You\'re testing a resonator that is to be used at 107.9 MHz. It doesn\'t pass th

Question

You're testing a resonator that is to be used at 107.9 MHz. It doesn't pass the test, because it actually resonates at 108.2 MHz, which totally sucks. The best way to bring down the resonant frequency is to add parallel capacitance. A careful measurement of the inductance shows that L = 5.914x10-8 H (59.14 nH). (NOTE: Because this circuit is close to resonance and differences are small, you need to use all four significant figures. In actual calculation it might be good to keep at least five significant figures, then round to four at the end.)

a. Find the actual capacitance of the circuit.

b. Find the capacitance that the circuit must have in order to resonate at the proper frequency.

c. How much parallel capacitance must you add in order to bring the circuit into resonance?

d. The dc resistance of the circuit is R = 8.565x10-2[ohms]. Determine the capacitive and inductive reactances, and the impedance, when the circuit is driven at the correct 107.9 MHz.

e. If the circuit is driven with a peak voltage of 300.0V (still at the correct 107.9 MHz), find the peak current in the circuit, and from it the peak voltages across all three elements.

Explanation / Answer

f = w/2pi = (1/2pi)*1/sqrt(LC)

a) C = (1/4*pi^2)*1/L = 1/(4*3.13^2*5.914*10^-8*(108.2*10^6)^2)

C = 36.85*10^-12 F

b) C' = (1/4*pi^2)*1/L = 1/(4*3.13^2*5.914*10^-8*(107.9*10^6)^2)

C' = 37.06*10^-12 F

c) C' = C + C1

C1 = C' - C = 0.21*10^-12 F

d) Xc = 1/WC' = 1/(2*pi*f*C') = 1/(2*3.14*107.9*37.06*10^-12)

Xc = 39.82*10^6 ohms

Xl = wL = (2*3.14*107.9*5.914*10^-8)= 40.073*10^?6ohms

impedance Z = = sqrt(R^2 - (Xl -Xc)^2) = 39.82*10^6 ohms

e) I = V/Z = 7.53 810^-6 A

across C = Vc = I*Xc = 299.84 V

across L Vl = I*XL = 301.74969e?12 V

across R = VR = I*R = 644.9445e?9 V

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