In electrostatics, we have $$\nabla \times \vec{E} = 0$$. Hence, we can define a scalar potential $V$, where $$\vec{E} = -\nabla V$$. We know from Faraday's law that $$\nabla \times \vec{E} = -\frac{\partial \vec{B}}{\partial t}$$. In situations involving constant currents(D.C.) the rate of change of the magnetic field is zero, so we can continue to use the concept of scalar potential, also known as voltage in circuit theory, for analysis of DC circuits. However,in practice, in the literature, we find the concept of voltage being used to study alternating current circuits as well. Further, it is also used in the study of high frequency alternating current circuits used for generation of electromagnetic waves. In such situations $$ \oint \vec{E}.\vec{dl} \neq 0$$. So strictly speaking, we cannot use the concept of voltage to study alternating current circuits, especially in the high frequency applications. Yet, in practice we find that it continues to be used in the literature. So is it really justified to use this concept outside of its realm of validity, and why or in what way?


3 Answers 3


Yes, it is possible to use the concept of voltage and related tools in AC circuits. The fact that the contour integral of $\vec E$ isn't zero isn't a problem. It would only be a problem if you first, "gradient" equation were right in this context. However, it's not. The correct equation needed for more general electromagnetic setups is $$ \vec E = -\nabla \Phi - \frac{\partial \vec A}{\partial t} $$ The identity for $\vec E$ implied by this definition isn't simply ${\rm curl}\,\vec E=0$ but the full Faraday's law Maxwell's equation you wrote down $$ \nabla \times \vec{E} = -\frac{\partial \vec{B}}{\partial t} $$ because $\vec B = \nabla\times \vec A$. The equations above are fully consistent with one another and compatible with the fact that $\Phi$, the vector potential, is a well-defined field even in general situations involving time-dependent fields and magnetic fields.

There is an ambiguity in the choice of $\Phi,\vec A$, the gauge invariance, but useful conventions may be assumed to fix this ambiguity in the case of AC circuits. When it's done, the non-magnetic parts of the circuits work just like before or in DC circuits. The voltage may be calculated as the differences of $\Phi$ just like in the DC case.

For coils, one has to appreciate that the integrated electric field isn't the only relevant contribution. Instead, one finds the electromotive force, EMF, a modified sibling of "ordinary" voltage which has to be added to coils (and batteries) for the total sums of voltages over closed loops to vanish.

One may also study harmonic time-dependence in which all quantities depend on time as $\cos\omega t$ which is usually complexified to $\exp(i\omega t)$.

  • $\begingroup$ hi, you mentioned that 'useful conventions may be assumed to fix this ambiguity in the case of AC circuits. When it's done, the non-magnetic parts of the circuits work just like before or in DC circuits.'. Can you specify exactly what these conventions are? In other words, are you suggesting choice of a specific gauge, and if so, then which particular gauge are you suggesting? $\endgroup$
    – guru
    Commented Aug 8, 2013 at 10:50
  • $\begingroup$ Yes, @guru. Vasiliy's answer below sketches some of the logic. For ordinary circuit, the pieces are separated, so the magnetic field is confined to the part of the circuits with coils (except for the negligible electromagnetic waves that are being emitted). So the gauge choice has to agree with the $\vec A=0$ outside the vicinities of the coils. This doesn't really fix the gauge completely and it's completely unfixed inside the coils but it doesn't really matter because the voltage of the coils is treated separately by the magnetic flux, a quantity that is gauge-invariant and gauge-insensitive $\endgroup$ Commented Aug 9, 2013 at 11:41
  • $\begingroup$ Hi @LubosMotl, if $\vec{E} = -\nabla V$ we know that any two potentials can atmost differ by a constant. Hence, potential difference is an absolute and hence the voltage specification in any D.C. power suppy makes sense. However, the situation is more complicated for changing magnetic fields. Now there is no guarantee that any two potentials will differ by atmost a constant because $\vec{E} = -\nabla V-\frac{\partial \vec{A}}{\partial t}$ So any gauge transformation $$\vec{A} ' = \vec{A} + \nabla \psi$$ and $$\phi' = \phi - \frac{\partial \psi}{\partial t}$$ is valid. $\endgroup$
    – guru
    Commented Aug 10, 2013 at 14:04
  • $\begingroup$ So this means that any two potentials need not differ by just a constant so when you say that a particular power supply is say 100V A.C., then shouldn't you also mention which gauge? $\endgroup$
    – guru
    Commented Aug 10, 2013 at 14:07
  • $\begingroup$ If not, then the choice of gauge for the potential would be different in different parts of the circuit which would lead to a complicated situation. $\endgroup$
    – guru
    Commented Aug 10, 2013 at 14:16

