Consider the LR circuit shown in the figure.

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 :

  1. (1) $\frac{2.7 E L}{R^{2}}$

  2. (2) $\frac{E L}{2.7 R^{2}}$

  3. (3) $\frac{7.3 E L}{R^{2}}$

  4. (4) $\frac{E L}{7.3 R^{2}}$


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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