A tuning circuit in an antique radio receiver has a fixed inductance of 0.20 mH and a variable capacitor (Fig.

A tuning circuit in an antique radio receiver has a fixed inductance of 0.20 mH and a variable capacitor (Fig. 21.15). If the circuit is tuned to a radio station broadcasting at 880 kHz on the AM dial, what is the capacitance of the capacitor?

A tuning circuit in an antique radio receiver has a fixed inductance of 0.20 mH and a variable capacitor (Fig.
The coil and cap form a resonant circuit in order to tune the station. The formula is



f = 1/(2PI(LC)^0.5)



Solving for C, we get

(LC)^0.5 = 1/2PI(f) =%26gt;

LC = 1/(4PI^2 (f^2)) =%26gt;

C = 1/(4pi^2(f^2)L) =%26gt;

1/(4(9.87)(7.74)(10^8)(0.20))



= 163pF, if I entered it right - check, ok?
Reply:show the Fig. 21.15



try 10pf with air isolation
Reply:Frequency of resonant circuit (whether series or parallel) is



f = 1 / (2pi x sqrt(LC)) Hz

C = 1 / ((2pi x f)^2 x L) F



Here we have L = 0.2 x 10^(-3) and f = 880 x 10^3, so



C = 1 / ((2pi x 880 x 10^3)^2 x 0.2 x 10^(-3)) F = 0.16355 x 10^(-9) F = 0.164 nF.