#### In the last article, we have derived the equation for EMF induced in the alternator. The equation for per-phase induced emf E_{ph} in an alternator is given as,
*E*_{ph} = 4 K_{f} K_{c} K_{d} f Ï†* T volt*

When an alternator is loaded the whole induced emf doesn't appear across the output terminals. As the load on the alternator is varied, its terminal voltage V_{ph} also varies due to the following drops,
- Voltage drop I
_{a}R_{a} due to armature resistance R_{a}.
- Voltage drop I
_{a}X_{s} due to armature leakage reactance X_{L}.
- Voltage drop due to armature reaction.

The induced emf in the alternator has to supply the above drops while supplying the load. Therefore the equation for terminal voltage of an alternator is given as,
*E*_{ph} = V_{ph} + I_{a}R_{a} + I_{a}X_{s} volt

From the above voltage equation, let us draw the phasor diagram of a synchronous generator operating at different load power factors. The relation between terminal voltage and current for power factor analysis can be done by the phasor diagram.

Let,- E
_{ph} = Induced emf on load per phase. - V
_{ph} = Terminal voltage per phase - I
_{a} = Armature current - Ï† = Phase angle between I
_{a} and V_{ph} (i.e., p.f.) - R
_{a} = Armature resistance per phase - X
_{s} = Synchronous reactance ( leakage reactance + armature reaction reactance )

Taking V_{ph} as the refernce phasor. The phase relationship between armature induced emf E due to field flux Ï†_{f} and the current flowing through the armature I_{a} depends upon the power factor of the load.

*E*

_{ph}= 4 K_{f}K_{c}K_{d}f Ï†

*T volt*

_{a}R_{a}due to armature resistance R_{a}._{a}X_{s}due to armature leakage reactance X_{L}.*E*

_{ph}= V_{ph}+ I_{a}R_{a}+ I_{a}X_{s}volt_{ph}= Induced emf on load per phase._{ph}= Terminal voltage per phase_{a}= Armature current_{a}and V_{ph}(i.e., p.f.)_{a}= Armature resistance per phase_{s}= Synchronous reactance ( leakage reactance + armature reaction reactance )## Phasor Diagram at Unity Power Factor Load :

#### When the alternator is driving a unity power factor load ( resistive ) i.e., cos Ï† = 1. The armature current I_{a} will be in phase with V_{ph} as shown below.

#### From the triangle OBC, the expression for induced emf E_{ph} is given as,

#### At unity power factor cos Ï† = 1 and sin Ï† = 0. The equation is modified as,

## Phasor Diagram at Lagging Power Factor Load :

#### For lagging power factor loads the current I_{a} will lag the terminal voltage V_{ph} with an angle Ï†. At zero lagging power factor ( pure inductive ) the current I_{a} lags the voltage V_{ph} exactly by 90°. The below shows the phasor diagram for the lagging power factor.

#### The armature resistance drop I_{a}R_{a} is due to armature current I_{a}. Hence it always lies in phase with current I_{a} i.e., DE. Therefore, from the triangle OCE,

## Phasor Diagram at Leading Power Factor Load :

#### Similar to the lagging power factor the current I_{a} leads the voltage V_{ph} by Ï† at leading power factor loads as shown below. At zero leading power factor ( pure capacitive ) current I_{a} lead V_{ph} exactly by 90°.

#### From triangle OCE,

### From the above, we can conclude that,

- For unity and lagging p.f. the sign of I
_{a}X_{s} will be positive. Because the armature reaction effect in X_{s} will be demagnetizing and cross-magnetizing effect at unity and lagging power factor. - For leading p.f. the sign of I
_{a}X_{s} will be negative. Thus V_{ph} is more than E_{ph} due to the magnetizing effect of armature reaction.

_{a}X_{s}will be positive. Because the armature reaction effect in X_{s}will be demagnetizing and cross-magnetizing effect at unity and lagging power factor._{a}X_{s}will be negative. Thus V_{ph}is more than E_{ph}due to the magnetizing effect of armature reaction.