Re: Fan Resistors

[Igor Kessel <i6941TB@gnn.com>]

>Igor Kessel says:
>
>>> The voltage drop on the fan was Vf=6v. Since total voltage
>>> on the running engine Vt=14v, then the voltage drop on the
>>> resistor R2 is Vr=8v. Therefore the current flowing through
>>> R2 (and Rf) I=Vr/R2=40a! The power this poor sucker has to
>>> dissipate P=IČR2=320w.
>
>Forty amps strikes me as an awful lot of current to run a fan. Doesn't a 6
>volt drop (across the fan) at 40 amps mean the fan is using 240 watts?
>Continuing along this line, won't 6 volts at 40 amps gives a resistance of
>0.15 ohms (6/40) for the fan motor.  If switched to the high speed position
>(i.e., no other resistance), wouldn't current would jump to over 90 amps
>(14/.15) and power used to over 1200 watts (14*90)?

That is why the hi power relay was suggested by Audi AG in their service bulletin.
Should you ever lose this resistor, you'll notice it right away: your Audi would 
sound like a helicopter trying to take off. The fan would be silent until the 
T°-re becomes high enough to kick in the 3rd speed of the fan. Boy, is it loud! 
When I had this problem on my old 5000s I made so much noise on the red lights in 
the summer time that I could rival the so-called "muscle cars". Also, you'll lose 
the AC action @ idle because it kicks in the 2nd speed every time you put the AC 
on. If my memory serves me correctly, that Bosch fan is rated @ 1.4kw.
 
>Is it possible to measure the voltage drop across the resistor instead of
>the fan?

Yes, but I was too lasy to take the plastic cover off, when I had the fan's 
terminals exposed. It would be within your calculations. We are dealing with a 
very simple series DC circuit here.

Igor Kessel
the sweetheart: 200TQ, chipped and MOMO'd through out,
in Tornado "arrest-me-officer" Red;
the ex: 5000s, the EE's nightmare
Phila PA, USA
i6941TB@gnn.com