Two identical capacitors $C_{1}$ and $C_{2}$ of equal capacitance are connected as shown in the circuit. Terminals a and b of the key k are connected to charge capacitor $C_{1}$ using battery of EMF V volt. Now disconnecting a and b the terminals b and c are connected. Due to this, what will be the percentage loss of energy?

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75%

25%

50%

answer is D.

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Detailed Solution

$U_{i} = \frac{1}{2} {CV}^{2}$

On switching key at point c

$\frac{q_{0} - q}{c} = \frac{q}{c} 2 q = q_{0} q = (\frac{q_{0}}{2}) U_{f} = \frac{1}{2} {(\frac{q_{0}}{2})}^{2} \times \frac{1}{C} + \frac{1}{2} {(\frac{q_{0}}{2})}^{2} \times \frac{1}{C} U_{f} = \frac{q_{0}^{2}}{4 C} U_{f} = \frac{1}{4} {CV}^{2} loss = (\frac{U_{i} - U_{f}}{U_{i}}) \times 100 = \frac{(\frac{1}{2} - \frac{1}{4}) {CV}^{2}}{\frac{1}{2} {CV}^{2}} \times 100 = 50 %$

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