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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}

 

Answers (1)

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}

Posted by

Safeer PP

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