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
(Re: A only)
If you’re filling a bucket, then the accumulated result is some litres/gallons of water. The analogous quantity in electricity is some coulombs of charge. And it’s important to understand that the amount of water/amount of charge, is not the same as measuring an amount of work done.
A barrel of water at the top of a big hill, can do a lot more work flowing down to a reservoir 3 miles below, than the same quantity can do if it’s already sitting in the lower reservoir and has nowhere lower to flow to.
(eta: the same reasoning applies to water flowing between reservoirs which are at different pressures for reasons other than gravity. Elevation just happens to provide a convenient, uniform pressure gradient in this analogy.)
And a coulomb of charge can do a lot more work flowing from a +48V source to 0V, than from +12 to 0.
Work, in kinematics, is an amount of force applied, multiplied by the distance over which the force moves. So the work that can be done by an amount of falling water is a certain number of newtons *times* a certain number of meters.
Analogously, the amount of work that a given amount of charge can do by “falling” from a high-voltage reservoir to a low-voltage one, is the amount of charge *times* the ‘distance’ it falls (measured in volts).
And, in both cases, power is work per unit time.
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.
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).
A) A simpler way to think of this is a ball on the bottom of a hill. The weight on the ball is current. At 0 Voltage it means the ball is at the bottom of the hill. If you let the ball go it does not move or do anything.
A better way you think of voltage is thinking how did the voltage get there? For our ball situation think of it as you roll the ball from the bottom of the hill to up the hill until you get tired. We will call this 1 unit of tiredness. The energy you used to roll it up the hill is equal to the weight of the ball and the unit of tiredness. If you let the ball go then all that energy you used to push it up the hill will make it roll down.
Voltage is potential energy, something pushed the ball up the hill and if you let it go then it will roll back down with that same energy.
For the water analogy the voltage is the pressure of the water or the height it falls from. If the water has no pressure or height then it is just a bucket of water on the ground. That will not do anything no matter how big the bucket is.
Electricity is special because there are many ways to generate that voltage and we are good at storing it in places like batteries. This is like rolling the ball up the hill and until we start the circuit it will stay up there.
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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