Quote:
Originally Posted by Joe Johnson
...you will have two current paths -- and since loses are an I^2 R problem, halving your current can effectively quarter your electrical heat dissipation.
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Formula is correct, application is almost correct.
The
total current didn't get halved, it just got split into two paths. So, if everything else is the same, the power dissipated
per path is a quarter of the original, but you have twice as many paths, so the total power dissipated (lost) is half the original (which is still a good thing).
To put it another way, putting another identical load in parallel with the first means your R is cut in half, which directly says that the power is cut in half.
Also, I just said "if everything else is the same", but if you're changing motors and/or gearing, then everything isn't the same. If new motors or gearing gives you a better match of motor to load, then that will be where you see the biggest improvement, and heat caused by losses becomes a smaller issue.