If possible, find reasons why the resistance of a resistor may have a different value when measured with a sine wave signal (a.c.) than when measured with a d.c. (dirrect current) source.
The Attempt at a Solution
Please correct me if anything of what I've said below could be wrong in some way:
In an AC circuit, the sinusoidal current through a resistor is given by: [tex]i = Isin \omega t[/tex]. From Ohm's law, the instantaneous potential [tex]v_R[/tex] is: [tex]v_R = iR = (IR)sin \omega t = V_R sin \omega t[/tex].
What I don't get here, is that, according to the equation [tex]v_R = iR[/tex], the relationship between the instantaneous voltage and current is constant (giving R), but if we graph this, since the two are in phase, both the current and the voltage across the resistor will be 0 once every cycle. Doesn't that mean that the resistance, R, of the resistor will also vary? In this illustration: http://www.electronics-tutorials.ws/resistor/res32.gif , we also see that the voltage has steeper curves than the current, and they're not proportional, so [tex]v_R = iR[/tex] isn't true? I find this really confusing...
If we only look at the peak voltages, the resistance would be the peak of the voltage divided by the peak of the current.. it's only when we're talking about instantaneous current and voltage I'm getting confused...
Aside from what I've written above, whenever the frequency of the changing current in a circuit is any other than 0 (DC), any circuit element will exhibit some combination of resistive, capacitive and inductive behaviour. May this mean that we might see a slight drop in resistance measured on the resistor because of the "imagined" inductor in series with it?
Lastly, I read about something called the "skin effect" and the proximity effect that might effect circuit elements when there's an alternating current. Is this something different than the above? Or is it simply another way of saying that each element exhibits some inductive, capaticive, etc. behaviour?