In the case of DC motors why is the mechanical power output equal to back emf multiplied by armature current.


2 Answers 2


You probably have no trouble with the concept of a simple rotating coil generator.
A coil is rotated in a magnetic field and an emf $\mathcal E$ is generated.

If there is a resistance $R$ across the terminals of the generator then a current $I$ flows in the circuit.

To simplify matters assume that the frictional forces are very small and the coil resistance is $r$.
Then $\mathcal E = I(R+r) \Rightarrow \mathcal EI = I^2R + I^2r$ where $I$ is the current flowing in the circuit.
The $I^2R$ term represents the useful power out and the $I^2r$ represents the wasted power.

So what is $\mathcal E I$?
It represents the mechanical power input which keeps the coil rotating at a constant speed.

Now suppose that the resistor $R$ is replaced by a battery of emf $E$ with $E < \mathcal E$
In this case the situation is not much different from that when the resistor was in place with $\mathcal EI - EI = I^2r \Rightarrow \mathcal EI = EI + I^2r$ where $EI$ now represents the power supplied to the battery (to recharge it?) whereas with a reistor in place that power is dissipated as heat.

If you increase the emf of the battery such that $E> \mathcal E$ you now have a reversal of the current thorough the coil and $ EI - \mathcal EI = I^2r \Rightarrow EI = \mathcal EI + I^2r$ where $EI$ is the power supplied by the battery and as before $I^2r$ represent the power losses in the coil.

And what of $\mathcal E I$?
Well you call $\mathcal E$ the back emf because it is is trying to drive current in the opposite direction to that produced by the battery and $\mathcal EI$ represents the mechanical power output from the motor.


Short version: with any motor/engine the total energy put in is equal to the total energy put out, averaged over a full cycle, always. In the case of an electric motor, the power put in will be the back emf times the current, because $P=IV$. If you assume that heat losses are small, then all of that power has to come out the other end of the motor as work.

It's no different, fundamentally, than if I lift something with a lever the work I do on the lever will be equal to the work done by the lever on the thing it lifts.


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