Question

The variable capacitor in the tuner of an AM radio has a capacitance of 2900 pF when the radio is tuned to a station at 550 kHz.

What must the capacitance be for a station near the other end of the dial, 1610 kHz?

Express your answer to three significant figures and INCLUDE the appropriate units.

C =

Answer 1

here,

when C1 = 2900 pF , f1 = 550 KHz

let the capacitance be C2 when frequency , f2 = 1610 KHz

as the inductance is constant

f1^2 * C1 = f2^2 * C2

550^2 * 2900 = 1610^2 * C2

solving for C2

C2 = 338 pF

the capacitance must be 338 pF