The FLA of a motor is the amperage that the motor operates at. So to answer the question, 100%.
About one tenth, but at a low power factor so that the no-load power might be only 3% of the on-load power.
approximately 67% of full load current
caca
2 to 5% of full load current
This is a good indication that the motor is being overloaded. If the motor load is belt driven, remove the belt and then check the motor amps. If it goes back to normal FLA then there is a problem with the driven load. If the motor amperage stays high check the motor bearings for the problem.
The ratio is between 30% to 40 %
Across the line starting of a motor can be as high as 300% of the full load amps.
The current flowing through a transformer's secondary is the current drawn by the load, so it will be exactly the same as the current flowing through your induction motor -assuming that is the load. Don't really understand the point of your question!
Full load starting current is typically in the region of 5or 6 times the full load motor current;.
2 to 5% of full load current
As the no load current is the current due to core losses of the motor which is very small .in no load terminals are open circuited no current flows through it, a small current flows which is due to core
While starting a motor from standstill to its rated speed, the motor has to overcome the inertia and generate enough torque to over come it. In the process the motor takes higher current during the starting. Once started and set in motion the current reduces ti its normal value. Full load current is lower than the starting current normally.
whenever the load increases,the current drawn by the motor to do or to fulfill the required energy to the load. so the current will increase generally.Increase in load will cause the full utilization of motor,so speed of rotor will decrease.
This is a good indication that the motor is being overloaded. If the motor load is belt driven, remove the belt and then check the motor amps. If it goes back to normal FLA then there is a problem with the driven load. If the motor amperage stays high check the motor bearings for the problem.
The ratio is between 30% to 40 %
Across the line starting of a motor can be as high as 300% of the full load amps.
The current flowing through a transformer's secondary is the current drawn by the load, so it will be exactly the same as the current flowing through your induction motor -assuming that is the load. Don't really understand the point of your question!
Depends on what sort of motor it was. If it was the sort you find in an ordinary electric drill, then it would slow down. But rememer, the only way you can reasonably reduce the current in to motor is to reduce the voltage in the supply. A motor takes what current it can, dependent on Ohm's Law. To reduce the current, reduce the voltage. But in a three-phase motor, the speed being dependent on the rotation of the phases, it would more or less stay the same. But this assumes it's not under load. A load would cause it to slow down.
starting current of 3 phase 75 KW induction motor
The starting current is high because the motor's rotor winding has very low resistance. It's similar to a transformer with a shorted secondary windings. As the motor accelerates,the back emf increases which resist the flow of current in the rotor winding. Hence,the current drop to the rated full-load value.