An inductance L and a resistance R are first | vedclass

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Question

(C) When the battery is disconnected,the current through the $RL$ circuit starts decreasing exponentially according to the equation $i = i_0 e^{-Rt/L}

Vedclass · Accepted Answer

$\frac{L}{R} \ s$

Vedclass · Answer

(C) When the battery is disconnected,the current through the $RL$ circuit starts decreasing exponentially according to the equation $i = i_0 e^{-Rt/L}$.Given that the current reduces to $37\%$ of its initial value,we have $i = 0.37 i_0$.Substituting this into the equation: $0.37 i_0 = i_0 e^{-Rt/L}$.Since $e^{-1} \approx 0.3678 \approx 0.37$,we can write $0.37 \approx e^{-1}$.Therefore,$e^{-1} = e^{-Rt/L}$.Comparing the exponents,we get $1 = \frac{Rt}{L}$.Solving for $t$,we find $t = \frac{L}{R}$ seconds.

An inductance $L$ and a resistance $R$ are first connected to a battery. After some time,the battery is disconnected,but $L$ and $R$ remain connected in a closed circuit. Then the current reduces to $37\%$ of its initial value in

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