permanent magnet alternator load problem

 Hi all,
I've made a simple permanent magnet alternator. For the sake of
experimentation I've attached the rotor to a D.C . motor, and the
motor to a constant voltage power supply. When I complete the circuit
with my wire coils, for some reason the current draw goes up  and the
rpms go down on the rotor.  This happens whether there is a load or no
load across the coils. The coils are in no way connected(intentionally
at least) to the rotor or motor. I don't know what is happening, or
how to prevent it. Any help would be appreciated.

Thanks,
Mark

  Mark brought forth on stone tablets:

Are you saying that when you short your coils that the motor driving
them slows down and draws more current?  Well, then your alternator is
working!

All the power it is producing is being wasted in the short, but the
motor has to provide it, and it slows down and draws more currnet to do so.

bob

 
The coils are stationary, and the magnets are moving because they are
connected externally to the motor shaft. You seemed to think that my
alternator coils were moving as well.  The coils being shorted are NOT
the coils inside the motor; the coils being shorted are the stationary
alternator coils.  The motor magnets and coils are not connected to
the alternator coils. The only reason the motor is involved is to turn
the magnets. Sorry for the confusion. Any ideas??

Thanks,
Mark

 

Mark, the responder understood EXACTLY how you have it set up, and
what you are doing, and his answer was CORRECT.

When you short the stationary coils you are consuming power,which is
dissipated as heat. This power comes from the "prime mover" which in
ypur case is the DC motor, which will draw more power and slow down id
the power supply is not "stiff" ENOUGH TO HANDLE IT.

--
Posted via a free Usenet account from http://www.teranews.com

 Try to operate your alternator into a big variable resistor. With a
voltmeter and ammeter attached. When the V drops to .707 of the Open
Circuit value, note that load resistance... power in the load will
decrease above and below that load