JEE Mains · Physics · STD 12 -Electromagnetic induction
- JEE Mains 2020
- Hard
- MCQ
A shown in the figure, a battery of \(emf \;\varepsilon\) is connected to an inductor \(L\) and resistance \(R\) in series. The switch is closed at \(t=0 .\) The total charge that flows from the battery, between \(\mathrm{t}=0\) and \(\mathrm{t}=\mathrm{t}_{\mathrm{c}}\;( \mathrm{t}_{\mathrm{c}}\) is the time constant of the circuit) is
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Answer & Solution
Correct Answer
(D) \(\frac{\varepsilon \mathrm{I}}{\mathrm{e} \mathrm{R}^{2}}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{i}=\mathrm{i}_{0}\left(1-\mathrm{e}^{-\mathrm{R}t / \mathrm{L}}\right)=\mathrm{i}_{0}\left(1-\mathrm{e}^{-t / T_{\mathrm{C}}}\right)\)
\(\mathrm{q}=\int_{0}^{\mathrm{T}_{\mathrm{C}}} \mathrm{i} \mathrm{dt}\)
\(=\int_{0}^{T_{c}} \frac{\varepsilon}{R}\left(1-e^{-t / T_{c}}\right)\)
\(=\left.\frac{\varepsilon}{R}\left(t-\frac{e^{-t / T_{c}}}{-1 / T_{c}}\right)\right|_{0} ^{T_{c}}\)
\(=\frac{e}{R}\left(T_{c}-T_{c} e^{-1}\right)-\frac{e}{R}\left(0+T_{c}\right)\)
\(\mathrm{q}=\frac{\mathrm{e}}{\mathrm{R}} \times \mathrm{T}_{\mathrm{C}} \mathrm{e}^{-1}\)
\(=\frac{\varepsilon}{\mathrm{R}} \times \frac{\mathrm{L}}{\mathrm{R}} \frac{1}{\mathrm{e}}\)
\(=\frac{\varepsilon \mathrm{L}}{\mathrm{e} \mathrm{R}^{2}}\)
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