A capacitor of capacitance C is charged using a cell of emf V and the cell is disconnected. The charged capacitor is now connected across another cell of emf 2V. The positive and negative plates are connected to the positive and negative terminals of the cell, respectively. Find the ratio of work done by the two cells used to charge the capacitor $(\frac{w_{2}}{w_{1}})$ is _________ .

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answer is 2.

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Detailed Solution

Charge on the capacitor after it is connected to the first cell is
   $Q_{1} = C V$
Work done by the first cell is
   $W_{1} = Q_{1} V = C V^{2}$
Charge on the capacitor after it is connected to the second cell is
$Q_{2} = C (2 V) = 2 C V$
Charge supplied by the second cell is
   $= Q_{2} - Q_{1} = C V$
Work done by the second cell is
$W_{2} = (C V) (2 V) = 2 C V^{2}$
$∴ \frac{W_{1}}{W_{2}} = \frac{C V^{2}}{2 C V^{2}} = \frac{1}{2}$