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In the given figure each plate of capacitance C has partial value of charge

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$\frac{{CE{R_1}}}{{{R_1} - r}}$

$\frac{{CE{R_1}}}{{{R_2} - r}}$

$\frac{{CE{R_2}}}{{{R_2} + r}}$

answer is C.

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Detailed Solution

$In{\text{ }}steady{\text{ }}state{\text{ }}current{\text{ }}drawn{\text{ }}from{\text{ }}the{\text{ }}battery\,\,i = \frac{E}{{({R_2} + r)}}\$

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In steady state capacitor is fully charged hence No current will flow through line (2)

Hence potential difference across line (1) is $V = \frac{E}{{({R_2} + r)}} \times {R_2}$ ,the same potential difference appears across the capacitor, so charge on capacitor

$Q = C \times \frac{{E{R_2}}}{{({R_2} + r)}}$

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