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

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 current through 5Ω resistor = V₃/5 = 18.11/5 = 3.62A. The power absorbed by 5Ω resistor = (3.62)²) × 5 = 65.52W.

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.