1. Consider the circuit shown below. Find the current I1 (A).
a) -1
b) -2
c) -3
d) -4
Explanation: Applying Super mesh analysis, the equations will be I1+I1+10+I2+I2=0. I1+I2=-5. I2-I1=1. On solving, I1=-3A.
2. Consider the circuit shown below. Find the current I2 (A).
a) -2
b) -1
c) 2
d) 1
Explanation: Applying Super mesh analysis, the equations will be I1+I1+10+I2+I2=0. I1+I2=-5. I2-I1=1. On solving, I2=-2A.
3. Find the power (W) supplied by the voltage source in the following figure.
a) 0
b) 1
c) 2
d) 3
Explanation: I3-I2=2. As I2=-2A, I3=0A. The term power is the product of voltage and current. So, power supplied by source = 10×0=0W.
4. Find the current i3 in the circuit shown below.
a) 8.18
b) 9.18
c) 10.18
d) 8.8
Explanation: For 2nd loop, 10 + 2(i2-i3) + 3(i2-i1) = 0. For 3rd loop, i3 + 2(i3-i2)=10. As i1=10A, On solving above equations, we get i3=8.18A.
5. Find the current i2 in the circuit shown below.
a) 6.27
b) 7.27
c) 8.27
d) 9.27
Explanation: For 2nd loop, 10 + 2(i2-i3) + 3(i2-i1) = 0. For 3rd loop, i3 + 2(i3-i2)=10. As i1=10A, On solving above equations, we get i2=7.27A.
6. 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
Explanation: 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.
7. Nodal analysis can be applied for non planar networks also.
a) true
b) false
Explanation: Nodal analysis is applicable for both planar and non planar networks. Each node in a circuit can be assigned a number or a letter.
8. In nodal analysis how many nodes are taken as reference nodes?
a) 1
b) 2
c) 3
d) 4
Explanation: 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.
9. Find the voltage at node P in the following figure.
a) 8V
b) 9V
c) 10V
d) 11V
Explanation: I1 = (4-V)/2, I2 = (V+6)/3. The nodal equation at node P will be I1+3=I2. On solving, V=9V.
10. Find the resistor value R1(Ω) in the figure shown below.
a) 10
b) 11
c) 12
d) 13
Explanation: 10=(V1-V2)/14+(V1-V3)/R1. From the circuit, V1=100V, V2=15×2=30V, V3=40V. On solving, R1=12Ω.