Whoa. That's comparing chalk and cheese. The problem with train wiring is usually volt drop, which depends on current and conductor resistance (which is basically the conductor size or awg. A 100W light bulb (normally, in the US) runs off 120V, so the current is less than 1 Amp, which requires a not very fat conductor. A layout at 100W is run off (lets say) 12V which means 10 Amps. So for the same volt drop in the wiring means using a conductor 10x fatter than the mains driven bulb. And it gets worse actually. If we say that the cable for the 100W bulb 'loses' 2 Volts then the bulb will only get 108V - not a big deal. But the assumptions above mean that the layout wiring will also lose 2V, meaning our 12V is now 10V - that is easily enough to notice. So in theory you need even more conductor at lower voltages to keep the volt losses at a small proportion of the system's proper voltage. The correct calculation is to find the max current required, decide on the acceptable volt drop, and then use the cable resistance for the different sizes to find out what the smallest size is that will not exceed the volt drop over the run length. (You may find on the web tables that give the 'resistance' of common cables in the form of V/A/m which is Volts(drop) per Amp per metre. Probably per foot or yard in US. Just be careful as to whether the figure is per metre (foot) of wire or run - run includes the return leg, otherwise you need to double the figure to allow for that.)
The salient point is that more thought needs to go into selecting wire than blindingly following the general misinterpreted axioms. Also, proper power modeling also includes the resistive losses of the wire and the rail.
