LC circuit and fixed inductance problem

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SUMMARY

The discussion centers on calculating the required capacitance to tune an LC circuit with a fixed inductance of 0.66 μH to a frequency of 8.11 MHz. The user initially applied the formula ω = 1/√(LC) and derived an incorrect capacitance value of 23000 pF. The correct approach involves using the formula F = 1/(2π√(LC)), where ω is defined as ω = 2πF, to accurately determine the capacitance needed for tuning.

PREREQUISITES
  • Understanding of LC circuit theory
  • Familiarity with the relationship between frequency, inductance, and capacitance
  • Knowledge of unit conversions, specifically between Farads and picofarads
  • Basic proficiency in algebraic manipulation of equations
NEXT STEPS
  • Study the derivation and application of the formula F = 1/(2π√(LC))
  • Learn about the significance of resonant frequency in LC circuits
  • Explore practical applications of LC circuits in radio tuning
  • Investigate the effects of varying capacitance on circuit performance
USEFUL FOR

Electronics students, radio frequency engineers, and hobbyists interested in understanding and designing LC circuits for tuning applications.

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A fixed inductance 0.66 [tex]\mu H[/tex] 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 [tex]\omega = 1/ \sqrt LC[/tex]
Solving for C gave me [tex](1/ \omega)^2 /L[/tex]
Then I converted all my numbers.
0.66[tex]\mu H[/tex] = 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.
 
Last edited:

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