The oscillating circuit has a natural frequency of 60 kHz. How many kilohertz will the frequency change if the inductance is increased 2.25 times and the capacitance is reduced by 2 times?
Initial data: ν1 (natural frequency of the oscillatory circuit) = 60 kHz; L2 = 2.25L1 (change in loop inductance); C2 = 0.5C1 (change in the capacitance of the circuit).
1) Determine the change in the oscillation frequency of the circuit: ν2 = 1 / T = 1 / (2 * Π * √ (L2 * C2)) = 1/2 * Π * √ (2.25L1 * 0.5C1)) = (1 / √ (1.125)) * ν1 = 0.9428ν1.
2) The value of the vibration frequency: ν2 = 0.9428 * 60 * 10 ^ 3 = 56.568 kHz.
3) Change in frequency: Δν = ν1 – ν2 = 60 – 56.568 = 3.432 kHz.
Answer: The oscillation frequency will decrease by 3.432 kHz.
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