# How do I determine the resonance frequency of this circuit

I have a circuit that consists of a voltage source in series with a resistor and a parallel of an inductor with a capacitor. I have to determine the resonance frequency of this circuit. I am very confused about this. First of all the impedance of the circuit is given by

$$Z=R+j(\frac{\omega L}{1-\omega^2 L C})$$

Now my thought was immediately to proceed as I did in the RLC series circuit i.e. to make the imaginary part of Z zero.

I obtain $$\\omega=0\$$. However the correct answer should be: $$\omega=\frac{1}{\sqrt{LC}}$$

But that would make the imaginary part of the impedance infinity and therefore the whole impedance infinity and the current zero. But isn't impedance when the current reaches its peak?

I'm so confused about all of this. I read that there is series impedance and parallel resonance but what should I peak and why are there 2 types of resonance. Shouldn't they be equivalent? Isn't resonance just the circuit behaving as a resistor?

Can someone explain me what is going on? Thanks!

• For the parallel resonant circuit, you get resonance when the imaginary part goes to infinity, not to zero. – The Photon Nov 26 '18 at 1:29
• What does the "l" stand for? It's usually used for length, but I don't suppose that that's the case here.... – CoolKoon Nov 26 '18 at 1:33
• @The Photon But why should I consider parallel resonance? I have a L//C in series with a resistance? Do you know where I can find an explanation for this? – Granger Obliviate Nov 26 '18 at 1:36
• @CoolKoon sorry i meant "L" not "l". I will fix it, thanks – Granger Obliviate Nov 26 '18 at 1:37
• The imag. part of your expression for Z is wrong. – LvW Nov 26 '18 at 7:56

Resonance frequency (in this case) is when the impedance of the inductor $$\L\$$ and capacitor $$\C\$$ are equal to each other. That should be your starting point. The rest of your assumptions are only true for RLC series and cannot be extended to parallel RLC.