8. (15 points) In an ideal circuit consisting of a capacitor with capacitance \( C=1 \) μF and an inductor with inductance \( L_{2}=2 \) mH, undamped free harmonic oscillations of current occur with an amplitude of \( I_{\max }=10 \) mA. At the moment when the current through the inductor \( L_{2} \) is maximal, the switch \( K \) is closed. Determine the maximum voltage across the capacitor after this. The inductance of the inductor \( L_{1}=4 \) mH.

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Answer: $516 \mathrm{mB}$.

Solution. After the key is closed, Kirchhoff's rule for the loop consisting of the coils: $L_{1} I_{1}^{\mid}+L_{2} I_{2}^{\mid}=0$.

We get that $L_{1} I_{1}+L_{2} I_{2}=$ const.

When the voltage across the capacitor is maximum, the current in the circuit segment with the capacitor is zero. Therefore, the currents through the coils are the same.

We get: $L_{2} I_{\max }=\left(L_{1}+L_{2}\right) I$.

We obtain that $I=\frac{L_{2} I_{\max }}{L_{1}+L_{2}}=\frac{2 \cdot 10^{-3} \cdot 10 \cdot 10^{-3}}{6 \cdot 10^{-3}}=\frac{10}{3} \cdot 10^{-3} \mathrm{~A}$.

Conservation of energy $\frac{L_{2} I_{\max }^{2}}{2}=\frac{\left(L_{1}+L_{2}\right) I^{2}}{2}+\frac{C U_{\max }^{2}}{2}$.

We get:

$$
U_{\max }=\sqrt{\frac{L_{2} I_{\max }^{2}}{C}-\frac{\left(L_{1}+L_{2}\right) I^{2}}{C}}=
$$

$\sqrt{\frac{2 \cdot 10^{-3}\left(10 \cdot 10^{-3}\right)^{2}}{1 \cdot 10^{-6}}-\frac{\left(4 \cdot 10^{-3}+2 \cdot 10^{-3}\right)\left(\frac{10}{3} \cdot 10^{-3}\right)^{2}}{1 \cdot 10^{-6}}}=0.516 \mathrm{~V}=516 \mathrm{mV}$.

(3 points)