LC circuit and fixed inductance problem

AI Thread Summary
A fixed inductance of 0.66 μH is used with a variable capacitor to tune into a radio station broadcasting at 8.11 MHz. The initial calculations using the formula ω = 1/√(LC) led to an incorrect capacitance value of 23000 pF. The user suggests using the formula F = 1/(2π√(LC)) for a more accurate solution. They also note that ω can be calculated as 2πF. The discussion seeks clarification on the correct approach to determine the capacitance in picofarads (pF).
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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?
 
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Might be easier to use

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

then solve for C.

omega = 2 PI * F.
 
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