I feel this is a bit of a misinterpretation of Ohm’s Law and might be confusing to some. The relation is **V=I*R**, which is the voltage drop across a resistor. So yes, for a constant R, higher current equates to a higher voltage drop

But what we really care about is power. Power gets things done; resistance does not.

The important equation for our purposes is: **P=V*I**

Simply put, less current is required at a higher voltage to achieve the same electrical power power output:

- 2000W / 37.0V =
**54A** *@10s*
- 2000W / 44.4V =
**45A** *@12s*

But really… higher voltages just give you the ability to produce more power. 12S can produce 20% more power than 10S at the same current. Using the same example, 54A on 12S produces 2400W

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*Back to the topic at hand:* *10s4p vs 12s3p*

**10s4p will have better range and less voltage sag than 12s3p**. Because at the end of the day you have 4 more cells and about 10% more watt-hours, as others have said

*Example:* At 2000W, 10s4p would draw 13.5A/cell, while 12s3p would draw 15A/cell

So the current draw *per cell* is actually less on 10s4p than 12s3p, which means slightly less voltage sag. And it has higher capacity (higher watt-hours), which means slightly better range.

If you wanted a fair comparison, assume both battery configurations are limited to 15A/cell. In that case, the 10s4p battery can produce 2222W, while the 12s3p battery can only do 2000W. So the power difference between the 10s and 12s (from above) is actually offset in this case, since you are really comparing 40 batteries to 36

Long story short, 12S can produce more power than 10S for the same [total] current. But a 10s3p battery can produce more power than 12s4p for the same per-cell current

*Overall the difference is not worth stressing about… go ride your skateboard*

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**TL;DR:** More batteries = More better