The simplest way to start a three-phase induction motor is to connect its terminals to the line. In an induction motor, the magnitude of the induced EMF in the rotor circuit is proportional to the stator field and the slip speed (the difference between synchronous and rotor speeds) of the motor, and the rotor current depends on this EMF. When the motor is started, the slip speed is equal to the synchronous speed, as the rotor speed is zero (slip equal to 1), so the induced EMF in the rotor is large. As a result, a very high current flows through the rotor. This is similar to a transformer with the secondary coil short circuited, which causes the primary coil to draw a high current from the mains. When an induction motor starts DOL, a very high current is drawn by the stator, on the order of 5 to 9 times the full load current. This high current can, in some motors, damage the windings; in addition, because it causes heavy line voltage drop, other appliances connected to the same line may be affected by the voltage fluctuation. To avoid such effects, several other strategies are employed for starting motors.
Across the line starting of a motor can be as high as 300% of the full load amps.
armature
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.
Large DC motors with field windings instead of permanent magnets present a very heavy load when starting. To prevent fuses blowing, a series of ever smaller resistors are sequentially switched in series with the motor. As the motor picks up speed, a back EMF in opposition to the applied voltage limits the maximum current. When the motor reaches it's running speed the 'starter' is out of circuit.
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.
because of starting current of induction motor is very high and it damages the insulation of motor
Across the line starting of a motor can be as high as 300% of the full load amps.
armature
How do you calculate voltage drop for starting motor current
Yes. Circuit breakers are designed to accommodate for a short-lived current spike. The motor does not draw high current for long at starting and hence it's possible.
Motor starting current is typically 5-7 times the rated current of the motor. (For three phase induction motors)
load is heavier so starting torque is requiredAnswerBecause the same current is passing through both the armature and field windings, the torque is proportional to the squareof the current. Since the starting current is alway high (no back emf), the torque will be very high indeed.
Series motors have the highest starting torque. The torque is proportional to the square of the current, and the starting current is so high it has to be limited by a resistor called a rheostat. Series motos are used mainly on trams and trolley buses.
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.
what is the starting current of 2ton window ac voltas
It isn't. It is only kept at maximum resistance when the motor is not running. That is done to limit the starting current.