Globe amperage draw

[timjf]

Hi everyone can anyone please advise me of the amperage draw of
5 watt 12 volt globes / 10 watt 12 volt globes / and 15 watt 12 volt globes or can anyone advise me of the formulas to be able to calculate this problem All advise most welcome
Thanks Tim

[hellosailor]

Ohm's Law:

5 watts = 12 volts x 0.41666 amps.

That is, watts divided by volts equals amps.

Ten watts? same voltage? Twice as many amps. Fifteen watts? Three times as many amps.

[timjf]

Thanks Hellosailor does the same formula apply to incandesant globes as well as halogen globes I have halogen globes on my boat and that is my point of interest thanks
Tim

[btrayfors]

Tim,

The formula VOLTS X AMPS = WATTS applies to all electrical circuits, ac or dc.

Bill

[Lurch]

Originally Posted by btrayfors

Tim,

The formula VOLTS X AMPS = WATTS applies to all electrical circuits, ac or dc.

Bill

The above is true for all dc loads, and for ac resistive loads, like lighting. It isn't true, however, for inductive ac loads, like motors. (Or for ac capacitive loads either, but you shouldn't have to worry about those.) In the case of inductive or capacitive loads, P(watts)=I(amps) x E(volts) x PF(power factor).

Power factor calculation can get a little hairy, so it's best to avoid it. Fortunately, this isn't difficult. Motor nameplates should give you a value for "amps" (or "I", or "FLA") which can just be used directly. If you can't find that, there may be a power given in "VA"(volt-amps), which is "apparent power", that can be used just like dc power for calculations.

In the worst case, if you're stuck with ac watts, go ahead and calculate it like dc, then figure that the motor will really draw about 1.5-1.6 times the current you calculated.

[yotphix]

please excuse the brief hijack Tim but...Lurch, does 1.5 or 1.6 describe the operating load or the startup load?

[Lurch]

Originally Posted by yotphix

please excuse the brief hijack Tim but...Lurch, does 1.5 or 1.6 describe the operating load or the startup load?

It doesn't really describe ANYTHING very well It's just a "fudge factor" to convert true load to apparent load for an "average" case. If you used it on "nameplate" watts, it would be a guess for operating current. Inrush can be 6 to 10 times that level, but only for a few seconds. Overload devices are already calculated to take that into consideration. It isn't significant for capacity purposes.

However, that's just a "worst case" workaround for an old piece of equipment with a worn out nameplate. You should always have a full load current (or equivalent) value to use. I just tossed it in for completeness in case Tim was figuring his system requirements, and had an ac motor in the mix.

[hellosailor]

Tim, basically yes.

Inductive loads, surge loads, AC circuits and conversion factors (VoltAmps rather than Watts) all can make it complicated but even then, using Ohm's Law gives you a reasonable ballpark for normal use. Generally, anything that makes a "boink" noise when you turn it on (like a big TV set or stereo) or has to start a motor (air conditioners and vacuum cleaners) will be an inductive load that uses a great deal more power when it first turns on. That surge can be 2x-3x as much more as it routinely uses, but you probably don't need to worry about that stuff if you are just trying to figure out your power budget on a "small" craft. The nav lights, the radio, the intruments, all can be figured by Ohm's Law. If you have any electrical winches, an anchor winch, or electrical refrigeration on board, THOSE motors all have a surge, so they need fuses/breakers which can withstand the starting pulse, and they will consume way more power for the first second of operation, than they normally consume. Because the time is so short, you can usually just use a "fudge factor" for your calculations, if the name plate on them doesn't list the surge current.