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The motor “A” stator is supplied with voltage at a certain frequency. The power available at the stator can be divided and utilized as two parts.
- The power is consumed to overcome I^2 * R and core losses.
- The second part is given to the rotor for its rotation.
The power supplied to the rotor is further consumed into two parts. They are as follows:
- The rotor has to rotate. So, first part is utilized for the rotation of the rotor. “N” is converted into the mechanical power.
- The remaining (Nsa-N) is the power available at slip frequency and this is passed to the slave or auxiliary motor “B.”
Shifting the focus to Slave motor “B” or the auxiliary motor “B”, the power (Nsa-N) supplied to the stator of the motor “B”, is again utilized into two parts.
- To overcome the stator I^2 * R losses and
- Passed to the rotor for its rotation. (Mechanical power).
N — actual speed of the cascaded set.
Nsa — synchronous speed of motor A, being independent of N.
Now when the cascaded set is supplied with the voltage at a frequency to the stator of motor “A”, an e.m.f is induced in the rotor “A” at same frequency. This is supplied to motor “B”, thus both the motors start to develop a forward torque. As the shaft speed increases, the rotor frequency of motor “A” falls thus making the synchronous speed of motor “B” to fall. The cascaded set settles down at a speed which is stable, when the shaft speed equals the speed of rotation of the field of motor “B.”
It can be seen that the mechanical outputs of the two cascaded motor set are in the ratio similar to the ratio N: (Nsa-N).
It is also evident that the mechanical outputs are in the ratio of number of poles of the motors.