Determine the range of capacitance

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    Capacitance Range
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Homework Help Overview

The problem involves determining the range of capacitance values required for an LC-tuned circuit to effectively pick up FM radio signals, given a specific inductance value and frequency range.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of the formula for resonant frequency in LC circuits and the necessary algebraic manipulations to derive capacitance. There are attempts to calculate capacitance for both low and high frequencies, with some questioning the correctness of earlier algebraic steps.

Discussion Status

There is ongoing discussion about the correct application of formulas and the need to perform calculations for both ends of the frequency range. Some participants have provided alternative expressions for capacitance, indicating a productive exploration of the problem.

Contextual Notes

Participants note the importance of correctly interpreting the relationship between frequency and capacitance, particularly that higher frequencies correspond to lower capacitance values. There is also mention of potential requirements from a homework system regarding the format of the answer.

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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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