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# Help me solve this In the given figure each plate of capacitance C has partial value of charge

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

• Option 1)

CE

• Option 2)

$\frac{CER_1}{R_2-r}$

• Option 3)

$\frac{CER_2}{R_2+r}$

• Option 4)

$\frac{CER_1}{R_1-r}$

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steady state current drawn from the battery $i=\frac{E}{R_2+r}$

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

Hence potential difference across line (1) is  $V=\frac{ER_{2}}{R_2+r}$

the same potential difference appears across the capacitor, so charge on capacitor $Q=\frac{CER_2}{R_2+r}$

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