# Modelling the equivalent series resistance's voltage in a circuit

Say a 2H inductor has an equivalent series resistance of 2000 ohm at a particular frequency. Logically to me, that inductor would be modeled as an inductor in series with a resistor. However, that doesn't quite work in my mind. For example, that would imply that the resistor has its own voltage across it, when (I believe) it is the same as the voltage across the inductor itself. If that's true, then it would be a parallel resistance, not a series resistance.

I obviously misunderstand something. How do you model it?

• The inductor has wires, which has resistance. Thus the losses, the resistance part, is spread through the wiring. Another "loss" involves core iron heating. – analogsystemsrf Apr 12 at 13:27
• Keep in mind that when we use a model of something, we generally expect that it is a faithful model only at its interface with the rest of the world. We do not require that everything inside the model be exactly like the real world. – Elliot Alderson Apr 12 at 14:02
• Also, if an inductor has resistance then that resistance is independent of frequency. If you want to say "at a particular frequency" then you should refer to the inductor's impedance or reactance. – Elliot Alderson Apr 12 at 14:13

The resistor does have the same voltage as the inductor. The only way it would have a different voltage accross it would be for the inductor to have a resistive component. In the model of a series inductor and resistor the inductor is ideal (ie only reactive impedance no resistive) therefore the voltage at the resistor is the voltage at the inductor.

The "equivalent series resistance" in the model is not physically localized like a real resistor, so it doesn't have its own voltage drop. The point of including it in the model is that, if you treat the inductor as a black box (i.e., you can only access the terminals, not the internals) it behaves the same way as an ideal inductor in series with a resistor. It makes the circuit analysis easier, but that doesn't mean it's literally true...it just seems that way until you look inside.

Consider an inductor in the real world it is a coil of wire and this wire has resistance distributed along its entire length.

The best model is therefore to consider a tiny inductor in series followed by a tiny resistor and this is repeated infinitely many times such that if you were to add up all the inductances you get the total inductance. Similarly if you add up all the resistances you get the ESR (Equivalent Series Resistance).

You therefore can never measure the inductor voltage or the resistor voltage because there is no point in the circuit to measure it. We can only measure $$\ V_R + V_L \$$.

Mathematically we can model this as a single inductor in series with a single resistor. For almost all cases this is good enough; only if we are trying to model the H and E fields around a component to we need any more detail.

The voltage across the ideal single inductor $$\ V_L \$$ and the ideal single resistor $$\ V_R \$$ will be different and frequency dependant.

We must not confuse this with the AC impedance however $$\ X_L = 2 \cdot \pi \cdot f \cdot L = \omega \cdot L \$$. This is sometimes modelled as a resistor but does not account for the $$\ 90^o \$$ phase shift.