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ExamsJEE MainPhysics

The circuit shown has been running for a long time. In it a battery drives a steady current through R₁ and an inductor L, while R₂ is connected across the inductor branch through switch S. At the instant just after switch S is opened, find the voltage across the inductor V_L and state which labelled point (A or B) is at the higher potential. Take R₁ = 4.0 ohm, R₂ = 8.0 ohm, L = 2.5 H and the battery EMF = 6.0 V.

  1. V_L = 12 V; Point B is at the higher potential
  2. V_L = 12 V; Point A is at the higher potential
  3. V_L = 6 V; Point A is at the higher potential
  4. V_L = 6 V; Point B is at the higher potential

Correct answer: V_L = 12 V; Point B is at the higher potential

Solution

In steady state V_L = 0, so the inductor current is set only by the battery and R₁: I₀ = EMF/R₁ = 6/4 = 1.5 A. The inductor cannot change this current instantly. When S is opened, the only path left is through R₂, so 1.5 A is forced through the 8 ohm resistor giving V_L = I₀ * R₂ = 1.5 * 8 = 12 V. The inductor acts as a source maintaining the current direction, which raises point B above point A.

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