A capacitor of capacitance C has initial char | vedclass

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Question

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

Vedclass · Accepted Answer

$\frac{Q_0}{2 \sqrt{L C}}$

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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: $E_{	ext{total}} = \frac{Q_0^2}{2C}$.Let $E_c$ be the energy in the capacitor and $E_i$ be the energy in the inductor at any time $t$.We are given that $E_c = 3 E_i$.Since $E_{	ext{total}} = E_c + E_i$,we can write $E_{	ext{total}} = 3 E_i + E_i = 4 E_i$.Substituting the total energy: $4 E_i = \frac{Q_0^2}{2C}$.Therefore,$E_i = \frac{Q_0^2}{8C}$.Since the energy in the inductor is given by $E_i = \frac{1}{2} L i^2$,we have $\frac{1}{2} L i^2 = \frac{Q_0^2}{8C}$.Solving for $i$: $i^2 = \frac{Q_0^2}{4LC}$.Taking the square root: $i = \frac{Q_0}{2 \sqrt{LC}}$.

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

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