**Subject :**Electrical Machines II (AC Machines)

**Unit :**Alternators

## Modified phasor diagram by two reaction theory

**Phasor diagram:**

The equivalent circuit of a salient-pole synchronous generator is shown in Fig. 1 (a). The component currents I_{d} and I_{q} provide component voltage drops *jI _{d} X_{d}* and

*jI*as shown in Fig. 1(b) for a lagging load power factor. The armature current I

_{q}X_{q}_{a}has been resolved into its rectangular components with respect to the axis for excitation voltage E

_{0}.The angle ψ between E

_{0}and I

_{a}is known as the internal power factor angle. The vector for the armature resistance drop I

_{a}R

_{a}is drawn parallel to I

_{a}. Vector for the drop I

_{d}X

_{d}is drawn perpendicular to I

_{d}whereas that for I

_{q}× X

_{q}is drawn perpendicular to I

_{q}. The angle δ between E

_{0}and V is called the power angle. Following phasor relationships are obvious from Fig. 1 (b)

*E _{0} = V I_{a}R_{a} jI_{d} X_{d} jI_{q} X_{q} and I_{a} = I_{d} I_{q}*

If R_{a} is neglected the phasor diagram becomes as shown in Fig. 2 (a). In this case,

*E _{0} = V jI_{d} X_{d} jI_{q} X_{q}*

*Fig: 1 *

Incidentally, we may also draw the phasor diagram with terminal voltage V lying in the horizontal direction as shown in Fig. 2 (b). Here, again drop I_{a}R_{a} is || I_{a} and I_{d} X_{d} is ⊥ to I_{d} and drop I_{q} X_{q} is ⊥ to I_{q} as usual.

**Calculations from Phasor Diagram**

*Fig: 2*

In Fig. 3, dotted line AC has been drawn perpendicular to Ia and CB is perpendicular to the phasor for E0. The angle ACB = ψ because angle between two lines is the same as between their perpendiculars. It is also seen that