Capacitors

JEE Main level, single correct, medium

175 of 180

Question 175 of 180

In the given circuit diagram, when the current reaches a steady-state in the circuit, the charge on the capacitor of capacitance CC will be:

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Answer: (D) CEr2(r+r2)CE\dfrac{{r}_{2}}{\left(r+{r}_{2}\right)}

In steady state, current through capacitor will be zero, so current through r1{r}_{1} will be zero.
Hence, the net current flowing in the circuit will be,
I=Er+r2I=\dfrac{E}{r+{r}_{2}}
The potential difference across the capacitor will be,
V2=I×r2=Er2r+r2{V}_{2}=I\times {r}_{2}=\dfrac{E{r}_{2}}{r+{r}_{2}}
Therefore, charge on the capacitor plate will be,
q=CV2=CEr2r+r2q=C{V}_{2}=\dfrac{CE{r}_{2}}{r+{r}_{2}}

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