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Is impedance and resistance the same thing in a simple resistor?

Thinking of just this case using an AC current with a frequency of 1k Hz, thru a simple non-inductive resistor.

Is the impedance of the resistor the same thing as the resistance measured using DC?  Is it equal to the resistance in value or different altogether? (sentence edited to change inductance to impedance)

I think it's the same thing and so equal in value.  Or maybe I've still got a bunch of reading to do.

gmoon1 year ago
You're mixing terms a little bit here. Inductance isn't impedance.

Impedance is basically "resistance to AC", so it takes into account other effects like inductance and capacitance. And impedance changes with frequency (and is dynamic, not static, when an AC current is applied).

Resistors (assuming they aren't wire wound) have nether inductance nor capacitance, so their DC and AC resistance is the same. I.E., the resistance equals the impedance--for the resistor alone.
Re-design (author)  gmoon1 year ago
gmoon answered first but all great answers and I'm satisfied.
Re-design (author)  gmoon1 year ago
You caught my mistake. I did not mean to use Inductance in the second sentence, I meant to use impedance. It was early in my morning when I typed this and had not had my caffeine yet.
Been there... ;-)
Yes. I mean that's the quick answer.  I am assuming the resistor you're writing about is the ideal kind, not the real kind.

The impedance of an ideal resistor is:
XR = R, where R is resistance in ohms (Ω)

The impedance of an ideal inductor is:
XL = j*ω*L, where ω is angular frequency in rad/s and ω=2*π*f, L is self-inductance in henrys (H)

The impedance of an ideal capacitor is:
XC = (1/(j*ω*C)), where ω is angular frequency in rad/s and ω=2*π*f, C is capacitance in farads (F)

I also claim that impedors "add" the same way that resistors do.  For example if you've got X1 and X2 in series, the total impedance is just their sum:
X1and2series = X1+X2. 

If X1 and X2 are in parallel, it is:
X1and2parallel = (X1^-1 + X2^-1)^-1

The only thing that makes it more complicated is all those j*omegas, which in general will make the answer a complex number, and dependent on omega.

Some easy examples:

Q:What's the impedance of an ideal 100 Ω resistor, at f=60 Hz.
A:  (100.00 + 0*j) Ω = 100 Ω

Q:What's the total impedance of an ideal 100 Ω resistor in series with an ideal 100 μF capacitor, again at f=60 Hz.
A:  (100.00 - 26.53*j) Ω
(Since XR= 100 , and XC= 1/(j*2*pi*60*100e-6), and the total impedance is XR+XC)

mrmerino1 year ago
Ideally, a perfect resistor resists DC and AC with the same impedance. Realistically, every electronic element has some kind of parasitic capacitance and/or inductance that would affect AC differently than DC. That is usually so small that it only really becomes a problem when you have like really really high frequencies (megahertz and such).

For one kilohertz, I wouldn't worry about that. The AC impedance of the resistor will be so close to the DC resistance that you might as well call them the same.
If you have a specifically NON-inductive resistor, as you said, then the DC resistance dominates any other effects, up to some frequency limit.

You can't measure the inductance of a non-inductive resistor, because it doesn't have any, by definition...