This set of Electric Circuits Questions and Answers for Experienced people focuses on “Thevenin and Norton Equivalents, More on Deriving a Thevenin Equivalent”.

1. Find the voltage across 24Ω resistor by using Thevenin’s theorem.

a) 8V

b) 9V

c) 1V

d) 6V

View Answer

Explanation: 1. Remove 24Ω resistor and calculate the voltage across the open circuit.

2. Calculate the thevenin’s resistance and by using it, the thevenin’s current.

3. V

_{24Ω}=I*R (can also verify by using Nodal analysis).

2. Calculate Thevenin’s voltage for the network shown below where the voltage source is 4V.

a) 6V

b) 4.71V

c) 5V

d) 1V

View Answer

Explanation: In the circuit given, thevenin’s voltage is nothing but the open circuit voltage which is V

_{x}. Applying KVL, it is obtained.

3. Find the Thevenin’s resistance for the network given.

a) 6.75Ω

b) 5.85Ω

c) 4.79Ω

d) 1.675Ω

View Answer

Explanation: Remove all the voltage/current sources and calculate the equivalent resistance.

4. Find the current through (5+j4) Ω resistor.

a) 0.9-j0.2 A

b) 0.78-j0.1 A

c) 2.7-j0.5 A

d) 1A

View Answer

Explanation: 1. Remove the 5+j4 Ω branch and calculate thevenin’s voltage.

(V= v across 6Ω resistor- v across 8Ω resistor)

2. Calculate Z

_{th}. (10//6 and 8//8)

3. Current= (V

_{th}/ (Z

_{th}+Z).

5. The voltage across 6Ω resistor is __________

a) 7.5V

b) 6.78V

c) 20V

d) 8.5V

View Answer

Explanation: Remove the resistor across which voltage is to be calculated and short circuit it. By using short circuit current and resistance calculate the current across 6Ω resistor and thereby voltage. (In this 10Ω resistor is also short-circuited since 10//0).

6. Find the Norton’s current for the circuit given below.

a) 5A

b) 3.33A

c) 4A

d) 1.66A

View Answer

Explanation: I

_{N}= (20/10) + (10/5).

7. Calculate IN for the given network.

a) 0A

b) 1A

c) 4.37A

d) 0.37A

View Answer

Explanation: Using nodal analysis V

_{x}is calculated. I

_{N}=V

_{x}/4.

8. Calculate R_{Th} for the network given.

a) 8Ω

b) 7Ω

c) 2Ω

d) 1Ω

View Answer

Explanation: 5//20 and then in series with 3Ω resistor.

9. Thevenin’s equivalent circuit consists of a ____________

a) Voltage source in series with a resistor

b) Current source in parallel with a resistor

c) Voltage source in parallel with a resistor

d) Current source in series with a resistor

View Answer

Explanation: Thevenin’s equivalent circuit contains a Voltage source in series with a resistor.

10. Norton’s equivalent circuit consists of a _____________

a) Voltage source in series with a resistor

b) Current source in parallel with a resistor

c) Both voltage and current sources

d) Current source in series with a resistor

View Answer

Explanation: Norton’s equivalent circuit consists of a Current source in parallel with a resistor.

11. Thevenin’s voltage is equal to ____________

a) Short circuit voltage

b) Open circuit current

c) Open circuit voltage

d) Short circuit current

View Answer

Explanation: Thevenin’s voltage is equal to open circuit voltage.

12. Norton’s current is equal to ____________

a) Short circuit voltage

b) Open circuit current

c) Open circuit voltage

d) Short circuit current

View Answer

Explanation: Norton’s current is equal to Short circuit current.

13. Thevenin’s resistance R_{Th} = ___________

a) V_{Th}/I_{SC}

b) VSC/I_{Th}

c) V_{Th}/I_{Th}

d) V_{SC} /I_{SC}

View Answer

Explanation: Thevenin’s resistance is defined as the ratio of open circuit voltage to the short circuit current across the terminals of the original circuit.

14. What is the expression forthe thevenin’s current if there is an external resistance in series with the R_{Th}?

a) V_{Th}/I_{Th}

b) V_{Th}/ (R_{Th}-R)

c) V_{Th}/ (R_{Th}+R)

d) V_{Th}/R_{Th}

View Answer

Explanation: I

_{Th}= V

_{Th}/ (R

_{Th}+R).

15. One can find the thevenin’s resistance simply by removing all voltage/current sources and calculating equivalent resistance.

a) False

b) True

View Answer

Explanation: Yes. One can find the thevenin’s resistance simply by removing all voltage/current sources and calculating equivalent resistance.

**Sanfoundry Global Education & Learning Series – Electric Circuits.**

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