## Mesh Analysis and Nodal Analysis MCQ (Interview-Exam) Question-Answer

**Q.1** Find the value of the currents I1, I2 and I3 flowing clockwise in the first, second and third mesh respectively.

**A.** 2.34A, -3.53A, -2.23A

**B.** -1.18A, -1.17A, -1.16A

**C.** 1.54A, -0.189A, -1.195A

**D.** 4.33A, 0.55A, 6.02A

**Ans : **1.54A, -0.189A, -1.195A

**Q.2** Find the value of the currents I1 and I2 flowing clockwise in the first and second mesh respectively.

**A.** 0.96A, -1.73A

**B.** -0.96A, -1.73A

**C.** 0.96A, 1.73A

**D.** -0.96A, 1.73A

**Ans : **0.96A, -1.73A

**Q.3** Find the value of V1 if the current through the 1 ohm resistor=0A.

**A.** 78.89V

**B.** 33.33V

**C.** 83.33V

**D.** 87.87V

**Ans : **83.33V

**Q.4** Find the value of V if the current in the 3 ohm resistor=0.

**A.** 6.5V

**B.** 8.5V

**C.** 7.5V

**D.** 3.5V

**Ans : **7.5V

**Explanation: ** Taking the mesh currents in the three meshes as I1, I2 and I3, the mesh equations are:

3I1+0I2+0V=5

-2I1-4I2+0V=0

0I1+9I2+V=0

Solving these equations simultaneously and taking the value of I2=0, we get V=7.5V.

**Q.5** I1 is the current flowing in the first mesh. I2 is the current flowing in the second mesh and I3 is the current flowing in the top mesh. If all three currents are flowing in the clockwise direction, find the value of I1, I2 and I3.

**A.** 10.67A, 7.67A, 2A

**B.** 3.67A, 6.67A, 2A

**C.** 7.67A, 10.67A, 2A

**D.** 7.67A, 8.67A, 2A

**Ans : **7.67A, 10.67A, 2A

**Q.6** Calculate the mesh currents I1 and I2 flowing in the first and second meshes respectively.

**A.** 0.5A, 2.5A

**B.** 3.2A, 6.5A

**C.** 1.75A, 1.25A

**D.** 2.3A, 0.3A

**Ans : **1.75A, 1.25A

**Q.7** Mesh analysis employs the method of ___________.

**A.** KCL

**B.** Neither KVL nor KCL

**C.** KVL

**D.** Both KVL and KCL

**Ans : **KVL

**Q.8** Calculate the mesh currents.

**A.** 2A, 1A, 0.57A

**B.** 6A, 7A, 8.99A

**C.** 7A, 6A, 6.22A

**D.** 3A, 4A, 5.88A

**Ans : **2A, 1A, 0.57A

**Explanation: ** I=V/R. Total resistance R = 20+40=60ohm. V=120V. I=120/60=2A.

**Q.9** Mesh analysis can be used for __________.

**A.** Non-planar circuits

**B.** Neither planar nor non-planar circuits

**C.** Planar circuits

**D.** Both planar and non-planar circuits

**Ans : **Planar circuits

**Q.10** Mesh analysis is generally used to determine _________.

**A.** Current

**B.** Power

**C.** Voltage

**D.** Resistance

**Ans : **Current

**Explanation: ** I=V/R. Total resistance R = 20+40=60ohm. V=120V. I=120/60=2A.

**Q.11** Calculate the node voltages V1 and V2.

**A.** 26.67V, 11.33V

**B.** 13V, 12V

**C.** 12V, 13V

**D.** 11.33V, 26.67V

**Ans : **11.33V, 26.67V

**Explanation: ** The nodal equations are:

2V1-V2=-4

-4V1+5V2=88

Solving these equations simultaneously, we get V1=11.33V and V2=26.67V.

**Q.12** Find the value of the node voltage V.

**A.** 60V

**B.** -40V

**C.** -60V

**D.** 40V

**Ans : **-60V

**Explanation: ** The node equation is:

-2+8+V/10=0 => 6 + v/10 = 0 => v = -10*6 = -60V

Solving this equation, we get V = -60V.

