, SOLUTION MANUAL
Electronic Principles 9th Edition by Albert Paul Malvino,
Patrick Hoppe
Part 1
Electronic Principles
Chapter 1 Introduction us more insight into how changes in load resistance affect
the load voltage.
SELF-TEST 12. It is usually easy to measure open-circuit voltage and
1. a 7. b 13. c 19. b shorted-load current. By using a load resistor and measuring
2. c 8. c 14. d 20. c voltage under load, it is easy to calculate the Thevenin or
3. a 9. b 15. b 21. b Norton resistance.
4. b 10. a 16. b 22. b
5. d 11. a 17. a 23. c PROBLEMS
6. d 12. a 18. b
1-1. Given:
V = 12 V
JOB INTERVIEW QUESTIONS
RS = 0.1 Ω
Note: The text and illustrations cover many of the job interview
Solution:
questions in detail. An answer is given to job interview questions
only when the text has insufficient information. RL = 100RS
RL = 100(0.1 Ω)
2. It depends on how accurate your calculations need to be. If RL = 10 Ω
an accuracy of 1 percent is adequate, you should include the
source resistance whenever it is greater than 1 percent of the Answer: The voltage source will appear stiff for values
load resistance. of load resistance of ≥10 Ω.
5. Measure the open-load voltage to get the Thevenin voltage 1-2. Given:
VTH. To get the Thevenin resistance, reduce all sources to
zero and measure the resistance between the AB terminals to RLmin = 270 Ω
get RTH. If this is not possible, measure the voltage VL across RLmax = 100 kΩ
a load resistor and calculate the load current IL. Then divide Solution:
VTH – VL by IL to get RTH. RS < 0.01 RL (Eq. 1-1)
6. The advantage of a 50 Ω voltage source over a 600 Ω voltage RS < 0.01(270 Ω)
source is the ability to be a stiff voltage source to a lower RS < 2.7 Ω
value resistance load. The load must be 100 greater than the
internal resistance in order for the voltage source to be Answer: The largest internal resistance the source can
considered stiff. have is 2.7 Ω.
7. The expression cold-cranking amperes refers to the amount 1-3. Given: RS = 50 Ω
of current a car battery can deliver in freezing weather when
it is needed most. What limits actual current is the Thevenin Solution:
resistance caused by chemical and physical parameters inside RL = 100RS
the battery, not to mention the quality of the connections RL = 100(50 Ω)
outside. RL = 5 kΩ
8. It means that the load resistance is not large compared to the Answer: The function generator will appear stiff for
Thevenin resistance, so that a large load current exists.
values of load resistance of ≥5 kΩ.
9. Ideal. Because troubles usually produce large changes in
voltage and current, so that the ideal approximation is ade- 1-4. Given: RS = 0.04 Ω
quate for most troubles. Solution:
10. You should infer nothing from a reading that is only RL = 100RS
5 percent from the ideal value. Actual circuit troubles will RL = 100(0.04 Ω)
usually cause large changes in circuit voltages. Small
changes can result from component variations that are still RL = 4 Ω
within the allowable tolerance. Answer: The car battery will appear stiff for values of
11. Either may be able to simplify the analysis, save time when load resistance of ≥ 4 Ω.
calculating load current for several load resistances, and give
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McGraw-Hill Education. 1-1
, 1-5. Given: Solution:
RS = 0.05 Ω RL = 0.01RS (Eq. 1-4)
I=2A RL = 0.01(250 kΩ)
Solution: RL = 2.5 kΩ
V = IR (Ohm’s law) IL = IT [(RS)/(RS + RL)] (Current divider formula)
V = (2 A)(0.05 Ω) IL = 5 mA [(250 kΩ)/(250 kΩ + 10 kΩ)]
V = 0.1 V IL = 4.80 mA
Answer: The voltage drop across the internal resistance is Answer: The load current is 4.80 mA, and, no, the current
0.1 V. source is not stiff since the load resistance is not less than
1-6. Given: or equal to 2.5 kΩ.
