# How to find the frequency of the AC source in RL circuit?

I' m asked to find the frequency of my source with the given data (RL-serial circuit):

$$(R_L=3 \Omega;L=0.04H), R=30\Omega, cos(\phi)=0,819 (ind), X_L>X_C$$

This looks easy but somehow I cannot get the result offered in the result section: The only thing that I did is this:

• $$R_L=\omega L=2\pi fL \Rightarrow f=\frac{R_L}{2\pi}\frac{1}{L}= 11.93 Hz$$
• $$cos\phi=\frac{R}{Z}\Rightarrow Z=\frac{R}{cos\phi}= 36.63\Omega$$ I also don't understand how I get different results for impedance (Z) when I find it through the usual formula: $$Z=\sqrt{R^2 + R_L^2}=30.14\Omega; Z\neq Z ?$$

Why isn't the frequency which I got from the first equation correct? What am I doing wrong?

I also referred here but could't find anything relevant: http://www.electronics-tutorials.ws/accircuits/series-resonance.html Can I at least get a hint? I've been starring at this for 2 hours and can't find out the correct way to solve it.

Original question: EDIT I'm given $$R_L \quad not\quad X_L$$

• Something seems to be missing. Where does Xc come from? – Decapod Sep 14 '16 at 9:37
• Xc is not given, only Xl, I'll now post the original question it is in croatian but you will see the data... – Eugen Sunic Sep 14 '16 at 9:45
• What is RL? Is it an extra load resistor? Because given that data RL is not the inductive reactance. Things dont add up. – crowie Sep 14 '16 at 10:34
• Actually it must be the series resistance of the inductor so total resistance of the circuit would be 33ohms plus the XL.. – crowie Sep 14 '16 at 10:37

## 1 Answer

Finally I found the answer: $$R_f=R_L+R=33\Omega$$ $$\phi=arcos(0.819)=35.01$$ $$tg\phi=\frac{X_L}{R_f}\Rightarrow X_L=tg\phi* R_f=23.11$$ $$X_L=\omega L=2\pi fL \Rightarrow f=\frac{X_L}{2*\pi}\frac{1}{L}=\frac{23.11}{2\pi 0.04}= 91.95 Hz$$

Correct answer is e) 92Hz

• What is tg? Because I was going to say $$\phi = arctan(X/R)$$ – crowie Sep 14 '16 at 10:43
• Tangent as tg... – Eugen Sunic Sep 14 '16 at 10:45