Two identical capacitors are in parallel and are connected to a battery of emf 3000V, store 1120J of energy. When they are connected in series to the same battery, the energy stored in each capacitor is

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530 J

378 J

140 J

70 J

answer is B.

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Detailed Solution

$\begin{array}{l} E_{1} = \frac{1}{2} C_{P} V^{2} = 1120; E_{2} = \frac{1}{2} C_{S} V^{2} = ? \\ \Rightarrow \frac{E_{1}}{E_{2}} = \frac{C_{P}}{C_{S}} = \frac{2 C}{C / 2} = 4 = \frac{1120}{E_{2}}; E_{2} = 280 J \end{array}$
Energy stored in each capacitor=140J

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