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(B) For a short-circuited $RL$ circuit,the current at time $t$ is given by $I = I_0 e^{-t/	au}$,where $	au$ is the time constant.Given that at $t =

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$	au = \frac{t_0}{\log_e \alpha}$

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(B) For a short-circuited $RL$ circuit,the current at time $t$ is given by $I = I_0 e^{-t/	au}$,where $	au$ is the time constant.Given that at $t = t_0$,the current becomes $I = \frac{I_0}{\alpha}$.Substituting these values into the equation:$\frac{I_0}{\alpha} = I_0 e^{-t_0/	au}$$\frac{1}{\alpha} = e^{-t_0/	au}$Taking the natural logarithm on both sides:$\ln(1/\alpha) = -t_0/	au$$-\ln(\alpha) = -t_0/	au$$\ln(\alpha) = t_0/	au$Therefore,the time constant is $	au = \frac{t_0}{\ln(\alpha)}$ or $	au = \frac{t_0}{\log_e \alpha}$.

$A$ coil carrying a steady current is short-circuited. The current in it decreases $\alpha$ times in $t_0$. The time constant of the circuit is:-

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