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I don't understand how energy is being transferred in a DC circuit. For example here I have a simple circuit powered by a constant voltage. Some form of energy is being transferred from the source to the light bulb and the light bulb lights up.

A constant voltage means there is a constant electric field between the terminals of the light bulb. A constant current means there exists a magnetic field of constant magnitude looping around the wire. If both the electric field and magnetic field is constant, there cannot be EM waves propagating along the wire to deliver energy to the light bulb.

If the energy is delivered by the constant electron drift, something has to move a lot faster than the drift current itself to signal or deliver the energy to the load (otherwise the light bulb wouldn't light up instantaneously). But again, both the E-field and M-field are constant, there cannot be waves propagating.

So what is delivering energy in a DC circuit and how does it work?

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  • $\begingroup$ A voltage difference and current… $\endgroup$
    – Jon Custer
    Jul 5, 2022 at 15:19
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    $\begingroup$ there is a reason why electricity is taught with voltage and current and not with EM waves. "Actually the energy is transferred by waves" is frequently a way to score internet debate points, and rarely a useful way to understand the situation. $\endgroup$
    – user253751
    Jul 5, 2022 at 15:50
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    $\begingroup$ @user253751 Unless, for example, you're trying to track down EMI or suppress ringing in your cables. $\endgroup$
    – John Doty
    Jul 5, 2022 at 16:23
  • $\begingroup$ Related: Why is Power = Voltage × Current?, How does energy flow in a circuit? Which is correct? $\endgroup$
    – Puk
    Jul 5, 2022 at 17:40

2 Answers 2

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E and B are static, from Poyntings theorem the energy that is being used is transfered by the poynting vector

$$\vec{S} =\frac{1}{\mu_0} \vec{E} × \vec{B}$$

$\vec{E},\vec{B} ≠ 0$ and are not parrallel, meaning there must be a poynting vector

Static fields can also have a poynting vector, The difference, is that for static fields in freespace:

$$\nabla \cdot \vec{S} = 0$$

Meaning there is no net energy gain or loss in any region of space, the amount of energy going into a region, equals the amount of energy leaving that region.

Hence energy being transfered, without "seeing" energy move.

Edit: imagine the initial flipping of a switch, the em wave will initially propagate, hitting the charges and do work, its just continual constant work once the wave has passed.

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  • $\begingroup$ The energy balance would rather be: $\vec{\nabla}\vec{S}=-\vec{j}\vec{E}$ The electromagnetic energy is not conserved but is converted into kinetic energy and transferred to the ohmic conductor. $\endgroup$
    – Vincent Fraticelli
    Jul 5, 2022 at 20:11
  • $\begingroup$ Yup! Which is why I said freespace $\endgroup$
    – jensen paull
    Jul 6, 2022 at 5:55
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First, the E field is perpendicular to the wire, not parallel to it. And you should really draw the return conductor, it's a necessary part of the circuit.

You may, if you like, analyze this using electromagnetic waves. The wire, together with the return, supports electromagnetic wave propagation in a TEM mode. This mode has no low frequency cutoff: it can transmit energy at zero frequency, the case here.

Note that there is no electric field around the wire to the right of the light bulb (everything is at ground potential). That portion of the circuit has, in this abstraction, waves propagating in both directions, phased so that the electric field cancels.

Of course, we don't usually model this sort of circuit this way: other approaches are easier.

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