Cable Current Carrying capacity for Pulses Power

I think that, generally speaking, the issue is one of heating of the wire. You can use your duty cycle to obtain the average current and choose your wire accordingly.

However - you do need to be a little careful. As an example, I was using a resistor in a circuit. I had a voltage spike of 100 volts with a duty cycle of about 5%. The average power was very small < 1/8 watt. A standard resistor worked fine, but when I replaced it with a surface mount resistor of the same average power rating, the pulse vaporized the resistor film. Even though the power dissipated was still < 1/8 watt, the peak power dissipated during the pulse was 20 times that, and the thin film of the surface mount resistor couldn't dissipate the heat.

So - very thin wires with very large pulses of very small duty cycle is something to watch out for. But, for reasonable pulses and duty cycles with fairly substantial wire - wire that can dissipate the instantaneous heat, you should be OK.

Notice that I put a lot of qualifiers in there. You can start with the current carrying capacity of wire:

http://www.powerstream.com/Wire_Size.htm

and go from there. I would be comfortable with an average current of less than half of the wire's rating, and a peak current of less than 5 times the wire's rating.

You will want to test, of course. Run your max current through the wire and measure its temperature rise. If it takes several seconds (10, say) for the wire to heat up appreciably (get hot to the touch), then that wire is OK to use , as long as your average current is within its limits.

You have to consider that the power goes up as the square of the current.

Consider a length of wire with resistance 1Ω. With a 1A continuous current, the power dissipated = 1²/1 = 1W.

Try 2A for ¹/₂ the time (50% duty cycle). Average power = 2²/2 = 2W.

Try 100A at 1% duty cycle. Average power = 100²/100 = 100W!

If it can't cool down during the OFF period, you can get into serious trouble.

ORLY?

You mean peak power, not average, right?

2A, 1Ω, 50% duty cycle. Peak power = I² x R = 4W. Average = 2W