# Transmit DC power with one cord?

My DC devices have two wires connecting them to the power source, but what would be needed to transfer DC power using just one? I depict it as a kind of headphone, which can work with just one wire.

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What do you mean by "one cord" or "two cords"? All electric devices need an input and an output (high voltage and low voltage) because they're all part of a current loop. –  Brandon Enright Dec 13 '13 at 22:29
If you mean one cable, then that's what most devices are powered by –  Jan Dvorak Dec 13 '13 at 22:30
Well, but why can't you transmit electric energy with just one? Imagine it were heat. You could be able to transmit it with just one cord. –  Quora Feans Dec 13 '13 at 22:31
If you mean one wire, then no. The device would be accumulating (or losing) electrons. Even one wire / AC is very inefficient (the environment has to serve as the second wire, but it's not connected in a conductive way) –  Jan Dvorak Dec 13 '13 at 22:32
@QuoraFea Headphones use two (actually three for stereo headphones) conductors. It just looks like one wire because they're bundled up together. –  Brandon Enright Dec 13 '13 at 22:38

You could deliver a short burst, but not continuous power. Not without having a sink of some kind.

In fact the example you gave about the heat also needs a sink of some kind. If you transmitted heat down a rod, the rod would heat up until the thermal gradient would disappear. Then the heat flow would cease.

The exact same thing happens with electricity - the charge moves while there is a gradient, stops when there is none. With no sink or return path the electrons won't flow.

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To build on Jan Dvorak's comment and also user1512321's answer:

If you mean one wire, then no. The device would be accumulating (or losing) electrons. Even one wire / AC is very inefficient (the environment has to serve as the second wire, but it's not connected in a conductive way)

In true DC conditions, the electron flow is wholly analogous to flowing water. If there truly is only one conductive path to whatever appliance is being powered, then, by conservation of charge, this means that charge must be building in the appliance, and this cannot happen indefinitely. The continuity equation $\partial_t \rho + \nabla\cdot\vec{J} = 0$ where $\rho$ is the charge density field and $\vec{J}$ the current density is, in true DC ($\partial_t=0$):

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

which means that the flux lines of current (the integral curves of $\vec{J}$) must be closed loops, hence there is always a return path.

In time varying situations, charge can accumulate for short times and, in particular, it can drain from and gather in places periodically. The rhythmically gathering and draining charge $\partial_t \rho$ gives rise to the generalised idea of conserved current which includes the displacement current (see my answer here).

So Single Wire Earth Return works by a combination of:

1. Pure conduction current, if the system truly is D.C., with the return path being charge flowing through the literal earth, so this is very inefficient if the soil is poorly conductive (although you do have a huge cross sectional area on your side!);

2. Periodical shuttling to and fro of charge and cyclic gathering and draining of charge at each point in an A.C. system. Ground return telegraph is partly through this mechanism. The best way to think of energy transfer is through the Poynting vector: energy transfer happens through the freespace around the conductors by way of propagating electromagnetic waves and the conducting wires let charge shuffle around so that these waves are guided along them.

However, even in 2. closed conductive loops must exist if the energy transfer is by way of transverse electromagnetic waves (see my answer here). Power transmission along one conductor even in A.C. with no conductive return path can happen, but only for higher order waveguide modes. What this means is that the waves always have Fourier components directed at nonzero angles to the conductor, so this method, known as a Goubau line is lossy and inefficient if the waves travel outside the conductor. If we use a hollow waveguide, however, they are bounced along inside the conductor. So hollow conductor waveguides are an A.C. transmission means that can theoretically be 100% efficient (in practice there are losses at the waveguide's surface owing to the skin effect) that truly doesn't use any return conduction path. The charges simply oscillate in back and forth along the conductors surface to guide the wave. If you simulate a pulse along the waveguide and visualise the charge motion as the pulse passes, the charge motion is highly "peristaltic", i.e. there is a good analogy between the charge and the muscles in your gullet as you swallow food.

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