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An inductor of inductance $L=400 \mathrm{mH}$ and resistors of resistance $R_{1}=2 \Omega$ and $R_{2}=2 \Omega$ are connected to a battery of emf $12 \mathrm{~V}$ as shown in the figure. The internal resistance of the battery is negligible. The switch $S$ is closed at $t=0 .$ The potential drop across $L$ as a function of time is
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Verified Answer
The correct answer is:
$12 \mathrm{e}^{-5 t} \mathrm{~V}$
Growth in current in $L R_{2}$ branch when switch is closed is given by
$$
\begin{array}{l}
i=\frac{E}{R_{2}}\left[1-e^{-R_{2} t / L}\right] \\
\Rightarrow \frac{d i}{d t}=\frac{E}{R_{2}} \cdot \frac{R_{2}}{L} e^{-R_{2} t / L}=\frac{E}{L} e^{-\frac{R_{2} t}{L}}
\end{array}
$$
Hence, potential drop across $\mathbf{L}=\left(\frac{E}{L} e^{-R_{2} t / L}\right) L=E e^{-R_{2} t / L}$
$=12 e^{-\frac{2 t}{400 \times 10^{-3}}}=12 \mathrm{e}^{-5 t} \mathrm{~V}$
$$
\begin{array}{l}
i=\frac{E}{R_{2}}\left[1-e^{-R_{2} t / L}\right] \\
\Rightarrow \frac{d i}{d t}=\frac{E}{R_{2}} \cdot \frac{R_{2}}{L} e^{-R_{2} t / L}=\frac{E}{L} e^{-\frac{R_{2} t}{L}}
\end{array}
$$
Hence, potential drop across $\mathbf{L}=\left(\frac{E}{L} e^{-R_{2} t / L}\right) L=E e^{-R_{2} t / L}$
$=12 e^{-\frac{2 t}{400 \times 10^{-3}}}=12 \mathrm{e}^{-5 t} \mathrm{~V}$
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