# Why is an electric motor more efficient at higher loads?

My question is driven by the plot below. We see that acceptable operating range of a motor is between 50-100% of the rated load. Below 40% or so the efficiency of the motor drops off dramatically.

What is the cause of this phenomenon?

• Because at low loads you're still spending the same energy just getting the armature to turn, but you are drawing out less energy for useful work. Dec 6, 2012 at 18:21
• It depends on the type of electric motor. AC, DC, and there are plenty of subtypes of both. You'll be more likely to get a good answer if you can provide this information. Dec 6, 2012 at 20:07

First, efficiency of an electric motor is just output power divided by input power. Input power is your electrical input power, which is VI. Output power is your mechanical output power, which is speedtorque.

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 role in the overall efficiency. Granted, other inefficiencies begin to occur at larger loads ($$I^2R$$ losses, copper losses, stray load losses, etc.) but in a well-designed motor the efficiency will peak in the 80-100% load range.

• The first use of energy is to generate rotational energy and overcome static friction, once upto speed, dynamic friction, dominates, at constant power input. This continues until I2R heating and eventually eventually cooling losses dominate. Oct 22, 2018 at 14:54

Its the mass of the rotor relative to the mass of the load.

No matter what a motor must drive its own rotor. Lets say that rotor weighs 1 kilogram. If you were to drive a load that also weighs a kilogram, then 50% of the energy is driving the load and 50% driving the rotor, so it cannot be more than 50% efficient. If you increase to 2 kilograms as a load, then the rotor is now only 33% of the total mass, and so it can be up to 66% efficient.

As you increase the load the rotor becomes a smaller and smaller fraction of the total mass being driven, meaning a greater portion of total electricity spent is actually doing useful work.

The reason is cost driven by Physics. A motor with a larger torque rating requires larger magnetic fields to be generated.

Magnetic fields require either more turns and/or higher current. Higher current requires larger wires. In both cases the motor is heavier. If those magnetic fields are not used to capacity the energy to drive the extra mass is wasted.

So motors are designed to work at specific load range to save cost.

The most common motor, then induction motor has a lot of its losses dependent on slip. Slip is the speed ratio of the motor's rotor compared the the speed of its stator's magnetic field. That means that the slower the rotor turns, the larger the slip, and the closer it approaches the stator's field, the smaller slip is. Losses such as copper loss have an inverse relationship with slip so as slip increases, the losses decrease.

How does a load affect this? Loading the motor induces a torque in the opposite direction slowing down the rotor's speed. Therefore slip increases and losses are minimized.