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

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The homework involves calculating the applied voltage in a resonant circuit with a 2mH inductor, a 20-ohm resistor, and a 0.3nF capacitor, where a resonant current of 50mA flows. The voltage is determined to be 1V at this current. When the current is reduced to 30mA, the voltage remains the same, but the frequency changes, leading to confusion in calculating the new frequency and phase difference. The initial attempt to find the new frequency resulted in an incorrect impedance calculation, mistakenly interpreting the impedance value as frequency. Ultimately, the correct frequency is identified as 206kHz after realizing the need to divide by 2π.
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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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