There are 3 different effects that damage wires.
1: Electron- hole pressure ionic break down
2: Electron-hole resistance, thermal break down .
3: physical damage.
1: High voltage pushes deeper into the conductor strands, and if too thin the electron pressure vs electron ionic bonding coherence of the strand conductor does bad things.
Where as voltages, the pressure shoves electron current into the conductor paths, and this is where ohms law comes into play.
2:High current (I) and less strands = more resistance. The voltage has to divide into less current pathways. V/I=R
However if your conductor strands are up to the task, the higher the voltages actually aids current as more electrons and holes is pushed through a single strand because of higher pressure, this causes lower resistance to current naturally..
1:So say if your wire is rated at 40v 100 amps and your pushing 50v through it at 1 amp. Not all conductors are going to be used efficiently each strand being utilized to max potential becomes bombarded by energy and slowly each strand will break down and react with its environment.. Slowly each conductor strand fails until you have next to none left. Slowly cutting current pathways until the last few remain, the resistance raises at 50v. Not after too long you wont get 1 amp through this wire little only 100 amps.
2: Say your wire is rated at 60v 10 amps. And youre pushing 50v @ 50 amps. There isnt enough voltage to interfere with the ion bonds in the conductor, but the current will be forced through all the strands less efficiently as the electron are not being pushed so deep into conductor and the strands are resistance resisting both the pressure and current permitted to flow though the conductor from point A to B . Current (I) is multiplied by the resistance equals voltage. I*R=V
Both current and volts shall be wasted using both. 50v/50A=1R and 50A*1R=50v, tells us if we short negative to positive we will have a 50v drop between Point A- and b+ and 50 amps shall pass..
The Volts by the amount of Amps being forced and impeded * to a 50v voltage drop and 50amps passage its indicative to the symbol * between point A and B of the conductor. This causes heat, Watts.
I*V=W Causes thermal break down.
Volt rating is rated by how much volts each strand can take,
For current it's more of a guide to get you desired amps, Consider cut off limit to V and the resistance of device, then the resistance per meter or the desired SI unit shall equal amps allowed.
We must equate the entire voltage drop out of the fets to get an idea how much our phase wire will suffer under load.
Windings resistance are approx .01R
4 foot of 8awg is approx 0.004 ohms.
20 amp controller approx 2.4 ohms ohms
Cheap 1C battery .12
Power leads .001
Lets just say about 2.5 ohms full circuit. 50v/2.5 ohms = 20 amps.
We have all the stuff needed to calculate the voltage drop across the phase wires.. @ 20 amps.
2.5R - 0.004 =2.496R
20A*2.496R= 49.92v-
50v-49.92v =.8v
We have a .8v drop across our phase wire @ 20 amps.
16 watts per 4feet= 4 watts per foot.
Consider PWM and 3 phase/4*3 = 3 watts per foot. Gawd I don't know. Depends on the frequency,
Imagine a foot long by 4mm diameter copper resistor burning 3 watts on max load. Or .3% of the heat a 1000 watt stove element spread over a 4mm * 1 foot wire.
I have no idea how hot a stove element gets, lets plug in 500 deg C. Generous figure at best I guess.
We are approximately adding 1.5 degC @ 20 amps to our phase wire,
Add another shunt to our controller @ 40 amps. Total resistance now 1.3 ohms we can do 38 amps. A better pack gets us closer to 40 amps,
Save me the v drop calculation and we will see about 7 watts/foot and 3.5 deg C on our phase conductor. 38 amps, That's close enough to 3Eman estimate. Add a blob of solder over the shunt and we can do 50 amps easy.
Also consider that we are only dealing with constant peak current values and at 50 amps and really mostly probably doing 20 amps average..