V=9V 1-12. Solution:
RS = 0.4 Ω VTH = VR2
Solution: VR2 = VS[(R2)/(R1 + R2)] (Voltage divider formula)
I = V/R (Ohm’s law) VR2 = 36 V[(3 kΩ)/(6 kΩ + 3 kΩ)]
I = (9 V)/(0.4 Ω) VR2 = 12 V
I = 22.5 A RTH = [R1R2/R1 + R2] (Parallel resistance formula)
Answer: The load current is 22.5 A. RTH = [(6 kΩ)(3 kΩ)/(6 kΩ + 3 kΩ)]
RTH = 2 kΩ
1-7. Given:
IS = 10 mA Answer: The Thevenin voltage is 12 V, and the
Thevenin resistance is 2 kΩ.
RS = 10 MΩ
Solution:
RL = 0.01 RS
RL = 0.01(10 MΩ) 6 kΩ R1
RL = 100 kΩ
Answer: The current source will appear stiff for load 36 V
resistance of ≤100 kΩ.
1-8. Given: 3 kΩ R2 VTH
RLmin = 270 Ω
RLmax = 100 kΩ
Solution:
RS > 100 RL (Eq. 1-3)
RS > 100(100 kΩ)
RS > 10 MΩ 36 V 6 kΩ R1
Answer: The internal resistance of the source is greater
than 10 MΩ.
1-9. Given: RS = 100 kΩ
Solution: 3 kΩ R2 RTH
RL = 0.01RS (Eq. 1-4)
RL = 0.01(100 kΩ)
RL = 1 kΩ
Answer: The maximum load resistance for the current (a) Circuit for finding VTH in Prob. 1-12. (b) Circuit
source to appear stiff is 1 kΩ. for finding RTH in Prob. 1-12.
1-10. Given:
IS = 20 mA 1-13. Given:
RS = 200 kΩ VTH = 12 V
RL = 0 Ω RTH = 2 kΩ
Solution: Solution:
RL= 0.01RS I = V/R (Ohm’s law)
RL= 0.01(200 kΩ) I = VTH/(RTH + RL)
RL= 2 kΩ I0Ω = 12 V/(2 kΩ + 0 Ω) = 6 mA
Answer: Since 0 Ω is less than the maximum load I1kΩ = 12 V/(2 kΩ + 1 kΩ) = 4 mA
resistance of 2 kΩ, the current source appea rs stiff; thus I2kΩ = 12 V/(2 kΩ + 2 kΩ) = 3 mA
I3kΩ = 12 V/(2 kΩ + 3 kΩ) = 2.4 mA
the current is 20 mA.
I4kΩ = 12 V/(2 kΩ + 4 kΩ) = 2 mA
1-11. Given: I5kΩ = 12 V/(2 kΩ + 5 kΩ) = 1.7 mA
I = 5 mA I6kΩ = 12 V/(2 kΩ + 6 kΩ) = 1.5 mA
RS = 250 kΩ Answers: 0 Ω 6 mA; 1 kΩ, 4 mA; 2 kΩ, 3mA; 3 kΩ, 2.4
RL = 10 kΩ mA; 4 kΩ, 2 mA; 5 kΩ, 1.7 mA; 6 kΩ, 1.5 mA.
Copyright 2021 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of
McGraw-Hill Education. 1-2
, R Solution:
TH
RN = RTH (Eq. 1-10)
RTH = 10 kΩ
V R IN = VTH/RTH (Eq. 1-12)
TH L
VTH = INRN
VTH = (10 mA)(10 kΩ)
Thevenin equivalent circuit for Prob. 1-13. VTH = 100 V
Answer: RTH = 10 kΩ, and VTH = 100 V
1-14. Given:
VS = 18 V R
TH
R1 = 6 kΩ
10 kΩ
R2 = 3 kΩ VTH
Solution:
VTH = VR2 100 V
VR2 = VS[(R2)/(R1 + R2)] (Voltage divider formula)
VR2 = 18 V[(3 kΩ)/(6 kΩ + 3 kΩ)] Thevenin circuit for Prob. 1-17.
VR2 = 6 V
RTH = [(R1 × R2)/(R1 + R2)] (Parallel resistance formula) 1-18. Given (from Prob. 1-12):
RTH = [(6 kΩ × 3 kΩ)/(6 kΩ + 3 kΩ)] VTH = 12 V
RTH = 2 kΩ RTH = 2 kΩ
Answer: The Thevenin voltage decreases to 6 V, and the Solution:
Thevenin resistance is unchanged.
