Does resistance felt by the parallel resonant circuit because of electromagnetic induction inside the ferritic material influence the resonant frequency calculation? enter image description here

The circuit is like this, but the coil is transfering power to the ferritic material. This increases resistance of the circuit. Does that resistance also effect the resonant frequency, or do I have to take into account only the resistance of the coil, but not the load itself?


I found this equation for the resonance calculation (which implies that the resistor next to capacitor is infinetely small). But the problem is that when I use that equation, even at 10 Ohms I get that the resonant frequency is almost 0. This can't be right. Is that equation even remotely right?

EDIT: When I test this circuit in a simulator I see that resistance does have an effect on the resonant freuqency. Increasing the resistance of L branch decreases it's current, so I have to reduce the frequency untill both branches have the same current. People say that resistance plays no role in determining the resonance of the RLC circuit, but that is only true for a series RLC circuit. I have also confirmed this with the RLC meter while setting up a resonant circuit. But the problem is that I don't get the same values. If I add the resistor the resonance decreases more in an actual setup, than in the simulation software. But I can find no equation that take resistance into account. The one I provided is wrong, since if I add a 6 Ohm resistor in series with inductor I calculate practicaly 0 Hz resonance, which is not the case. Does anyone have the working equation for it?

edit 2: I have found the solution. In parallel the resonance point is shifted and is calculated by that equation. I get imaginary numbers if I increase the resistance past a certain points because the circuit becomes overdamped so it is not resonating anymore. But the interesting thing is that resonant frequency is not the frequency at which this circuit has the lowest current flowing trough the power source. Nor is it a frequency at which the L and C currents are the same for reasons I don't understand. But if looking for the lowest source current draw is what we're after, than it can for the circuit here

enter image description here

be calculated by

Umax= (x - y)^1/2

x = (a + b)^1/2

a = 1/(LC)^2(1+2*RL/R)

b = (RL/L)^2*(2/(L*C))

y = (RL/L)^2

In my case I don't have the resistor R, only RL, which means that the equation a becomes

a = 1/(L*C)^2

When I simulate at which frequency the generator current is lowest, it exactly matches this frequency.

  • \$\begingroup\$ This increase in impedance is already included in the \$X_L = \omega*L \$ which is exactly what determines the point of resonance \$\endgroup\$ – Joren Vaes Sep 12 '17 at 14:15
  • \$\begingroup\$ What are you calling the "load" in your circuit? \$\endgroup\$ – The Photon Sep 12 '17 at 16:20
  • \$\begingroup\$ I do not understand the first comment. The impedance is determined by R and by XL, not just XL. The load in my example is a ferritic material which has eddy currents induced in it because of changing magnetic field caused by the coil. This is felt on the primary circuit as increased resistance. I am asking, does that increased resistance also need to be included in calculation of the resonant frequency, or do I have to only include the resistance of the coil, but ignore the increase of resistance because of the ferritic material in the presence of the coil? \$\endgroup\$ – MaDrung Sep 13 '17 at 5:50
  • \$\begingroup\$ When I do tests with a rlc meter I see that resistance does in fact change the resonant frequency as it is expected, because resonance accurs when both parallel currents are equal. But when I add resistance I see that the resonance shifts by 20 % instead of falling to 0 as the equation suggests. \$\endgroup\$ – MaDrung Sep 13 '17 at 7:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.