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depending on kva class of the motor between 4-8 times the running current
The start up current should be listed on the motor nameplate as FLA , full load amps.
To create rotating magnetic field inside motor stator and it is done by capacitor. current drawn by motor running winding is lagging in nature when capacitor is connected in series with starting winding then the phase angle of running winding current and starting winding currents changes which creates a rotating magnetic field and motor is able to run.
How do you calculate voltage drop for starting motor current
A series-wound commutator motor has the best starting torque because the torque is proportional to the square of the current, and the starting current is set by a current-limiting resistor which is switched out as the motor builds up speed.
depending on kva class of the motor between 4-8 times the running current
When you first turn on a motor it is starting from a static position and more current is required to get the motor up to speed (Starting current) than to keep it running (running current). Since watts equals amps times voltage you can see the difference in wattage is related to current. If you look at watts as work being done it is obvious that it requires more work to get the motor running than to keep it running.
The start up current should be listed on the motor nameplate as FLA , full load amps.
To create rotating magnetic field inside motor stator and it is done by capacitor. current drawn by motor running winding is lagging in nature when capacitor is connected in series with starting winding then the phase angle of running winding current and starting winding currents changes which creates a rotating magnetic field and motor is able to run.
How do you calculate voltage drop for starting motor current
It isn't. It is only kept at maximum resistance when the motor is not running. That is done to limit the starting current.
At a steady state (when up and running) Watts = Volts x Amps so you just need to solve the equation 75,000 / 400 = ? However, you add the caveat "starting". If your application involves a motor, for example, the starting current will be higher than the steady state current. One rule of thumb says the starting current may be as high as 6 times the running current. This higher current starts high and then decreases as the motor comes up to speed.
Motor starting current is typically 5-7 times the rated current of the motor. (For three phase induction motors)
what is the starting current of 2ton window ac voltas
The starting current is high because when the motor is not rotating no back-emf is generated, leaving the starting current to be determined by the armature resistance, which should be low.
A series-wound commutator motor has the best starting torque because the torque is proportional to the square of the current, and the starting current is set by a current-limiting resistor which is switched out as the motor builds up speed.
A motors locked rotor current is the same as the motors starting current. This is the point where the voltage is applied to a non rotating motor, time cycle zero. Because the motor is not rotation and generating a back EMF to oppose the inrush current, the current will go as high as 6 times the running current of the motor.