# For a synchronous motor, the maximum value of torque developed at an angle of

### Right Answer is:

90 degree

#### SOLUTION

**For a synchronous motor, the maximum value of torque is developed at an angle of 90 degrees.**

When a synchronous motor operates under load it draws active power from the line. The active power absorbs by the motor depends on the supply voltage E and excitation Voltage E_{o}.

Therefore power is given as

**P = ( E × E _{o}/X_{s})sinδ.**

Where

** P** = mechanical power of the motor per phase

**X _{s}** = synchronous reactance per phase

**δ**= Torque angle between E

_{o}and E.

This equation shows that the mechanical power increase with the torque angle and its maximum value is reached when **δ = 90°**.

Therefore peak power per phase is given as

**P _{max} = E_{o}E/X_{s}**.

**The maximum torque occurs at a torque angle of 90 electrical degrees.** This is called the pull-out torque, which indicates the maximum value of torque that a synchronous motor can develop without pulling out of synchronism. In general, its value varies from 1.25 to 3.5 times the full-load torque. In actual practice, however, the motor will never operate at a torque angle close to 90 electrical degrees because the stator current will be many times its rated value at this operation.

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