In a series RL circuit a constant emf $E = 5 V$ is applied across a resistor $R = 10 Ω$ and an inductor $L$ (unknown). The current after $t = 0.20 s$ is measured to be $0.30 A$ . Find the self‑inductance $L$ of the circuit.

easy 2 marks Class 12 · Physics · Electromagnetic Induction

Answer

L \approx 2.2 H

The correct answer is $L \approx 2.2 H$ — Finally $L = τ R = 0.218 \times 10 \approx 2.18 H \approx 2.2 H$ .

Step-by-step solution

The steady‑state current is $I_{ss} = E / R = 5/10 = 0.5 A$ .
For an RL circuit $I (t) = I_{ss} (1 - e^{- t / τ})$ where $τ = L / R$ .
Insert the given values: $0.30 = 0.5 (1 - e^{- 0.20/ τ})$ .
Solve for the exponential term: $1 - e^{- 0.20/ τ} = 0.6 \Rightarrow e^{- 0.20/ τ} = 0.4$ .
Take natural logarithm: $- 0.20/ τ = ln 0.4$ so $τ = - 0.20/ ln 0.4$ .
Compute $τ = 0.20/0.9163 \approx 0.218 s$ .
Finally $L = τ R = 0.218 \times 10 \approx 2.18 H \approx 2.2 H$ .

Common mistake: Using

I (t) = I_{ss} e^{- t / τ}

instead of

I (t) = I_{ss} (1 - e^{- t / τ})

Concept tested: Self‑induction (time constant)