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Ohm's law is often motivated by the microscopic relation \begin{align} \vec{j} = \sigma \vec{E} \end{align} From this you can easily derive \begin{align} U = RI \end{align} , given that \begin{align} U = \int_{\text{along the resistor}} \vec{E} \vec{ds} \end{align}

However, there are different definitions of "voltage", for example the "voltage" $U$ used in the circuit analysis for inductors, \begin{align} U = L \frac{d I}{dt} \end{align} uses $U$ to be the difference in the lorentz-gauge scalar potential, and not the line integral of the electric field (which would be 0 in a conducting inductor).

Hence the question:

  • What is the standard way to deal with this ambiguity of "voltage" when it comes to resistors?

  • Is it common to enhance Ohm's law to account for the additional electric field that a time varying magnetic field would create?

  • Or is it instead common to use the definition of "voltage" being a line integral?

  • Or does one simply not bother, because in the approximation of the lumped element model, resistors are never exposed to time varying magnetic fields / rotational electric fields at all?

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2 Answers 2

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When dealing with circuits one usually uses quasistatic approximation, that is treating the electric and magnetic fields, as if they were static fields, neglecting the radiation terms. In this case the electric field is fully described by its scalar potential and the line integral along the field equals to the potential difference at the end points.

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  • $\begingroup$ That doesn't seem to add up with the example of the inductor, which I mentioned. In an inductor, treating the fields as quasistatic would mean that there isn't any induction happening. Also, in an inductor, the electric field is not described by the scalar potential alone. $\endgroup$
    – Quantumwhisp
    Commented Jun 30, 2020 at 12:32
  • $\begingroup$ @Quantumwhisp in circuit theory one uses a condensed description in terms of resistance, voltage, current, inductance and capacitances. All other phenomena are assumed embedded in these quantities and the equations relating them. It is a bit like Lego - the pieces are made from a real material, but they connect together with simple junctions. $\endgroup$
    – Roger V.
    Commented Jun 30, 2020 at 12:36
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    $\begingroup$ Yep, more or less, it's the final case here - > "in the approximation of the lumped element model, resistors are never exposed to time varying magnetic fields / rotational electric fields at all." As Vadim mentions, if you have time varying magnetic fields / rotational electric fields on your resistor, to use circuit theory, you need to separate out these effects as additional inductance and capacitance elements. $\endgroup$
    – Tom Feng
    Commented Jun 30, 2020 at 13:25
  • $\begingroup$ If you put a closed conducting loop in a varying magnetic field, the induced emf will drive the current through each small element of resistance, but there will no measurable voltage drop between any two points in the loop. $\endgroup$
    – R.W. Bird
    Commented Jun 30, 2020 at 14:40
  • $\begingroup$ @R.W.Bird I agree with you that there won't be a measurable voltage drop in the sense that I use the term "voltage", which is difference in scalar potential. It get's interesting however at the term "measurable": Of course you can measure "something" when you use a voltmeter. If there's no flux change in between the volt meter's path and the path between the two points of interest, you can even measure the line integral of the electric field. So ... can you ellaborate what you mean by "measurable" exactly? $\endgroup$
    – Quantumwhisp
    Commented Jun 30, 2020 at 15:07
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What is the standard way to deal with this ambiguity of "voltage" when it comes to resistors?

When the rates of change of the voltage across and current through are large enough, physical resistors are better represented as an ideal resistor combined with other ideal circuit elements that model non-ideal phenomena due to the changing electric and magnetic fields. For example: Resistors are not resistors

For example, here is a chip resistor model from the Vishay technical note Frequency Response of Film Chip Resistors:

enter image description here

High frequency measurements from 0.1 GHz to 40 GHz were performed on industry standard flip chip thin film resistors from Vishay Thin Film. The results of these measurements are reported in this paper. A lumped circuit model is presented that accurately predicts the response of various part values and case sizes.

Clearly, Ohm's law alone does not give the terminal voltage $v_{R'}(t)$ across the resistor due to a time varying current $i_{R'}(t)$ through:

$$LC\frac{d^2}{dt}v_{R'} + RC\frac{d}{dt}v_{R'} + v_{R'}(t) = Ri_{R'}(t) + L\frac{d}{dt}i_{R'}$$

But, for rates of change small enough, Ohm's law alone is sufficient.

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  • $\begingroup$ Do you know if there are cases where the combinations of inductor, capacitor and resistor (and others...) are used to describe the effect of an "external" time varrying magnetic field (for example one generated by another element in the same circuit) $\endgroup$
    – Quantumwhisp
    Commented Jun 30, 2020 at 15:09
  • $\begingroup$ @Quantumwhisp, are you thinking about, e.g., non-zero coupling (mutual inductance) between the inductor in the resistor model above to other inductors (whether desired or parasitic) in the circuit? $\endgroup$
    – Alfred Centauri
    Commented Jul 1, 2020 at 15:23
  • $\begingroup$ yes, that's exactly what I mean. $\endgroup$
    – Quantumwhisp
    Commented Jul 1, 2020 at 15:46

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