The inductance in the alternating current circuit is 1 mH, the electrical capacity is 0.4 μF. The oscillation period of the circuit is?

These tasks: L (inductance in the alternating current circuit) = 1 mH (1 * 10-3 H); C (electrical capacity) = 0.4 μF (0.4 * 10-6 F).

The oscillation period of the circuit under consideration is determined by the Thompson formula: T = 2 * Π * √ (L * C).

Calculation: T = 2 * 3.14 * √ (1 * 10-3 * 0.4 * 10-6) = 0.0001256 s (0.1256 ms).

Answer: The circuit under consideration has an oscillation period of 0.1256 ms.



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