A radio inductor.

[Cernie]

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 gotta 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 \omega is Z = \omega L. Ohm's law is V = IZ.

Substituting and rearranging,

\frac{V}{L I} = \omega

Solving for \omega, I get ~188 kHz.

- Warren

[Cernie]

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?

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

Thus:

http://www.google.com/search?hl=en&ie=UTF-8&oe=UTF-8&q=%2812+%2F+%284.6+*+10%5E-3+*+2.2+*+10%5E-3%29%29+%2F+2+pi&btnG=Google+Search

- Warren