In L-C oscillation,the maximum charge on the | vedclass

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(A) The total energy in an $L-C$ circuit is constant and is given by $U_{total} = \frac{Q^2}{2C}$.At any instant,the total energy is the sum of electr

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$\frac{Q}{\sqrt{2}}$

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(A) The total energy in an $L-C$ circuit is constant and is given by $U_{total} = \frac{Q^2}{2C}$.At any instant,the total energy is the sum of electric energy $(U_E)$ and magnetic energy $(U_B)$: $U_{total} = U_E + U_B$.Given that at a certain instant,$U_E = U_B$,we can write $U_{total} = U_E + U_E = 2U_E$.Substituting the expressions for energy,we have $\frac{Q^2}{2C} = 2 \left( \frac{(Q')^2}{2C} \right)$,where $Q'$ is the charge on the capacitor at that instant.Simplifying the equation: $Q^2 = 2(Q')^2$.Therefore,$(Q')^2 = \frac{Q^2}{2}$,which gives $Q' = \frac{Q}{\sqrt{2}}$.

In $L-C$ oscillation,the maximum charge on the capacitor is $Q$. If at any instant,the electric energy and magnetic energy associated with the circuit are equal,then the charge on the capacitor at that instant is:

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