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Originally Posted by LauraN
*drags out physics*
... power=I2R. So the amount of power you'd be drawing would decrease because you've decreased the resitance, but then increase proportionally the the square of the corresponding increase in current... I think I've totally confused myself...
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Actually, all the stuff about increasing yet decreasing could be explained a lot clearer and in an easy-to-understand way.
Given V=IR and P=I
2R; we know that IR is always constant because your voltage will remain constant, so IR can equal K. Essentially P=I(IR) [I just rearranged the variables under the commutative property], so P=KI. This shows that you need not worry about decreasing resistance, compensating in amperage, and increasing due to squaring; rather, there's just a simple
linear relationship.
You could even graph it if you wanted to, whereas P=y, K=m, and I=x under the function y=mx+b! (b=0, obviously, because when your voltage is zero your IR HAS to be zero as well)