Solve Radio Inductor Problem: Find Frequency for 12V, 2.2mA, 4.6mH

[Cernie]

A radio inductor. Please help!

Hello.
I am seeking immediate help with a problem I have. Here is the problem:

A Radio Inductor. You want the current amplitude through a inductor with an inductance of 4.60 mH (part of the circuitry for a radio receiver) to be 2.20 mA when a sinusoidal voltage with an amplitude of 12.0 V is applied across the inductor.
What frequency is required?

What I tried to do was use this equation:
w = 1/(square root of(LC)) and combining it with w = I/Q which is also w0 = I/(CV), and then I would have C which I could use to find w. But that's got to be the wrong way. I've tried some other ways which I think would make more sense, but it just doesn't seem to come out right. Is there any chance you can help me with this?
Thank you,
Cernie.

 

[HallsofIvy]

I don't see how we can do this without knowing the frequency of the applied 12 volt current.

 

[chroot]

The (real) impedance of an inductor of inductance L at frequency [itex]\omega[/itex] is [itex]Z = \omega L[/itex]. Ohm's law is V = IZ.

Substituting and rearranging,

[tex]\frac{V}{L I} = \omega[/tex]

Solving for [itex]\omega[/itex], I get ~188 kHz.

- Warren

 

[Cernie]

How?

How do you get that answer?
I've already tried this method and I always get the same answer, w = 1185771 Hz = 1.19MHz.

 

[chroot]

Cernie,

Perhaps you're making some arthmetic mistake?

[tex]\frac{V}{L I} = \frac{12}{4.6 \cdot 10^{-3} \cdot 2.2 \cdot 10^{-3}} = \omega = 2 \pi f[/tex]

Thus:

http://www.google.com/search?hl=en&...^-3+*+2.2+*+10^-3))+/+2+pi&btnG=Google+Search

- Warren