Effect of frequency of a.c. circuit, resonance and temperature in resistor

[The_Lobster]

In an L-R-C circuit, if the reactance of the capacitor equals that of the inductor, X_L = X_C, and consequently the current is at it's max and we have resonance, will the temperature in a resistor with a positive temperature coefficient increase faster in this situation, and thus get increasing resistance, since more current is going back and forth through the resistor (though less voltage)?

Hope my question made sense:P Sorry for the long sentence...

 

[sophiecentaur]

Yes- it makes sense.
Are you assuming a series resonance with a source and R in series with the whole lot? (Diagrams often help in this sort of question, however crude - the diags not the question, I mean .

 

[The_Lobster]

I'm considering any circuit I think, but would that have any effect (I assume it will since you asked! :) ) I'm basically wondering if, in an a.c. circuit, a measured resistance of a resistor could be affected by temperature increasing in the resistor due to higher frequency, or at resonance frequency since current is largest at this point... (however subtle the effect...)

J:)

 

[sophiecentaur]

You are basically asking about the changes in current. The resistance / temperature thing would be a consequence of the change in current and would modify the response.

First step would be to say that the current will be highest at resonance - when the reactance is zero. That would mean that if the circuit were driven with lots of volts, there would be more power dissipated in the resistor at resonance - so it would get hot. That wouldn't alter the resonant frequency but it would mean that the peak of the resonance response (the bell shape) would be less if the resistance / temperature coefficient were positive. This effect would be progressive as you approach resonance so the normal 1/√(1-x²) shape of curve would not apply but be flattened at the top due to the thermal changes.

You'd need actual values to go any further but it wouldn't be too hard, I think. Any good circuit simulation prog would do it.

 

[The_Lobster]

Thank you so much, sophiecentaur! That makes all the sense in the world! I learned something tonight :)

 

[The_Lobster]

There's only one thing I still don't get: If the impedance is at its lowest at resonance frequency, why doesn't the voltage go down as the current goes up? Effectively negating the effect? Since there is less impedance in the circuit? This way it seems as if the power should stay the same... I know this is not true, but I don't understand properly why.. Does the voltage stay the same?

I think I need a good circuit simulation program!:P

Edit: the voltage would stay the same, unless there are other elements in series with it that the voltage will divide across (voltage divider rule) ? In the latter case, will the power still go up at resonance frequency?

 

[sophiecentaur]

If you are feeding the series RLC from a voltage source, then the voltage will, by definition, be unchanged. If there is significant source resistance, then the RMS voltage will dip near resonance. The effect of source resistance will be less than the dramatic reduction of reactance near resonance for a circuit with a significant Q factor. The source resistance is, of course, just as relevant to the Q as the R you put in the RLC.
This is why I said that a diagram is useful. Where are you actually measuring these quantities in your thought experiment?

PS. I think that too much emphasis is put on simulation programs at the expense of understanding what the analysis is telling us. The algebra of this is pretty simple and it tells you much more than you can get by putting arbitrary actual numbers into someone else's application. If you can possibly manage it, I'd recommend the analytical approach.