RN = RTH (Eq. 1-10)
1-15. Given: RN = 2 kΩ
VS = 36 V IN = VTH/RTH (Eq. 1-12)
R1 = 12 kΩ IN = 12 V/2 kΩ
R2 = 6 kΩ IN = 6 mA
Solution: Answer: RN = 2 kΩ, and IN = 6 mA
VTH = VR2
VR2 = VS[(R2)/(R1 + R2)] (Voltage divider formula)
R
VR2 = 36 V[(6 kΩ)/(12 kΩ + 6 kΩ)] IN N
VR2 = 12 V
RTH = [(R1R2)/(R1 + R2)] (Parallel resistance formula) 6 mA 2 kΩ
RTH = [(12 kΩ)(6 kΩ)/(12 kΩ + 6 kΩ)]
RTH = 4 kΩ Norton circuit for Prob. 1-18.
Answer: The Thevenin voltage is unchanged, and the
Thevenin resistance doubles. 1-19. Shorted, which would cause load resistor to be connected
1-16. Given: across the voltage source seeing all of the voltage.
VTH = 12 V 1-20. a. R1 is open, preventing any of the voltage from reaching
RTH = 3 kΩ the load resistor. b. R2 is shorted, making its voltage drop
zero. Since the load resistor is in parallel with R2, its
Solution: voltage drop would also be zero.
RN = RTH
1-21. The battery or interconnecting wiring.
RN = 3 kΩ
IN = VTH/RTH 1-22. RTH = 2 kΩ
IN = 12 V/3 kΩ Solution:
IN = 4 mA RMeter = 100RTH
Answer: IN = 4 mA, and RN = 3 kΩ RMeter = 100(2 kΩ)
RMeter = 200 kΩ
Answer: The meter will not load down the circuit if the
IN RN
meter impedance is ≥ 200 kΩ.
4 mA 3 kΩ
CRITICAL THINKING
1-23. Given:
Norton circuit for Prob. 1-16. VS = 12 V
IS = 150 A
1-17. Given:
IN = 10 mA Solution:
RN = 10 kΩ RS = (VS)/(IS)
RS = (12 V)/(150 A)
RS = 80 mΩ
Copyright 2021 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of
McGraw-Hill Education. 1-3
Electronic Principles 9th Edition by Albert Paul Malvino,
Patrick Hoppe
Part 1
Electronic Principles
Chapter 1 Introduction us more insight into how changes in load resistance affect
the load voltage.
SELF-TEST 12. It is usually easy to measure open-circuit voltage and
1. a 7. b 13. c 19. b shorted-load current. By using a load resistor and measuring
2. c 8. c 14. d 20. c voltage under load, it is easy to calculate the Thevenin or
3. a 9. b 15. b 21. b Norton resistance.
4. b 10. a 16. b 22. b
5. d 11. a 17. a 23. c PROBLEMS
6. d 12. a 18. b
1-1. Given:
V = 12 V
JOB INTERVIEW QUESTIONS
RS = 0.1 Ω
Note: The text and illustrations cover many of the job interview
Solution:
questions in detail. An answer is given to job interview questions
only when the text has insufficient information. RL = 100RS
RL = 100(0.1 Ω)
2. It depends on how accurate your calculations need to be. If RL = 10 Ω
an accuracy of 1 percent is adequate, you should include the
source resistance whenever it is greater than 1 percent of the Answer: The voltage source will appear stiff for values
load resistance. of load resistance of ≥10 Ω.
5. Measure the open-load voltage to get the Thevenin voltage 1-2. Given:
VTH. To get the Thevenin resistance, reduce all sources to
zero and measure the resistance between the AB terminals to RLmin = 270 Ω
get RTH. If this is not possible, measure the voltage VL across RLmax = 100 kΩ
a load resistor and calculate the load current IL. Then divide Solution:
VTH – VL by IL to get RTH. RS < 0.01 RL (Eq. 1-1)
6. The advantage of a 50 Ω voltage source over a 600 Ω voltage RS < 0.01(270 Ω)
source is the ability to be a stiff voltage source to a lower RS < 2.7 Ω
value resistance load. The load must be 100 greater than the
internal resistance in order for the voltage source to be Answer: The largest internal resistance the source can
considered stiff. have is 2.7 Ω.
