Question:
Consider the LR circuit shown in the figure. If the switch S is closed at $\mathrm{t}=0$ then the amount of charge that passes
through the battery between $\mathrm{t}=0$ and $t=\frac{L}{R}$ is :
Correct Option: , 2
Solution:
(2)
We have, $i=i_{0}\left(1-e^{-t / c}\right)=\frac{\varepsilon}{R}\left(1-e^{-t / c}\right)$
Charge, $q=\int_{0}^{\tau} i d t$
$=\frac{\varepsilon}{R} \int_{0}^{\tau}\left(1-e^{-t / \tau}\right) d t=\frac{E}{R} \frac{\tau}{e}=\frac{E}{R} \times \frac{(L / R)}{e}$
$=\frac{E L}{2.7 R^{2}}$
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