I couldn't find the explanation for this query of mine.
Why back emf in dc motor is nearly equal to the applied voltage on no-load?
I tried to search on the internet but all in vain.
Electronic – Back Emf of dc motor
back-emfdc motor
back-emfdc motor
I couldn't find the explanation for this query of mine.
Why back emf in dc motor is nearly equal to the applied voltage on no-load?
I tried to search on the internet but all in vain.
Best Answer
You can think of the motor as behaving like a resistor connected to a generator (the latter corresponding to the back EMF). It produces a torque that is proportional to the current through the resistor.
The higher the difference between the back EMF and the input voltage, the higher the current. The higher the current, the higher the torque.
If the motor is stalled, the entire input voltage appears across the resistor and a lot of torque is produced. Since it only takes a bit of torque to overcome losses such as bearing friction and windage and magnetic losses when there is no load, the motor spins up until the back EMF is almost equal to the input voltage and the torque has dropped until the motor is no longer accelerating significantly. The current resulting from the remaining difference (multiplied by the input voltage) gives you the power consumption with no load. Efficiency is zero, of course, with no load since it is producing zero output power at the shaft.
The losses in the copper windings are the current squared times the resistance (which is the winding resistance) aka \$I^2R\$ losses.
If the windings had zero resistance the back EMF would always be equal to the input voltage, (even under load) since any difference would allow infinite current to flow, generating infinite torque.