# RLC circuits

Discussion in 'General Electronics Chat' started by kerny, Jun 4, 2011.

1. ### kerny Thread Starter New Member

Apr 16, 2011
6
0
I have a questioning about simplifying RLC circuits.

I have a series RLC (resonant) circuit. I rewrote the equations in terms of impedances i.e

Z = R + jwL + 1/jwC = R + jwL -jwC

For the sake of simplicity, lets assume that the resulting equation ended up being Z = 1+ j2. (If w = 1, L = 5, C = 3, R = 1)

Therefore, in my computer simulation, i just included a resistor and an inductor in my model. Since this is what is effectively a RL circuit, how can i calculate the resonant frequency? This doesn't seem to make sense to me since a regular RL circuit does not resonate.

I have included the angles (j), and the magnitude, so why doesn't this simplification work?

Thank you.

2. ### Maged A. Mohamed Member

Apr 6, 2011
18
1
Z = R + jwL + 1/jwC = R + jwL -j/wC
to find the resonance you do NOT substitute for values as assuming (If w = 1, L = 5, C = 3, R = 1) means you stated a single case could be in or out of resonance
To find the resonance you calculate the total impedance vs source frequency , then you find to lowest impedance value
I hope that helps
best regards

3. ### kerny Thread Starter New Member

Apr 16, 2011
6
0
Sorry, I should have been more careful when typing my equation. Thanks for pointing that out.

I also know that the resonance frequency has the lowest impedance.

However, I'm still not sure if my method is correct or not. Can I model a RLC as a RL circuit? My simulation doesn't seem to be agreeing with me, but mathematically, i don't really understand why it wouldn't.

4. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
782
No you can't. The series impedance varies with frequency. Below resonance the circuit is capacitive in nature and inductive above resonance. At resonance the circuit is purely resistive.

As Maged A. Mohamed suggests you might consider what happens to the impedance as the frequency ω varies.

To find resonance, you determine the value of ω where the inductive reactance cancels the capacitive reactance. Where the magnitudes |XC| & |XL| are equal. |XC|=1/(ωC) & |XL|=ωL ........

5. ### Maged A. Mohamed Member

Apr 6, 2011
18
1
Exactly as t n k says, another way to see it, in series resonance, the current is unified i.e. forced by AC source, which will cause voltage to be either +90 or - 90 on each. That also cause Higher voltages on L,C than source
varying the f will increase one's impudence and decrease the other till resonance, they are equal. the total 90 - (-90) degrees result in 180 degree at resonance cancelling the voltage sum on L+C which cause the drop in impudence at resonance

R on the other hand does not the phase shift C provides, that is why it can not be eliminated.

Last edited: Jun 4, 2011