A body of capacity 4,mu F is charged to 80,V | vedclass

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(A) Initial energy of the capacitor with capacitance $C_1 = 4\,\mu F$ charged to $V_1 = 80\,V$ is given by $U_i = \frac{1}{2} C_1 V_1^2 = \frac{1}{2}

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$7.8$

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(A) Initial energy of the capacitor with capacitance $C_1 = 4\,\mu F$ charged to $V_1 = 80\,V$ is given by $U_i = \frac{1}{2} C_1 V_1^2 = \frac{1}{2} 	imes (4 	imes 10^{-6}) 	imes (80)^2 = 0.0128\,J = 12.8\,mJ$.When the two capacitors are connected,the common potential $V$ is given by $V = \frac{C_1 V_1 + C_2 V_2}{C_1 + C_2} = \frac{4 	imes 80 + 6 	imes 30}{4 + 6} = \frac{320 + 180}{10} = 50\,V$.The final energy of the $4\,\mu F$ capacitor at potential $V = 50\,V$ is $U_f = \frac{1}{2} C_1 V^2 = \frac{1}{2} 	imes (4 	imes 10^{-6}) 	imes (50)^2 = 0.005\,J = 5.0\,mJ$.The energy lost by the $4\,\mu F$ capacitor is $\Delta U = U_i - U_f = 12.8\,mJ - 5.0\,mJ = 7.8\,mJ$.

$A$ body of capacity $4\,\mu F$ is charged to $80\,V$ and another body of capacity $6\,\mu F$ is charged to $30\,V$. When they are connected,the energy lost by the $4\,\mu F$ capacitor is.......$mJ$.

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