**Q.13** Calculate the node voltages.

**A.** 32.34V, 7.87V, 8.78V

**B.** 45.44V, 6.67V, 7.77V

**C.** 30.77V, 7.52V, 18.82V

**D.** 34.34V, 8.99V, 8.67V

**Ans : **30.77V, 7.52V, 18.82V

**Explanation: ** The nodal equations, considering V1, V2 and V3 as the first, second and third node respectively, are:

-8+(V1-V2)/3-3+(V1-V3)/4=0

3+V2+(V2-V3)/7+(v2-V1)/3=0

-2.5+(V3-V2)/7+(V3-V1)/4+V3/5=0

Solving the equations simultaneously, we get V1=30.77V, V2=7.52V and V3=18.82V.

**Q.14** Find the node voltage V.

**A.** 2V

**B.** 4V

**C.** 1V

**D.** 3V

**Ans : **4V

**Explanation: ** The nodal equation is:

(V-10)/2+(V-7)/3+V/1=0

Solving for V, we get V=4V.

**Q.15** Nodal analysis is generally used to determine_______.

**A.** Current

**B.** Power

**C.** Voltage

**D.** Resistance

**Ans : **Voltage

**Q.16** Find the value of V1 and V2.

**A.** 23.32V, 46.45V

**B.** 56.32V, 78, 87V

**C.** 87.23V, 29.23V

**D.** 64.28V, 16.42V

**Ans : **64.28V, 16.42V

**Explanation: ** The nodal equations are:

0.3V1-0.2V2=16

-V1+3V2=-15

Solving these equations simultaneously, we get V1=64.28V and V2=16.42V.

**Q.17** Nodal analysis can be applied for________.

**A.** Non-planar networks

**B.** Neither planar nor non-planar networks

**C.** Planar networks

**D.** Both planar and non-planar networks

**Ans : **Both planar and non-planar networks

**Q.18** If there are 10 nodes in a circuit, how many equations do we get?

**A.** 9

**B.** 7

**C.** 10

**D.** 8

**Ans : **9

**Q.19** How many nodes are taken as reference nodes in a nodal analysis?

**A.** 2

**B.** 4

**C.** 1

**D.** 3

**Ans : **1

## Mesh Analysis:

**Definition:**Mesh analysis is a method used to analyze electrical circuits by applying Kirchhoff’s Voltage Law (KVL) to loops or meshes in the circuit.**Mesh:**A mesh refers to a closed loop in the circuit that does not contain any other loops within it.**Currents:**Mesh analysis assigns mesh currents to each mesh in the circuit. These currents circulate within the loop and can be either clockwise or counterclockwise.**KVL Application:**Applying KVL to each mesh in the circuit allows the derivation of equations relating the voltages across circuit elements to the mesh currents.**Solution:**Once the equations are derived, they can be solved simultaneously to determine the values of the mesh currents and consequently, the voltages and currents in the entire circuit.**Advantages:**Mesh analysis is particularly useful for circuits with multiple current sources and only a few voltage sources.**Disadvantages:**Mesh analysis can become complex in circuits with a large number of meshes, and it may not be suitable for circuits with many voltage sources.

## Nodal Analysis:

**Definition:**Nodal analysis is a method used to analyze electrical circuits by applying Kirchhoff’s Current Law (KCL) to the nodes (connection points) in the circuit.**Node:**A node is a point in the circuit where two or more circuit elements are connected.**Node Voltages:**Nodal analysis assigns voltages to each node in the circuit with respect to a reference node (usually the ground).**KCL Application:**Applying KCL at each node in the circuit allows the derivation of equations relating the currents entering and leaving each node.**Solution:**Once the equations are derived, they can be solved simultaneously to determine the node voltages and consequently, the currents and voltages in the entire circuit.**Advantages:**Nodal analysis is particularly useful for circuits with multiple voltage sources and only a few current sources. It also simplifies the analysis of circuits with many parallel branches.**Disadvantages:**Nodal analysis can become complex in circuits with a large number of nodes, and it may not be suitable for circuits with many current sources.