An RL circuit consists of a resistor $R=20\Omega$ and an inductor $L=0.50\text{H}$. At $t=0$ a constant emf $\mathcal{E}=10\text{V}$ is applied across the series combination. Find the time required for the current to reach $90\%$ of its steady‑state value.

Step-by-step solution

1. The steady‑state (long‑time) current is given by Ohm's law: $I_{\text{ss}}=\mathcal{E}/R=10/20=0.5\text{A}$.
2. For an RL circuit, the current grows as $I(t)=I_{\text{ss}}(1-e^{-t/\tau })$ where $\tau =L/R$ is the time constant.
3. Calculate the time constant: $\tau =L/R=0.50/20=0.025\text{s}$.
4. Set $I(t)=0.9I_{\text{ss}}$ and solve: $0.9=1-e^{-t/\tau }$ $\Rightarrow$ $e^{-t/\tau }=0.1$.
5. Take natural logarithm: $-t/\tau =\ln 0.1=-\ln 10$ $\Rightarrow$ $t=\tau \ln 10$.
6. Insert $\tau$: $t=0.025\times 2.3026\approx 0.0576\text{s}$.
7. Round to two significant figures: $t\approx 5.8\times 10^{-2}\text{s}$.

Concept tested: Time constant of an RL circuit

Published 14 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked