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Effect of excitation on armature current and pf

Short answer

With constant load, varying DC excitation changes Eb. Under-excitation (Eb<V) yields lagging pf and high current; normal excitation (Eb≈V) gives unity pf and minimum current; over-excitation (Eb>V) yields leading pf and increased current. Armature current follows V-curves; pf follows inverted V-curves.

Explain the variation of armature current and power factor for different values of excitation current (normal excitation, under-excitation and over-excitation) in a synchronous motor with suitable figures and phasor diagrams.

In a synchronous motor with constant load (constant power P=3VIcos⁡ϕP = 3VI\cos\phi), varying the DC field current changes the back emf EbE_b. Since VV and PP are fixed, Icos⁡ϕI\cos\phi stays constant, so the tip of the current phasor moves along a line perpendicular to VV, while Ebsin⁡δE_b\sin\delta also stays constant.

(a) Normal excitation

  • Field current gives EbE_b such that the current is in phase with VV: unity pf.
  • Armature current is minimum for that load.

(b) Under-excitation

  • EbE_b is smaller; the resultant voltage Er=V−EbE_r = V - E_b swings so that the current lags VV.
  • Armature current is larger than at normal excitation; the motor draws lagging reactive power (magnetising current) from the supply.

(c) Over-excitation

  • EbE_b is larger; the current leads VV.
  • Armature current again increases; the motor supplies reactive power to the supply (synchronous condenser action).
 Phasors at constant load (V reference, Rs = 0)

   I3 (lead, over-excited)
     \
      \     I1 (upf, normal)
  O----+------> V      I tip moves on the
      /                vertical line I cos(phi)
     /                 = constant
   I2 (lag, under-excited)

V-curves and inverted V-curves

  Ia                         pf
  |  \       /  full load    |      _
  |   \     /                |    /   \  (inverted V)
  |  \ \   / /               |   /     \
  |   \_\_/_/  no load       |  /       \
  +------------> If          +--------------> If
  lag | upf | lead          lag  upf  lead
  • V-curves: IaI_a vs IfI_f at constant load, shaped like a "V". The minimum point of each curve is at unity pf. The left side (under-excited) is lagging, the right side (over-excited) is leading. Higher load gives a higher curve.
  • Inverted V-curves: pf vs IfI_f; the peak is unity pf. The locus of minimum points (unity pf) shifts to the right as load increases, since more excitation is needed at higher load.
ExcitationpfArmature current
UnderLaggingHigh
NormalUnityMinimum
OverLeadingHigh

This property lets an over-excited synchronous motor drive its load and also correct the plant power factor.

✓ Answer checked by Claude Sonnet 5.5

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