# LC circuit and fixed inductance problem

1. Apr 24, 2006

### Punchlinegirl

A fixed inductance 0.66 $$\mu H$$ is used in series with a variable capacitor in the tuning section of a radio. What capacitance tunes the circuit into the signal from a station broadcasting at 8.11 MHz? Answer in units of pF.

I used the equation $$\omega = 1/ \sqrt LC$$
Solving for C gave me $$(1/ \omega)^2 /L$$
Then I converted all my numbers.
0.66$$\mu H$$ = 6.6e-7 H
8.11 MHz= 8110000 Hz
So plugging in, (1/8110000)^2 / 6.6e-7
gave me 2.30e-8 F
then converting to pF, gave me 23000, which isn't right.. can someone tell me what I'm doing wrong?

2. Apr 24, 2006

### Hammie

Might be easier to use

F = 1/(2PI * Sqrt (LC))

then solve for C.

omega = 2 PI * F.

Last edited: Apr 24, 2006