Questions about electric motor power consumption

P = VI

since V = IR, the following also work

P = I^2/R
P = V^2/R

everything follows ohms law, if something didn't, it wouldn't be a law

Expanding upon Stuh's expaination a bit.

P, p = power in Watts, or Joules if you use the right units
V, v = voltage in Volts
I, i = current in Amperes.
R, z = resistance in Ohms, reactance in Ohms

Watt's and Ohm's laws are correct, but you must properly phrase them using small letters. The large letters apply for DC currents with only pure resistive components in the ciruit. The small letters apply with reactive components and AC or DC current.

This is very much more than your question, but basically the copper windings of the motor cause the voltage and current to get out of phase with each other. If you were just to put a a wattmeter in the circuit you would not get the true power consumed by the motor. There is power consumed and released as heat, friction, etc, which is not monitored directly by a watt meter because of the phase difference between the current and the voltage. To get the true powe consumed you need a seperate volt meter and ammeter and then multiply their readings.

This gets nicely complicated, and I find it lots of fun, and you can learn about it in a basic electrical engineering course. That's what really kept me in college to get my degree is that kind of stuff.

Kelly

Actually, there are a few excpetions. Most notably semi-conductors. Only ohmic devices follow ohm's law. Also, you left out the affects of temperature on the power consumption of a device.

Wow, you spend one night reading and researching a topic, and suddenly you know alot about it The suns starting to come up, and my research paper is almost complete. Thanks for the response though, I was confused about whether or not motor's were ohmic devices (when i posted that). Now I'm not.

A.c. motors involve impedence, but you did ask about d.c.