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!

  • \$\begingroup\$ For the parallel resonant circuit, you get resonance when the imaginary part goes to infinity, not to zero. \$\endgroup\$
    – The Photon
    Commented Nov 26, 2018 at 1:29
  • \$\begingroup\$ What does the "l" stand for? It's usually used for length, but I don't suppose that that's the case here.... \$\endgroup\$
    – CoolKoon
    Commented Nov 26, 2018 at 1:33
  • \$\begingroup\$ @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? \$\endgroup\$ Commented Nov 26, 2018 at 1:36
  • \$\begingroup\$ @CoolKoon sorry i meant "L" not "l". I will fix it, thanks \$\endgroup\$ Commented Nov 26, 2018 at 1:37
  • 2
    \$\begingroup\$ The imag. part of your expression for Z is wrong. \$\endgroup\$
    – LvW
    Commented Nov 26, 2018 at 7:56

2 Answers 2


Regarding series versus parallel resonance: only when a complete path exists, for the circulating currents, can resonance occur. In parallel resonance, the L and C are obviously in a tight loop. In series resonance, in some oscillators, the "circulating currents" take a path that included the VDD wiring, and the oscillator will fail **unless* a low-loss capacitor is used in VDD bypassing.


But isn't impedance when the current reaches its peak?

I'm assuming you mean resonance. That statement is true only for series RLC circuits, but is not the case in general.

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.

  • 2
    \$\begingroup\$ Your "definition" of resonance (equal impedances for L and C) is correct for the described circuit only!! The general definition requires that the phase shift between voltage and current is zero (mag. part of the whole impedance to be real). \$\endgroup\$
    – LvW
    Commented Nov 26, 2018 at 7:58
  • \$\begingroup\$ You're right, I've edited my answer to reflect that this is a special case. \$\endgroup\$
    – Sean M
    Commented Nov 26, 2018 at 16:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.