Current Resonance: Explaining I1 Existence After Switch Closure

[Jiggy-Ninja]

Yet another PF contributor is questioning the basic theory on evidence from a simulator. Tut Tut, when will you boys learn?

This would not be a problem if you were actually to build the circuit in analogue components because you would get a 'very high and very sharp' peak as you swept the frequency of your analogue generator and you'd 'believe' the theory.

There is no question that, at resonance, there is infinite impedance. BUT your simulation is dealing with idealised components. You didn't include any series resistance in your resonant loop so the resonance peak is very very sharp (to the limit of the accuracy of your digital calculations). It cannot be surprising when a calculation which is, in effect, subtracting a massive number from another massive number, fails to give a non-zero answer. Your simulation (or snare and delusion, as I refer to them) has, in fact, given you the slightly off-resonance result. btw, it would be interesting to know the phase of the current - which would tell yo which side of resonance the simulator thought we were.

Anyone would think I was a technophobe!

The components might not be ideal "behind the scenes", as I mentioned in my last post. Jst a small winding resistance is enough to produce the measured deflection.

By my eyes, the red wave is slightly lagging the green wave. The question now is what the green wave is measuring.

 

[sophiecentaur]

You would really be better to put a known resistance of reasonable size in series, then calculate the response and the simulator will give you a sensible result.
btw, why do you have to use your "eyes"? The simulator will probably tell you phase value.

Why the question "what is the green wave measuring?" Doesn't it tell you that it's measuring the current through C? Without the L, it's the same as for the R (covered up) and then it gets all that reactive current from L,. I am surprised that there's not a bigger 'magnification factor' at resonance. Why not nudge the frequency or a component value and see whether the simulator can give you more for another condition.

But we're only discussing the nuts and bolts of a simulation, surely. Not the basic theory.

 

[Jiggy-Ninja]

You would really be better to put a known resistance of reasonable size in series, then calculate the response and the simulator will give you a sensible result.
btw, why do you have to use your "eyes"? The simulator will probably tell you phase value.

Why the question "what is the green wave measuring?" Doesn't it tell you that it's measuring the current through C? Without the L, it's the same as for the R (covered up) and then it gets all that reactive current from L,. I am surprised that there's not a bigger 'magnification factor' at resonance. Why not nudge the frequency or a component value and see whether the simulator can give you more for another condition.

But we're only discussing the nuts and bolts of a simulation, surely. Not the basic theory.

I've never used that particular simulator before, so I don't quite know how it works. That why I only speculated that the parts are nonideal, instead of trying to confirm it. And I need to use my eyes since I'm just looking at the picture posted, I don't have the setup to tweak.

Are you getting me confused with BobS?

I don't know what sort of "magnification factor" you are talking about, though. At that frequency, the capacitor has a reactance of about 314 ohms. With 12V peak put across it (seems to be what the SINE generator is set at), that's a peak current of 38 mA, which is what's shown with the green graph. The inductor would have almost the same amount, but 180 degrees out of phase, and cancel out the capacitor current.

 

[sophiecentaur]

By magnification factor, I was referring to the green curve to red curve ratio. At resonance, with no loss in the resonator, there should be no current in the resistor.
Re - the "by eye" thing: I mixed up who had written what, actually. Wake up, that confused man!