Question

Колебательный контур состоит из конденсатора ёмкостью 1,2 мкФ и катушки индуктивностью 42 мГн. Определите период и частоту свободных электромагнитных колебаний в этом контуре.

Answer 1

....................................................

Answer 2

Calculating the Period and Frequency of Free Electromagnetic Oscillations

The period and frequency of free electromagnetic oscillations in a circuit can be determined using the formula:

Period (T) = 2π√(LC) Frequency (f) = 1/T

Where: - L = inductance of the coil in henries (H) - C = capacitance of the capacitor in farads (F)

Given: - Capacitance (C) = 1.2 µF = 1.2 × 10^-6 F - Inductance (L) = 42 mH = 42 × 10^-3 H

Calculation

Using the given values of capacitance and inductance, we can calculate the period and frequency of the free electromagnetic oscillations in the circuit.

Period (T) = 2π√(LC) Frequency (f) = 1/T

Substituting the given values: T = 2π√(1.2 × 10^-6 × 42 × 10^-3) f = 1/(2π√(1.2 × 10^-6 × 42 × 10^-3))

Result

After calculating the above expressions, the period and frequency of the free electromagnetic oscillations in the given circuit are as follows:

Period (T) ≈ 0.0029 seconds Frequency (f) ≈ 344 Hz

Therefore, the period of the free electromagnetic oscillations in the circuit is approximately 0.0029 seconds, and the frequency is approximately 344 Hz.