In this article let us draw the phasor diagram of synchronous motor and also derive the expression for back emf E_{b} and load angle α for various factors.

### Analysis of Phasor Diagram Under Normal Conditions :

#### Let,- V = Supply voltage per phase
- I
_{a} = Armature current per phase - Φ = p.f. angle or angle between V and I
_{a} - cos Φ = p.f. at which motor is working
- α = Load angle or Torque angle corresponding to the load on the motor

The phasor diagram with all the above details at normal excitation is shown below.

_{a}= Armature current per phase_{a}When the value of back emf is equal to applied voltage E_{b} = V then the synchronous motor is said to be at normal excitation. The angle θ is called an internal machine angle or impedance angle. It is constant for a motor.

The significance of θ is that it tells us that I_{a} lags behind E_{r} by an angle θ. Practically armature resistance R_{a} is very small compared to reactance X_{s} and hence θ tends to 90°. It is expressed as,

From the phasor diagram, armature current is given by I_{a} = E_{r} / Z_{s} and, synchronous impedance Z_{s} = R_{a} + jX_{s}. The vector difference of E_{b} and V gives the resultant emf E_{r} which represents I_{a} Z_{s}. The resultant emf E_{r} is expressed as,

The nature of the power factor is lagging if I_{a} lags V by angle Φ while it is leading if I_{a} leads V by angle Φ. Let us see the phasor diagram and expression for back emf and load angle at different power factor loads.

### Phasor Diagram at Lagging PF :

When field excitation is made in such a way that back emf is less than the applied voltage (E_{b} < V) then the motor is said to be 'Under-Excited'. Here the torque angle α is small and I_{a} lags behind V with poor power factor angle Φ. The phasor diagram is shown below.

#### Applying cosine rule to triangle OAB,

#### Applying sine rule to triangle OAB,

#### Hence load angle α can be calculated once E_{b} is known.

### Phasor Diagram at Leading PF :

When excitation is increased in such a way that E_{b} > V, the motor is said to be 'Over-excited'. Here the current I_{a} is comparatively larger and leads the voltage V by an angle Φ as shown below.

#### Applying cosine rule to triangle OAB,

#### Applying sine rule to triangle OAB,

#### Hence load angle α can be calculated once E_{b} is known.

### Phasor Diagram at Unity PF :

The change in excitation for which the armature current will be in phase with voltage so that power factor becomes unity, this occurs when E ≅ V. Therefore Φ = 0 and cos Φ = 1.