A body of capacity $4 \mu F$ is charged to 8 0 \nabla and another body of capacity $6 \mu F$ is charged to 30V. When they are connected the energy lost by $4 \mu F$ capacitor is
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Initial energy of body of capacitance 4 μ μ F is $U_i = \frac{1}{2} \times (4 \times 10^{-6})(80)^2 = 0.0128 \, J$
Final potential on this body after connection is $v = \frac{4 \times 80 + 6 \times 30}{4 + 6} = 50 \, \mathrm{V}.$ So final energy on it
$U_f = \frac{1}{2} \times 4 \times 10^{-6} (50)^2 = 0.005 \text{ J}$
Energy lost by this body = U i – U f = 7.8 mJ
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