All LEDs can be modelled as zener diodes with a colour/substrate specific forward voltage Vf and a series resistive Rs, where they combine both to give the Vf at rated current.
Rs tends to be small so you can neglect it for approximations of adding current limiting series resistors. (see below)
Therefore the current is non-linear and proportional to the voltage difference between the supply and the Vf drop at desired current.
Batteries with low voltage variation are ideal such as Lithium primary cells. Most White flashlights using 3V per LED use these without series resistors as the Li cell is also 3V. However they may be specify a sorted bin of LED's to achieve this.
My Rule of Thumb is to string arrays of LED's such that the voltage difference is ~1V for the current limiting Resistor for a fixed regulator. If the supply range has a wide range, e.g. 10 ~15V then a constant current sink circuit is best.
Additional Info
For more accuracy over a wider range of currents, you can determine the Rs value from the specsheets for a given temperature. The Vf forward voltage also is a function of temperature which affects the results slightly. THe Rs of LED's is much lower than the dynamic Rs of Zeners using silicon junctions.
- 20mA HB devices are <20 ohms.
- 300mA HB devices are < 2 ohms.
- 1Amp power modules are ~ 0.3 ohm.
- Rs for LED arrays , add in series, and divide in parallel.
- Old technology LEDs were much higher Rs values.
- Rs will reduce as the current increases but you can approximate it at the 10% of rated current value and extrapolate if the device stays at constant temp.
- Because of the Shockley effect with voltage variationmyou can actually calculate the junction temperature from the voltage drop of a calibrated LED.
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
10W is the maximum power that the supply can provide. The actual power (and current) will depend on the load connected to the supply.
In your example, the smallest resistor that can be safely connected to the supply is 2.5 ohms, which will result in a current of 2A and power of 10W. If a resistor smaller than that is used, it will attempt to draw more than 2A, with a power greater than 10W. What happens next will depend on the power supply, but the supply's output voltage will fall below the rated 5V, and the supply may overheat or the protection fuse may trip.