# Input power & Torque to a DC Motor

I have an DC motor connected across a 12V supply, I got the rpm around 200 rpm. Armature Terminal Voltage of 9V. Winding resistance = 1ohm How can I find the torque of the motor and the Input power without using an external device. EDIT:The External Device is stalling the motor in this case.

• There isn't really a direct relation between input power and torque of the motor: You have no means of knowing what resistive heating losses the motor has, nor the core losses, nor stiction parameters, nor bearing and mounting losses. Nov 10, 2013 at 8:37

You would have an easier time if you knew the motor constants as I describe in detail to my answer to this question: How to improve torque and RPM of a DC motor? You should be able to get those from the motor manufacturer and they would take into account the losses mentioned by the others.

But you can get a ballpark estimate given what you have. Given the voltage, armature resistance and Ohm's law, you are able to find the current and therefore get the input power.

You can then determine torque at a known speed with the following equation:

$P_{in} = P_{out}$
$V \cdot I = \tau \cdot \omega$

Where:

$\omega = \text{angular velocity in radians per second}$

So you would just need to convert RPM to rad/sec using:

$\dfrac{RPM \cdot 2 \cdot \pi}{60} = rad/sec$

The input power is volts multiplied by amps. Without an external device such as a mechanical load on your motor you can't determine torque.

With the motor unloaded there is torque but this is due to bearing friction and windbag. It's impossible to determine this by measuring power because there are also copper losses in the stator and armature.

Knowing the armature dc resistance helps to get a better estimate but realistically if you want to determine torque you need to explain "external device"

• I am sorry to not have mentioned that. The motor is being stalled by a brake. Nov 10, 2013 at 17:35
• @cheeky please modify your question to describe the full picture. Nov 10, 2013 at 18:07

$T = K\phi I_a$ .....(1)

where $K$ is a constant, $Ia$ is armature current.

Back-emf $E = KØω$ ....(2)

Dividing (1) and (2),

$\frac{T}{E} = \frac{I_a}{ω}$

$T = E(\frac{I_a}{ω})$.....(3)

Now, $ω = 2π(200/60)~~rad/sec = 20π/3~~~rad/sec$

$I_a = (V -E)/R_a$ where $V$ = Terminal voltage, $R_a$ = Armature Resistance

$I_a= (12 - 9) / 1 = 3 A$

Thus from (3), Torque of DC Motor $= 9(3\times3 /20π ) = 81/20π = 1.3~N$ (Ans.)