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

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

Quantrex Academy · Accepted Answer

Correct answer: D. CE r _ 2 (r + r _ 2 )

In steady state, current through capacitor will be zero, so current through ${r}_{1}$ will be zero.

Hence, the net current flowing in the circuit will be,
$I = \dfrac{E}{r + {r}_{2}}$
The potential difference across the capacitor will be,
${V}_{2} = I \times {r}_{2} = \dfrac{E{r}_{2}}{r + {r}_{2}}$

Therefore, charge on the capacitor plate will be,
$q = C{V}_{2} = \dfrac{CE{r}_{2}}{r + {r}_{2}}$

In the given circuit diagram, when the current reaches a...

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