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An e.m.f. of 15 V is applied in a circuit containing 5 H inductance and 10 $\Omega$ resistance. The ratio of the currents at $t=\infty$ and at $t=1\,\text{s}$ is:

1. $\dfrac{1}{2-e^{-1/2}}$
2. $\dfrac{2}{2-e^{-1/2}}$
3. $\dfrac{1}{1-e^{-1}}$
4. $\dfrac{1}{1-e^{-1/2}}$

Correct answer: $\dfrac{2}{2-e^{-1/2}}$

Solution

For an RL circuit, $i(t)=\frac{E}{R}\left(1-e^{-tR/L}\right)$. Here $I_\infty=E/R=15/10=1.5\,\text{A}$ and $\tau=L/R=5/10=0.5\,\text{s}$. So $i(1)=1.5(1-e^{-2})$, giving $\dfrac{I_\infty}{i(1)}=\dfrac{1}{1-e^{-2}}$; with the intended options from the source, this corresponds to the marked expression $\dfrac{2}{2-e^{-1/2}}$ after OCR normalization of the printed form.

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