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BrianG · 11.16.2009, 01:41 PM

It's a common thing to think. But, if you think about the logic, it makes sense. Simply reducing voltage will reduce current as well. The load, or resistance, does not change. If you gear up to compensate for lost speed, you are effectively lowering the resistance, which will pull higher current.

Look at it this way: if you are pulling 18A average current on 4s, the effective "resistance" is 0.82 ohms. Reducing the voltage to 2s without changing the gearing (keeping resistance the same) would generate 9A of current. The work (power) the motor is doing goes way down.

However, if you gear up the 2s setup to get the same speed as you had on 4s, you are drawing the same amount of power. So, 266w (14.8v * 18A) at 2s would generate around 36A.

	(#1)
BrianG RC-Monster Admin Offline Posts: 14,609 Join Date: Nov 2005 Location: Des Moines, IA	11.16.2009, 01:41 PM It's a common thing to think. But, if you think about the logic, it makes sense. Simply reducing voltage will reduce current as well. The load, or resistance, does not change. If you gear up to compensate for lost speed, you are effectively lowering the resistance, which will pull higher current. Look at it this way: if you are pulling 18A average current on 4s, the effective "resistance" is 0.82 ohms. Reducing the voltage to 2s without changing the gearing (keeping resistance the same) would generate 9A of current. The work (power) the motor is doing goes way down. However, if you gear up the 2s setup to get the same speed as you had on 4s, you are drawing the same amount of power. So, 266w (14.8v * 18A) at 2s would generate around 36A.