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Synchronous motor drawing lagging and leading current

Short answer

A synchronous motor's power factor is controlled by field excitation at constant load. Under-excitation (Eb < V) gives lagging current and inductive behaviour; over-excitation (Eb > V) gives leading current, acting as a capacitor. This is demonstrated by V-curves, where lowest point is unity pf.

A synchronous motor can operate as both inductive and capacitive characteristics. Justify this statement with relevant phasor diagrams. (Explain how a three-phase synchronous motor can be operated to draw lagging as well as leading current from the source.)

A synchronous motor runs at constant speed whatever its excitation. For a constant load (constant input power VIcos⁡ϕVI\cos\phi), changing the field current changes the back emf EbE_b, and hence the magnitude and power factor of the armature current. So the same motor can behave as an inductive (lagging) or a capacitive (leading) load.

Motor equation (per phase, RaR_a neglected):

Vˉ=Eˉb+jXsIˉorIˉ=Vˉ−EˉbjXs\bar V = \bar E_b + jX_s\bar I \qquad\text{or}\qquad \bar I = \frac{\bar V - \bar E_b}{jX_s}

The resultant voltage Eˉr=Vˉ−Eˉb\bar E_r = \bar V - \bar E_b drives the current, which lags ErE_r by 90°.

1. Under-excitation (Eb<VE_b < V): lagging pf (inductive)

           V
   O-------------->
    \`-.        /
     \  `-.   / Er = jIXs
      \    `/
   I   \   Eb (short)
  (lags V)
  • Ebcos⁡δ<VE_b\cos\delta < V; the motor draws a lagging current.
  • The field is too weak, so the armature takes a lagging (magnetising) current from the supply to make up the missing flux. The motor then behaves like an induction motor or an inductor.

2. Normal excitation (Eb≈VE_b \approx V): unity pf

  • Field current is just enough for the required flux; current is in phase with VV and is minimum for that load.
   O--------------> V
   O-------->  I (in phase)
     \       /
      Eb    / Er

3. Over-excitation (Eb>VE_b > V): leading pf (capacitive)

     I (leads V)
      ^       V
      |  O---------->
      |    \        .
      |      \    . Er
      |   Eb (long)
  • Eb>VE_b > V; the resultant ErE_r makes the current lead VV.
  • The motor supplies reactive power to the system and acts as a capacitor. Running at no load like this, it is called a synchronous condenser, used for power-factor improvement and voltage control.

Summary

ExcitationEbE_b vs VV (approx.)CurrentpfActs as
UnderEb<VE_b < VLargeLaggingInductor
NormalEb≈VE_b \approx VMinimumUnityResistor
OverEb>VE_b > VLargeLeadingCapacitor

These variations are shown by the V-curves (armature current vs field current at constant load), whose lowest point is at unity pf.

✓ Answer checked by Claude Sonnet 5.5

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