Network Theory Multiple Choice Questions on “Nodal Analysis”.
1. If there are 8 nodes in network, we can get ____ number of equations in the nodal analysis.
A. 9
B. 8
C. 7
D. 6
Answer: C
Clarification: Number of equations = N-1 = 7. So as there are 8 nodes in network, we can get 7 number of equations in the nodal analysis.
2. Nodal analysis can be applied for non planar networks also.
A. true
B. false
Answer: A
Clarification: Nodal analysis is applicable for both planar and non planar networks. Each node in a circuit can be assigned a number or a letter.
3. In nodal analysis how many nodes are taken as reference nodes?
A. 1
B. 2
C. 3
D. 4
Answer: A
Clarification: In nodal analysis only one node is taken as reference node. And the node voltage is the voltage of a given node with respect to one particular node called the reference node.
4. Find the voltage at node P in the following figure.
A. 8V
B. 9V
C. 10V
D. 11V
Answer: B
Clarification: I1 = (4-V)/2, I2 = (V+6)/3. The nodal equation at node P will be I1+3=I2. On solving, V=9V.
5. Find the resistor value R1(Ω) in the figure shown below.
A. 10
B. 11
C. 12
D. 13
Answer: C
Clarification: 10=(V1-V2)/14+(V1-V3)/R1. From the circuit, V1=100V, V2=15×2=30V, V3=40V. On solving, R1=12Ω.
6. Find the value of the resistor R2 (Ω) in the circuit shown below.
A. 5
B. 6
C. 7
D. 8
Answer: B
Clarification: As V1=100V, V2=15×2=30V, V3=40V. (V1-V2)/14+(V1-V3)/R2=15. On solving we get R2 = 6Ω.
7. Find the voltage (V) at node 1 in the circuit shown.
A. 5.32
B. 6.32
C. 7.32
D. 8.32
Answer: B
Clarification: At node 1, (1/1+1/2+1/3)V1-(1/3)V2 = 10/1. At node 2, -(1/3)V1+(1/3+1/6+1/5)V2 = 2/5+5/6. On solving above equations, we get V1=6.32V.
8. Find the voltage (V) at node 2 in the circuit shown below.
A. 2.7
B. 3.7
C. 4.7
D. 5.7
Answer: C
Clarification: At node 1, (1/1+1/2+1/3)V1-(1/3)V2 = 10/1. At node 2, -(1/3)V1+(1/3+1/6+1/5)V2 = 2/5+5/6. On solving above equations, we get V2=4.7V.
9. Find the voltage at node 1 of the circuit shown below.
A. 32.7
B. 33.7
C. 34.7
D. 35.7
Answer: B
Clarification: Applying Kirchhoff’s current law at node 1, 10 = V1/10+(V1-V2)/3. At node 2, (V2-V1)/3+V2/5+(V2-10)/1=0. On solving the above equations, we get V1=33.7V.
10. Find the voltage at node 2 of the circuit shown below.
A. 13
B. 14
C. 15
D. 16
Answer: B
Clarification: Applying Kirchhoff’s current law at node 1, 10 = V1/10+(V1-V2)/3. At node 2, (V2-V1)/3+V2/5+(V2-10)/1=0. On solving the above equations, we get V2=14V.