Synchronous Motor - Effect of Excitation on Armature Current & Power Factor

     When synchronous motor is loaded and load is kept constant, the input power drawn by motor will remain constant i.e., √3 V I Cos φ is constant, because input voltage, V and input power are constant. so I Cos φ i.e., active component of current is constant, for constant load.

     When excitation is changed, the magnitude of induced emf changes. The load angle, is also constant for a constant load.

Let,
 
   E_r  =  Resultant voltage between V̅ and E̅_b  =  I_a Z_a

   θ  =  Internal angle  =  Angle between E̅_r and I̅_a  =  Tan -1 ( X_s  / R_a )

   I_a  =  Armature current 

   φ  =  Power factor angle  =  Angle between V̅ and I̅_a

                                         

(i) When the excitation is reduced in such a way that E_b  =  V. Shown in below figure. I_a lags ' V ' by an angle φ.

(ii) When the excitation is reduced in such a way that E_b  <  V, the motor is said to under excited. Resultant  ' E_r ' advanced in clockwise direction and an angle ' θ ' is constant, I_a also moves in clockwise direction.

     As seen from the above figure angle ' φ ' is increased and power factor is decreased. To maintain I_a Cos φ constant, I_a is increased. Therefore for low values of excitation, I_a increases and power factor is lagging in nature.

(iii) When excitation is increased in such a way that E_b  >  V, the motor is said to be over excited. Resultant  ' E_r ' is moved in anti-clockwise direction and so the ' I_a ' also moves in anti-clockwise direction ( as angle between E_r and  I_a is  θ ). Current I_a leads voltage by an angle ' φ '.

     As excitation goes on increasing. E_r and I_a go on moving in anti-clockwise direction, angle φ increases. power factor decreases and I_a increases as I_a Cos φ  =  constant. Thus for high values of excitation. current I_a increases and power factor is leading in nature.

(iv) For unity power factor. E_b is slightly greater than V ( E_b  ≈  V ). This is shown in below figure. The excitation for which the power factor of the motor is unity is called as normal excitation. Then I_a is in phase with V. 

     Now I_a Cos φ  =  Constant, Cos φ  = 1 is at its maximum hence I_a, is minimum for normal excitation.

The Above Conditions Clearly Indicates the Following 

(i) A synchronous motor draws a lagging current, when it is under excited.

(ii) At normal excitation, it draws minimum armature current and power factor is unity. 

(iii) The motor draws a leading current, when it is over excited.