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(B) Initial charge on the capacitor is $q_i = CV$.Final charge on the capacitor is $q_f = C(2V) = 2CV$.Since the battery is connected with opposite po

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$2.25$

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(B) Initial charge on the capacitor is $q_i = CV$.Final charge on the capacitor is $q_f = C(2V) = 2CV$.Since the battery is connected with opposite polarity,the charge flowing through the battery is $\Delta q = q_f - (-q_i) = 2CV + CV = 3CV$.The work done by the battery is $W = \Delta q 	imes (2V) = (3CV)(2V) = 6CV^2$.Initial energy stored is $U_i = \frac{1}{2}CV^2$.Final energy stored is $U_f = \frac{1}{2}C(2V)^2 = 2CV^2$.By the work-energy theorem,$W = (U_f - U_i) + H$,where $H$ is the heat generated.$H = W - (U_f - U_i) = 6CV^2 - (2CV^2 - 0.5CV^2) = 6CV^2 - 1.5CV^2 = 4.5CV^2$.The ratio of heat generated to the final energy stored is $\frac{H}{U_f} = \frac{4.5CV^2}{2CV^2} = 2.25$.

$A$ 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

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