AC Motor

When the ratio of mechanical power delivered to electrical power consumed is highest.

When it is turned off.

Are you trying to start that "zero divided by zero" up roar again? Go ask Siri and see what you get.

Just trying to match the quality of the answer to the quality of the question.

He won't!!

When running at or near its full load capability.

Depends on the motor...

"First, efficiency of an electric motor is just output power divided by input power. Input power is your electrical input power, which is V*I. Output power is your mechanical output power, which is speed*torque.

Given that, we can see that efficiency for every motor is going to be 0% at no load (i.e., maximum speed at 0 torque). Efficiency will then increase as torque increases until it reaches a maximum and then it will start to drop off until stall torque is reached. At this point, the efficiency is 0% again because the speed will be zero.

The other way to ask your question is why is efficiency low at low loads? Friction is the main cause of inefficiency at low loads. Losses due to friction are essentially constant with respect to load so at low loads, the majority of your input power may be used to overcome friction. As the load increases, friction plays a smaller and smaller roll in the overall efficiency. Granted, other inefficiencies begin to occur at larger loads (I2RI2R losses, copper losses, stray load losses, etc.) but in a well-designed motor the efficiency will peak in the 80-100% load range."

http://physics.stackexchange.com/questions/46113/why-is-an-electric-motor-more-efficient-at-higher-loads

Dear Mr. Rabaacca,

During our college days, this topic was taught during 3rd year Engg.Studies.

For DC MOTOR the Max. Efficiency is obtained for a particular Load at which the constant loss is equal to the copper loss which is (I^2) xR

Similarly a Transformer will deliver maximum efficieny at a particular load where constant loss is equal to the copper loss which is (I^2) xR

For AC Motor will deliver maximum efficieny at a particular load where constant loss is equal to the copper loss which is (I^2) xR

The above statement is derived by an equation involving Integration.

DHAYANANDHAN.S

Without knowing the type of AC motor (induction, universal) and the coil resistances, inductances and core reactances [not to mention the mechanical loads] to generate the necessary equations for a mathematical analysis, any answer will be a guess. My guess is that some point in the starting process will be where things will be most efficient. At this point the torque will be at a maximum. If this is an AC induction motor then the design of the coupling rotor will determine at what slip this maximum will occur. The NEMA torque curves for common rotor designs means that this theoretical work does not have to be repeated anywhere but a classroom. Operating a motor close to this peak is not advised.

The question needs a verb before it can be answered concisely.

LOL!!

Happy New Year!!

I think it needs more than verb, but then, Life is!

And the question is?

How can anyone attempt to answer this statement presented by the OP?

What are you talking about? The OP has presented neither a genuine statement nor a genuine question.

Generally in the motor itself at a speed and torque when fixed losses equal variable losses - but also in priority terms of what...?

Mechanical power, electrical power, installation cost, maintenance cost, running cost - they all get caught in the 'efficiency' net....