Discussion
In the given figure, the generator is delivering 1.0 pu power to infinite bus (V = 1 pu), with the generator terminal voltage Vt = 1 pu. What is the power angle?
Answer: D
The power angle asked for is the angle of the generator terminal voltage \(V_t\) ahead of the infinite bus. The reactance between them is the transformer in series with the two parallel lines of j0.5 each.
\[ X = 0.1 + \frac{0.5}{2} = 0.35\ \text{pu} \]
\[ \begin{aligned} P = \frac{V_tV}{X}\sin\delta \\ &= \frac{1 \times 1}{0.35}\sin\delta \\ &= 1 \Rightarrow \sin\delta = 0.35 \end{aligned} \]
\[ \delta = 20.5^\circ \]
The generator reactance of 0.25 pu would only be used for the angle of the internal voltage behind it, so it is not needed here. The answer is option D.
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