In a d.c. generators, we have seen that due to the armature resistance drop and brush drop it is not possible to have all the induced e.m.f. available across the load. The voltage available to the load is called terminal voltage. The concept is same in case of alternators. The entire induced e.m.f. can not be made available to the load due to the various internal voltage drops. So the voltage available to the load is called terminal voltage denoted as. In case of three phase alternators as all the phases are identical, the equations and the phasor diagrams are expressed on per phase basis.

So if E

_{ph }is the induced e.m.f. per phase in the alternator, there are following voltage drops occur in an alternator.i) The drop across armature resistance I

_{a }R_{a}both I_{a }and R_{a}are per phase values.ii) The drop across synchronous reactance I

_{a }X_{s}_{}, both I_{a }_{}and X_{s}are per phase values. After supplying these drops, the remaining voltage of E

_{ph }is available as the terminal voltage V_{ph}.**Note**: Now drop I

_{a }R

_{a}is always in phase with I

_{a }due to a resistive drop while current I

_{a }lags by 90

^{o}with respect to drop I

_{a }X

_{s}as it is a drop across purely inductive reactance.

Hence all these quantities can not be added or subtracted algebraically but must be added or subtracted vectorially considering their individual phases. But we can write a voltage equation in its phasor from as,

This is called voltage equation of an alternator.

From this equation, we can draw the phasor diagram for various load power factor conditions and establish the relationship between E

_{ph }and V_{ph}, in terms of armature current i.e. load current and the power factor cos(Φ).

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