Maximum and minimum values of a variable capacitor

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
2 replies · 5K views
rlc
Messages
128
Reaction score
1

Homework Statement


upload_2015-2-25_11-3-1.png

A parallel circuit like that in the figure forms the tuning circuit for an AM radio. If the inductor has a value of 8.6×10-6 H, what must be the maximum and minimum values of the variable capacitor if the radio receives frequencies from 590k Hz to 1670k Hz?
a) Maximum C?
b) Minimum C?

Homework Equations


w=2*pie*f
f=1/(2*pie*sqrt(LC))

sqrt(LC) = 1 / (2 pi f)
LC = 1 / (2 pi f)^2
C = 1 / [ (2 pi f}^2 L)

The Attempt at a Solution


Is the inductor value 8.6E-6 H the variable f in the equation? Are those equations even the ones I use?
590k Hz= 5.9E5 Hz
How do I begin this problem? I'm just really confused
 
Physics news on Phys.org
rlc said:
Is the inductor value 8.6E-6 H the variable f in the equation? Are those equations even the ones I use?
By convention the variable name used for inductance is L. f is frequency. C is capacitance.
 
  • Like
Likes   Reactions: rlc
C = 1 / [(L)(4pi^2)(f^2)]
Cmax= 1 / [(8.6E-6)(4pi^2)(590E3)^2]
Cmax=8.461E-9 F

Cmin=1 / [(8.6E-6)(4pi^2)(1670E3)^2]
Cmin=1.056E-9 F

*I kept getting this wrong because the equation in the Relevant Equations part had a 2 instead of a 4*