Re: Fan Resistors

[John Hobson <johnh@haht.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)?

Is it possible to measure the voltage drop across the resistor instead of
the fan?

Am I missing something? It just seems like a great deal of current and power
to run a fan. Thanks.

John