I have my '55 2nd series GMC heater out of the truck for a rebuild so I did a little measuring and a little figuring. This is a 6 volt system and has a single resistor in the fan speed control to give two fan speeds.
The resistance of the motor measures out at around 0.9 ohms. The single resistor measures about 3.7 ohms.
Since the resistors job is to drop the voltage to the motor to something less than 6 volts, I calculated the current flowing in the resistor for varying voltages across the resistor:
E = volts
R = resistance (ohms)
I = current (amps)
P = IxIxR (power in watts)
I = E/R (current through resistor
1 volt drop: 1/3.7 = 0.27 amps ; .269 watts
2 volt drop: 2/3.7 = 0.54 amps ; 1.07 watts
3 volt drop: 3/3.7 = 0.81 amps ; 2.43 watts
4 volt drop: 4/3.7 = 1.08 amps ; 4.32 watts
5 volt drop: 5/3.7 = 1.35 amps ; 6.74 watts
6 volt drop: 6/3.7 = 1.62 amps ; 9.71 watts
Lets take the case with the 4 volt drop and calculate the voltage dropped in the motor.
I = 1.08 amps throughout the circuit.
Motor resistance = .9 ohm (measured)
Motor voltage drop due to resistance
Mvd = 1.08 amp x .9 ohm = .972 volt
If we add up all of the voltage drops in the circuit, the should all sum to zero.
Battery = + 6.0 volts
Resistor = - 4.0 volts
Motor = - .972 volts
-----------------------
?????? = + 1.028 volts
Looks like we have a little voltage left over!
We know the current is steady at 1.08 amps lets
redo this calculation in power Watts
Battery = 6.0 volts x 1.08 amps = 6.48 watts (energy in)
Resistor = 4.0 volts x 1.08 amps = 4.4 watts (heat from resistor)
Motor = 0.972 volts x 1.08 amps = 1.05 watts (heat in motor windings)
Looks like we have 1.028 volts x 1.08 amps = 1.11 watts of energy actually turning the fan!
Stuart, in your experiment, your assumption that the fan would always draw 6v/.95ohm = 6.3 amps, only applies when the rotor is locked and the torque produced in the shaft is at it's maximum.
If you release the shaft and allow it to rotate, the current draw and the torque in the shaft will drop as the motor speeds up. The more mechanical power used by the driven device, the higher the current draw.
John
