If a capacitor of capacity 900 mu F is charge | vedclass

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(C) The energy stored in a capacitor is given by the formula $U = \frac{1}{2} CV^2$.Given for the first capacitor: $C_1 = 900 \mu F$ and $V_1 = 100 \

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$300$

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(C) The energy stored in a capacitor is given by the formula $U = \frac{1}{2} CV^2$.Given for the first capacitor: $C_1 = 900 \mu F$ and $V_1 = 100 \ V$.The energy stored in the first capacitor is $U_1 = \frac{1}{2} C_1 V_1^2$.When this energy is transferred to a second capacitor of capacity $C_2 = 100 \mu F$, let the new potential be $V_u$.Since the total energy is transferred, $U_1 = U_2$, where $U_2 = \frac{1}{2} C_2 V_u^2$.Equating the energies: $\frac{1}{2} C_1 V_1^2 = \frac{1}{2} C_2 V_u^2$.Substituting the values: $\frac{1}{2} 	imes 900 	imes 10^{-6} 	imes (100)^2 = \frac{1}{2} 	imes 100 	imes 10^{-6} 	imes V_u^2$.Simplifying the equation: $900 	imes 100^2 = 100 	imes V_u^2$.$V_u^2 = \frac{900 	imes 10000}{100} = 90000$.$V_u = \sqrt{90000} = 300 \ V$.

If a capacitor of capacity $900 \mu F$ is charged to $100 \ V$ and its total energy is transferred to a capacitor of capacity $100 \mu F$, then its potential will be (in $V$)

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