In an L-R decay circuit,the initial current a | vedclass

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(B) The energy stored in an inductor is given by $U = \frac{1}{2} L i^2$. Initially,$U_0 = \frac{1}{2} L I^2$. When the energy reduces to one-fourth o

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$LI/2R$

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(B) The energy stored in an inductor is given by $U = \frac{1}{2} L i^2$. Initially,$U_0 = \frac{1}{2} L I^2$. When the energy reduces to one-fourth of its initial value,$U = \frac{1}{4} U_0 = \frac{1}{4} (\frac{1}{2} L I^2)$. Thus,$\frac{1}{2} L i^2 = \frac{1}{4} (\frac{1}{2} L I^2)$,which implies $i^2 = \frac{1}{4} I^2$,so $i = \frac{I}{2}$. In an $L-R$ decay circuit,the current at time $t$ is $i = I e^{-t/	au}$,where $	au = L/R$. Setting $i = I/2$,we get $I/2 = I e^{-t/	au}$,so $e^{-t/	au} = 1/2$,which means $t/	au = \ln 2$,or $t = 	au \ln 2$. The total charge $q$ flown through the resistor is $q = \int_{0}^{t} i dt = \int_{0}^{	au \ln 2} I e^{-t/	au} dt$. $q = I [-	au e^{-t/	au}]_{0}^{	au \ln 2} = -I 	au (e^{-\ln 2} - e^0) = -I 	au (1/2 - 1) = -I 	au (-1/2) = \frac{I 	au}{2}$. Substituting $	au = L/R$,we get $q = \frac{IL}{2R}$.

In an $L-R$ decay circuit,the initial current at $t = 0$ is $I$. The total charge that has flown through the resistor until the energy in the inductor has reduced to one-fourth of its initial value is:

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