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
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.