In the framework of EE theory there is widely accepted assumption of "lumped circuit".

Lumped circuit is the circuit where all the elements are "infinitesimal" (size less), and the whole circuit is "infinitesimal" too (zero area). The above means that even in presence of time varying magnetic field, the magnetic flux through circuit elements and through the area of the circuit itself is assumed to be zero.

Just like any other physical model, the lumped model has its limitations too. The rule of thumb is that lumped model breaks when the wavelengths involved are comparable to circuit's dimensions (or shorter). However, even in these high frequency cases, the lumped model may still be applied if the high frequency effects are accounted for by adding to the circuit additional components (transmission lines, parasitic inductances and capacitances, etc...).

When the lumped model may not be applied (due to very high frequencies involved), the simple notations of EE theory may not be used anymore. In these cases one must build a physical model of the circuit and solve Maxwell's equations. However, to my best knowledge, these cases are treated by physics doctors in a specialized facilities, not by electrical engineers.

  • $\begingroup$ This doesn't answer the question, which asked about the applicability of voltage in circuits where there's changing magnetic fields $\endgroup$ Commented Aug 8, 2013 at 10:51
  • 1
    $\begingroup$ @LarryHarson, If you assume zero magnetic flux through all circuit's components, why do you care about changing magnetic fields? $\endgroup$
    – Vasiliy
    Commented Aug 8, 2013 at 11:32
  • $\begingroup$ All circuits possess large finite areas with a changing magnetic field through it. Don't forget the interconnecting wires between components. $\endgroup$ Commented Aug 8, 2013 at 11:45
  • $\begingroup$ @LarryHarson, I don't understand what you're trying to say. Do you? $\endgroup$
    – Vasiliy
    Commented Aug 8, 2013 at 11:49
  • $\begingroup$ You need to link the components with interconnecting wires, so you'll end up with a circuit loop. And as the current in this loop changes, so will the magnetic flux in the loop. $\endgroup$ Commented Aug 8, 2013 at 12:10

we cannot use the concept of voltage to study alternating current circuits, especially in the high frequency applications.

This is a common misconception, usually on part of physics teachers and students, electrical engineers seem to get this right more often.

What is going on here is that voltage, in its basic meaning used in AC/DC circuits and Kirchhoff's voltage law, is difference of the Coulomb potential, or equivalently, line integral of the Coulomb electric field from one point to another. The Coulomb field is a function of all charges magnitudes and positions' and is conservative; any line integral of it over closed curve is zero. Thus the integral from one point to another does not depend on path, only on the endpoints. All this is true regardless of what total electric field looks like; voltage only cares about the Coulomb field.

When you see electric field written as

$$ \mathbf E = -\nabla \Phi -\partial_t \mathbf A $$

this does not by itself fix $\Phi$, but in practice, in working with AC/DC circuits where radiation is negligible, the most natural way to fix $\Phi$ is to fix it to the electrostatic definition

$$ \Phi(\mathbf x, t) = \int K\frac{\rho(\mathbf x',t)}{|\mathbf x - \mathbf x'|}d^3\mathbf x'. $$

Then, $-\nabla \Phi$ is the Coulomb field, and $-\partial_t \mathbf A$ is the rest of the field; in DC circuits this contribution is zero, but in AC circuits, it is not, and we call it induced electric field.


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