Worked examples: Loop (mesh) current method

WORKED EXAMPLES

Example 1 Write the mesh equations for the circuit shown in the figure and solve for the current in the 12Ω resistor.

Solution: The loop currents are assumed as shown. It is required to find i3. By inspection we can write,

= 11 (204) + 4(-140-16) -4 (28 + 44)

= 2244 – 1092 - 288

= 864

= 480 (72) + 600 (-44-28)

= - 8640

I₃= Δ₃/Δ = - 8640 / 864

= - 10A

 (-) indicates that the assumed direction of I₃ is not the actual direction I₃ The actual direction of I3 is anti-clockwise. i.e., the current through 12 Ω is 10A flowing from the right to the left terminal of the resistor.

Example 2 Write and solve the equations for the mesh currents in the network in figure.

Solution: For solving the problems on the mesh current (loop current) method, the preferable method is to convert the practical current sources into practical voltage sources.

The original circuit is redrawn as in the figure below after transforming the current sources into voltage sources.

By Inspection putting the above circuit equations in the matrix form,

I₁ = Δ₁/ Δ =116/29 = 4A

I₂ = Δ₂/ Δ = 261/29 = 9A

Example 3 Write the mesh equations for the network shown in figure by inspection and find the power absorbed by 8 Ω resistor.

Solution: Let the loop currents be as shown. By inspection

Power absorbed by 8 Ω resistor VI = 100 × 8.44 8.44 W

The mesh equations are as expressed below:

I₂ I₁ - 4 I₂ = 100  ....(i)

-4I₁ + 24 I₂-10 I₃ = 0 ...(ii)

- 10 I₂ + 14 I₃ = -40... (iii)

Example 4 Write the mesh equations for the network shown in figure below by inspection and find the power absorbed by the 3-ohm resistor.

Solution: Applying KVL to the mesh ABCD, we get

-5I₁-(I₁-I₂) 1 + I₃ =0

⇒ 6I₁-I₂ = 13....(i)

Aapplying KVL to the mesh BEFC,

-3I₃-(I₂-I₁) 1 = 0

⇒ I₁-I₂-3I₃ = 0 ... (ii)

The source current is given by

(I₃-I₂) = 2 ... (iii)

Multiplying (iii) by 3 and adding to (ii)

I₁-4I₂ = 6....(iv)

Multiplying (i) by 4 and subtracting from (iv)

23 I₁ =46

 I₁= 2A

Therefore, the power observed in 3 Ω resistor

= I₁² × 3 = 2² × 3 = 12 watt

Example 5 In the Circuit shown in the figure, IB = 10 μA. Find the value of resistance R_in

Solution:

Step 1: The given circuit can be drawn as below.

Step 2: Assume that the loop currents I, and I2 are as indicated.

Step 3:

Given that I₁-I₂= I_B = 10 x 10^-6 A... (1)

Applying KVL to the left loop, we get