A series RL circuit is driven by a constant emf of 12 V. The resistor has resistance 8 Ω and the inductor has self‑inductance L (unknown). At t=0.10 s after the emf is applied, the current through the circuit is measured to be 1.2 A. Find the value of L.

Step-by-step solution

1. Calculate the steady‑state current: $I_{\infty }=V/R=12/8=1.5\text{ A}$.
2. Use the RL current expression: $I(t)=I_{\infty }(1-e^{-t/\tau })$, where $\tau =L/R$.
3. Set $I(0.10)=1.2$ A: $1.2=1.5(1-e^{-0.10/\tau })$.
4. Solve for the exponential: $e^{-0.10/\tau }=1-0.8=0.2$. Take natural log: $-0.10/\tau =ln0.2=-1.6094$.\ Thus $\tau =0.10/1.6094=0.0621\text{ s}$.
5. Find $L$ from $\tau =L/R$: $L=0.0621\times 8=0.4968\text{ H}\approx 0.50\text{ H}$.

Concept tested: Inductive time constant and current growth

Published 18 September 2026 · How this solution was prepared and checked