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

  • Thread starter Thread starter Cernie
  • Start date Start date
  • Tags Tags
    Inductor Radio
AI Thread Summary
To determine the frequency required for a radio inductor with an inductance of 4.6 mH to allow a current of 2.2 mA when a 12V sinusoidal voltage is applied, the impedance formula Z = ωL and Ohm's law V = IZ are used. The calculation leads to the angular frequency ω being approximately 188 kHz. However, another participant, Cernie, reported consistently calculating a frequency of 1.19 MHz, suggesting a potential arithmetic error in their approach. The correct formula for frequency involves substituting values into ω = V/(LI), which can be further converted to frequency using f = ω/(2π). The discussion emphasizes the importance of careful arithmetic in solving such electrical engineering problems.
Cernie
Messages
2
Reaction score
0
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.
 
Last edited:
Physics news on Phys.org
I don't see how we can do this without knowing the frequency of the applied 12 volt current.
 
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
 
Last edited:
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.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top