250 watt HQI ballast.

[ron101]

ron101 wrote:
Overall power draw on the circuit is voltage x amperage; which should remain the same for both a 110 and a 220v setup. However power lost to heat is [correction : resistance x current squared] .[/i]


Ron, you realize thoes are both formula to figure out the same thing.. the second one assumes you don't know the voltage, and the first one assumes you don't know the resistance of the load.

so both will give you the total power consumed by a load and if you know all the factors and do it both ways you should come up with a number that is very close to the other.

Steve

Yes and no, it depends on which context in which they are used. If you are evaluating systems where the power output is only heat (ie space heater) and at a constant voltage then that formula is simply interchangeable with the others.

However if you are evaluating a system where not all the energy put into the system is converted to heat (ie lighting system or electric car) then current x voltage is the sum of I^2 x line resistance + useable power output + non-thermal losses .

This is why it is more efficient to transfer power at high voltage/low current then the other way around.

Anyways the answer to my original question depends on how the DC restistance of the MH ballast changes with the different voltage taps on it. If it is constant and the power input is constant there should be less heat from the ballast and more energy to the lamp (heat and/or light), as miniscule as it may be. When I do my upgrade Ill break out the multimeter and find out...