Question:
In the given circuits (a) and (b), switches \(S_1\) and \(S_2\) are closed at \(t=0\) and kept close for a long time. The variation of currents in the two circuits for \(t\ge0\) are shown in the options. (Figures are schematic and not drawn to scale.)
In the given circuits (a) and (b), switches \(S_1\) and \(S_2\) are closed at \(t=0\) and kept close for a long time. The variation of currents in the two circuits for \(t\ge0\) are shown in the options. (Figures are schematic and not drawn to scale.)
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Circuit (a) is an RC charging circuit (current decays from E/R); circuit (b) is an LR circuit (current grows from 0 to E/R).
Updated On: Nov 11, 2026
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The Correct Option is B
Solution and Explanation
Circuit (a): series C, R and battery E. Initially the uncharged capacitor acts as a short circuit, so \(i(0)=\frac{E}{R}\); then \(i=\frac{E}{R}e^{-t/RC}\) decays to zero.
Circuit (b): series L, R and battery E. Initially the inductor opposes the change, so \(i(0)=0\); then \(i=\frac{E}{R}\left(1-e^{-Rt/L}\right)\) rises to the steady value \(\frac{E}{R}\).
So curve (a) starts at E/R and decays, curve (b) starts at 0 and rises to E/R. Graph (2) shows this with correct labels (graph (3) has the labels swapped, (1) and (4) have non-monotonic curves).
Answer: (2) (a) decays from E/R, (b) rises to E/R
Circuit (b): series L, R and battery E. Initially the inductor opposes the change, so \(i(0)=0\); then \(i=\frac{E}{R}\left(1-e^{-Rt/L}\right)\) rises to the steady value \(\frac{E}{R}\).
So curve (a) starts at E/R and decays, curve (b) starts at 0 and rises to E/R. Graph (2) shows this with correct labels (graph (3) has the labels swapped, (1) and (4) have non-monotonic curves).
Answer: (2) (a) decays from E/R, (b) rises to E/R
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