Trumpy, the formula you used only applies if the full 120V is across the 1 ohm. The formula you want in this case is P = I^2*R. To make the numbers round, suppose the lamp had a 120 watt light bulb (try to find that one on the shelf somewhere!). Then:
The load current is 120W/120V = 1 amp.
The power dissipated at the bad receptacle is (1 amp)^2 * 1 ohm = 1 watt.
I would certainly expect that a box, whether plastic or steel, could contain that.
I suspect that the resistance of the receptacle was higher than 1 ohm. Still, if you assume that the resistance of a light bulb is a constant (it isn't), and solve for the worst case at the box, you will find that the maximum power that the receptacle could be dissapating with our hypothetical 120 watt light bulb is 30 watts. Which is quite a bit of power in a box. And which brings me back to my question of the relative performance of plastic and steel boxes.
[This message has been edited by SolarPowered (edited 08-24-2006).]
