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LC-цепь

1  ?? Найдите максимальный ток через катушку $I_\max$.

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$q_{30}=0$ 0.20
$q_{10}=q_{20}$ 0.20
$\cfrac{q_{10}}{C_1}+\cfrac{q_{20}}{C_2}=U_0$ 0.20
$q_{10}=q_{20}=\cfrac{C_1C_2}{C_1+C_2}U_0$ 0.20
$$W_{0}=\cfrac{q_{10}^{2}}{2 C_{1}}+\cfrac{q_{20}^{2}}{2 C_{2}}=\cfrac{C_{1} C_{2} U_{0}^{2}}{2\left(C_{1}+C_{2}\right)}$$ 0.20
$q_1=q_2+q_3$ 0.20
$\cfrac{q_2}{C_2}=\cfrac{q_3}{C_3}$ 0.20
$\cfrac{q_1}{C_1}+\cfrac{q_2}{C_2}=U_0$ 0.20
$q_2=\cfrac{C_1C_2}{C_1+C_2+C_3}U_0$ 0.20
$$W=\cfrac{C_{1}\left(C_{2}+C_{3}\right) U_{0}^{2}}{2\left(C_{1}+C_{2}+C_{3}\right)}+\cfrac{L I_{\max }^{2}}{2}$$ 0.20
$A=(q_1-q_{10})U_0$ 0.20
$W_0+A=W$ 0.20
$$I_{\max }=\sqrt{\cfrac{C_{3}}{\left(C_{1}+C_{2}\right)\left(C_{1}+C_{2}+C_{3}\right) L}} C_{1} U_{0}$$ 0.20
2  ?? Определите минимальное напряжение $U_\min$ на конденсаторе $C_2$.

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$U_\min=2U_2-U_{20}$ 0.20
$$U_{\min }=\cfrac{C_{1}\left(C_{1}+C_{2}-C_{3}\right)}{\left(C_{1}+C_{2}\right)\left(C_{1}+C_{2}+C_{3}\right)} U_{0}$$ 0.20