GATE 2025 EE Q.35: solution and answer

Official answer: 66.25 to 66.75Source: Official answer key, GATE 2025 EE (IIT Roorkee)

1. Phase voltage: $V=\frac{3300}{\sqrt{3}}=1905.26\text{V}$.

2. At no load, the induced emf equals the terminal voltage in both magnitude and phase: $|E|=|V|$.

3. Pole conversion: for a 4-pole machine, electrical angle $=\frac{P}{2}\times$ mechanical angle $=2\times 2^{\circ }=4^{\circ }$. So the torque angle is $\delta =4^{\circ }$ electrical.

4. Take $V=1905.26\angle 0^{\circ }\text{V}$ and $E=1905.26\angle (-4^{\circ })\text{V}$, and since stator resistance is negligible, $I_a=\frac{V-E}{j{X}_s}$.

5. For two equal-magnitude phasors separated by angle $\delta$: $$|V-E|=2V\sin \left(\frac{\delta }{2}\right)=2(1905.26)\sin (2^{\circ })\approx 132.99\text{V}$$

6. Magnitude of current: $$|I_a|=\frac{|V-E|}{X_s}=\frac{132.99}{2}\approx 66.49\text{A}$$