# Power Question



## Thomas Cross (Nov 14, 2016)

Will I be tripping the FUSE BOX at my apartment?? I plan on buying two 500watt lights connected to 202 AC. I live in France and the voltage in my apartment goes up to 240. Thanks


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## 480sparky (Nov 14, 2016)

Depends on the rating of the fuse itself, as well as whatever else that might be drawing power that's on the same circuit.

While it's possible you may have issues, it's not likely.


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## Designer (Nov 14, 2016)

Thomas Cross said:


> Will I be tripping the FUSE BOX at my apartment?? I plan on buying two 500watt lights connected to 202 AC. I live in France and the voltage in my apartment goes up to 240. Thanks


If the two lights are the only use on the breaker at that time, no.  Check the amperage rating on the breaker for that circuit, and I'm sure you will see that it is more than 5 amps.  Probably more like 15 amps.  That is what it is for most residential lighting circuits in the U.S.  

If there is additional load on the circuit, and the breaker does trip, then you can plug one light into a different circuit.


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## evancamp13 (Nov 14, 2016)

Hello, I'm new to this forum, but I am an electrical engineer by trade, so I think this is an appropriate first post for me. Lights are purely resistive load, so your lights will be pulling just over 2A each at 500W. Your breakers are likely rated for well over 5A each, so you should be fine

Sent from my XT1254 using Tapatalk


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## davisphotos (Nov 15, 2016)

Watts are volts x amps, my circuits here in the US are all 20 amp, at 120 volts, which means I can pull 2400 watts per circuit, with some caveats.  not super familiar with UK wiring, but I would imagine you would be able to support a similar load.


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## evancamp13 (Nov 15, 2016)

davisphotos said:


> Watts are volts x amps, my circuits here in the US are all 20 amp, at 120 volts, which means I can pull 2400 watts per circuit, with some caveats.  not super familiar with UK wiring, but I would imagine you would be able to support a similar load.



This is correct at an elementary level, but once you enter into AC loads and Power Factor with Capacitive and Inductive loads, the power relationship begins to look a little different. Which is why I added the caveat that lights are typically completely resistive loads, which makes the P=VI accurate.


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