# DC motor electrical and mechanical power relationship

## Main Question or Discussion Point

Hey guys, I know I still have a tread unanswered but I got a project in which I have to include some physical theory about the working of motors. I know P=V*I and also P=T*W where V is voltage, I is current, P is power, T is torque and W is angular speed. My question is, can I say V*I=T*W?. I believe it is not, I think V*I is input power and T*W is output power. I got a datasheet from the motor I´m gonna use and the graph seems to confirm my hypothesis. However, as always, your professional answer will be really appreciated.
Thank you :)

Related Electrical Engineering News on Phys.org
I forgot to add the datasheet, if I divide 0.073 w which is the output power on the chart and divide it by the power obtained by multiplying V*I=0.1525 w I get 0.478 which is efficiency and looking in the graph below I get 0.32 mN*m of torque which seems correct graphically. These data where obtained from the maximum efficiency section.
http://www.jameco.com/Jameco/Products/ProdDS/238465.PDF

billy_joule
I know P=V*I and also P=T*W where V is voltage, I is current, P is power, T is torque and W is angular speed. My question is, can I say V*I=T*W?.
Yes you can say that, but what that implies is that input power = output power. That may be a valid assumption in some cases.
But, of course, in many cases it's not valid and we must consider the fact that no motor is 100% efficient.

jim hardy
Gold Member
2019 Award
Dearly Missed
Draw in your mind a box enclosing said motor.

Energy in = energy out
it must come out as either mechanical work via shaft
or as heat lost to inefficiency.

Unloaded, the I^2R losses in the armature's copper are smaller than at full load
but the friction and windage losses, a function of speed, don't care about load.

So your assumption that input power X efficiency = output power is correct.
Check efficiency at very low , intermediate and very high loads
since load and current I are roughly proportional
you'll probably find a y=ax2+b, efficiency = (some constant m) X load2 + (some amount b) of friction and windage.
You can refine it further since speed affects windage, and you should just for the academic interest.

that's my 2 cents.

CWatters
Homework Helper
Gold Member
The efficiency of the motor = 100 * power out/power in.

This can vary depending on the load on the motor. Eg no load means the motor torque is zero and so the efficiency is zero.

Close to max load the efficiency is usually between 60 and 95% depending on the type of motor. Some manufacturers will put the efficiency in their data sheets.

Thank you for your answers :) I'm understanding this a lot more. I got another question. Is it possible to measure the voltage drop in a dc motor directly with a multimeter? The last time I tried I got random values in the multimeter. A friend of mine told me I had to white until the reading was stable but I still had my doubts.

jim hardy
Gold Member
2019 Award
Dearly Missed
Is it possible to measure the voltage drop in a dc motor directly with a multimeter?
IN or ACOSS ? How are you measuring it?
Sparking at the brushes makes electrical noise, aka "static"
an analog meter may work better for you.

Answer is "Yes and it's not difficult."
You can even touch a meter lead gently to the commutator to measure voltage "inside" if that's what you meant... i use a graphite drafting pencil lead held in an alligator clip so as to not scratch the soft copper commutator.

Merlin3189
Homework Helper
Gold Member
I'm not sure why you're having problems with measuring the voltage. It certainly can be done.
I guess the problem is the noisy voltage. Perhaps if you put a capacitor (say 1uF) across the motor, that would be enough to smooth the voltage.

Waiting for the voltage to be stable should just mean waiting until the speed is constant. If the speed is varying, the voltage is varying. If the variation is slow enough, you should be able to take measurements, but it is difficult to record voltage and current simultaneously unless you have some automated system, because you can't look at two meters at once.
Digital multimeters add to your problems when voltages are varying. Mine is a pretty poor one and only takes about one reading per second, but better meters which take 3 or 4 readings per second still make the display hard to read. Some do have a hold function which is very helpful. Locking any autoranging function helps. I would generally use my analogue meter for this sort of task, as it gives a continuous reading.

Coming back to the OP a few details which don't seem to have been noted yet:
The stall current is given, 0.22 A, so you can calculate the winding resistance. If the applied voltage is 2.5 V (I think it is, though the sheet is less than clear), R=2.5 V / 0.22 A about 11.4Ω
Then at any current you can calculate how much voltage is used overcoming this resistance. The remaining voltage x the current is the power that is used to overcome other losses (usually small) and generating mechanical power to drive the load.
The non-resistive part of the voltage is proportional to the speed.
The torque should be proportional to the current. (Though it clearly isn't according to the data here.)

But a caveat about all these calculations: in my experience the calculations and the declared figures in the datasheets rarely match exactly. For example, if you look at this motor at peak efficiency, the power out is given as 73mW but the torque 0.00033 Nm and speed 2190 rpm give a power of 75mW. Only 2.7% difference, but when all your data can have this level of uncertainty, it's easy to end up with significant errors when you try to do calculations based on 3 or 4 data.