In an ideal oscillatory circuit with a natural frequency v1 = 20 kHz, when the capacitor was replaced

| April 21, 2021 | education

In an ideal oscillatory circuit with a natural frequency v1 = 20 kHz, when the capacitor was replaced with another frequency, it became equal to v2 = 30 kHz. What will be the oscillation frequency of the circuit if you connect these two capacitors in parallel?

To find the oscillation frequency of the specified circuit after the parallel connection of capacitors, we apply the formula: ν = 1 / (2 * Π * √ (L * C1 ^ 2)) = 1 / (2 * Π * √ (L * (C1 + C2))) = 1 / (2 * Π * √ (L * (1 / (ν1 ^ 2 * (2Π) ^ 2 * L) + 1 / (ν2 ^ 2 * (2Π) ^ 2 * L)))) = 1 / (2 * Π * √ (1 / (ν1 ^ 2 * (2Π) ^ 2 + 1 / (ν2 ^ 2 * (2Π) ^ 2)))) = 1 / √ (1 / ν1 ^ 2 + 1 / ν2 ^ 2) = 1 / (√ (ν2 ^ 2 + ν1 ^ 2) / (ν1 ^ 2 * ν2 ^ 2))) = ν1 * ν2 / √ (ν2 ^ 2 + ν1 ^ 2).

The values ​​of the variables: ν1 – the oscillation frequency at the initial capacitor (ν1 = 20 kHz); ν2 is the oscillation frequency after replacing the capacitor (ν2 = 30 kHz).

Calculation: ν = ν1 * ν2 / √ (ν2 ^ 2 + ν1 ^ 2) = 20 * 30 / √ (30 ^ 2 + 20 ^ 2) = 16.64 kHz.

Answer: The oscillation frequency of the indicated circuit will be equal to 16.64 kHz.



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