Question:
In the given circuit diagram when the current reaches steady state in the circuit, the charge on the capacitor of capacitance \(C\) will be:
In the given circuit diagram when the current reaches steady state in the circuit, the charge on the capacitor of capacitance \(C\) will be:
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In steady state no current flows in the capacitor branch, so the capacitor voltage equals the voltage across \(r_2\) in the battery loop \(E, r, r_2\).
Updated On: Nov 11, 2026
- \(CE\,\frac{r_1}{(r_1 + r)}\)
- \(CE\)
- \(CE\,\frac{r_1}{(r_2 + r)}\)
- \(CE\,\frac{r_2}{(r + r_2)}\)
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The Correct Option is D
Solution and Explanation
Circuit (from the figure): the top branch has the battery \(E\) in series with \(r\); the middle branch has \(C\) in series with \(r_1\); the bottom branch has \(r_2\); all three are connected in parallel between the same two nodes.
In steady state the capacitor blocks current, so no current flows in the \(C\)-\(r_1\) branch and there is no drop across \(r_1\).
The current flows only in the loop \(E\), \(r\), \(r_2\): \(I = \frac{E}{r + r_2}\).
The capacitor branch is in parallel with \(r_2\), so the potential difference across \(C\) is \(V = I r_2 = \frac{E r_2}{r + r_2}\).
Charge on the capacitor: \(Q = CV = CE\,\frac{r_2}{r + r_2}\).
Answer: (4) \(CE\,\frac{r_2}{(r + r_2)}\)
In steady state the capacitor blocks current, so no current flows in the \(C\)-\(r_1\) branch and there is no drop across \(r_1\).
The current flows only in the loop \(E\), \(r\), \(r_2\): \(I = \frac{E}{r + r_2}\).
The capacitor branch is in parallel with \(r_2\), so the potential difference across \(C\) is \(V = I r_2 = \frac{E r_2}{r + r_2}\).
Charge on the capacitor: \(Q = CV = CE\,\frac{r_2}{r + r_2}\).
Answer: (4) \(CE\,\frac{r_2}{(r + r_2)}\)
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