Well you’re in good company 
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Two words
Synchronous Rectification
Current flows through the battery only half the time, while current flows through the LED 100% of the time regardless of duty cycle, so the average current through the battery is lower.
At this point people should just give up trying to understand how it works and just remember these two concepts
Motor Current = Maximum torque - how fast you’ll accelerate
Battery Current = Maximum power - maximum speed/climbing rate
Power conversion black magics handles the rest.
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when gesc? 
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VESC big buck with fast switches and compuder
The average current through the battery and through the LED would be the same, and the current would only flow through the LED 100% of the time if you have some sort of filter circuit on the switching. Am I not right? This is assuming a simple DC circuit with a single switching transistor.
I strongly disagree with this statement.
Take a circuits class.
This is not how inverters work. The load is inductive and there is either a catch diode or synchronous complimentary transistor. This is also not how 99% of LED drive circuits work either. They use a buck converter to deliver a continuous current.
In the case of something resistive, like an LED, during the on time the LED sees rated current, the value of this current is the analog of “motor current” in this example. During the off time the led sees no current. The battery sees half the rated current. Inductive loads do not behave this way. Interestingly enough most LED circuits use a buck converter to deliver a constant current to the LED because the forward voltage never matches the supply voltage. Current through an LED determines brightness. Current through a motor winding determines torque. LED forward voltage determines power output. Motor BEMF voltage determines power output. LED forward voltage is always lower than supply voltage. Motor BEMF voltage is always lower than supply voltage. Supply voltage divided by power output is supply current. LEDs are driven by a buck converter. Motors are driven by a buck converter.
Current cannot change by large values during a switching cycle due to the inductance.
Draw the current paths during different parts of the switching cycle for an inductive load.
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There’s got to be a good water analogy illustration or video about how voltage/current/duty cycle in magic boxes works. I think I get it, but it still feels like a mind bender to try to visualize the relationships between the battery side and motor side. I do also want to understand it better though.
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RC explained on YouTube has some great explanations on all this stuff
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The motor inductance acts as the filter. Inductors resist changes in current flow, so the inductance of the motor tries to keep the current flowing constantly, even during the off-periods of the PWM waveform.
That assumes the battery voltage and LED forward voltage are the same. If the battery is say 30 volts, and the LED is 3 volts, then with an inductor in series with the LED and a freewheeling diode, the LED current could be ten times the battery current without breaking the law of conservation of energy.
Current is not conserved, nor is voltage. Power is conserved (power = volts x amps), but that equation can be shifted to one side or the other easily, trading lower voltage for increased current, and vise versa.
There’s no free lunch, and the number of watts you get from the battery is the same number (excluding losses) that you get at the wheels. It just might be in a different ratio of voltage and current.
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I was talking hypothetically. I was comparing the switching of an LED in a simple circuit with the switching on the motor from the ESC, to illustrate that intuitively, if you reduce the duty cycle, you reduce the current flow. I wasn’t talking about ESCs or buck converters at that point.
Imagine an LED connected to an microcontroller (say, an Arduino) through a resistor, with the MC outputting a PWM signal. With that setup you get a dimmer LED due to persistence of vision, but in addition, the current flow is reduced on average. That’s literally all I meant.
In the case of having no RC or other filter, I would think the battery sees near-zero current during the off time of the LED.
I’m pretty sure I roughly understand inductance. When a voltage is applied to an inductor*, a magnetic field builds around the inductor. This generates a voltage opposing the first voltage so little current flows. As the magnetic field increases, the opposing voltage decreases until current flows freely. When the original voltage is removed, the magnetic field collapses and in doing so induces a voltage such that current continues to flow for some time at a rapidly decreasing rate.
*Or any conductor to some extent, since I presume there’s no perfect non-inductive conductor.
I guess when I said “Imagine I have an LED…” I thought it was obvious that meant “with only a switching voltage source and nothing else”. Because I completely understand that if I add inductor into the mix I then have an RL filter.
Totally understand all of that 
So to summarise, I was confused because I wasn’t thinking of the motor as inductor. I was thinking
- Battery == Voltage source
- ESC == Very expensive switch
- Motor == Resistive load
And with that in mind, I couldn’t understand how there could be a higher current flow at the motor than was coming out of the battery. Obviously the key was that the motor is an inductor and thus in combination with the ESC creates a buck converter.
Which is honestly kind of puzzling, since I fully and completely knew that motors are inductive loads, but for whatever reason I wasn’t thinking of them as such 
Meh, sometimes I’m just really thick.
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What really still has me confused is that the motor takes power when supplied by the VESC, but it also generates power in the form of back EMF. How does the BEMF power relate to the power output of the motor, of the VESC and of the whole system? Like does the VESC just push whatever it’s allowed to into the battery and give off the rest as heat or is some of that power recycled back into the system? And does the BEMF power help the motor spin at all?
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As we’ve seen, I’m no expert, but I believe the key is that voltage is not power. The BEMF of the motors isn’t “forcing” current back into the battery. It’s generating a voltage. If the ESC allowed the current to flow back into the battery (by turning on the FETs), then current will flow. And actually, that’s precisely how regen braking works. As soon as current flows, the motor is actually doing work and generating power, and thus it converts your kinetic energy into electric current energy.
When you’re applying throttle, the voltage from the ESC to the motor is normally higher than the BEMF that the motors generate, and as such current still wants to flow from the battery to the motor. As the BEMF voltage from the motors gets close to the battery voltage, current wants to flow less and less. That’s what determines the maximum speed of the electric motors. If the BEMF voltage was higher than the battery voltage, current would want to flow back into the battery. If the ESC allows it, you get regen braking.
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Does this mean “regenerative braking” isn’t actually braking? The BEMF is only generated while the motor is spinning but not receiving a voltage from the ESC.
Doesn’t it take more power for electric resistive braking than the BEMF would produce? Doesn’t the motor have to match or overcome the BEMF to even apply the brake?
All braking is regenerative. Dynamic braking is merely shunting that power to load resistors.
The voltage from BEMF directly opposes the voltage coming from the ESC. The voltage from the ESC will drive a current in one direction, while the BEMF will drive current in the opposite direction, like a tug of war.
Forward torque is produced when the ESC wins.
Braking torque is produced when the BEMF wins.
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this made everything click, thanks.
isn’t it really inefficient to have the two voltages opposing each other while driving the motor? isn’t there a way to recycle the power instead of just giving it off as heat?
They’re in series, so they mostly cancel out and you get power out at the shaft.
The power does not turn into heat, it turns into either electric power in the battery or mechanical power at shaft.
I mean when the BEMF is higher than the allowed regenerative current sent to the battery, doesn’t the VESC just give off the rest as heat? or is it putting that leftover power back into the motor?
There is no leftover power, it all goes into the battery.
even if I set the maximum regen current to 0? or 0.1?
Then you don’t get any regen or braking.
What will happen is the ESC will match the voltage of the BEMF so that the BEMF never wins the tug of war.
If your motor overspeeds and exceeds battery voltage then it will blow up the ESC
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