# Is Power consumed by a motor under various load conditions constant?

Does a motor (dc /Induction/Synchronous ) consumes more power when operating at higher load than when operating at some lower load? Or power consumption remains constant/same in both cases?

What I think is that, Power drawn by motor should increase as we increase the load on motor.

But, someone told me that Power drawn remains constant.For example a 10 kW motor will always consume 10 kW irrespective of load on it.He said power consumed is P=3 VIpf (pf=power factor) for a 3 phase induction motor.As load increases current I increases but Power consumption remains constant as V ,pf change to compensate the increase in I in order to make power remain constant.

• Quick thought: How does the speed of a motor influence the consumption? The faster the rotation the faster the commutation of currents through the inductive parts, which results in higher impedance and less current. A motor under load runs slower, the impedance is lower, and higher current, i.e. more power, can flow. – JimmyB Jan 15 '15 at 14:28
• In general, a motor consumes more power when under load. This goes for most AC and DC motors. (And your friend doesn't understand power factor very well.) – Hot Licks Jan 15 '15 at 17:19
• If what Someone told you was true then where does all that power go when there is no load? It would have to be lost as heat but motors only get hot under load. Don't make life decisions based on Someone's advice! – Transistor Feb 2 '16 at 21:41

Don't let inductive current lag (apparent power) confuse the issue. Ohm's law still applies.

When you apply a voltage to a motor, the resulting motion generates a "back ElectroMotive Force" against the coils; its rotation feeds back a resistance that regulates the current, in proportion to the motor speed.

If an external load reduces the speed of the motor, that "back emf" is reduced (the resistance of the motor decreases), increasing current in order to maintain the voltage, and thus power increases.

Your friend is incorrect. The rating on a motor is the rated power that the motor produces at the shaft. It isn't the power that the motor draws. Also note that this is the rated power, which means that it doesn't always produce this power, only that this is the maximum power that the motor can safely run at continuously. The input power of the motor will vary depending on the load.

"For example a 10 kW motor will always consume 10 kW irrespective of load on it. He said power consumed is P=3 VIpf (pf=power factor) for a 3 phase induction motor".

If this would be the case, motors as we know now would probably not exist.

Since V is constant, the product (I x PF) must remain constant as well. As the power factor varies between 0 and 1, the motor would run at a ridiculously low PF at full load. You could only have solved this if motors would be incredibly over-dimensioned or PF would have come standard with every motor.

At steady state, the electrical power input to the motor is exactly equal to the output mechanical power at the motor shaft, minus any losses.

You can arrive at this conclusion by considering the conservation of energy. If the amount of power input to the motor were more than the amount of power being taken by the load, the extra power would have to be going somewhere. (In real life, excess power is stored as kinetic energy - the motor accelerates.)

To illustrate that power in is proportional to power out, see the example motor curves below. Note that power input to the motor (red curve) is roughly proportional to power output. Also note that power at no load is 12,100 watts - this represents the losses of the motor at no load, i.e. friction losses, iron losses, copper losses. You are right and someone is wrong. a motor with no load draws little current and also has a low power factor. when loaded he power factor increases and current draw increase.

power consumed is a slippery term, for example the first law of thermodynamics puts it at 0

motors produce heat mostly by electric resistance, so heating is proportional to the current flowing through the motor, current is basically proportional to torque, so that increases with load too.

Put a large motor like a vacuum cleaner or a circular saw on a long enstension lead with an incandescent lamp. at night start the motor up and you'll see the lamp dim and then brighten again, this if proof that he motor poses a variable load. A similar experiement can be dome with a pocket fan motor flashlight globe and a 10 ohm resistor in series with the battery. (slow the fan by hand ans watch the lamp dim)

About the only thing likely to be constant is reactive power

Power consumed by a motor = power required by the load + losses. So you were correct.

The confusion that your "friend" has may be related to the fact that a PORTION of those losses in the motor are fixed, meaning there are losses associated with just making the chunks of iron and strands of copper act like a motor at all, and those losses remain the same regardless of loading on the motor. So as the load on the motor decreases, even though the QUANTITY of power lost decreases with the load decreasing, the PERCENTAGE of total load that those fixed losses represent does go up.

So for a rough rough example let's say you have full load on a motor that is drawing 1000W with 100W of total losses, and 10W of that is fixed losses in the motor, so 10% of the losses are the fixed losses. If the load on the motor drops to 500W, and the losses drop to maybe 60W, but that 10W loss in the motor itself remains constant, that 10W represents 17% of the total losses now, instead of 10%. That sort of issue gets some people confused into thinking that the actual total power doesn't change because "losses increase". It is a RELATIVE increase in PERCENTAGE, not an ACTUAL increase that takes place.