the rotating device can act as a motor and a generator. When you supply to that device, then it works as a motor, if you draw voltage that is generated from the rotating device by rotating it, you call it generator. So, when you supply voltage to the rotting device, you wil have the emf, that is the voltage generated to counter the your supply voltage
Thanks for the reply.
1 more things, the speed of the motor is based on a average current right?
then, how to relate it when we provide more voltage to the motor, then the motor speed will run more faster?
can u tell me how this relate to the current?
because as i understand : V(supply voltage)-CEMF(counter emf)= I(current)R(resistance) and the CEMF = w(speed).k(constant of motor)
now i dunno why when we increase V, then the speed will increase as well.
thank.
regards,
sysysy
---------- Post added at 15:04 ---------- Previous post was at 14:39 ----------
The shown equations clarify, that the statement is incorrect. Saying it's based on voltage would be almost correct in contrast.
The motor type hasn't been mentioned yet. To avoid misunderstandings, it should be said, that the equations are valid for a motor with constant excitation, e.g. a PMDC motor.
So what's the role of motor current? Basically it's proportional to the load torque. An unloaded motor is consuming only a small idle current, representing the torque of friction losses and additional electrical losses like armature eddy currents. They are in fact speed dependent, and in so far, the idle current is based on speed, rather than the other way round.
So i have tried to come out the equation w = V/k - TR/k square
so from equation above, when increase of V, then w will increase. ( there is ntg relate with I, am i correct)?
but i am not really understand with that idle current.
because when we reduce the gear of motor, it will increase torque of the motor but reduce speed of motor.
anyway, from what i have analysed in the equation
w (↑) = V (↑) /k- TR/k².... i assume v,k and r is constant. so when T is large, then w will decrease, and when T is small, w will be increase...am i correct?
this is becaus v is already fix...so when v minus large T....w become small and when V minus small T, then w become larger....
Inverse proportional is usually understood as f(x) ~ 1/x not f(x) ~ -x, that's why I'm asking.
I think, the discussed equation is pretty clear. Not considering generatoric operation, the motor has it's maximum speed with no load (T=0), the speed drops, when the load increases.
Some people call it a "theory", a practical man measures motor speed with different loads, put the points in a diagram, and draw a line through it.
The quoted literature doesn't apply to a PMDC motor, it's about variation of the field current of a DC motor with electrical exictation (shunt wound motor).
Furthermore, the load configurations isn't specified, at least not on the shown page. Thus it's unclear what sets the torque in this experiment.
Can you elaborate on torque and its relation to speed?? Because, In some motors, I have read of Four quadrant operations, where speed and torque maybe directly or indirectly proportional?? Does that relation have anything to do with constant excitation?