This set of Basic Electrical Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Mesh Analysis”.
1. Find the value of the currents I1, I2 and I3 flowing clockwise in the first, second and third mesh respectively.
a) 1.54A, -0.189A, -1.195A
b) 2.34A, -3.53A, -2.23A
c) 4.33A, 0.55A, 6.02A
d) -1.18A, -1.17A, -1.16A
Explanation: The three mesh equations are:
Solving the equations, we get I1= 1.54A, I2=-0.189 and I3= -1.195A.
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
Explanation: The two mesh equations are:
Solving the equations simultaneously, we get I1=0.96A and I2=-1.73A.
3. Find the value of V if the current in the 3 ohm resistor=0.
Explanation: Taking the mesh currents in the three meshes as I1, I2 and I3, the mesh equations are:
Solving these equations simultaneously and taking the value of I2=0, we get V=7.5V.
4. Find the value of V1 if the current through the 1 ohm resistor=0A.
Explanation: Taking I1, I2 and I3 as the currents in the three meshes and taking I3=0 since it is the current across the 1 ohm resistor, the three mesh equations are:
Solving these equations simultaneously we get V1= 83.33V.
5. Calculate the mesh currents I1 and I2 flowing in the first and second meshes respectively.
a) 1.75A, 1.25A
b) 0.5A, 2.5A
c) 2.3A, 0.3A
d) 3.2A, 6.5A
Explanation: In this circuit, we have a super mesh present.
Let I1 and I2 be the currents in loops in clockwise direction. The two mesh equations are:
Solving these equations simultaneously, we get I1 = -1.75A and I2 = 1.25A.
Since no specific direction given so currents in loop 1 and loop 2 are 1.75A and 1.25A respectively.
6. 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) 7.67A, 10.67A, 2A
b) 10.67A, 7.67A, 2A
c) 7.67A, 8.67A, 2A
d) 3.67A, 6.67A, 2A
Explanation: The two meshes which contain the 3A current is a super mesh. The three mesh equations therefore are:
Solving these equations simultaneously we get:
I1=7.67A, I2=10.67A and I3=2A.
7. Calculate the mesh currents.
a) 7A, 6A, 6.22A
b) 2A, 1A, 0.57A
c) 3A, 4A, 5.88A
d) 6A, 7A, 8.99A
Explanation: The two meshes which contain the 3A source, act as a supper mesh. The mesh equations are:
Solving these equations simultaneously, we get the three currents as
I1=-1A, I2=0.57A, I3=2A
So currents are 2A, 1A, 0.57A.
8. Mesh analysis employs the method of ___________
c) Both KVL and KCL
d) Neither KVL nor KCL
Explanation: KVL employs mesh analysis to find the different mesh currents by finding the IR products in each mesh.
9. Mesh analysis is generally used to determine _________
Explanation: Mesh analysis uses Kirchhoff’s Voltage Law to find all the mesh currents. Hence it is a method used to determine current.
10. Mesh analysis can be used for __________
a) Planar circuits
b) Non-planar circuits
c) Both planar and non-planar circuits
d) Neither planar nor non-planar circuits
Explanation: If the circuit is not planar, the meshes are not clearly defined. In planar circuits, it is easy to draw the meshes hence the meshes are clearly defined.
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