Why terminal voltage exceeds emf on capacitive load

For a synchronous generator,

$$\bar E = \bar V + \bar I R_a + j\bar I X_l + j\bar I X_a = \bar V + \bar I(R_a + jX_s)$$

where $X_a$ represents armature reaction and $X_l$ the leakage reactance. With a capacitive (leading pf) load, $V$ becomes greater than $E$ for two reasons.

1. Armature reaction is magnetising

• With a purely capacitive load, the armature current leads the emf by 90°. The current in a conductor is maximum before the pole axis reaches it, so the armature mmf acts along the field axis and aids the main flux.
• The resultant air-gap flux increases, so the voltage actually generated in the air gap rises.
• For a practical leading pf (e.g. 0.8 lead), one component of armature mmf is magnetising and the other cross-magnetising; the net effect still raises the voltage.

2. Reactance drop opposes the terminal voltage (phasor view)

The drop $jIX_s$ is perpendicular to $I$. When $I$ leads $V$, $jIX_s$ has a component pointing against $V$, so the vector sum $E = V + I(R_a + jX_s)$ can be shorter than $V$:

             I      E
              \    /'.
               \  /   '.  j I Xs
                \/      '.
                O----------+ V
                   I R  (small, along I)

From the components (taking current as reference):

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

Because the reactive term is $V\sin\phi - IX_s$ (minus sign for leading pf), $E$ can be less than $V$, i.e. $V > E$.

Result

• Terminal voltage rises as the capacitive load increases; the external characteristic rises.
• Voltage regulation $=\dfrac{E - V}{V}\times 100\%$ is negative.
• A smaller field current is needed to keep rated voltage. Example: an alternator feeding a long unloaded cable or a capacitor bank (Ferranti-like rise).