Network Theory Questions and Answers – Compensation Theorem

This set of Network Theory Interview Questions and Answers for freshers focuses on “Compensation Theorem”.

1. Reciprocity Theorem is applied for _____ networks.
a) Linear
b) Bilateral
c) Linear bilateral
d) Lumped
View Answer

Answer: c
Explanation: Reciprocity Theorem is applied for linear bilateral networks, not for linear or for linear bilateral or for lumped networks.

2. Reciprocity Theorem is used to find the change in _______ when the resistance is changed in the circuit.
a) Voltage
b) Voltage or current
c) Current
d) Power
View Answer

Answer: b
Explanation: Reciprocity Theorem is used to find the change in voltage or current when the resistance is changed in the circuit. If reciprocity theorem is satisfied the ratio of response to excitation is same for the two conditions.

3. Find the current through 3Ω resistor in the circuit shown below.
Find the current through 3Ω resistor in the circuit in given figure
a) 1
b) 2
c) 3
d) 4
View Answer

Answer: a
Explanation: Total resistance in the circuit = 2+[3||(2+2||2)] = 3.5Ω. The total current drawn by the circuit =10/(4+6||3) = 1.67A. Current through 3Ω resistor = 1.11A ≅1A.
advertisement
advertisement

4. Determine the current flowing in the ammeter having 1Ω internal resistance connected in series with the 3Ω resistor as shown in the below circuit.
Find the current through 3Ω resistor in the circuit in given figure
a) 0.91
b) 0.92
c) 0.93
d) 0.94
View Answer

Answer: c
Explanation: Current through 3Ω resistor = 1.11A. So voltage drop across 1Ω resistor = 1.11×1 = 1.11V. Now the circuit can be modified as
The total current drawn by the circuit =10/(4+6||3) = 1.67A in figure
Now current through 3Ω resistor = 0.17A. This current is opposite to the current calculated before. So ammeter reading = (1.11-0.17) = 0.94A.

5. Find the current through 6Ω resistor in the circuit shown below.
Find current flowing in ammeter having 1Ω internal resistance connected in series
a) 0.33
b) 0.44
c) 0.55
d) 0.66
View Answer

Answer: c
Explanation: Total resistance in the circuit = 4+6||3Ω. The total current drawn by the circuit = 10/(4+6||3)=1.67A. Current through 6Ω resistor = 0.55A.
Sanfoundry Certification Contest of the Month is Live. 100+ Subjects. Participate Now!

6. Determine the current flowing in the ammeter having 1Ω internal resistance connected in series with the 6Ω resistor as shown in the below circuit.
Find current flowing in ammeter having 1Ω internal resistance connected in series
a) 0.4
b) 0.45
c) 0.9
d) 0.95
View Answer

Answer: b
Explanation: Current through 3Ω resistor = 0.55A. So voltage drop across 1Ω resistor = 0.55×1 = 0.55V.
Now the circuit can be modified as
Current is opposite to current calculated before ammeter reading = (0.55-0.0.94) = 0.45A
Now current through 6Ω resistor = 0.094A. This current is opposite to the current calculated before. So ammeter reading = (0.55-0.0.94) = 0.45A.

7. Find the current through 6Ω resistor in the circuit shown below.
Find the current through 6Ω resistor in the circuit shown
a) 0.11
b) 0.22
c) 0.33
d) 0.44
View Answer

Answer: c
Explanation: Total current in the circuit = 10/(4+3||2||6)=2A. Current through 6Ω resistor = 2×(3||2)/(6+3||2)=0.33A.
advertisement

8. Consider the following circuit. Determine the current flowing in the ammeter having 1Ω internal resistance in series with the 6Ω resistor.
Find the current through 6Ω resistor in the circuit shown
a) 0.1
b) 0.2
c) 0.3
d) 0.4
View Answer

Answer: c
Explanation:
Current flowing in ammeter having 1Ω internal resistance in series with 6Ω resistor is 0.3
New total current = 0.33/(7+4||2||3)=0.04A. Now reading of ammeter = 0.33-0.04=0.29A ≅ 0.3A.

9. Find the current through 3Ω resistor in the circuit shown below.
Find the current through 3Ω resistor in the circuit shown
a) 0.45
b) 0.56
c) 0.67
d) 0.78
View Answer

Answer: c
Explanation: Total current = 10 / (4 + (6||2||3) = 2A. Current through 3Ω resistor= 2 x (6||2)/(3 + (6||2)) = 0.67A.
advertisement

10. Consider the following circuit. Determine the current flowing in the ammeter having 1Ω internal resistance in series with the 3Ω resistor.
Find the current through 3Ω resistor in the circuit shown
a) 0.6
b) 0.7
c) 0.8
d) 0.9
View Answer

Answer: a
Explanation: The current flowing in the ammeter having 1Ω internal resistance in series with the 3Ω resistor shown in the circuit is 0.6 A.
Current flowing in the given circuit is 0.6 A
Current through 3Ω resistor = 0.67/(7 + (4||6||2)) = 0.08A. Ammeter reading = 0.67 – 0.08 = 0.59 ≅ 0.6A.

Sanfoundry Global Education & Learning Series – Network Theory.

To practice all areas of Network Theory for Interviews, here is complete set of 1000+ Multiple Choice Questions and Answers.

If you find a mistake in question / option / answer, kindly take a screenshot and email to [email protected]

advertisement
advertisement
Subscribe to our Newsletters (Subject-wise). Participate in the Sanfoundry Certification contest to get free Certificate of Merit. Join our social networks below and stay updated with latest contests, videos, internships and jobs!

Youtube | Telegram | LinkedIn | Instagram | Facebook | Twitter | Pinterest
Manish Bhojasia - Founder & CTO at Sanfoundry
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

Subscribe to his free Masterclasses at Youtube & discussions at Telegram SanfoundryClasses.