It's true that P=VxI. But you cant just calculate the amps this way 200 / 7.2 = 27.8 amps.
I cant remember the reason but the right way to do it is like this:
The 200 watt is at 12 volt. So 200/12 = 16,7.
Now we now the volt and the amps and thus we can calculate the resistance: U=R*I, that is 12/16,7=0,72 ohm.
So at 7,2 volt, the amps will be: 7,2/0,72=10 amps.
There will however be som inaccuracy: While the bulb gets hotter, the resistance will decrease. So at a lower voltage, the resistance will be a little higher. Besides, when the pack discharge, the voltage will decrease, thus the amps will also decrease.
Anyway, I think there's too muc hype about these discharge rates. The mathcing companies discharge at high rates, simply because higher numbers sell better.
So average Joes like me, do fine with 10 amps discharging. At practising, just drive the car till the pack dumps. At races, discharging between heats at 10 amps, will do fine.