Radio Receiver

  • Thread starter purduegirl
  • Start date
  • #1
74
0

Homework Statement



A radio receiver contains an LC circuit whose natural frequency of oscillation can be adjusted, or tuned, to match the frequency of the incoming radio waves. The adjustment is made by means of a variable capacitor. Suppose that the inductance of the circuit is 11.00 μH. What capacitance must the capacitor be adjusted to if the circuit is to span the 540.00 kHz range?


Homework Equations



[tex]\omega[/tex] = [tex]\sqrt{L/C}[/tex]

The Attempt at a Solution



[tex]\omega[/tex] = [tex]\sqrt{L/C}[/tex]
[tex]\omega[/tex] = 540.00 kHz
L = 11.00 microH
C = what we're looking for

I solved for C getting C = [tex]\frac{1}{\omega^2 L}[/tex]
C = [tex]\frac{1}{(540000Hz)^2 * .000011 H}[/tex]
C = [tex]\frac{1}{3.20E6}[/tex]
C = 3.1176E-7 F

Where am I going wrong?
 

Answers and Replies

  • #2
74
0
C = 1/3207600
 
  • #3
marcusl
Science Advisor
Gold Member
2,792
450
You are mixing up frequency with angular frequency. Remember that

[tex]\omega=2\pi f[/tex]
 

Related Threads on Radio Receiver

  • Last Post
Replies
1
Views
8K
Replies
5
Views
3K
Replies
2
Views
369
  • Last Post
Replies
3
Views
3K
Replies
5
Views
8K
  • Last Post
Replies
17
Views
2K
Replies
3
Views
315
  • Last Post
Replies
10
Views
9K
  • Last Post
Replies
6
Views
18K
  • Last Post
Replies
1
Views
2K
Top