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I'm designing a system to power up my vehicle with 12vDC and I have 8 motors each draws 25A at max and they are not expected to work always at the same time, meaning that in the maximum case 6 of them working 70% and 2 of them works 100%, so my question is that what power supply should I choose to be enough to power the vehicle up like what power rating should it be ? 1kw or 2kw ? And why ? I mean if there is any equation or calculation method to use ? Please help me with that.

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  • $\begingroup$ You've got a good answer already, so I'll just point out that you're looking at a possible current load of 8*25 A = 200 amps. Stranded wire rated for that current is "double aught," written 00 or 2/0. That kind of wire sells for about \$2.50 per foot (about \$7 per meter) where I am, and you'll need big lugs and a large crimper to work with it. You might be able to get by with undersizing the power supply, but I would not recommend undersizing the wire. Also, a typical car battery is about 40 Ah, so a design case of 155 A would drain a car battery in (40/155) h ~= 15 minutes. $\endgroup$ – Chuck Jan 3 '19 at 12:38
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Power equals voltage times current. One motor therefore has maximum power of 12V * 25A = 300W

6 x 0.7 x 300W = 1260W (1.26kW)

2 x 1.0 x 300W = 600W (0.6kW)

Total: 1860W (1.86kW)

So the maximum power is equal to 1.86kW meaning you need to have a supply power being able to output at least that much (though in reality, at least a bit more).

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  • $\begingroup$ @AbdelRahman - Please upvote answers that you found helpful and accept answers (using the check mark to the left of the answer) if it solved your problem. $\endgroup$ – Chuck Jan 4 '19 at 21:42

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