Basically, when the load is increased on the dc motor, the speed of the motor reduces. In the dc motor, armature develops an emf which opposes the supply voltage V called back emf E_{b}. The resultant voltage across the armature circuit is V - E_{b}. If R_{a} is the armature resistance the expression for armature current is given by,

Therefore the back emf in a d.c motor depends on the armature speed. Similarly, in a synchronous motor, a back emf is set up in the stator (armature) by the rotor flux which opposes the supply voltage. This back emf depends on the excitation given to the field winding i.e., rotor excitation but not on the speed.

When the synchronous motor is loaded, as the name implies it is a constant speed motor. Its speed cannot decrease (i.e., irrespective of load) as the synchronous motor operates at a constant speed known as synchronous speed.

Similar to the d.c motor, armature current for synchronous motor is the voltage across armature divided by impedance of armature circuit. But now voltage difference is the vector difference of V and E_{b}. Therefore armature current per-phase,

#### Where,- Z
_{s} = Synchronous impedance - R
_{a} = Stator winding resistance per-phase - X
_{s} = Synchronous reactance of stator per-phase - V = Supply voltage per-phase
- E
_{b} = Back emf per-phase

_{s}= Synchronous impedance_{a}= Stator winding resistance per-phase_{s}= Synchronous reactance of stator per-phase_{b}= Back emf per-phase## Synchronous Motor on No-Load (without losses) :

When the motor running on no-load with field excitation. Since the load is absent and no losses are assumed in the motor the rotor is free to rotate. Therefore there exists a strong magnetic locking between the stator and rotor axis as shown below.

The back emf E_{b} is equal and opposite to the supply voltage V as shown in the below phasor diagram.

But the vector difference of V & E_{b}is zero. Hence, I_{a} = 0. The motor takes no power as there is neither load nor losses. In practice, this is impossible, as we knew every motor has mechanical losses and iron losses along with small copper losses on it.

## Synchronous Motor on No-Load (with losses) :

Now let us assume the synchronous motor on no-load with losses. If we consider losses in the motor also there exists magnetic locking between stator and rotor. But due to losses, the magnetic locking is in such a way that the rotor axis lies behind the stator axis with a small angle as shown below.

So the rotor axis falls back with respect to the stator axis by angle Î±. This angle Î± decides the amount of current required to produce the torque on the rotor and to supply various losses. Hence this angle is called a load angle or power angle or coupling angle or angle of retardation.

Now through | E_{b} | = | V |. Back emf E_{b} will not be located in exact opposition with V, displaced from its initial position by angle Î± as shown in the above phasor diagram.

Hence the vector difference between E_{b} and V is not zero but gives rise to a resultant phasor E_{r}. But V̅ - E̅_{b} = I̅_{a} Z_{s} = E̅_{r}, this resultant decides the amount of current I_{a} is drawn to produce the torque. Therefore armature current is,

Under the no-load condition, Î± is very small and hence E̅_{r} is also very small. Hence at no-load, the current drawn by the motor is very small.

## Synchronous Motor on Load :

When a synchronous motor is loaded, though the speed remains constant but the load angle Î± increases i.e., the angle by which the rotor pole axis lies behind the stator pole axis increases.

As load angle Î± increases, the resultant voltage E_{r} increases, and the current drawn by the motor also increases. The current drawn by the synchronous motor when there is an increase in the loaded demand has to supply the required torque to drive the load and losses in the motor.

In the phasor diagram, the angle Î± goes on increasing as the load increases. The maximum permissible limit of Î± is 90° electrical, if it increases beyond 90°, magnetic locking between stator and rotor breaks and the motor comes out of synchronism and stops.

Hence, we can say that the maximum amount of torque produced in the motor is at load angle Î± = 90°. The value of maximum torque is called pull out torque. The characteristics drawn between torque produced and load angle Î± is shown below.

### Angle of Retard and Angle of Advance :

The angle with which E_{b} phasor moves from its initial position so as to gives rise resultant emf is called Angle of Retard or Load angle Î±.

*Angle of advance = 180° - Angle of retard*