An inductor (L = 100 mH),a resistor (R = 100 | vedclass

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Question

(A) Initially,when steady state is achieved,the current in the circuit is $i_0 = \frac{E}{R} = \frac{100 \ V}{100 \ \Omega} = 1 \ A$.When the battery

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$\frac{1}{e} \ A$

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(A) Initially,when steady state is achieved,the current in the circuit is $i_0 = \frac{E}{R} = \frac{100 \ V}{100 \ \Omega} = 1 \ A$.When the battery is disconnected and points $A$ and $B$ are short-circuited,the circuit becomes an $LR$ decay circuit.The equation for the decay of current is given by $i(t) = i_0 e^{-\frac{R}{L}t}$.Given $R = 100 \ \Omega$,$L = 100 \ mH = 0.1 \ H$,and $t = 1 \ ms = 10^{-3} \ s$.The time constant of the circuit is $	au = \frac{L}{R} = \frac{0.1 \ H}{100 \ \Omega} = 10^{-3} \ s$.Substituting the values into the decay equation:$i(t) = 1 	imes e^{-\frac{10^{-3}}{10^{-3}}} = 1 	imes e^{-1} = \frac{1}{e} \ A$.

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