Super Node Analysis - Free MCQ Practice Test with solutions, GATE EE Network

The value of the current (A) in the equivalent current source of the voltage source the short circuit current at the terminals A and B is I = 60/30 = 2A.

The equation at node 1 is 10 = V₁/3 + (V₁ - V₂)/2
According to super Node analysis, (V₁ - V₂)/2 = V₂/1 + (V₃ - 10)/5 + V₃/2V₂ - V₃ = 20.
On solving, we get, V₂ = 11.5V.

The term power is defined as the product of voltage and current and the power delivered by the source (6A) = V₂ x 6 = 3.72 × 6 = 22.32W.

Applying Super Node Analysis, the combined equation of node 1 and node 2 is (V₁ - V₃)/3 + 3 + (V₂ - V₃)/1 - 6 + V₂/5 = 0.
At node 3, (V₃ - V₁)/3 + (V₃ - V₂)/1 + V₃/2 = 0.
Also V₁ - V₂=10.
On solving above equations, we get V₂ = 3.72V ≈ 4V.

The value of the voltage (V) in the equivalent voltage source of the current source the voltage across the terminals A and B is (6)(5) = 30V.

At node 1, (V₁ - 40 - V₃)/4 + (V₁ - V₂)/6 - 3 - 5 = 0.
Applying Super Node Analysis at node 2 and 3, (V₂ - V₁)/6 + 5 + V₂/3 + V₃/5 + (V₃ + 40 - V₁)/4 = 0.
Also, V₃ - V₂ = 20.
On solving above equations, V₃ = 18.11V ≈ 18V.

The equation at node 1 is 10 = V₁/3+(V₁ - V₂)/2.
According to super Node analysis, (V₁ - V₂)/2 = V₂/1 + (V₃ - 10)/5 + V₃/2V₂ - V₃ = 20.
On solving, we get, V₁ = 19V.

Applying Super Node Analysis, the combined equation of node 1 and node 2 is (V₁ - V₃)/3 + 3 + (V₂ - V₃)/1 - 6 + V₂/5 = 0.
At node 3, (V₃ - V₁)/3 + (V₃ - V₂)/1 + V₃/2 = 0.
Also V₁ - V₂ = 10.
On solving above equations, we get V₃ = 4.5V.

Applying Super Node Analysis, the combined equation of node 1 and node 2 is (V₁ - V₃)/3 + 3 + (V₂ - V₃)/1 - 6 + V₂/5 = 0.
At node 3, (V₃ - V₁)/3 + (V₃ - V₂)/1 + V₃/2 = 0.
Also V₁ - V₂ = 10.
On solving above equations, we get V₁ = 13.72V ≈ 14V.