A 20,F capacitor is charged to 5,V and isolat | vedclass

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(D) The initial energy stored in the $20\,F$ capacitor is given by $U_i = \frac{1}{2} C_1 V_1^2 = \frac{1}{2} 	imes 20 	imes 5^2 = 250\,J$.When conn

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$150$

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(D) The initial energy stored in the $20\,F$ capacitor is given by $U_i = \frac{1}{2} C_1 V_1^2 = \frac{1}{2} 	imes 20 	imes 5^2 = 250\,J$.When connected in parallel,the common potential $V$ is given by $V = \frac{C_1 V_1 + C_2 V_2}{C_1 + C_2} = \frac{20 	imes 5 + 30 	imes 0}{20 + 30} = \frac{100}{50} = 2\,V$.The final energy of the system is $U_f = \frac{1}{2} (C_1 + C_2) V^2 = \frac{1}{2} 	imes (20 + 30) 	imes 2^2 = \frac{1}{2} 	imes 50 	imes 4 = 100\,J$.The decrease in energy is $\Delta U = U_i - U_f = 250\,J - 100\,J = 150\,J$.Alternatively,using the formula $\Delta U = \frac{C_1 C_2}{2(C_1 + C_2)} (V_1 - V_2)^2 = \frac{20 	imes 30}{2(20 + 30)} (5 - 0)^2 = \frac{600}{100} 	imes 25 = 6 	imes 25 = 150\,J$.

$A$ $20\,F$ capacitor is charged to $5\,V$ and isolated. It is then connected in parallel with an uncharged $30\,F$ capacitor. The decrease in the energy of the system will be.......$J$

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