I'm making this post mostly to check things for another post I'll make (if I don't find any error up to this point).

I'm trying to calculate the efficiency for an ideal DC series motor as a function of the speed, without any load attached to it.

Once I'm considering it's ideal (in classical mechanics), then there's no friction and it will accelerate indefinitely for any given input voltage U. The mechanical output is going to be entirely converted into rotational kinetic energy at any moment.

The expressions I developed for the series motor are:

T = KtI²

Torque is proportional to the square of the current with a constant Kt

I = (U-Eb)/R

The current is equal to the resulting voltage (applied voltage minus back emf) divided by the total series resistance

Eb = KeNI

The back emf is proportional to the speed and the current (once it's the same for the field winding) with a constant Ke

Solving this the current and the torque will be:

I = U/(R+KeN)


The mechanical power (Pm), electrical power (Pe) and Efficiency (E) will be:


Pe = U²/(R+KeN)


The efficiency varies only with the speed, and as N tends to infinity, the expression tends to E=Kt/Ke

I'm unsure of the meaning of this result.

I know that at least until the expression for efficiency the dimensional analysis checks, both sides of the fraction have the unit Ohm (N·m·s/C²), so the result is dimensionless as efficiency should be.

This also shows that the efficiency is independent of the applied voltage or the torque its exerting, the only variable in there is the speed.

Is this true for a situation with load?

I mean, if I find the equilibrium speed for a given load and applied voltage, then I can find the efficiency directly through that expression?

  • \$\begingroup\$ "I'm trying to calculate the efficiency for an ideal DC series motor as a function of the speed, without any load attached to it. - Efficiency is power out / power in, so with no load the answer is easy - 0%. \$\endgroup\$ Apr 16, 2021 at 3:51
  • \$\begingroup\$ @BruceAbbott There is mechanical power, as I said, it's accelerating the rotor, converting electrical energy into rotational kinetic energy indefinitely (once I'm considering an ideal motor with no friction and no limit to its speed). And I'm talking about the efficiency of that conversion as a function of N. \$\endgroup\$ Apr 16, 2021 at 3:53
  • \$\begingroup\$ The energy is stored inside the motor, so by definition the efficiency (output power / input power ) is zero. But then you ask whether it is true with a load at 'equilibrium speed', which is a totally different situation. The mechanical energy stored as the motor gets up to speed has no impact on efficiency while running at 'equilibrium speed'. \$\endgroup\$ Apr 16, 2021 at 4:07
  • \$\begingroup\$ @BruceAbbott I know they're different situations, but all in all I'm considering a purely inertial load (the rotor itself) and this doesn't necessarily change how the power conversion works. I just chose this load for simplicity, but the mechanical power expression was derived solely knowing the expression for current and torque as a function of N and U. \$\endgroup\$ Apr 16, 2021 at 4:18
  • \$\begingroup\$ Kinetic energy stored in the motor is irrelevant. You might as well just call it zero, since an ideal motor has no inertia and reaches 'equilibrium speed' instantaneously. \$\endgroup\$ Apr 16, 2021 at 4:36

2 Answers 2


Given that, for an ideal motor, \$K_t = K_e\$, I think you've just found that even when you include armature resistance, as the current goes to zero the efficiency goes to 1.

Or, in other words, the more that you can neglect the one and only non-ideal part in your model, the more the motor resembles the ideal motor.


If you consider ideal motor inertia=0, then all electrical energy input will be transformed in mechanical energy and no losses. Efficiency will be 100%. If not, some losses have been dissipated in R during acceleration (think than, in the start (speed=0), current will I=U/R, so power loss is U^2/R @ t=0). Current will decrease during acceleration until 0, so integrating this power loss (I^2*R) with inertia data you can calculate acceleration losses.


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