Electrical Resonance and Complex Numbers

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SUMMARY

The discussion focuses on determining the resonance angular frequency \(\omega\) in a circuit where the complex impedance \(Z\) is real. The circuit consists of a resistor \(R\) and inductor \(L\) in series, with a capacitor \(C\) in parallel. The total impedance is expressed as \(Z_{tot}=\frac{R-i(\omega C R^2 + \omega^3 L^2 C - \omega L)}{(\omega C R)^2+(\omega^2 L C - 1)^2}\). To find \(\omega\), the imaginary part of \(Z\) must be set to zero, leading to the equation \(\omega C R^2 + \omega^3 L^2 C - \omega L=0\), which simplifies to a quadratic formula for \(\omega\).

PREREQUISITES
  • Understanding of complex impedance in electrical circuits
  • Familiarity with resonance in RLC circuits
  • Knowledge of Euler's relation in complex numbers
  • Ability to manipulate quadratic equations
NEXT STEPS
  • Study the derivation of resonance conditions in RLC circuits
  • Learn how to apply Euler's relation in electrical engineering contexts
  • Explore the implications of complex impedance on circuit behavior
  • Investigate the significance of the phase angle in AC circuits
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Electrical engineering students, circuit designers, and anyone involved in analyzing RLC circuits and resonance phenomena.

thirdchildikari
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Homework Statement



A circuit is said to be in resonance if it's complex impedance Z (in terms of R, L, and C (being the resistance, inductance, and capacitance)) is real. We are to determine the resonance angular frequency \omega in terms of R, L, and C.

Homework Equations



The circuit is setup so that the resistor and inductor are in series, and the capacitor in parallel with them. I've solved for he total impedance as,

Z_{tot}=\frac{R-i(\omega c R^2 + \omega^3 L^2 C - \omega L)}{(\omega C R)^2+(\omega^2 L C - 1)^2}

And this agrees with the answer supplied in the text. I feel that Euler's relation should be playing a role here as well.

The Attempt at a Solution



The question mentions that the circuit would be in resonance if Z is real. I'm thinking that must mean

Z_{real}=\frac{R}{(\omega C R)^2+(\omega^2 L C - 1)^2}

But even if that's true, how would I then find \omega? This real component of Z isn't equal to anything. Using Euler's relation seems promising, but for some reason I just don't know what steps to take here. Do I use Euler's relation and set it equal to the real component of Z?

Also, another question of mine (about a similar, yet different circuit setup) mentions the angle of Z being 45 degrees? That would mean real and imaginary components of Z are equal, but where does that get me? And do I need an angle for my original question.

This stuff is driving me nuts...any help would be appreciated.
 
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thirdchildikari said:

Homework Statement



A circuit is said to be in resonance if it's complex impedance Z (in terms of R, L, and C (being the resistance, inductance, and capacitance)) is real. We are to determine the resonance angular frequency \omega in terms of R, L, and C.

Homework Equations



The circuit is setup so that the resistor and inductor are in series, and the capacitor in parallel with them. I've solved for he total impedance as,

Z_{tot}=\frac{R-i(\omega c R^2 + \omega^3 L^2 C - \omega L)}{(\omega C R)^2+(\omega^2 L C - 1)^2}

And this agrees with the answer supplied in the text. I feel that Euler's relation should be playing a role here as well.

The Attempt at a Solution



The question mentions that the circuit would be in resonance if Z is real. I'm thinking that must mean

Z_{real}=\frac{R}{(\omega C R)^2+(\omega^2 L C - 1)^2}

But even if that's true, how would I then find \omega? This real component of Z isn't equal to anything. Using Euler's relation seems promising, but for some reason I just don't know what steps to take here. Do I use Euler's relation and set it equal to the real component of Z?

Just set the imaginary part of Z equal to zero! In other words set
\omega c R^2 + \omega^3 L^2 C - \omega L=0
You may divide by omega throughout (since we assume omega is not zero itself) and that leaves you with a quadratic formula for omega

Also, another question of mine (about a similar, yet different circuit setup) mentions the angle of Z being 45 degrees? That would mean real and imaginary components of Z are equal, but where does that get me? And do I need an angle for my original question.

This stuff is driving me nuts...any help would be appreciated.
It leaves you with an equation. Just set the imaginary part equal to the real part.

Hope this helps

Patrick
 
nrqed said:
Just set the imaginary part of Z equal to zero! In other words set
\omega c R^2 + \omega^3 L^2 C - \omega L=0
You may divide by omega throughout (since we assume omega is not zero itself) and that leaves you with a quadratic formula for omega

...wow...why did I not see that hours ago?

Thanks a ton...I'm really sort of awestruck at how simple that was and how I didn't see it sooner.
 
thirdchildikari said:
...wow...why did I not see that hours ago?

Thanks a ton...I'm really sort of awestruck at how simple that was and how I didn't see it sooner.

:-) :-)

Sometimes we miss the obvious.
 

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