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ExamsJEE MainPhysics

In a series L-R circuit connected to a battery of constant emf E, the switch is closed at t = 0. Let e be the magnitude of the emf induced across the inductor and i the circuit current at time t. Which graph correctly represents e versus i?

  1. A straight line of e decreasing linearly with i, from a positive intercept down to zero
  2. A curve where e decreases nonlinearly and approaches zero asymptotically
  3. A curve where e increases nonlinearly from the origin and saturates
  4. A straight line of e increasing linearly with i from the origin

Correct answer: A straight line of e decreasing linearly with i, from a positive intercept down to zero

Solution

By KVL around the L-R loop, E = i*R + L*di/dt, and the inductor emf magnitude is e = L*di/dt = E - i*R. This is linear in i: at t = 0, i = 0 so e = E (maximum); as i grows to its steady value E/R, e falls to 0. So the e-vs-i graph is a straight line decreasing from intercept E to zero.

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