A capacitor of 4,mu F charged to 50,V is conn | vedclass

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(A) The total energy before connection is given by $U_i = \frac{1}{2} C_1 V_1^2 + \frac{1}{2} C_2 V_2^2$.$U_i = \frac{1}{2} 	imes 4 	imes 10^{-6}

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$1.5$ and $1.33$

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(A) The total energy before connection is given by $U_i = \frac{1}{2} C_1 V_1^2 + \frac{1}{2} C_2 V_2^2$.$U_i = \frac{1}{2} 	imes 4 	imes 10^{-6} 	imes (50)^2 + \frac{1}{2} 	imes 2 	imes 10^{-6} 	imes (100)^2 = 50 	imes 10^{-4} + 100 	imes 10^{-4} = 1.5 	imes 10^{-2}\,J$.When connected in parallel with like charges,the common potential $V$ is given by $V = \frac{C_1 V_1 + C_2 V_2}{C_1 + C_2} = \frac{4 	imes 50 + 2 	imes 100}{4 + 2} = \frac{400}{6} = \frac{200}{3}\,V$.The total energy after connection is $U_f = \frac{1}{2} (C_1 + C_2) V^2$.$U_f = \frac{1}{2} 	imes (4 + 2) 	imes 10^{-6} 	imes (\frac{200}{3})^2 = 3 	imes 10^{-6} 	imes \frac{40000}{9} = \frac{40000}{3} 	imes 10^{-6} = 1.33 	imes 10^{-2}\,J$.

$A$ capacitor of $4\,\mu F$ charged to $50\,V$ is connected to another capacitor of $2\,\mu F$ charged to $100\,V$ with plates of like charges connected together. The total energy before and after connection in multiples of $10^{-2}\,J$ is:

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