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A capacitor of capacitance C is initially charged to a potential difference of V volt. Now it is connected to a battery of 2V with opposite polarity. The ratio of heat generated to the final energy stored in the capacitor will be n/4 where ‘n’ is

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answer is 11.

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Detailed Solution

$\begin{array}{l} \Rightarrow ΔU = \frac{1}{2} C [(2 V)^{2} - V^{2}] = \frac{3}{2} {CV}^{2} \\ W_{B} = 2 V [2 CV - (- CV)] = 6 {CV}^{2} \end{array}$

$H e a t g e n e r a t e d = 6 C V^{2} - 1.5 C V^{2} = 4.5 C V^{2}$

$r a t i o = \frac{H e a t g e n e r a t e d}{F i n a l e n e r g y s t o r e d} = \frac{4.5 C V^{2}}{2 C V^{2}} = 2.25$

$\frac{n}{4} = 2.25 \Rightarrow n = 11$

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