The inductance of the coil and the capacitance of the capacitor of the ideal oscillatory circuit are 6 mH and 4 nF, respectively. The amplitude of charge oscillations on the capacitor plates is 12 nC. Find the amplitude of the current fluctuations in the circuit.
Let’s use the formula that relates the charge to the voltage across the capacitor:
q = CU, where q is the charge, C is the capacitance of the capacitor, U is the voltage, then:
U = q / C.
The amplitude of voltage fluctuations will be:
4 * 10 ^ (- 9) / 12 * 10 ^ (- 9) = 1/3 V.
Since the circuit is ideal, that is, there are no losses, then the energy stored on the capacitor is completely transferred to the coil, we get the equality:
CU ^ 2/2 = LI ^ 2/2;
I ^ 2 = CU ^ 2 / L;
I = U * √ (C / L).
I = 1/3 * √4 * 10 ^ -9 / 6 * 10 ^ (- 3) = 1/3 * √2 / 3 * 10 ^ -3 = 0.27 mA.
Answer: 0.27 mA.
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