>Extra resistance in the system will not cause a problem with the fan
>resistor.  Quite the contrary, it will make it run cooler.

Reason is that because of the extra resistance there's less total
power dissipated in the system, and of that lower power
proportionally less of it is dissipated in the resistor.

Your radiator fan is rated 450 watts at (probably) 13.2 V, giving it
a draw of 34 A and an effective resistance of

P = I^2R
--> R = P / I^2
= 450 W / 34 A ^2
= ~400 milliohm (its actual resistance is much less but when the
motor is spinning it generates a back voltage that simulates
resistance for our purposes).

So to slow it down you stick let's say one ohm in series with
it.  And we'll also stick in a disclaimer:  when the motor changes
speed the back voltage is going to change and its effective
resistance will, too.  But we're going to pretend that the motor has
a fixed characteristic because dealing with that theoretically would
go beyond my ability without helping the example.

Ok, we now have 1.4 ohms in the circuit and the 13.2 volts is
distributed across it proportionally, so the motor sees 4/14 of the
voltage or 3.8 V and the resistor sees 10/14 or 9.4 V.

Total power in the circuit has gone down to
P = E^2 / R
= 13.2^2 V / 1.4 R
= ~124 watts

and of that the resistor is dissipating
P = E^2 / R
= 9.4^2 V / 1.0 R
= ~88 watts  with the balance in the fan.

Now throw in a bad connection that adds a tenth of an ohm.  Total
power is now down to
P = E^2 / R
= 13.2^2 V / 1.5 R
= ~116 watts.

The bad connection is seeing 1/15 the total voltage or 880 mV and is
dissipating
P = E^2 / R
= .880^2 V / 0.1 R
= 7.75 watts, so it's acting like a night-light bulb or about half a
small soldering iron.  It's maybe in trouble.

But the resistor is now seeing 10/15 of 13.2 V instead of 10/14, so
it's coasting along at
P = E^2  R
= 8.8^2 V / 1.0 R
= ~77 watts, cool as a roasted cucumber.

Yours,
David

--
David Beierl -- dbeierl@attglobal.net