Consider the LR circuit shown in the figure. If the switch S is closed at $t = 0$ then the amount of charge that passes through the battery between $t = 0$ and $t = \frac{L}{R}$ is:

Question

Consider the LR circuit shown in the figure. If the switch S is closed at t = 0 then the amount of charge that passes through the battery between t = 0 and t = L R is:

TestHub · Accepted Answer

The current in the LR circuit at any time $t$ is
$I = \frac{E}{R} (1 - e^{- \frac{R t}{L}})$

Where $E, R, L, I$ are the EMF, resistance, inductance and the current respectively.
$∴$ Charge flown through the circuit from time 0 to $\frac{L}{R}$ is
(charge) $∆ q = \int_{0}^{\frac{L}{R}} I . d t = \int_{0}^{\frac{L}{R}} \frac{E}{R} (1 - e^{- \frac{R t}{L}}) d t$
$\Rightarrow ∆ q = \frac{E}{R} {[t + \frac{L}{R} e^{- \frac{R t}{L}}]}_{0}^{\frac{L}{R}}$
$\Rightarrow ∆ q = \frac{E}{R} [\frac{L}{R} + \frac{L}{R} e^{- 1} - \frac{L}{R}]$
$\Rightarrow ∆ q = \frac{E}{R} \times \frac{L}{e R}$
$\Rightarrow ∆ q = \frac{E L}{e R^{2}} = \frac{E L}{2.7 R^{2}}$

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