# The inductance of the coil and the capacitance of the capacitor of the ideal oscillatory circuit are 6 mH

**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.