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Q.

Two capacitors are in parallel and when connected to a source of 3000 V, store 250 J of energy. When they are connected in series to the same source, the energy stored decreases by 190 J for the same potential. Their capacities are in the ratio

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a

3 : 5

b

3 : 2

c

4 : 3

d

2 : 7

answer is A.

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Detailed Solution

$\begin{array}{l} \frac{1}{2} C_{P} V^{2} = 250; \frac{1}{2} C_{s} V^{2} = 250 - 190 = 60 \\ \frac{C_{P}}{C_{s}} = \frac{250}{60} = \frac{25}{6}; \frac{C_{1} + C_{2}}{(\frac{C_{1} C_{2}}{C_{1} + C_{2}})} = \frac{25}{6} \\ \frac{{(C_{1} + C_{2})}^{2}}{C_{1} C_{2}} = \frac{25}{6} \\ Take \frac{C_{1}}{C_{2}} as x; 6 x^{2} + 13 x + 6 = 0; x = \frac{2}{3} or \frac{3}{2} \end{array}$

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