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 Î¦ 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.

_{a}R_{a}due to armature resistance R_{a}._{a}X_{s}due to armature leakage reactance X_{L}.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}voltFrom 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 - Î¦ = 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)

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

## Phasor Diagram at Unity Power Factor Load :

When the alternator is driving a unity power factor load (resistive) i.e., cos Î¦ = 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 Î¦ = 1 and sin Î¦ = 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 Î¦. 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 Î¦ 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.