Two identical capacitors C₁ and C₂ of equal capacitance are connected as shown in the circuit terminals a and b of the key k are connected to charge capacitor C₁ 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 C.

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Detailed Solution

$\frac{\frac{1}{2} \frac{c_{1} c_{2}}{c_{1} + c_{2}} {(v_{1} - v_{2})}^{2}}{\frac{1}{2} c_{1} v_{1}^{2} + \frac{1}{2} c_{2} v_{2}^{2}} \times 100 = 50 %$

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