The variable capacitor in the tuner of an AM radio has a capacitance of 2700 pF...?

Question

when the radio is tuned to a station at 570 kHz.
(a) What must be the capacitance for a station at 1600 kHz?
pF
(b) What is the inductance (assumed constant)?
µH

Answer 1

when L is constant the fo (resonance freq varies as sqrt (1/C)

as LC = 1/w^2= 1 /4 pi^2 f0^2
w = 2 pi f0
when tuned at 570 kHz then resonance is struck so

L C1 = 1 / 4*pi^2 (570*10^3)^2 ----(1)

a) at 1600 kHz tuning

L C2 = 1 / 4*pi^2(1600*10^3)^2 ----(2)

divide (1) and (2)
C2/C1= (570/1600)^2 = 0.12691

C2 = C1 * 0.12691
C2 = 2700 pF * 0.12691 =342.667 pico farad
b) L is

L C2 = 1 / 4*pi^2 (1600*10^3)^2

L =1 / 4*pi^2 * 342.667 *10^-12 (1600*10^3)^2

L = 28.90 *10^-6 Henry

L = 28.90 micro Henry

Answer 2

For a resonance circuit,

w^2 = 1/(LC)
where w = 2*pi*f

Substituting C=2700pF and f=570kHz into the equation, you get L=28.88uH.

Substituting L=28.88uH and f=1600kHz into the equation, you get C=342.67pF.