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
A: Filling a bucket is not power–it is energy storage (if the bucket is high up, I mean). What you want is not just the rate at which water flows, but how hard the water can push against something to make it move, e.g. a turbine. I think you would agree that a bigger turbine can generate more power than a smaller turbine, if the flow of water is exactly the same in both systems. That is because, for the same amount of flow, a big turbine has more surface area for the water to push against. Meaning, it experiences more total force due to the pressure. That means pressure matters for doing work. We already know flow matters (a closed pipe with stationary high-pressure water in it won’t turn the turbine, because the pressure is the same on both sides), so that must mean that both pressure *and* flow matter. In electricity, “pressure” is voltage, and “flow” is current.
B: Batteries cannot maintain a perfectly constant voltage, and because they are chemical in nature, they will always have some variation due to temperature (colder temperatures slow down most chemical reactions). If that is the rating for your bike’s battery, then under most reasonable conditions (I assume “SOC” means “Standard Operating Conditions”), it should consistently produce a voltage between 30 and 42 volts. As the battery is used, it can fall off in both voltage and current. *In general,* so long as the load on the battery is much higher than the battery’s (very small) internal resistance, it will act almost exactly like a constant voltage source, and thus the current will depend only on how much resistance is present in the circuit. In the water-pipe analogy, for a pump that always pushes water at a fixed pressure, more water will flow through big pipes than through really tiny pipes.
C: Your error is that the “Power(heat loss)” equation presumes that *all* power is lost as heat, which is not true in this case. Instead, some energy is lost as heat due to the internal resistance of the motor, and some energy is used to turn the motor. A motor’s efficiency is how much energy it actually does turn into useful work, divided by the total energy put into it. In this case, your total power output would be P=IV+I²R, where R is the resistivity of the motor itself (not the rotation, just the motor as an electrical circuit element). The *useful* power output would thus be only IV, current through the motor times voltage drop across the motor.
D: In a theoretical ideal motor, where you can supply any current and voltage you like, this would be true. Real motors have quirks and foibles which make the simplifications behind these formulae fail. It *might* be the case that doubling the current flowing through the motor would quadruple the heat output of the motor, but it might also be significantly less or significantly more–or it might destroy the motor to double the current.
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