IS echanical efficiency of an electric motor lifting a load W at linear speed v is (Weight X speed)/ (Source voltage E x Current I) or is it (Weight X speed)/ (Terminal voltage V X Current I) ? Or something else? Pls explain your answer.

[danelectrical]

I think it is the first, because E - back emf  (call it B here) = V (terminal voltage) = IR ( R = resistance of coil)

So EI - BI = I^2 R  Giving EI = BI + I^R

The EI part gives the input electrical power

I^2 R part is the power dissipated as heat

Therefore BI must be the mechanical power developed which must be the same as force x speed = Wv

So efficiency = BI/EI  = Wv/EI

What do you think?

[GerdW]

Hi Dan,

"What do you think?"

I think this is a Mechanics homework

Also this question is not LabView related, so why not look for wikipedia or some fundamental Physics forum?

You have not explained what source voltage refers to...

Message Edited by GerdW on 12-16-2009 10:24 AM