Test: Super Mesh Analysis - Electrical Engineering (EE) MCQ

For 2nd loop, 10 + 2(i₂ - i₃) + 3(i₂ - i₁) = 0. For 3rd loop, i₃ + 2(i₃ - i₂) = 10. As i₁ = 10A, On solving above equations, we get i₃ = 8.18A.

The current in the first loop is equal to 10A. So the current i₁ in the circuit is i₁ = 10A.

Applying Super mesh analysis, the equations will be I₁ + I₁ + 10 + I₂ + I₂ = 0.
I₁ + I₂ = -5. I₂ - I₁ = 1.
On solving, I₂ = -2A.

Applying Super mesh analysis, the equations will be I₂ - I₁ = 2
-10 + 2I₁ + I₂ + 4 = 0.
On solving the above equations, I₂ = 3.33A.

Applying Super Mesh analysis, (10 + 5)I₁ – 10(I₂) – 5(I₃) = 50.
2(I₂) + I₃ + 5(I₃-I₁) + 10(I₂-I₁) = 0.
I₂ – I₃ = 2.
On solving above equations, we get I₁ = -8A.

For 2nd loop, 10 + 2(i₂ - i₃) + 3(i₂ - i₁) = 0. For 3rd loop, i₃ + 2(i₃ - i₂) = 10.
As i₁ = 10A, On solving above equations, we get i₂ = 7.27A.

I₃ - I₂ = 2.
As I₂ = -2A, I₃ = 0A.
Th term power is the product of voltage and current. So, power supplied by source = 10 × 0 = 0W.

Applying Super mesh analysis, the equations will be I₁ + I₁ + 10 + I₂ + I₂ = 0.
I₁ + I₂ = -5.
I₂ - I₁ = 1.
On solving, I₁ = -3A.

By applying mesh analysis in loop 1,

-V₁ + (I₁ – I₂) + 2(I₁ – I₃) = 0

⇒ 3I₁ – I₂ – 2I₃ = V₁

By applying mesh analysis in loop 2,

(I₂ – I₁) – V₂ + I₂ + (I₂ – I₃) = 0

⇒ -I₁ + 3 I₂ – I₃ = V₂

By applying mesh analysis in loop 3,

2(I₃ – I₁) + (I₃ – I₂) + V₃ = 0

⇒ -2I₁ – I₂ + 3I₃ = -V₃

By representing all the mesh equations in the matrix form, we get