Test: Mesh & Nodal Analysis - Electrical Engineering (EE) MCQ

The two mesh equations are:
5I1-3I2=10
-3I1+7I2=-15
Solving the equations simultaneously, we get I1=0.96A and I2=-1.73A.

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.

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:
15I1-5I2=V1
-5I1+10I2+0V1=0
0I1-3I2+0V1=10
Solving these equations simultaneously we get V1= 83.33V.

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:
I₂-I₁=3
-5I₁-3I₂=5
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.

Explanation: The two meshes which contain the 3A current is a super mesh. The three mesh equations therefore are:
I₃=2A
I₂-I₁=3
-2I₁-I₂=-26
Solving these equations simultaneously we get:
I₁=7.67A, I₂=10.67A and I₃=2A.

The two meshes which contain the 3A source, act as a supper mesh. The mesh equations are:
I1-I2=3
-11I1-4I2+14I3=-10
10I1+4I2-28I3=0
Solving these equations simultaneously, we get the three currents as 2A, 1A and 0.57A.

KVL employs mesh analysis to find the different mesh currents by finding the IR products in each mesh.

Mesh analysis uses Kirchhoff’s Voltage Law to find all the mesh currents. Hence it is a method used to determine current.

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.

The node equation is:
-2+8+V/10=0 => 6 + v/10 = 0 => v = 10*6=>60v
Solving this equation, we get V=60V.

The nodal equations are:
2V1-V2=-4
-4V1+5V2=88
Solving these equations simultaneously, we get V1=11.33V and V2=26.67V.

The nodal equation is:
(V-10)/2+(V-7)/3+V/1=0
Solving for V, we get V=4V.

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=24.32V, V2=4.09V and V3=7.04V.

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.

Nodal analysis uses Kirchhoff’s Current Law to find all the node voltages. Hence it is a method used to determine voltage.

KCL employs nodal analysis to find the different node voltages by finding the value if current in each branch.

The number of equations we get is always one less than the number of nodes in the circuit, hence for 10 nodes we get 9 equations.

Nodal analysis can be applied for both planar and non-planar networks since each node, whether it is planar or non-planar, can be assigned a voltage.

In nodal analysis one node is treated as the reference node and the voltage at that point is taken as 0.

Applying nodal analysis at node V_A -

Solving eqn (i) We get,
V_A = 4.29 volts
But in the given options only the value near to 4.29 is 4.25 volts.