Why Squirrel cage rotor is short circuited?

[niko_bellic]

In a 3-phase squirrel cage induction machine, why the rotor terminals are short circuited?

 

[vk6kro]

This is because we want a large induced current to flow in these windings.

This current flow produces the magnetic field which makes the rotor turn.

 

[Bob S]

Unlike a dc brush motor, there are no windings on the rotor. The only way the stator magnetic field can produce a torque on the rotor is when there are induced currents in the squirrel cage in the rotor. The are no induced currents in the rotor itself because it is laminated. The currents are induced in the squirrel cage only when the rotor RPM slips relative to the synchronous RPM. For example, for 60 Hz ac power, there is no induced current and no torque at 1800 RPM (synchronous), but when the rotor RPM drops to ~1740 RPM, there are induced currents in the squirrel cage and a resultant torque. If the squirrel cage were not shorted, there would be no currrent and no torque.

Bob S

 

[m.s.j]

Shorting of rotor bars in form of squirrel cage cause the suitable low impedance path for rotor induced currents. The impedance characteristics (R & X) of rotor winding infects the motor performance, starting torque, and total losses consequently motor efficiency.
When the rotor is stationary, the revolving magnetic field cuts the short-circuited secondary conductors at synchronous speed and induces in them line-frequency currents. To supply the secondary IR voltage drop, there must be a component of voltage in time phase with the secondary current, and the secondary current, therefore, must lag in space position behind the revolving air-gap field. A torque is then produced corresponding to the product of the air-gap field by the secondary current times the sine of the angle of their space-phase displacement.
At standstill, the secondary current is equal to the air-gap voltage divided by the secondary impedance at line frequency. As the rotor speeds up, with a given air-gap field, the secondary induced voltage and frequency both decrease in proportion to s. Thus, the secondary voltage becomes sE₂, and the secondary impedance R₂+jsX₂ .
The maximum torque is approximately equal to E²/2X. This gives the basic rule that the percent maximum torque of a low-slip polyphase motor at a constant impressed voltage is about half the percent starting current.
By choosing the value of R₂, the slip at which maximum torque occurs can be fixed at any desired value. The maximum-torque value itself is affected, not by changes in R₂, but only by changes in X and to a slight degree by changes in X_M.

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