Originally Posted by skellyo (Post 254866)

I've run across this a few times here and on other forums. I keep seeing folks comment in various forums that running a higher voltage setup will yield lower current draw. This is only true when you drop the kv of the motor!

Ohm's law is quite simple and proves this easily:

P = V^2/R or P = I^2*R

Assume a motor with R = 0.2 ohm on 4S:
P = 14.8^2/0.2
P = 1095W
Now solve for I.
1095 = I^2*R
I = 74A

Now take the same motor on 6S:
P = 22.2^2/0.2
P = 2464W
Again, solve for I.
2464 = I^2*R
I = 111A

Now let's make an assumption that you run a lower kv motor with R = 0.3 on 6S:

P = 22.2^2/0.3
P = 1642W
Solve for I.
1642 = I^2*R
I = 74A

Now, this is simplified a bit assuming a purely resistive load in the functions. However, it proves that it is possible to get more power with maintaining the same current (or even less if you further increase resistance and voltage input). But, it also proves that if you think current will be lower because you increase voltage on the same motor, you're quite incorrect.

That is all. I'll step off my soapbox now.