This set of Network Theory Multiple Choice Questions & Answers (MCQs) focuses on “Mesh Analysis”.
1. Mesh analysis is applicable for non planar networks also.
Explanation: Mesh analysis is applicable only for planar networks. A circuit is said to be planar if it can be drawn on a plane surface without crossovers.
2. A mesh is a loop which contains ____ number of loops within it.
d) no loop
Explanation: A loop is a closed path. A mesh is defined as a loop which does not contain any other loops within it.
Explanation: We know if there are n loops in the circuit, n mesh equations can be formed. So as there are 2 loops in the circuit. So 2 mesh equations can be formed.
4. In the figure shown in the question 2, the current through loop 1 be I1 and through the loop 2 be I2, then the current flowing through the resistor R2 will be?
Explanation: Through the resistor R2 both the currents I1, I2 are flowing. So the current through R2 will be I1-I2.
5. If there are 5 branches and 4 nodes in graph, then the number of mesh equations that can be formed are?
Explanation: Number of mesh equations= B-(N-1). Given number of branches = 5 and number of nodes = 4. So Number of mesh equations = 5-(4-1) =2.
Explanation: Consider current I1 (CW) in the loop 1 and I2 (ACW) in the loop 2. So, the equations will be Vx+I2-I1=0. I1=5/2=2.5A. I2=4Vx/4= Vx. Vx+Vx-2.5=0. Vx =1.25V.
Explanation: According to mesh analysis, (1+3+6)I1 – 3(I2) – 6(I3) =10
-3(I1) + (2+5+3)I2 =4 -6(I1) + 10(I3) = -4 +20 On solving the above equations, I1=4.3A.
8. Find the current I2 (A) in the figure shown in the question 7.
Explanation: According to mesh analysis, (1+3+6)I1 – 3(I2) – 6(I3) = 10. -3(I1) + (2+5+3)I2 = 4. -6(I11) + 10(I3) = -4 + 20 On solving the above equations, I2 =1.7A.
9. Find the current I3 (A) in the figure shown in the question 7.
Explanation: According to mesh analysis, (1+3+6)I1 – 3(I2) – 6(I3) = 10. -3(I1) + (2+5+3)I2 = 4. -6(I1) + 10(I3) = -4 + 20. On solving the above equations, I3 = 4.7A.
Explanation: Applying mesh analysis, 5(I1) + 2(I1-I2) = 10. 10(I2) + 2(I2-I1) + 40 = 0. On solving, I1 = 0.5A, I2 = -3.25A. So current through R2 resistor is 0.5-(-3.25) = 3.75 A.
Sanfoundry Global Education & Learning Series – Network Theory.
To practice all areas of Network Theory, here is complete set of 1000+ Multiple Choice Questions and Answers.