120The idea is that a different load is used with the different voltages. As a simple example:

- on 120V: a 120V/120W lightbulb is 120 ohms (when hot) and draws 1 amp.
- on 240V: a 240V/120w lightbulb is 480 ohms (when hot) and draws 1/2 amp.

In both cases, the load is 120 watts but by doubling the voltage, we get away with 1/2 the current, allowing smaller wires.

Similarly, we could start out with the 240V 1/2A, put that into a 2:1 transformer, and get out 120V 1A for the 120V lightbulb. For the 240V section of the transmission, we can reduce the wire size due to the lower current. Now scale this up to 10KV transmission lines going into 120V/240V residental wiring.

Your estimate is off by several orders of magnitude. Wikipedia gives the resistivity of air as being around \$10^{16}\ \Omega \cdot m\$. I'd guess an actual resistance between two points would be at least on the order of teraohms. Assuming \$1\ T\Omega\$, that gives a current of 5 picoamps, which is far too small to measure easily. As pointed out in an answer to another EE.SE question, the material the battery is made of is probably a better conductor than air.

To actually figure out what's going on in extreme situations, you need a more detailed model of the materials involved. How many electrons and/or ions are available for conduction? An ideal dielectric (insulator) has no free electrons, but a real dielectric might. What's the strength of the electric field? If you have a 40 kilovolt voltage source, you can rip apart air molecules, creating lots of free electrons! A less extreme example would be a vacuum tube, which "conducts" through empty space \$(R = \infty)\$ using electrons liberated from a piece of metal.

Ohm's law is an approximation that works for many materials at low voltages, frequencies, and temperatures. But it is far from a complete description of electrodynamics and physical chemistry, and should not be treated as such.

To answer your question more directly, regardless of whether a tiny current flows through the air, there can definitely be a voltage between the terminals. Voltage is another way of describing the electric field. Wherever there is an electric field, there is a voltage difference, even in a vacuum with no matter at all! HyperPhysics shows what this looks like.

Specifically, the *gradient* of the voltage field gives you the magnitude and direction of the electric field:

$$\vec E = -\nabla V$$

I don't know whether a tiny current actually flows through the air, but hopefully now you have a better appreciation for the physics of the situation. :-)

## Best Answer

The best reason I've heard is to avoid this: -

V = 2 V (which of course is meant to say "voltage = 2 volts")

U = 2 V sounds more sensible after all we use a different symbol for current (I) and also amps. Voltage is a bit on its own - we wouldn't say "amps = 2 amps" or "current = 2 currents".

It seems to me this is the sensible reason for choosing U over V but having said that I never use "U"! Maybe I should?