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(A) The current in a decaying $L-R$ circuit is given by $i = i_0 e^{-t/	au}$,where $	au = L/R$ is the time constant.The energy stored in the inducto

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$(\ln 2) \frac{L}{R}$

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(A) The current in a decaying $L-R$ circuit is given by $i = i_0 e^{-t/	au}$,where $	au = L/R$ is the time constant.The energy stored in the inductor is $U = \frac{1}{2} L i^2$.Let $U_0$ be the initial energy,$U_0 = \frac{1}{2} L i_0^2$.We are given that the energy reduces to one-fourth of its initial value,so $U = \frac{1}{4} U_0$.Substituting the expressions for energy: $\frac{1}{2} L i^2 = \frac{1}{4} (\frac{1}{2} L i_0^2)$.This simplifies to $i^2 = \frac{1}{4} i_0^2$,or $i = \frac{1}{2} i_0$.Substituting $i = i_0 e^{-t/	au}$,we get $i_0 e^{-t/	au} = \frac{1}{2} i_0$.$e^{-t/	au} = \frac{1}{2} \Rightarrow e^{t/	au} = 2$.Taking the natural logarithm on both sides: $t/	au = \ln 2$.Therefore,$t = 	au \ln 2 = \frac{L}{R} \ln 2$.

In a decaying $L-R$ circuit,the time after which energy stored in the inductor reduces to one-fourth of its initial value is

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