A motor running at its rated power provides 2.6 lb-ft torque at its rated speed of 2020 rpm.?
The motor draws 5.6-A from a 230-V supply at a phase angle of 37 degrees. Would the effciency of the motor be 82.7% or 1.6%

Question

A motor running at its rated power provides 2.6 lb-ft torque at its rated speed of 2020 rpm.?
The motor draws 5.6-A from a 230-V supply at a phase angle of 37 degrees. Would the effciency of the motor be 82.7% or 1.6%

anonymous · Answer

can u calculate the output power from the given data?!

anonymous · Answer

I don't understand your question.

anonymous · Answer

to find efficiency.. u need to calculate what is the output power and the input power !

anonymous · Answer

I've already done that but i'm not sure if the input power is 901.6 W(5.6x230x0.7) or 47656 W(5.6x230x37)

anonymous · Answer

power input = VI cos (phase angle)

anonymous · Answer

besides doesn't 1.6 percent look a little off to you? :P

anonymous · Answer

You can't just go off of the way an answer looks in math...

anonymous · Answer

well thats physics :P.. 

also.. no electrical circuit can give u a power HIGHER than VI.. right? .. so its always VIcos(phase cangle)

anonymous · Answer

actually shouldn't it be (5.6x230x0.8)

anonymous · Answer

and then the eff=72.4%

anonymous · Answer

i dunno how much cos(37) is :-/

anonymous · Answer

If i did it right on the calculator by putting in 37 then pressing cos it =0.79 which I rounded of to 0.8

anonymous · Answer

its 0.77.... googled :D