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 108.6 MHz, which totally sucks. So it's up to you to fix it. The best way to bring down the resonant frequency is to add parallel capacitance, leaving the inductance alone. A careful measurement of the inductance shows that L= 6.215x10-8 H. Use four significant figues in your calculations.

(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 to the circuit in order to make it resonate at the proper frequency?

Explanation / Answer

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

108.6*10^6 = 1/2*pi * sqrt ( 6.125*10^-8 *C)

solve , c= 3.47*10^-9 -------ANSWER a

b)

107.9*10^6 = 1/2*pi * sqrt ( 6.125*10^-8 *C)

solving , c =c= 3.55*10^-9 -------ANSWER b

C) extra addition of capcitance in paralle = 3.55*10^-9 - 3.47*10^-9 = 0.08 *10^-9 --------Answer c

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