If the charge on the capacitor is increased b | vedclass

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(C) The energy stored in a capacitor is given by $U = \frac{Q^2}{2C}$,where $Q$ is the charge and $C$ is the capacitance.Let the original charge be $Q

Vedclass · Accepted Answer

$30$

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(C) The energy stored in a capacitor is given by $U = \frac{Q^2}{2C}$,where $Q$ is the charge and $C$ is the capacitance.Let the original charge be $Q$. The original energy is $U_1 = \frac{Q^2}{2C}$.When the charge is increased by $3 \ C$,the new charge is $Q' = Q + 3$.The new energy is $U_2 = \frac{(Q+3)^2}{2C}$.Given that the energy increases by $21 \%$,we have $U_2 = U_1 + 0.21 U_1 = 1.21 U_1$.Substituting the expressions for $U_1$ and $U_2$:$\frac{(Q+3)^2}{2C} = 1.21 	imes \frac{Q^2}{2C}$.$(Q+3)^2 = 1.21 Q^2$.Taking the square root on both sides:$Q+3 = 1.1 Q$.$3 = 1.1 Q - Q$.$3 = 0.1 Q$.$Q = \frac{3}{0.1} = 30 \ C$.Therefore,the original charge on the capacitor is $30 \ C$.

If the charge on the capacitor is increased by $3 \ C$,the energy stored in it increases by $21 \%$. The original charge on the capacitor is (in $C$)

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