A 3-phase, $11\mathrm{kV},10\mathrm{MVA}$ synchronous generator is connected to an inductive load of power factor $(\sqrt{3}/2)$ via a lossless line with a per-phase inductive reactance of 5Ω. The per-phase synchronous reactance of the generator is 30Ω with negligible armature resistance. If the generator is producing the rated current at the rated voltage, then the power factor at the terminal of the generator is

The correct answer is 0.63 lagging.

To find the power factor at the terminal of the generator, we need to analyze the phasor relations of the system on a per-phase basis.

Step 1: Calculate the rated terminal voltage (per-phase) and rated current of the generator
The generator is operating at rated conditions, so the line-to-line terminal voltage is:
V_L-L = 11 kV
The rated per-phase terminal voltage of the generator is:
$V_t=\frac{11000}{\sqrt{3}}\approx 6350.85\text{ V}$

The rated apparent power of the 3-phase generator is S = 10 MVA. The rated current (which is also the line current and the phase current in a star-connected system) is:
$I=\frac{S}{\sqrt{3}\times V_{\text{L-L}}}=\frac{10\times {10}^6}{\sqrt{3}\times 11000}\approx 524.86\text{ A}$

Step 2: Express the current and line voltage drop using the load voltage as reference
Let the load phase voltage V_L be the reference phasor:
V_L = V_L ∠ 0° V
Since the load is inductive and has a power factor of cos(φ_L) = $\sqrt{3}/2$ lagging, the phase angle of the load is:
φ_L = arccos(0.866) = 30°
So, the load current lags the load voltage by 30°:
I = I ∠ -30° A = 524.86 ∠ -30° A

The generator is connected to the load through a transmission line with a per-phase inductive reactance of X_line = 5 Ω.
The generator terminal voltage (per-phase) is related to the load voltage by:
V_t = V_L + j I X_line

First, calculate the voltage drop across the line reactance:
j I X_line = (1 ∠ 90°) × (524.86 ∠ -30°) × 5
j I X_line = 2624.3 ∠ 60° V
j I X_line = 2624.3 × (cos 60° + j sin 60°) V
j I X_line = 2624.3 × (0.5 + j 0.866) = 1312.15 + j 2272.71 V

Thus, the terminal voltage phasor is:
V_t = (V_L + 1312.15) + j 2272.71 V

Step 3: Solve for the load voltage magnitude V_L
Since the magnitude of the generator terminal voltage is rated at |V_t| = 6350.85 V, we can write:
$|V_t{|}^2=(V_L+1312.15{)}^2+(2272.71{)}^2$
$(6350.85{)}^2=(V_L+1312.15{)}^2+5165211$
$40333296=(V_L+1312.15{)}^2+5165211$
$(V_L+1312.15{)}^2=35168085$
V_L + 1312.15 = 5930.27 V
V_L = 4618.12 V

Step 4: Find the power factor at the terminal of the generator
Substitute V_L back into the expression for V_t:
V_t = (4618.12 + 1312.15) + j 2272.71 = 5930.27 + j 2272.71 V

The angle θ_t of the terminal voltage phasor with respect to the load voltage reference is:
${\theta }_t=\arctan \left(\frac{2272.71}{5930.27}\right)=\arctan (0.3832)\approx 20.97{°}^{}$

The current phasor has an angle of θ_i = -30°.
The phase difference φ between the generator terminal voltage and the current is:
φ = θ_t - θ_i = 20.97° - (-30°) = 50.97°

Thus, the power factor at the terminals of the generator is:
cos(φ) = cos(50.97°) ≈ 0.63
Since the current lags the terminal voltage (θ_i < θ_t), the power factor is 0.63 lagging.