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
Principles of Electric Circuits, Electron Flow, 9th ed.Floyd
© 2010 Pearson Higher Education,Upper Saddle River, NJ 07458. • All Rights Reserved
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
Principles of Electric Circuits, Electron Flow, 9th ed.Floyd
© 2010 Pearson Higher Education,Upper Saddle River, NJ 07458. • All Rights Reserved
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
Principles of Electric Circuits, Electron Flow, 9th ed.Floyd
© 2010 Pearson Higher Education,Upper Saddle River, NJ 07458. • All Rights Reserved
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
Principles of Electric Circuits, Electron Flow, 9th ed.Floyd
© 2010 Pearson Higher Education,Upper Saddle River, NJ 07458. • All Rights Reserved
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
Principles of Electric Circuits, Electron Flow, 9th ed.Floyd
© 2010 Pearson Higher Education,Upper Saddle River, NJ 07458. • All Rights Reserved
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
Principles of Electric Circuits, Electron Flow, 9th ed.Floyd
© 2010 Pearson Higher Education,Upper Saddle River, NJ 07458. • All Rights Reserved
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
Principles of Electric Circuits, Electron Flow, 9th ed.Floyd
© 2010 Pearson Higher Education,Upper Saddle River, NJ 07458. • All Rights Reserved
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
Principles of Electric Circuits, Electron Flow, 9th ed.Floyd
© 2010 Pearson Higher Education,Upper Saddle River, NJ 07458. • All Rights Reserved
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
Principles of Electric Circuits, Electron Flow, 9th ed.Floyd
© 2010 Pearson Higher Education,Upper Saddle River, NJ 07458. • All Rights Reserved
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
Principles of Electric Circuits, Electron Flow, 9th ed.Floyd
© 2010 Pearson Higher Education,Upper Saddle River, NJ 07458. • All Rights Reserved
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
Principles of Electric Circuits, Electron Flow, 9th ed.Floyd
© 2010 Pearson Higher Education,Upper Saddle River, NJ 07458. • All Rights Reserved
SummarySummary
V T H
R T H
VTH is defined as
Thevenin’s theorem
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
Principles of Electric Circuits, Electron Flow, 9th ed.Floyd
© 2010 Pearson Higher Education,Upper Saddle River, NJ 07458. • All Rights Reserved
SummarySummary
Thevenin’s theorem
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
Principles of Electric Circuits, Electron Flow, 9th ed.Floyd
© 2010 Pearson Higher Education,Upper Saddle River, NJ 07458. • All Rights Reserved
SummarySummary
Thevenin’s theorem
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
Principles of Electric Circuits, Electron Flow, 9th ed.Floyd
© 2010 Pearson Higher Education,Upper Saddle River, NJ 07458. • All Rights Reserved
SummarySummary
Thevenin’s theorem
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
Principles of Electric Circuits, Electron Flow, 9th ed.Floyd
© 2010 Pearson Higher Education,Upper Saddle River, NJ 07458. • All Rights Reserved
SummarySummary
Thevenin’s theorem
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
Principles of Electric Circuits, Electron Flow, 9th ed.Floyd
© 2010 Pearson Higher Education,Upper Saddle River, NJ 07458. • All Rights Reserved
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
Principles of Electric Circuits, Electron Flow, 9th ed.Floyd
© 2010 Pearson Higher Education,Upper Saddle River, NJ 07458. • All Rights Reserved
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
Principles of Electric Circuits, Electron Flow, 9th ed.Floyd
© 2010 Pearson Higher Education,Upper Saddle River, NJ 07458. • All Rights Reserved
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
Principles of Electric Circuits, Electron Flow, 9th ed.Floyd
© 2010 Pearson Higher Education,Upper Saddle River, NJ 07458. • All Rights Reserved
SummarySummary
Norton’s theorem
RNIN
17.9 mA 359
The Norton circuit (without the load) is:
Chapter 8Chapter 8
Principles of Electric Circuits, Electron Flow, 9th ed.Floyd
© 2010 Pearson Higher Education,Upper Saddle River, NJ 07458. • All Rights Reserved
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
Principles of Electric Circuits, Electron Flow, 9th ed.Floyd
© 2010 Pearson Higher Education,Upper Saddle River, NJ 07458. • All Rights Reserved
SummarySummary
Maximum power transfer
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
Principles of Electric Circuits, Electron Flow, 9th ed.Floyd
© 2010 Pearson Higher Education,Upper Saddle River, NJ 07458. • All Rights Reserved
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
Principles of Electric Circuits, Electron Flow, 9th ed.Floyd
© 2010 Pearson Higher Education,Upper Saddle River, NJ 07458. • All Rights Reserved
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
Principles of Electric Circuits, Electron Flow, 9th ed.Floyd
© 2010 Pearson Higher Education,Upper Saddle River, NJ 07458. • All Rights Reserved
Current source
Maximum power transfer
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
Principles of Electric Circuits, Electron Flow, 9th ed.Floyd
© 2010 Pearson Higher Education,Upper Saddle River, NJ 07458. • All Rights Reserved
Terminal equivalency
Thevenin’s theorem
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
Principles of Electric Circuits, Electron Flow, 9th ed.Floyd
© 2010 Pearson Higher Education,Upper Saddle River, NJ 07458. • All Rights Reserved
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
Principles of Electric Circuits, Electron Flow, 9th ed.Floyd
© 2010 Pearson Higher Education,Upper Saddle River, NJ 07458. • All Rights Reserved
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
Principles of Electric Circuits, Electron Flow, 9th ed.Floyd
© 2010 Pearson Higher Education,Upper Saddle River, NJ 07458. • All Rights Reserved
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
Principles of Electric Circuits, Electron Flow, 9th ed.Floyd
© 2010 Pearson Higher Education,Upper Saddle River, NJ 07458. • All Rights Reserved
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
Principles of Electric Circuits, Electron Flow, 9th ed.Floyd
© 2010 Pearson Higher Education,Upper Saddle River, NJ 07458. • All Rights Reserved
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
Principles of Electric Circuits, Electron Flow, 9th ed.Floyd
© 2010 Pearson Higher Education,Upper Saddle River, NJ 07458. • All Rights Reserved
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
Principles of Electric Circuits, Electron Flow, 9th ed.Floyd
© 2010 Pearson Higher Education,Upper Saddle River, NJ 07458. • All Rights Reserved
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
Principles of Electric Circuits, Electron Flow, 9th ed.Floyd
© 2010 Pearson Higher Education,Upper Saddle River, NJ 07458. • All Rights Reserved
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
Principles of Electric Circuits, Electron Flow, 9th ed.Floyd
© 2010 Pearson Higher Education,Upper Saddle River, NJ 07458. • All Rights Reserved
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
Principles of Electric Circuits, Electron Flow, 9th ed.Floyd
© 2010 Pearson Higher Education,Upper Saddle River, NJ 07458. • All Rights Reserved
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
Principles of Electric Circuits, Electron Flow, 9th ed.Floyd
© 2010 Pearson Higher Education,Upper Saddle River, NJ 07458. • All Rights Reserved
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
Principles of Electric Circuits, Electron Flow, 9th ed.Floyd
© 2010 Pearson Higher Education,Upper Saddle River, NJ 07458. • All Rights Reserved
Quiz
Answers:
1. a
2. c
3. b
4. d
5. b
6. b
7. c
8. d
9. a
10. b