What happens under over excitation and under-excitation?
Consider SM having constant mechanical load (output constant, if losses neglected)
Case-1: Eb=VEb=V known as 100% excitation
armature II lags VV by small angle ϕϕ
θθ with ERER is fixed by stator constants i.e. tanθ=Xs/Ratanθ=Xs/Ra
Case-2: Eb<VEb<V excitation is less than 100%
ERER advanced clockwise and so is II (because it lags behind ERER by fixed angle θθ ).
|I||I| is increased but its PF is decreased (ϕϕ has increased).
Because input as well as VV are constant, hence the power component of II i.e., IcosϕIcosϕ remains same, but watt-less component II sinϕsinϕ is increased.
Hence, as excitation ↓↓, I ↑I ↑ but p.f. ↓↓ so that power component Icosϕ=OAIcosϕ=OA remain constant.
Locus of extremity of current vector is a straight horizontal line
Incidentally, when If ↓If ↓, the pull-out torque is also ↓↓ in proportion.
Case-3: Eb>VEb>V
ERER is pulled anticlockwise and so is II.
Now motor is drawing a leading current.
For some value of excitation, that II may be in phase with V i.e., p.f. is unity
At that time, the current drawn by the motor would be minimum.
Two important points stand out clearly from the prevailing discussion:
IaIa Vs IfIf
|I||I| varies with excitation and has large value both for low and high values of excitation (though it is lagging for low excitation and leading for higher excitation).
In between, it has minimum value corresponding to a certain excitation.
The variations of II with excitation is known as V-curves because of their shape.
cosϕcosϕ Vs IfIf
For the same input, II varies over a wide range and so causes the power factor also to vary accordingly.
When over-excited, motor runs with leading p.f. and with lagging p.f. when under-excited. In between, the p.f. is unity.
The variations of p.fp.f. with excitation looks like inverted V-curve.