Capacity of a capacitor is 48 mu F. When it i | vedclass

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(A) The energy stored in a capacitor is given by $U = \frac{q^2}{2C}$.Given: $C = 48 \mu F = 48 	imes 10^{-6} \ F$,$q_1 = 0.1 \ C$,$q_2 = 0.5 \ C$.Th

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$2500 \ J$

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(A) The energy stored in a capacitor is given by $U = \frac{q^2}{2C}$.Given: $C = 48 \mu F = 48 	imes 10^{-6} \ F$,$q_1 = 0.1 \ C$,$q_2 = 0.5 \ C$.The change in energy $\Delta U$ is given by:$\Delta U = U_2 - U_1 = \frac{q_2^2}{2C} - \frac{q_1^2}{2C} = \frac{1}{2C} (q_2^2 - q_1^2)$Substituting the values:$\Delta U = \frac{1}{2 	imes 48 	imes 10^{-6}} ((0.5)^2 - (0.1)^2)$$\Delta U = \frac{1}{96 	imes 10^{-6}} (0.25 - 0.01)$$\Delta U = \frac{0.24}{96 	imes 10^{-6}}$$\Delta U = \frac{0.24 	imes 10^6}{96} = \frac{240000}{96} = 2500 \ J$.

Capacity of a capacitor is $48 \mu F$. When it is charged from $0.1 C$ to $0.5 C$,the change in the energy stored is:

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