In a series RL circuit a constant emf $E = 15 V$ is applied at $t = 0$ . The resistor has $R = 5 Ω$ and the current after $0.20 s$ is $2.0 A$ . What is the self‑inductance $L$ of the circuit?

medium MCQ 2 marks Class 12 · Physics · Electromagnetic Induction

(A) $0.25 H$
(B) $0.50 H$
(C) $0.75 H$
(D) $1.00 H$

Answer

(B)

0.50 H

The correct answer is (B) $0.50 H$ — Take natural logarithm: $- \frac{1}{L} = ln \frac{1}{3} = - ln 3$ . Hence $\frac{1}{L} = ln 3$ and $L = \frac{1}{l n 3} = 0.50 H$ (to two significant figures).

Step-by-step solution

For a series RL circuit the current grows as $I (t) = \frac{E}{R} (1 - e^{- tR / L})$ .
Insert the known values: $2.0 = \frac{15}{5} (1 - e^{- 0.20 \times 5/ L}) = 3 (1 - e^{- 1/ L})$ .
Divide both sides by 3: $\frac{2}{3} = 1 - e^{- 1/ L}$ , so $e^{- 1/ L} = 1 - \frac{2}{3} = \frac{1}{3}$ .
Take natural logarithm: $- \frac{1}{L} = ln \frac{1}{3} = - ln 3$ . Hence $\frac{1}{L} = ln 3$ and $L = \frac{1}{l n 3} = 0.50 H$ (to two significant figures).

Common mistake: Forgetting the negative sign in the exponent or solving

e^{- tR / L} = \frac{2}{3}

instead of

\frac{1}{3}

Concept tested: Transient current in RL circuit

Published 21 September 2026 · How this solution was prepared and checked

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In a series RL circuit a constant emf E=15V is applied at t=0. The resistor has R=5Ω and the current after 0.20s is 2.0A. What is the self‑inductance L of the circuit?

Step-by-step solution

In a series RL circuit a constant emf $E = 15 V$ is applied at $t = 0$ . The resistor has $R = 5 Ω$ and the current after $0.20 s$ is $2.0 A$ . What is the self‑inductance $L$ of the circuit?