0 V. The Thevenin voltage of the generator is

a. 1.0 V

b. 2.0 V

c. 4.0 V

d. not enough information to tell.

Chapter 8Chapter 8



Quiz

7. A Wheatstone bridge is shown with the Thevenin circuit as viewed with respect to ground. The total Thevenin resistance (RTH + RTH’) is

a. 320

b. 500

c. 820

d. 3.47 k.

R3 R4

R2

RL

R1

VS

-

+A B

1.0 k 1.0 k

1.0 k 470

100

VTH VTH

RTH RL RTH'

'

Chapter 8Chapter 8



Quiz

8. The Norton current for the circuit is

a. 5.0 mA

b. 6.67 mA

c. 8.33 mA

d. 10 mA

R2

R1

RL

VS +10 V

1.0 k

1.0 k 1.0 k

Chapter 8Chapter 8



Quiz

9. The Norton resistance for the circuit is

a. 500

b. 1.0 k

c. 1.5 k

d. 2.0 k

R2

R1

RL

VS +10 V

1.0 k

1.0 k 1.0 k

Chapter 8Chapter 8



Quiz

10. Maximum power is transferred from a fixed source when

a. the load resistor is ½ the source resistance

b. the load resistor is equal to the source resistance

c. the load resistor is twice the source resistance

d. none of the above

Chapter 8Chapter 8



Quiz

Answers:

1. a

2. c

3. b

4. d

5. b

6. b

7. c

8. d

9. a

10. b

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Chapter 8 Principles of Electric Circuits, Electron Flow, 9 th ed. Floyd © 2010 Pearson Higher...

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