Scientific Theory

I am working on a scientific theory and I need a little help.
I have 2 car batteries in series to create 24v need to power an electric motor.
The motor is 14.5 amps and I need a way to control the amperage from the car batteries to the motor.
Any help would be greatly appreciated.

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Ice Dragon (author) 4 years ago
to me it is a theory even if it is engineering not science, but to meeven engineering is science based.
OK here is what i need to know: I need to know how/what i can use to control the speed of the motor cuz a potentiometer doesnt work or it blew out one or the other. any more help would be appreciated.
Lets put it this way if i am successful in this theory then i can advance it to power my house without outside power and hopefuly my car without outside power source.
Look up rheostats, you are talking mucho power here which would blow out normal electronics potentiomenters without control circuits.
Ice Dragon (author)  caitlinsdad4 years ago
ok the motor i am using is a unitemotor model# MY1016 i believe it is a 300 watts motor whould i match the rheostat to the same wattage or more or would a lesser one work?
Ice Dragon (author)  caitlinsdad4 years ago
Thanx
kelseymh4 years ago
I'm not exactly sure what you mean by "working on a scientific theory;" putting together an electric motor is engineering.

You should not need to "control the amperage" (at least, not if you actually believe in current, well known science). The motor is a resistive load. It will draw ("pull") exactly enough current (14.5 A) to perform its function. The important thing is to be sure that your current source is capable of supplying that current at the required voltage.

In your case, a standard auotmotive battery can certainly supply more than 14.5 A, so your engineering setup should work properly.
The motor ISN'T a resistive load......
No, it's more complicated (if I'm not completely mistaken, the load goes like I2, not like I). However, I believe that it is still true that the motor will draw current from the source, not that current gets pushed through, which is what the main confusion seemed to be. Given the level of sophistication of the question, I was trying to keep the discussion as simple as possible.
Current depends on load. The lower the speed, the lower the back EMF. As I said to Cdad V=IR+Vb.
Thanks. As the load and speed increase, so does the back EMF, which increases the current draw. faster than just I = V/R.
As the load increases, back EMF FALLS, so current increases.
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