Question
An inductor $$\left( {l = 100\,mH} \right),$$ a resistor $$\left( {R = 100\,\Omega } \right)$$ and a battery $$\left( {E = 100\,V} \right)$$ 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
An inductor $$\left( {l = 100\,mH} \right),$$ a resistor $$\left( {R = 100\,\Omega } \right)$$ and a battery $$\left( {E = 100\,V} \right)$$ 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
A.
$$\frac{I}{{eA}}$$
B.
$$eA$$
C.
$$0.1\,A$$
D.
$$1\,A$$
Answer :
$$\frac{I}{{eA}}$$
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 $$ - L\frac{{di}}{{dt}} - iR = 0$$
$$\eqalign{ & \Rightarrow \frac{{di}}{i} = - \frac{R}{L}dt \Rightarrow \int\limits_{{i_0}}^i {\frac{{di}}{i}} = \int\limits_0^t { - \frac{R}{L}} dt \cr & \Rightarrow {\log _e}\frac{i}{{{i_0}}} = - \frac{R}{L}t \Rightarrow i = {i_0}{e^{ - \frac{R}{L}t}} \cr & \Rightarrow i = \frac{E}{R}{e^{ - \frac{R}{L}t}} = \frac{{100}}{{100}}{e^{\frac{{ - 100 \times {{10}^{ - 3}}}}{{100 \times {{10}^{ - 3}}}}}} = \frac{1}{e} \cr} $$
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 $$ - L\frac{{di}}{{dt}} - iR = 0$$
$$\eqalign{ & \Rightarrow \frac{{di}}{i} = - \frac{R}{L}dt \Rightarrow \int\limits_{{i_0}}^i {\frac{{di}}{i}} = \int\limits_0^t { - \frac{R}{L}} dt \cr & \Rightarrow {\log _e}\frac{i}{{{i_0}}} = - \frac{R}{L}t \Rightarrow i = {i_0}{e^{ - \frac{R}{L}t}} \cr & \Rightarrow i = \frac{E}{R}{e^{ - \frac{R}{L}t}} = \frac{{100}}{{100}}{e^{\frac{{ - 100 \times {{10}^{ - 3}}}}{{100 \times {{10}^{ - 3}}}}}} = \frac{1}{e} \cr} $$
