Load characteristics and capacitive load voltage rise

Load (external) characteristics

The load or external characteristic of an alternator is the curve of terminal voltage $V$ against load (armature) current $I_a$, at constant speed and constant field current, for a given load power factor.

  Vt
  |                 ___.-- leading pf
  |          ___.--'
  |E ------'----------.___   unity pf
  |                       `--.
  |    `--.__                 lagging pf
  |          `--.__
  +--------------------------------> Ia
  0                  rated Ia

• All curves start at $V = E$ (no load).
• Lagging pf: voltage falls sharply because of the demagnetising armature reaction and the large $IX_s$ drop in phase with $V$.
• Unity pf: voltage falls slightly (cross-magnetising effect, $IR_a$ and small quadrature drop).
• Leading pf: voltage rises with load because armature reaction is magnetising.
• Hence a lower pf (lagging) gives poorer regulation; an automatic voltage regulator changes the field current to hold $V$ constant.

Why $V > E$ for capacitive load

Per phase, $\bar E = \bar V + \bar I(R_a + jX_s)$ and

$$E = \sqrt{(V\cos\phi + IR_a)^2 + (V\sin\phi - IX_s)^2}\quad\text{(leading pf)}$$

The reactive term $V\sin\phi - IX_s$ is reduced by the reactance drop, so $E$ can be smaller than $V$.

Armature reaction view: with a capacitive load the current leads the emf (by 90° for a pure capacitor). The armature mmf then acts along the axis of the field poles and aids the field flux (magnetising effect). The air-gap flux increases, so the voltage at the terminals is higher than the open-circuit emf $E$ produced by the field current alone.

 Leading pf phasor diagram
             I      E
              \    /'.
               \  /   '.  j I Xs
                \/      '.
                O----------+ V      (E < V)

Consequence: voltage regulation is negative; less excitation is needed at leading pf, and over-voltage can occur when a lightly loaded long line or capacitor bank is fed.