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Q.

A capacitor having capacitance ‘C’ is charged by a battery having e.m.f. ‘2V’ (Battery is disconnected after charging) . Now this charged capacitor is connected to another battery having e.m.f. 'V'. The positive plate of capacitor is connected with negative terminal of battery & vice-versa. The amount of heat produced after connecting it with battery is

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a

$1.5 {CV}^{2}$

b

$4.5 {CV}^{2}$

c

$4 {CV}^{2}$

d

$2.5 {CV}^{2}$

answer is A.

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Detailed Solution

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$Δ H = W_{b a t t e r y} - Δ U = V_{q (tansferred)} - Δ U q_{initial} = C (2 V) q_{final} = C V q_{transferred} = C V - (- 2 C V) = 3 C V Δ U = U_{f} - U_{i} = \frac{1}{2} C V^{2} - \frac{1}{2} C {(2 V)}^{2} = - \frac{1}{2} 3 C V^{2} Δ H = V (3 C V) - (- \frac{3}{2} C V^{2}) = 3 C V^{2} + \frac{3}{2} C V^{2} = \frac{9}{2} C V^{2} = 4.5 C V^{2}$

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