Question

Under steady state short circuit conditions, the armature reaction of a synchronous generator is:

• A. Cross-magnetizing
• B. Demagnetizing
• C. Magnetizing
• D. Partially cross-magnetizing

Hint:

Remember the quick reference for a generator: - Unity Power Factor = Cross-magnetizing - Zero Power Factor Lagging (Short-Circuit) = Purely Demagnetizing - Zero Power Factor Leading = Purely Magnetizing

Answer 1

The Correct Option is B

Concept: Armature reaction refers to the effect of the armature mmf (magnetic motormotive force) on the main field flux distribution of a synchronous machine. The nature of this armature reaction depends entirely on the power factor ($\cos\phi$) of the connected load, which determines the spatial phase angle between the internal induced EMF ($E_f$) and the stator armature current ($I_a$).

Step 1: Analyze the circuit parameters under short-circuit conditions.
During a steady-state short-circuit fault at the terminals of a synchronous generator, the external load impedance drops to zero. The circuit is limited only by the internal machine impedance: \[ Z_s = R_a + jX_s \] In standard large synchronous machines, the armature resistance $R_a$ is negligible compared to the synchronous reactance $X_s$ ($R_a \ll X_s$). Therefore, the short-circuit impedance is almost purely inductive: \[ Z_s \approx jX_s = X_s \angle 90^\circ \]

Step 2: Determine the phase relationship between voltage and current.
The short-circuit armature current $I_{sc}$ is given by: \[ I_{sc} = \frac{E_f}{Z_s} \approx \frac{E_f \angle 0^\circ}{X_s \angle 90^\circ} = \frac{E_f}{X_s} \angle -90^\circ \] This shows that under short-circuit conditions, the armature current $I_a$ lags the internal generated voltage $E_f$ by exactly $90^\circ$. This condition corresponds to a purely lagging zero power factor (ZPF lagging).

Step 3: Evaluate the spatial orientation of the magnetic fluxes.
* The main field flux ($\phi_f$) produces the maximum induced voltage $E_f$ when it is shifted $90^\circ$ ahead of it in space. * Since the current lags $E_f$ by $90^\circ$, the armature reaction flux ($\phi_a$) aligns directly opposite to the main field flux ($\phi_f$). \[ \phi_{\text{resultant}} = \phi_f - \phi_a \] Because the armature magnetic field acts directly against the main rotor field flux, it reduces the net flux, which is a purely demagnetizing effect. Consequently, the armature reaction is purely demagnetizing, which matches Option (B).