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(A) The total energy in the $LC$ circuit is constant and is equal to the initial energy stored in the capacitor: $U_{total} = \frac{Q_0^2}{2C}$.At any

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$\frac{Q_0}{2\sqrt{LC}}$

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(A) The total energy in the $LC$ circuit is constant and is equal to the initial energy stored in the capacitor: $U_{total} = \frac{Q_0^2}{2C}$.At any time $t$,the total energy is the sum of the energy in the capacitor $(U_C)$ and the energy in the inductor $(U_L)$: $U_{total} = U_C + U_L$.Given that $U_C = 3U_L$,we can substitute this into the energy conservation equation:$U_{total} = 3U_L + U_L = 4U_L$.Substituting the expressions for total energy and inductor energy:$\frac{Q_0^2}{2C} = 4 \left( \frac{1}{2} L i^2 \right) = 2 L i^2$.Solving for the current $i$:$i^2 = \frac{Q_0^2}{4LC} \Rightarrow i = \frac{Q_0}{2\sqrt{LC}}$.

$A$ capacitor of capacitance $C$ has an initial charge $Q_0$ and is connected to an inductor $L$ as shown. At $t = 0$,the switch $S$ is closed. What is the current through the inductor when the energy in the capacitor is three times the energy of the inductor?

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