7. The expression cold-cranking amperes refers to the amount 1-3. Given: RS = 50 Ω
of current a car battery can deliver in freezing weather when
it is needed most. What limits actual current is the Thevenin Solution:
resistance caused by chemical and physical parameters inside RL = 100RS
the battery, not to mention the quality of the connections RL = 100(50 Ω)
outside. RL = 5 kΩ
8. It means that the load resistance is not large compared to the Answer: The function generator will appear stiff for
Thevenin resistance, so that a large load current exists.
values of load resistance of ≥5 kΩ.
9. Ideal. Because troubles usually produce large changes in
voltage and current, so that the ideal approximation is ade- 1-4. Given: RS = 0.04 Ω
quate for most troubles. Solution:
10. You should infer nothing from a reading that is only RL = 100RS
5 percent from the ideal value. Actual circuit troubles will RL = 100(0.04 Ω)
usually cause large changes in circuit voltages. Small
changes can result from component variations that are still RL = 4 Ω
within the allowable tolerance. Answer: The car battery will appear stiff for values of
11. Either may be able to simplify the analysis, save time when load resistance of ≥ 4 Ω.
calculating load current for several load resistances, and give
Copyright 2021 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of
McGraw-Hill Education. 1-1
, 1-5. Given: Solution:
RS = 0.05 Ω RL = 0.01RS (Eq. 1-4)
I=2A RL = 0.01(250 kΩ)
Solution: RL = 2.5 kΩ
V = IR (Ohm’s law) IL = IT [(RS)/(RS + RL)] (Current divider formula)
V = (2 A)(0.05 Ω) IL = 5 mA [(250 kΩ)/(250 kΩ + 10 kΩ)]
V = 0.1 V IL = 4.80 mA
Answer: The voltage drop across the internal resistance is Answer: The load current is 4.80 mA, and, no, the current
0.1 V. source is not stiff since the load resistance is not less than
1-6. Given: or equal to 2.5 kΩ.
V=9V 1-12. Solution:
RS = 0.4 Ω VTH = VR2
Solution: VR2 = VS[(R2)/(R1 + R2)] (Voltage divider formula)
I = V/R (Ohm’s law) VR2 = 36 V[(3 kΩ)/(6 kΩ + 3 kΩ)]
I = (9 V)/(0.4 Ω) VR2 = 12 V
I = 22.5 A RTH = [R1R2/R1 + R2] (Parallel resistance formula)
Answer: The load current is 22.5 A. RTH = [(6 kΩ)(3 kΩ)/(6 kΩ + 3 kΩ)]
RTH = 2 kΩ
1-7. Given:
IS = 10 mA Answer: The Thevenin voltage is 12 V, and the
Thevenin resistance is 2 kΩ.
RS = 10 MΩ
Solution:
RL = 0.01 RS
RL = 0.01(10 MΩ) 6 kΩ R1
RL = 100 kΩ
Answer: The current source will appear stiff for load 36 V
resistance of ≤100 kΩ.
1-8. Given: 3 kΩ R2 VTH
RLmin = 270 Ω
RLmax = 100 kΩ
Solution:
RS > 100 RL (Eq. 1-3)
RS > 100(100 kΩ)
RS > 10 MΩ 36 V 6 kΩ R1
Answer: The internal resistance of the source is greater
than 10 MΩ.
1-9. Given: RS = 100 kΩ
Solution: 3 kΩ R2 RTH
RL = 0.01RS (Eq. 1-4)
RL = 0.01(100 kΩ)
RL = 1 kΩ
Answer: The maximum load resistance for the current (a) Circuit for finding VTH in Prob. 1-12. (b) Circuit
source to appear stiff is 1 kΩ. for finding RTH in Prob. 1-12.
1-10. Given:
IS = 20 mA 1-13. Given:
RS = 200 kΩ VTH = 12 V
RL = 0 Ω RTH = 2 kΩ
Solution: Solution:
RL= 0.01RS I = V/R (Ohm’s law)
RL= 0.01(200 kΩ) I = VTH/(RTH + RL)
RL= 2 kΩ I0Ω = 12 V/(2 kΩ + 0 Ω) = 6 mA
Answer: Since 0 Ω is less than the maximum load I1kΩ = 12 V/(2 kΩ + 1 kΩ) = 4 mA
resistance of 2 kΩ, the current source appea rs stiff; thus I2kΩ = 12 V/(2 kΩ + 2 kΩ) = 3 mA
I3kΩ = 12 V/(2 kΩ + 3 kΩ) = 2.4 mA
the current is 20 mA.
