GATE 2024 EE Q.41: solution and answer

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

1. With equal voltage magnitudes of $1$ pu, small angle differences and a lossless network, use the DC power flow: the real power flowing from bus $i$ to bus $k$ is $P_{ik}=\frac{{\theta }_i-{\theta }_k}{X_{ik}}$ pu, and the injection at a bus equals the sum of the flows leaving it.

2. Take bus 3 as the angle reference, ${\theta }_3=0$. Its injection is then fixed by the other two, $P_3=-(P_1+P_2)$, so it places no condition on $P_1$ and $P_2$.

3. The flow from bus 1 to bus 2 is zero: $\frac{{\theta }_1-{\theta }_2}{\alpha x}=0$. Since $\alpha x$ is finite and positive, ${\theta }_1={\theta }_2=\theta$. The $\alpha x$ line carries no power, so $\alpha$ does not appear in the result.

4. Bus 1 has two lines, to bus 2 (flow $0$) and to bus 3. Hence $P_1=0+\frac{{\theta }_1-{\theta }_3}{\beta x}=\frac{\theta }{\beta x}$.

5. Bus 2 likewise has the zero-flow line to bus 1 and the line to bus 3. Hence $P_2=0+\frac{{\theta }_2-{\theta }_3}{\gamma x}=\frac{\theta }{\gamma x}$.

6. Divide: $\frac{P_1}{P_2}=\frac{\theta /(\beta x)}{\theta /(\gamma x)}=\frac{\gamma }{\beta }$. This needs $\theta \ne 0$, which holds because $P_2\ne 0$ (as $m>0$, the injections are non-zero).

7. Using $P_1=m{P}_2$ gives $m=\frac{\gamma }{\beta }$, so $\gamma =m\beta$. This is option (A).