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: What you might be missing about the P = VI formula is that what the formula truly means is the power (P) consumed by a device is equal to the voltage drop across it (V) times the current flowing through it (I). So you have a device (like an electric motor) connected to a power supply (your battery) and no other devices connected in series. If your battery is supplying 40V, then the motor must have a 40V drop from one terminal to the other. The power consumption then does directly correlate to the current drawn by the motor, or in other words, more current = more power = more speed.
If you connect your battery to your motor, then to another device in series, and then complete the circuit, your voltage drop across the motor will no longer be 40V, so the actual power draw of the motor won’t be 40V * the current provided by the battery. There’ll be some extra math that depends on the other device as well.
If you want to use a water analogy, you can think of a water cutter is an example of high voltage but low current. It’s pushing the water extremely hard, so it has enough power to cut right through metal. A weir in a slow-flowing river can be an example of low voltage but high current. Even if the river isn’t flowing very fast, there’s still a huge amount of water pushing against the weir, which means a lot of power being dissipated.
B: A battery can discharge any amount of current up to its rated discharge capacity, which depend on the construction of the battery. For example, cold cranking amps for typical car batteries is the max current it can discharge for 30s at 32F. Temperature affects battery performance. The battery can also discharge near-0 amps, like when the car is off. Or it can discharge some current while the car is running.
C: Kindof. If you’re modelling your motor using a resister, not all of the power dissipated goes to heat. A brushless motor has about 85-90% efficiency according to google, which is much better than 50%. The 10-15% wasted power becomes heat and noise while the rest goes towards forward motion.
D: I think the answer to this depends on the efficiency of the motor, what voltage and currents it’s designed to work with, and the voltages and current actually being supplied to it.
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