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#### G.v.Mainberg

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Power draw is not a constant vs voltage in these motors... the example numbers are not correct. Don't confuse the rise in current draw when a motor stalls with running current versus voltage when the motor is running in the normal RPM range. Current in motors like these is related to torque which is related to fan blade, speed, etc. Increase the voltage (from the faster alternator as proposed) and the speed will increase....we all know this from hearing fans slow down when we drop and engine to idle and the fan(s) slow down due to alternator voltage dropping. Torque/load will increase with speed, and current has to increase; it may not increase much, but it won't go down. (This all comes out of the mathematical modeling for motors and involves back-EMF's, etc.)

With better voltage, the torque load on the engine is higher.
In other words:

Let's say, you have a 12V DC motor with 80 Watts. This means, the motor has a power consumption of 80W at 12V. Since power consumption P (Watt) = U (Voltage) x I (Current), as a result your current draw is P / U or 80W / 12V = 6.67A.
Also, the electrical resistance R (Ohm) = U / I, so your motor has a resistance R = 12V / 6.67A = 1.8Ohm. Since the resistance is constant (within our idealized example), you can see that you can rearrange the equation to I = U / R --> Since R is constant, if the voltage drops down, the current drops down in the same relation which will bring the power consumption down even further.
If the voltage drops by factor 2 (to 6V) and the current drops by factor 2 as well (to 3.33A), the power drops by factor 2[SUP]2[/SUP] = 4 from 80W to 20W (6V x 3.33A).

Dieter

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