JEE Mains · Physics · STD 12 -6. Electromagnetic induction
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
- A \(\frac{\varepsilon \mathrm{L}}{\mathrm{R}^{2}}\left(1-\frac{1}{\mathrm{e}}\right)\)
- B \(\frac{\varepsilon \mathrm{R}}{\mathrm{e} \mathrm{L}^{2}}\)
- C \(\frac{\varepsilon \mathrm{I}}{\mathrm{R}^{2}}\)
- D \(\frac{\varepsilon \mathrm{I}}{\mathrm{e} \mathrm{R}^{2}}\)
Answer & Solution
Correct Answer
(D) \(\frac{\varepsilon \mathrm{I}}{\mathrm{e} \mathrm{R}^{2}}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{i}=\mathrm{i}_{0}\left(1-\mathrm{e}^{-\mathrm{R}t / \mathrm{L}}\right)=\mathrm{i}_{0}\left(1-\mathrm{e}^{-t / T_{\mathrm{C}}}\right)\) \(\mathrm{q}=\int_{0}^{\mathrm{T}_{\mathrm{C}}} \mathrm{i} \mathrm{dt}\)…
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