Question

The figure shows the single-line diagram of a synchronous generator delivering \(P=50\) MW of power at unity power factor to an infinite bus.

\(I_S\) denotes the stator current phasor. If the field excitation is increased, which one of the following options correctly describes its effect on the stator current, power factor, and load angle of the machine?

• A. Stator current increases, power factor becomes lagging, load angle remains the same
• B. Stator current decreases, power factor becomes leading, load angle remains the same
• C. Stator current increases, power factor becomes lagging, load angle decreases
• D. Stator current increases, power factor becomes leading, load angle increases

Hint:

Use P = (Ef V / Xs) sin(delta) with P held fixed to see that delta must fall as Ef rises, and recall that overexciting a generator moves it to the lagging side of its V-curve, away from the unity-PF minimum-current point.

Answer 1

The Correct Option is C

Step 1: Write the real power equation of the generator.
For a synchronous generator connected to an infinite bus of voltage \(V\) through synchronous reactance \(X_s\), with excitation voltage \(E_f\) and load angle \(\delta\):
\[ P=\frac{E_fV}{X_s}\sin\delta \]
Since \(P\), \(V\), and \(X_s\) are all held constant here (the load and the bus are unchanged), increasing \(E_f\) means
\[ E_f\sin\delta=\text{constant} \]
so as \(E_f\) increases, \(\sin\delta\) must decrease, which means \(\delta\) decreases (for the normal range \(0<\delta<90^{\circ}\)).
Step 2: Recall the generator V-curve behavior.
At the starting operating point, the machine runs at unity power factor, which is the minimum-current point on its V-curve for this value of \(P\). Moving the excitation away from this minimum-current point in either direction (increasing OR decreasing \(E_f\)) increases the magnitude of the stator current \(|I_S|\), since the V-curve is roughly U-shaped with unity power factor at its minimum. Increasing \(E_f\) moves the machine to the overexcited side of this minimum.
Step 3: Determine the power factor after overexciting.
An overexcited synchronous generator supplies reactive power to the system (behaves like a source of lagging vars from the bus's point of view), which by standard generator convention means it operates at a LAGGING power factor. This is the mirror image of a generator's V-curve: overexcited (higher \(E_f\)) gives lagging power factor, underexcited (lower \(E_f\)) gives leading power factor.
Step 4: Combine the three effects.
Increasing the field excitation: increases \(|I_S|\) (Step 2), makes the power factor lagging (Step 3), and decreases the load angle \(\delta\) (Step 1).
Step 5: Rule out the other options.
Option (A) and (D) both claim the load angle stays the same or increases, contradicting Step 1, which shows \(\delta\) must decrease to keep \(P\) fixed as \(E_f\) rises. Option (B) claims the current decreases and the power factor becomes leading, but increasing excitation moves the machine toward the overexcited, lagging side, not the underexcited, leading side, and away from the unity-PF minimum-current point, so current cannot decrease.
Step 6: Final Answer.
\[ \boxed{\text{Stator current increases, power factor becomes lagging, load angle decreases}} \]