I4kΩ = 12 V/(2 kΩ + 4 kΩ) = 2 mA
1-11. Given: I5kΩ = 12 V/(2 kΩ + 5 kΩ) = 1.7 mA
I = 5 mA I6kΩ = 12 V/(2 kΩ + 6 kΩ) = 1.5 mA
RS = 250 kΩ Answers: 0 Ω 6 mA; 1 kΩ, 4 mA; 2 kΩ, 3mA; 3 kΩ, 2.4
RL = 10 kΩ mA; 4 kΩ, 2 mA; 5 kΩ, 1.7 mA; 6 kΩ, 1.5 mA.
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McGraw-Hill Education. 1-2
, R Solution:
TH
RN = RTH (Eq. 1-10)
RTH = 10 kΩ
V R IN = VTH/RTH (Eq. 1-12)
TH L
VTH = INRN
VTH = (10 mA)(10 kΩ)
Thevenin equivalent circuit for Prob. 1-13. VTH = 100 V
Answer: RTH = 10 kΩ, and VTH = 100 V
1-14. Given:
VS = 18 V R
TH
R1 = 6 kΩ
10 kΩ
R2 = 3 kΩ VTH
Solution:
VTH = VR2 100 V
VR2 = VS[(R2)/(R1 + R2)] (Voltage divider formula)
VR2 = 18 V[(3 kΩ)/(6 kΩ + 3 kΩ)] Thevenin circuit for Prob. 1-17.
VR2 = 6 V
RTH = [(R1 × R2)/(R1 + R2)] (Parallel resistance formula) 1-18. Given (from Prob. 1-12):
RTH = [(6 kΩ × 3 kΩ)/(6 kΩ + 3 kΩ)] VTH = 12 V
RTH = 2 kΩ RTH = 2 kΩ
Answer: The Thevenin voltage decreases to 6 V, and the Solution:
Thevenin resistance is unchanged.
RN = RTH (Eq. 1-10)
1-15. Given: RN = 2 kΩ
VS = 36 V IN = VTH/RTH (Eq. 1-12)
R1 = 12 kΩ IN = 12 V/2 kΩ
R2 = 6 kΩ IN = 6 mA
Solution: Answer: RN = 2 kΩ, and IN = 6 mA
VTH = VR2
VR2 = VS[(R2)/(R1 + R2)] (Voltage divider formula)
R
VR2 = 36 V[(6 kΩ)/(12 kΩ + 6 kΩ)] IN N
VR2 = 12 V
RTH = [(R1R2)/(R1 + R2)] (Parallel resistance formula) 6 mA 2 kΩ
RTH = [(12 kΩ)(6 kΩ)/(12 kΩ + 6 kΩ)]
RTH = 4 kΩ Norton circuit for Prob. 1-18.
Answer: The Thevenin voltage is unchanged, and the
Thevenin resistance doubles. 1-19. Shorted, which would cause load resistor to be connected
1-16. Given: across the voltage source seeing all of the voltage.
VTH = 12 V 1-20. a. R1 is open, preventing any of the voltage from reaching
RTH = 3 kΩ the load resistor. b. R2 is shorted, making its voltage drop
zero. Since the load resistor is in parallel with R2, its
Solution: voltage drop would also be zero.
RN = RTH
1-21. The battery or interconnecting wiring.
RN = 3 kΩ
IN = VTH/RTH 1-22. RTH = 2 kΩ
IN = 12 V/3 kΩ Solution:
IN = 4 mA RMeter = 100RTH
Answer: IN = 4 mA, and RN = 3 kΩ RMeter = 100(2 kΩ)
RMeter = 200 kΩ
Answer: The meter will not load down the circuit if the
IN RN
meter impedance is ≥ 200 kΩ.
4 mA 3 kΩ
CRITICAL THINKING
1-23. Given:
Norton circuit for Prob. 1-16. VS = 12 V
IS = 150 A
1-17. Given:
IN = 10 mA Solution:
RN = 10 kΩ RS = (VS)/(IS)
RS = (12 V)/(150 A)
RS = 80 mΩ
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McGraw-Hill Education. 1-3
