Solve Circuit Resonance Homework: 50mA Current & 30mA Frequency Change

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SUMMARY

The discussion revolves around solving a circuit resonance problem involving a 2mH inductor, a 20-ohm resistor, and a 0.3nF capacitor in series. The applied voltage at a resonant current of 50mA is calculated to be 1V. When the current is reduced to 30mA, the new frequency is determined to be 206kHz, with a phase difference of approximately 53 degrees. The user initially miscalculated the frequency by not dividing the impedance value by 2π, leading to confusion in their solution process.

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Homework Statement


A resonant current of 50mA flows in a circuit consisting of a 2mH inductor, 20 ohm resistor and 0.3nF capacitor all in series. What is the applied voltage? If the current is reduced to 30mA by changing the frequency but not the voltage, find the new frequency and the phase difference between the voltage and current.

Homework Equations



The j term is the imaginary term in all these equations.

Z_r = 20
Z_i = j \omega L
Z_c = \frac{1}{j \omega C}
\omega_0 = \frac{1}{\sqrt{LC}}

The Attempt at a Solution


Summing the impedances then using Ohm's law I find the voltage to be 1V at 50mA.

When the current is reduced, the voltage is still 1V but the frequency changes. What I've tried is to find the new total impedance using Ohm's law, then I rearrange the total impedance equation I used for the first part to make omega the subject, but I end up with a polynomial and am forced to use an equation to find the roots which I shouldn't need to use and I end up with imaginary terms I don't want. I end up with an answer of 1.3 megaohms, whereas the answer is suppose to be 206kHz (and approx 53 degrees phase).

I'd like some help with this please. I don't know whether I'm along the right lines by trying to find an equation for omega in terms of the component impedances.

Cheers.
 
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Never mind, I'm an idiot! I found the value for omega to be 1.3M and assumed this was the frequency, but I didn't divide it by 2 pi which would have given me the correct answer!

Thanks anyway.
 
You're making it harder than it is.
Zl = 2PifL
Zc = 1/2PifC.

Total impedance = 20 Ohms + Zl + Zc.
 

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