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Physics · PYQ NEET UG 2019(2)
हिन्दीTwo 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 e. m. f. \( V \) 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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Correct answer · Option D
50%
Explanation
Initial energy \( U_i = \frac{1}{2} C V^2 \). When connected to an identical uncharged capacitor \( C_2 \), the common potential becomes \( V/2 \). Final energy \( U_f = 2 \times \frac{1}{2} C (V/2)^2 = \frac{1}{4} C V^2 \). Loss percentage = \( \frac{U_i - U_f}{U_i} \times 100\% = 50\% \).
