GATE 2024 EE Q.40: solution and answer

Official answer: ASource: Official answer key, GATE 2024 EE (IISc Bengaluru)

1. Rated phase voltage: $V_{ph}=\frac{11000}{\sqrt{3}}=6350.85\text{V}$; rated current: $I=\frac{S}{\sqrt{3}{V}_L}=\frac{10\times 10^6}{\sqrt{3}\times 11000}=524.86\text{A}$.

2. Take the current as reference, $I=524.86\angle 0^{\circ }\text{A}$. The load power factor is $\sqrt{3}/2=0.866$ lagging, so for an inductive (lagging) load the load voltage leads the current by $30^{\circ }$: $V_{load}=v\angle 30^{\circ }$, with magnitude $v$ unknown.

3. Since the line is lossless (purely reactive, $X_{line}=5\Omega$), the generator terminal voltage is $V_{gen}=V_{load}+j{X}_{line}I=v\angle 30^{\circ }+(524.86\times 5)\angle 90^{\circ }=v\angle 30^{\circ }+2624.32\angle 90^{\circ }\text{V}$.

4. The generator is at its rated terminal (phase) voltage, so $|V_{gen}|=V_{ph}=6350.85\text{V}$. Writing $V_{gen}$ in rectangular form: $\text{Re}=0.866v$, $\text{Im}=0.5v+2624.32$.

5. Applying the magnitude condition: $(0.866v{)}^2+(0.5v+2624.32{)}^2=6350.85^2\Rightarrow v^2+2624.32v-33,446,283\approx 0\Rightarrow v\approx 4618.1\text{V}$ (the negative root is rejected as unphysical).

6. Generator terminal voltage components: $\text{Re}=0.866\times 4618.1\approx 4000\text{V}$, $\text{Im}=0.5\times 4618.1+2624.32\approx 4933\text{V}$.

7. Angle of the terminal voltage relative to the current: $\theta ={\tan }^{-1}\left(\frac{4933}{4000}\right)\approx 51^{\circ }$.

8. Generator terminal power factor $=\cos (51^{\circ })\approx 0.63$. Since the terminal voltage leads the current (positive angle), the current lags the voltage, so the power factor is lagging.