GATE 2023 EE Q.33: solution and answer

Official answer: 83 to 84Source: Official answer key, GATE 2023 EE (IIT Kanpur)

1. For a nominal-$\pi$ line, $V_S=A{V}_R+B{I}_R$ with $B=Z=(R+j\omega L)\ell$ (the series arm) and $A=1+\frac{YZ}{2}$.

2. The receiving-end active power is $P_R=\frac{V_S{V}_R}{|B|}\cos (\beta -\delta )-\frac{|A|V_R^2}{|B|}\cos (\beta -\alpha )$, where $\beta =\angle B$, $\alpha =\angle A$ and $\delta$ is the angle by which $V_S$ leads $V_R$.

3. $V_S$ and $V_R$ are both held at $275\text{kV}$, so only the first term depends on $\delta$. It is largest when $\cos (\beta -\delta )=1$, that is ${\delta }_{max}=\beta$.

4. Total series resistance: $R_{tot}=0.035\times 400=14\Omega$.

5. Total series reactance: $X_{tot}=2\pi (50)(1\times 10^{-3})(400)=125.66\Omega$.

6. $\beta ={\tan }^{-1}\left(\frac{X_{tot}}{R_{tot}}\right)={\tan }^{-1}\left(\frac{125.66}{14}\right)={\tan }^{-1}(8.976)\approx 83.64^{\circ }$.

7. The shunt capacitance appears only in $A$, which enters the second term. That term does not depend on $\delta$, so the capacitance does not change the angle. Hence $\theta =83.64^{\circ }$.