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A real inductor has resistance due to the wire. Should this resistance be modeled as in series or parallel with an ideal inductor? Assume the frequency of interest is so low as to make capacitive and skin effects negligible.

On one hand, it makes sense to model them in series, because if a DC voltage is applied across the real inductor, all current will still have to flow through the resistor (i.e. a short will not occur).

On the other hand, it also seems reasonable to model the LR components in parallel, since both will see the same voltage applied across them.

I would like to verify I am using the correct equations when experimentally determining inductance, and when selecting a capacitor to achieve resonance in an LC tank circuit (since the inductor's resistance will affect the ideal resonance frequency).

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    \$\begingroup\$ Series resistance for your first order equivalent. \$\endgroup\$ – winny Jun 7 '17 at 19:41
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An inductor's resistance is properly modeled as a resistor in series with an ideal inductor (as you said neglecting AC resistance and capacitive elements.)

It may seem like the resistance is in parallel with the inductor, since if you put a multimeter across the inductor you will read the resistance component across the two terminals.

However, if that were the case if you put a voltage step across the inductor you would get an immediate step of current (V/R) plus the ramp of current due to the build up of the flux.

In reality you do not see that behavior, the current starts at zero and ramps linearly until the inductor saturates or the drop on the series resistance becomes significant compared to the applied voltage.

So the resistance is effectively in series with the inductance.

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You use the appropriate model for the circuit.

A real inductor will show excess voltage for the current flowing. That's series R, which is essentially the winding resistance. It will also consume excess current for the voltage across its terminals. That's parallel R, which is essentially a model for the core losses.

In many cases, you can lump all the loss into a single position, particularly if you are using a resonant circuit with a capacitor. Choose the configuration that gives you the easier analysis.

Note both losses are frequency dependent. Don't forget that an inductor also has a stray capacitance, which makes its effective inductance frequency dependent as well.

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