0

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!

StainlessSteelRat
  • 7,996
  • 2
  • 18
  • 34
Granger Obliviate
  • 1,403
  • 11
  • 30
  • For the parallel resonant circuit, you get resonance when the imaginary part goes to infinity, not to zero. – The Photon Nov 26 '18 at 01:29
  • 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 01:33
  • @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 01:36
  • @CoolKoon sorry i meant "L" not "l". I will fix it, thanks – Granger Obliviate Nov 26 '18 at 01:37
  • @GrangerObliviate It's parallel resonance because the LC is in parallel. The resistor is irrelevant in this case (you can think of the actual parallel R as infinity, or open circuit). – Sean M Nov 26 '18 at 01:46
  • @SeanM and what if I had multiple parallel and series connections between inductors and capacitors? Should I consider what type of resonance? Also the resistor is in series with L//C, not in parallel. – Granger Obliviate Nov 26 '18 at 02:04
  • @GrangerObliviate The R in series with the L//C does not make it a series RLC circuit, it's just a resistor in series with a parallel LC. Regardless of multiple resistors in series/parallel, the formula for resonance remains the same. The rest of your assumptions cannot be extended to the general case. – Sean M Nov 26 '18 at 02:09
  • Please edit your question to include a schematic. Edit the question, then click on the schematic tool (a button with schematic symbols on it) and then draw your circuit. – user57037 Nov 26 '18 at 02:33
  • 2
    The imag. part of your expression for Z is wrong. – LvW Nov 26 '18 at 07:56
  • You've not worked out the characteristic equation, so far as I can tell. Also, the values of R, L, and C in your arrangement (as I gather it) can be critically damped, under-damped, or over-damped. In the case of under-damped there will be a damped frequency, as well. But in all cases $\omega_0=\frac{1}{\sqrt{L:C}}$ and $\alpha=\frac{1}{2:R:C}$. – jonk Nov 26 '18 at 08:00

2 Answers2

1

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.

analogsystemsrf
  • 34,253
  • 2
  • 19
  • 48
0

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.

Sean M
  • 136
  • 8
  • 2
    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). – LvW Nov 26 '18 at 07:58
  • You're right, I've edited my answer to reflect that this is a special case. – Sean M Nov 26 '18 at 16:41