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Q.

In a L-R decay circuit, the initial current is I. The total charge that has flown through the resistor till the energy in the inductor has reduced to one-fourth is initial value, is $\frac{L I}{X R}$ the value of ‘X’ is

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answer is 2.

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Detailed Solution

$E_{2} = \frac{E_{1}}{4} \begin{array}{l} I_{2} = \frac{I_{1}}{2} \\ I_{0} e^{- \frac{t}{τ}} = \frac{I_{0}}{2} \\ \frac{t}{τ} = \log 2 \\ μ = \int_{0}^{\log 2} I_{0} e^{- \frac{t}{τ}} \\ a f t e r s o l v i n g \\ μ = \frac{I L}{2 n} \end{array}$

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