Actually eli6 because its not like I know nothing about electricity at all: I understand that 1 amp = 6.24\*10\^18 electrons per second and voltage is the pressure behind it. I understand that P=IV and that volts = energy per coulomb.
What I don’t understand is: (A) why power is the current multiplied by voltage and not current alone. In terms of the water analogy, the current is how much water is flowing through a pipe per second. If I’m filling a bucket with water then regardless of how much pressure and resistance there is, the only thing that matters for the bucket is how much water flows through per second, aka power which would be watts. But based on P=IV not being P=I, I know I’m wrong but I don’t know why. Why does current have to be multiplied by voltage to get the power? Is the water analogy just flawed here?
(B) How an ebike battery and system works. My ebike has 30-42v depending on SOC, but what does that mean? Does it mean that there will always be 30-42v in the motor and the wires and everything else, with the only changing thing being amps? What about amps, do they change based on voltage or in a different way? What happens to both voltage and amps when using different power settings in the display?
(C) Does ohm’s law apply to heat loss in ebikes? If yes, my current understanding is the following:
Power(heat loss) = VI = (IR)I = I²R
Power(useful kinetic energy) = VI
Wait, that would mean since both are VI, that would mean there will always be 50% useful energy and the other 50% is loss. I know this is wrong but I don’t know why. Where am I mistaken?
(D) in conclusion, is it true that an electric motor running at 2x power will have 4x the heat loss over the same duration of time? If so, what are the formulas behind it?
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In: Physics
Don’t think of It like filling a bucket, think of it like turning a water wheel.
Like those water wheels that have the little waterfall going into them. if that water fall is very short (low voltage) there isn’t as much energy to be harvested, compared to the same amount of flow falling from a taller height (higher voltage).
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