This set of Network Theory Multiple Choice Questions & Answers (MCQs) focuses on “Supernode Analysis”.

1.Consider the figure shown below. Find the voltage (V) at node 1.

a) 13

b) 14

c) 15

d) 16

View Answer

Explanation: Applying Super Node Analysis, the combined equation of node 1 and node 2 is (V

_{1}-V

_{3})/3+3+(V

_{2}-V

_{3})/1-6+V

_{2}/5=0. At node 3, (V

_{3}-V

_{1})/3+(V

_{3}-V

_{2})/1+V

_{3}/2=0. Also V

_{1}-V

_{2}=10. On solving above equations, we get V

_{1}= 13.72V ≈ 14V.

2. In the figure shown in the question 1 find the voltage (V) at node 2.

a) 3

b) 4

c) 5

d) 6

View Answer

Explanation: Applying Super Node Analysis, the combined equation of node 1 and node 2 is (V

_{1}-V

_{3})/3+3+(V

_{2}-V

_{3})/1-6+V

_{2}/5=0. At node 3, (V

_{3}-V

_{1})/3+(V

_{3}-V

_{2})/1+V

_{3}/2=0. Also V

_{1}-V

_{2}=10. On solving above equations, we get V

_{2}= 3.72V ≈ 4V.

3. In the figure shown in the question 1 find the voltage (V) at node 3.

a) 4.5

b) 5.5

c) 6.5

d) 7.5

View Answer

Explanation: Applying Super Node Analysis, the combined equation of node 1 and node 2 is (V

_{1}-V

_{3})/3+3+(V

_{2}-V

_{3})/1-6+V

_{2}/5=0. At node 3, (V

_{3}-V

_{1})/3+(V

_{3}-V

_{2})/1+V

_{3}/2=0. Also V

_{1}-V

_{2}=10. On solving above equations, we get V

_{3}= 4.5V.

4. In the figure shown in the question 1 find the power (W) delivered by the source 6A.

a) 20.3

b) 21.3

c) 22.3

d) 24.3

View Answer

Explanation: The term power is defined as the product of voltage and current and the power delivered by the source (6A) = V

_{2}x6 = 3.72×6 =22.32W.

5. Find the voltage (V) at node 1 in the circuit

a) 18

b) 19

c) 20

d) 21

View Answer

Explanation: The equation at node 1 is 10= V

_{1}/3+(V

_{1}-V

_{2})/2. According to super Node analysis, (V

_{1}-V

_{2})/2=V

_{2}/1+(V

_{3}-10)/5+V

_{3}/2V

_{2}-V

_{3}=20. On solving, we get, V

_{1}=19V.

6. Find the voltage (V) at node 2 of the circuit shown in the question 5.

a) 11.5

b) 12

c) 12.5

d) 13

View Answer

Explanation: The equation at node 1 is 10= V

_{1}/3+(V

_{1}-V

_{2})/2

According to super Node analysis, (V

_{1}-V

_{2})/2=V

_{2}/1+(V

_{3}-10)/5+V

_{3}/2V

_{2}-V

_{3}=20. On solving, we get, V

_{2}=11.5V.

7. Find the voltage (V) at node 3 in the figure shown below.

a) 18

b) 20

c) 22

d) 24

View Answer

Explanation: At node 1, (V

_{1}-40-V

_{3})/4+(V

_{1}-V

_{2})/6-3-5=0. Applying Super Node Analysis at node 2 and 3, (V

_{2}-V

_{1})/6+5+V

_{2}/3+V

_{3}/5+(V

_{3}+40-V

_{1})/4=0. Also V

_{3}-V

_{2}=20. On solving above equations, V

_{3}= 18.11V ≈ 18V.

8. Find the power absorbed by 5Ω resistor in the figure shown in the question 7.

a) 60

b) 65.5

c) 70.6

d) 75

View Answer

Explanation: The current through 5Ω resistor = V

_{3}/5=18.11/5=3.62A. The power absorbed by 5Ω resistor = (3.62)

^{2})×5=65.52W.

9. Find the value of the voltage (V) in the equivalent voltage source of the current source shown below.

a) 20

b) 25

c) 30

d) 35

View Answer

Explanation: The value of the voltage (V) in the equivalent voltage source of the current source the voltage across the terminals A and B is (6)( 5) = 30V.

10. Find the value of the current (A) in the equivalent current source of the voltage source shown below.

a) 1

b) 2

c) 3

d) 4

View Answer

Explanation: The value of the current (A) in the equivalent current source of the voltage source the short circuit current at the terminals A and B is I=60/30=2A.

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