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?
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) =>
LC = 1/(4PI^2 (f^2)) =>
C = 1/(4pi^2(f^2)L) =>
1/(4(9.87)(7.74)(10^8)(0.20))
= 163pF, if I entered it right - check, ok?
show the Fig. 21.15
try 10pf with air isolation
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.