Quote:
Originally Posted by JayG
Now the fuse for each fan is 40 amp, if it was a full draw of 40 amps, it would be ~440 watts
12 volts through a .55 ohm resistor is ~260 watts or ~21 amps.
My guess is that the low speed resistor is ~ 300 watts rating if .55 ohm and ~250 watts if .8 ohm
What gets interesting is that the voltage drop on a .55 ohm resistor with a 12 v source is ~11.5 volts. That leaves ~.5 volt to the fan
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Sorry to revive an old thread, but the above is fundamentally wrong as it neglects the fact, that the electric motor has an impedance as well.
If we assume that the motor draws full 40A at 12V, it's impedance would be 0.3 Ohm. But that would be unreasonable, so if the motor drew 20A, the impedance it would be 0.6 Ohm. I would 'guess' that the actual impedance is somewhere in the middle - in fact, I'd say it's equal to the resistor, i.e. 0.55 Ohm.
Since the resistor for low speed operation is connected in series with the motor, the total impedance is 1.1 Ohm, meaning that the amperage draw is 11A and the resulting voltage drop is 6V, leaving another 6V for the fan, cutting the fan speed in half.
The power dissipation on the resistor is 66W, so a 100W 0.55 Ohm resistor should be just fine.