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A 50 volt potential difference is suddenly applied to a coil with $L=5 \times 10^{-3}$ henry and $R=180 \mathrm{ohm}$. The rate of increase of current after 0.001 second is
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The rate of increase of current
$=\frac{d i}{d t}=\frac{d}{d t} i_0\left(1-\mathrm{e}^{-R t / L}\right)$ $=\frac{d}{d t} i_0-\frac{d}{d t} i_0 \mathrm{e}^{-R t / L}$
$=0-i_0 \mathrm{e}^{-R t / L} \cdot \frac{d}{d t}\left(-\frac{R t}{L}\right)$ $=i_0 \frac{R}{L} e^{-R t / L}$
$=\frac{50}{180} \times \frac{180}{5 \times 10^{-3}} \times e^{-(180 \times 0.001) /\left(5 \times 10^{-3}\right)}$ $=10^4 \times \mathrm{e}^{-36} \mathrm{~A} / \mathrm{sec}$
$=\frac{d i}{d t}=\frac{d}{d t} i_0\left(1-\mathrm{e}^{-R t / L}\right)$ $=\frac{d}{d t} i_0-\frac{d}{d t} i_0 \mathrm{e}^{-R t / L}$
$=0-i_0 \mathrm{e}^{-R t / L} \cdot \frac{d}{d t}\left(-\frac{R t}{L}\right)$ $=i_0 \frac{R}{L} e^{-R t / L}$
$=\frac{50}{180} \times \frac{180}{5 \times 10^{-3}} \times e^{-(180 \times 0.001) /\left(5 \times 10^{-3}\right)}$ $=10^4 \times \mathrm{e}^{-36} \mathrm{~A} / \mathrm{sec}$
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