The slip is proportional to the load torque in an induction motor. This can also be seen as a reduction in speed causing extra current to be induced in the rotor to supply the increased torque.
When load increases, rotor speed decreases. When rotor speed decreases ,the difference (
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!
The ratio is between 30% to 40 %
Induction motors are designed for a certain power. If the power demand is lower than that, you still have to setup the magnetic fields as if you were going to supply that maximum power. As a result, the no-load power factor of an induction motor is quite small, i.e. poor.
The first thing you have to do is find the full load amps of the motor. The wire size feeding the motor has to be 125% of the full load current. The breaker is usually 250% of the full load current. If the voltage and amperage had been added to the question the exact breaker size could have been calculated.
Squirrel Cage Induction Motor Disadvantages:Some of disadvantages or demerits of squirrel cage induction motors are listed below:Main disadvantage of squirrel cage induction motor is that they have poor starting torque and high starting currents. Starting torque will be in the order of 1.5 to 2 times the full load torque and starting current is as high as 5 to 9 times the full load current. In slip ring induction motors, higher starting torque can be attained by providing an external resistance in the rotor circuits during starting of the slip-ring induction motor. This arrangement in slip-ring induction motors also reduces the high inrush currents during starting of induction motor.Squirrel cage induction motors are more sensitive to the supply voltage fluctuations. When the supply voltage is reduced, induction motor draws more current. During voltage surges, increase in voltage saturates themagnetic components of the squirrel cage induction motor.Speed control is not possible in squirrel cage induction motor. This is one of the major diadvantages of squirrel cage induction motors.The total energy loss during starting of squirrel cage motor is more compared to slip ring motors. This point is significant if the application involves frequent starting
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!
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.
rotor speed will decrease....the rotor current wil increase.....
rotor speed will decrease....the rotor current wil increase.....
90/3=30ma
The ratio is between 30% to 40 %
starting current of 3 phase 75 KW induction motor
1. Induction Motor has an air-gap but transformer has no air gap rather it is mutually link. 2. Induction Motor has high no-load current than transformer. 3. Induction Motor is a dynamic device. 4. Induction Motor has high power factor.
due to high reluctance air gap magnetisation current is very high as a result no load current is more
yes
A constant KVA load is an electrical load which has a constant apparent power consumption regardless of small changes in voltage. Constant KVA loads in industry are typically electric motor loads such as induction motors. The interesting thing about an induction motor and why it is called a constant KVA load is, for a constant level of torque output of the motor, if the terminal voltage decreases by say 5% the motor current will actually increase by about 5%. Note that this constant KVA effect will only occure withing an narrow range of voltage deviation. This is the opposite of a static (constant Z) load which if the voltage decreases the current decreases.
The magnetic flux that couples the rotor to the stator will weaken significantly reducing the motor's torque.