8. (15 points) In an ideal circuit consisting of a capacitor with capacitance \( C = 2 \) μF and an inductor with inductance \( L_{2} = 1 \) mH, undamped free harmonic oscillations of current with an amplitude \( I_{\max} = 5 \) mA occur. 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 \( L_{1} = 2 \) mH.

![](https://cdn.mathpix.com/cropped/2024_05_06_96efd573cfdf000c103ag-06.jpg?height=274&width=477&top_left_y=1476&top_left_x=869)
Answer: $90 \mathrm{mB}$.

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

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 $I=\frac{L_{2} I_{\max }}{L_{1}+L_{2}}=\frac{1 \cdot 10^{-3} \cdot 5 \cdot 10^{-3}}{3 \cdot 10^{-3}}=\frac{5}{3} \cdot 10^{-3} \mathrm{~A}$.

Law of 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:

$U_{\max }=\sqrt{\frac{L_{2} I_{\max }^{2}}{C}-\frac{\left(L_{1}+L_{2}\right) I^{2}}{C}}=\sqrt{\frac{1 \cdot 10^{-3}\left(5 \cdot 10^{-3}\right)^{2}}{2 \cdot 10^{-6}}-\frac{\left(2 \cdot 10^{-3}+1 \cdot 10^{-3}\right)\left(\frac{5}{3} \cdot 10^{-3}\right)^{2}}{2 \cdot 10^{-6}}}=0.09 \mathrm{~B}=$

$90 \mathrm{MB}$.

(3 points)

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