Electromagnetic Induction and Alternating Current 4 Question 1
1. Consider the $L-R$ 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
(Main 2019, 12 April II)
(a) $\frac{2.7 E L}{R^{2}}$
(b) $\frac{E L}{2.7 R^{2}}$
(c) $\frac{7.3 E L}{R^{2}}$
(d) $\frac{E L}{7.3 R^{2}}$
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Answer:
Correct Answer: 1. (b)
Solution:
- In an $L-R$ circuit, current during charging is given by
$$ I=I _0 \quad 1-e^{-\frac{R}{L} t} $$
where, $\quad I _0=\frac{E}{R}=$ saturation current
So, we have $\frac{d q}{d t}=I=I _0 \quad 1-e^{-\frac{R}{L} t}$
$\Rightarrow \quad d q=I _0 \quad 1-e^{-\frac{R}{L} t} d t$
So, charge $q$ that passes through battery from time $t=0$ to $t=\frac{L}{R}$ is obtained by integrating the above equation within the specified limits, i.e.
$$ \begin{aligned} q & =\int _0^{Q} d q=\int _{t=0}^{t=\frac{L}{R}} I _0 1-e^{-\frac{R}{L} t} d t \\ & =I _0 \quad t-\frac{1}{-\frac{R}{L}} \cdot e^{-\frac{R}{L} t} \\ & =\frac{E}{R} \frac{L}{R}+\frac{L}{R e^{\prime}}-0+\frac{L}{R}=\frac{E}{R} \times \frac{L}{R e}=\frac{E L}{R^{2} e} \\ \Rightarrow \quad & \quad q=\frac{E L}{2.7 R^{2}} \end{aligned} $$
