In the oscillatory circuit, a capacitor with a capacity of 1 μF and a coil with an inductance of 10 mH.

In the oscillatory circuit, a capacitor with a capacity of 1 μF and a coil with an inductance of 10 mH. The capacitor is charged to a maximum voltage of 200V. Determine the maximum charge of the capacitor and the maximum current.

These tasks: C (the capacity of the capacitor used) = 1 μF = 10 ^ -6 F; L (coil inductance) = 10 mH = 10 ^ -2 H; Um (maximum voltage across the capacitor) = 200 V.

1) Maximum current: C * Um ^ 2/2 = L * Im ^ 2/2, whence we express: Im = √ (C * Um ^ 2 / L) = √ (10 ^ -6 * 200 ^ 2/10 ^ -2) = 2 A.

2) The maximum charge of the capacitor: q * U / 2 = C * Um2 / 2, whence we express: q = C * Um = 10 ^ -6 * 200 = 0.2 * 10 ^ -3 Cl = 0.2 mC.

Answer: The maximum current is 2 A, the capacitor has a maximum charge of 0.2 mC.