A 3 phase synchronous motor is connected to AC mains, is running at full load and unity power factor. If its shaft load is reduced to half, with field current held constant, its new power factor will be:

In a 3-phase synchronous motor

\(V = E\ \cos δ\)

Where, 

V is terminal voltage

E = Back emf

δ is load angle

• With constant excitation, \(E\) is constant but \(δ\) decreases as load decrease.
• So that cosδ is increases.
• Then \(E\ \cos δ > V\), the machine operates at over-excitation.
• Thus operating at a leading power factor.

Note:

• Over-excited (\(E\ \cos δ > V\)) synchronous motor works at a leading power factor.
• Normal excited (\(E\ \cos δ = V\)) synchronous motor works at a unity power factor.
• The under-excited (\(E\ \cos δ < V\)) synchronous motor works at lagging power factor.