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An inductor $(L = 100\ mH)$, a resistor $(R = 100 \Omega)$ and a battery $(E = 100\ V)$ are initially connected in series as shown in the figure. After a long time the battery is disconnected after short circuiting the points $A $ and $ B$. The current in the circuit $1\ ms$ after the short circuit is
$\frac{1}{e}$ $A$
$e$ $A$
$0.1\ A$
$1$ $A$
Solution
Initially, when steady state is achieved,
$i=\frac{E}{R}$
Let $E$ is short circuited at $t=0 .$ Then
At $t=0, i_{0}=\frac{E}{R}$
Let during decay of current at any time the current
flowing is $-L \frac{d i}{d t}-i R=0$
$\Rightarrow \frac{d i}{i}=-\frac{R}{L} d t \Rightarrow \int_{i_{0}}^{i} \frac{d i}{i}=\int_{0}^{t}-\frac{R}{L} d t$
$\Rightarrow \log _{e} \frac{i}{i_{0}}=-\frac{R}{L} t \Rightarrow i=i_{0} e^{-\frac{R}{L} t}$
$\Rightarrow i=\frac{E}{R} e^{-\frac{R}{L} t}=\frac{100}{100} e^{\frac{-100 \times 10^{-3}}{100 \times 10^{-3}}}=\frac{1}{e}$
