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STANDARDPhysics-Current Electricity

STANDARD Physics Kirchhoff's Laws MCQ Question

Type: MCQ-conceptual-Medium-Class 12

A current of 6 A enters one corner P of an equilateral triangle PQR having 3 wires of resistances 2 Ω each and leaves by the corner R. Then the currents I₁ and I₂ are

A

2 A, 4 A

B

4 A, 2 A

C

1 A, 2 A

D

2 A, 3 A

Correct Answer

Option A

Detailed Explanation

Applying Kirchhoff's first law at the junction P, we get 6 = I₁ + I₂. Applying Kirchhoff's second law to the closed loop PQR, we get -2I₁ - 2I₁ + 2I₂ = 0 or, 4I₁ - 2I₂ = 0. By solving these equations, we get I₁ = 2 A, I₂ = 4 A.

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