Determine the range of capacitance

  • Thread starter Thread starter zeshkani_usa
  • Start date Start date
  • Tags Tags
    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.
zeshkani_usa
Messages
5
Reaction score
0

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)
 
Physics news on Phys.org
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
 
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Thread 'Correct statement about a reservoir with an outlet pipe'
The answer to this question is statements (ii) and (iv) are correct. (i) This is FALSE because the speed of water in the tap is greater than speed at the water surface (ii) I don't even understand this statement. What does the "seal" part have to do with water flowing out? Won't the water still flow out through the tap until the tank is empty whether the reservoir is sealed or not? (iii) In my opinion, this statement would be correct. Increasing the gravitational potential energy of the...
Back
Top