An 11 kV,50 Hz,two-pole star cylindrical synchronous motor has Xs=50 Ω/phase and negligible resistance. At fixed field it first draws 100 A at unity pf. Increased load makes current 120 A. Find the signed motor load angle in degrees.
Use phase terminal voltage V=11000/√3 V. At the initial unity-pf point, internal emf is E=V−j(50)(100), so its fixed magnitude is √(V²+5000²).
At the new current, |V−E|=XsI=6000 V. The phasor triangle gives $\cos\delta=(V^2+|E|^2-6000^2)/(2V|E|)$.
The magnitude is about 47.1°. For a motor the internal emf lags the terminal voltage, so the signed angle is **about −47.1°**, within −48 to −46°.
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