Network Theory Multiple Choice Questions on “Supernode Analysis”.
1. Consider the figure shown below. Find the voltage (V) at node 1.
A. 13
B. 14
C. 15
D. 16
Answer: B
Clarification: Applying Super Node Analysis, the combined equation of node 1 and node 2 is (V1-V3)/3+3+(V2-V3)/1-6+V2/5=0. At node 3, (V3-V1)/3+(V3-V2)/1+V3/2=0. Also V1-V2=10. On solving above equations, we get V1 = 13.72V ≈ 14V.
2. Consider the figure shown below. Find the voltage (V) at node 2.
A. 3
B. 4
C. 5
D. 6
Answer: B
Clarification: Applying Super Node Analysis, the combined equation of node 1 and node 2 is (V1-V3)/3+3+(V2-V3)/1-6+V2/5=0. At node 3, (V3-V1)/3+(V3-V2)/1+V3/2=0. Also V1-V2=10. On solving above equations, we get V2 = 3.72V ≈ 4V.
3. Consider the figure shown below. Find the voltage (V) at node 3.
A. 4.5
B. 5.5
C. 6.5
D. 7.5
Answer: A
Clarification: Applying Super Node Analysis, the combined equation of node 1 and node 2 is (V1-V3)/3+3+(V2-V3)/1-6+V2/5=0. At node 3, (V3-V1)/3+(V3-V2)/1+V3/2=0. Also V1-V2=10. On solving above equations, we get V3 = 4.5V.
4. Consider the figure shown below. Find the power (W) delivered by the source 6A.
A. 20.3
B. 21.3
C. 22.3
D. 24.3
Answer: C
Clarification: The term power is defined as the product of voltage and current and the power delivered by the source (6A. = V2x6 = 3.72×6 = 22.32W.
5. Find the voltage (V) at node 1 in the circuit shown below.
A. 18
B. 19
C. 20
D. 21
Answer: B
Clarification: The equation at node 1 is 10 = V1/3+(V1-V2)/2. According to super Node analysis, (V1-V2)/2=V2/1+(V3-10)/5+V3/2V2-V3=20. On solving, we get, V1=19V.
6. Consider the figure shown below. Find the voltage (V) at node 2.
A. 11.5
B. 12
C. 12.5
D. 13
Answer: A
Clarification: The equation at node 1 is 10 = V1/3+(V1-V2)/2
According to super Node analysis, (V1-V2)/2=V2/1+(V3-10)/5+V3/2V2-V3=20. On solving, we get, V2=11.5V.
7. Find the voltage (V) at node 3 in the figure shown below.
A. 18
B. 20
C. 22
D. 24
Answer: A
Clarification: At node 1, (V1-40-V3)/4+(V1-V2)/6-3-5=0. Applying Super Node Analysis at node 2 and 3, (V2-V1)/6+5+V2/3+V3/5+(V3+40-V1)/4=0. Also, V3-V2=20. On solving above equations, V3 = 18.11V ≈ 18V.
8. Find the power absorbed by 5Ω resistor in the following figure.
A. 60
B. 65.5
C. 70.6
D. 75
Answer: B
Clarification: The current through 5Ω resistor = V3/5=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
Answer: C
Clarification: 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
Answer: B
Clarification: 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.