Determine the range of capacitance

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    Capacitance Range
AI Thread Summary
To determine the capacitance range for an LC-tuned circuit with an inductance of 7.60 x 10^-7 H, the frequency formula f = 1/((2π)√(LC)) is used. Converting FM radio frequencies from MHz to Hz, calculations show that the capacitance for the lower frequency of 88 MHz yields a value of approximately 4.303 pF. It is important to perform calculations for both the low and high frequencies to ensure accurate results. The relationship indicates that higher frequencies correspond to lower capacitance values, as the capacitor discharges more quickly. Accurate calculations are essential for tuning the antenna to effectively receive FM signals.
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Homework Statement



FM radio stations use radio waves with frequencies from 88.0 to 108 MHz to broadcast their signals. Assuming that the inductance in an LC-tuned circuit has a value of 7.60 x 10-7 H, determine the range of capacitance values that are needed so the antenna can pick up all the radio waves broadcasted by FM stations. Let your answer to (a) be the smaller of the two values that define the limits of the range, and (b) be the larger of the two.

Homework Equations



f = 1/ ((2pi)sqrt(LC))

The Attempt at a Solution



i used the euqation above converted the MHz to Hz and tried to Solve first for the Low frequency 88 for C but its the wrong answer this is how it looks for C

C = (sqrt(1/f2pi))/(L)
 
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your equation:
C = (sqrt(1/f2pi))/(L) is wrong. You did some algebra errors... to get rid of the sqrt in your first equation, you need to square some things. I'd also do BOTH calculations (for high and low frequency)... just to be sure you get the low and range correct (if you are using a homework system that might ask for the high end and low end separately). High frequency actually corresponds with LOW capacitance -- the capacitor gets filled up and discharges more quickly... the system isn't as sluggish.
 
C = 1/((f^2)(2pi^2)L)
 
C = 1/((8.8e7)^2(2pi)2(7.60e-7))
C = 4.303e-12
 
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