Test: Network Equation (Mesh & Node) - 1 - Electrical Engineering (EE) MCQ

Applying the nodal analysis

1. Define nodes

• Left node = vL

• Right node = va

• Bottom node = ground

2. KCL at node va

• Currents entering va: 3 A (top source, left→right) + 1 A (bottom source, up) = 4 A

• Only path leaving va is through the 2 Ω resistor to the left node: I = (va − vL) / 2

• Set entering = leaving: (va − vL) / 2 = 4 → va − vL = 8 …… (1)

3. KCL at left node vL

• From step (1), 4 A enters vL from the 2 Ω branch

• 3 A leaves vL to the right through the top source

• So 1 A must go to ground through the 3 Ω resistor

• vL / 3 = 1 → vL = 3 V …… (2)

4. Solve for va

• From (1) and (2): va = vL + 8 = 3 + 8 = 11 V

Final answer: va = 11 V.

Correct Answer :- B

Explanation : Current due to voltage source is:

I' = 10/(64+36) 

= 0.1A.

Current due to current source is:

I = 0.5A.

Current ib = I' + I 

= 0.5+ 0.1

= 0.6A

3loops = 3KVL equations. Solving them gives currents flowing in the circuit. I₁ = 5A, I₂ = -1.47A, I₃ = 0.56A. Power supplied by dependent voltage source = 0.4V1 (I₁ - I₂).