A capacitor C is fully charged with voltage V _ 0 . After...

Heat Loss is given by,&

$H = \dfrac{{C}_{1}{C}_{2}}{2\left({C}_{1} + {C}_{2}\right)}\left({V}_{1} - {V}_{2}\right)2$

${V}_{1}$~is& initial voltage of capacitor& $C= {V}_{0}$

${V}_{2}$~is the initial voltage of capacitor& $\dfrac{C}{2}= 0$

$= \dfrac{c \times \dfrac{c}{2}}{2\left(c + \dfrac{c}{2}\right)}\left({V}_{0} - 0\right)2 = \dfrac{C}{6}{V}_{0}^{2}$

$H = \dfrac{1}{6}{C}_{0}{V}_{0}^{2}$

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