GATE 2024 EE – Question 41
For the three-bus lossless power network shown in the figure, the voltage magnitudes at all the buses are equal to 1 per unit (pu), and the differences of the voltage phase angles are very small. The line reactances are marked in the figure, where $\alpha$, $\beta$, $\gamma$, and $x$ are strictly positive. The bus injections $P_1$ and $P_2$ are in pu. If $P_1=mP_2$, where $m>0$, and the real power flow from bus 1 to bus 2 is 0 pu, then which one of the following options is correct?
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Correct answer: (A) $\gamma=m\beta$
Explanation
For small angle differences, $P_{ij}=\dfrac{\theta_i-\theta_j}{X_{ij}}$. Zero flow from bus 1 to bus 2 means $\theta_1=\theta_2$. Then bus 1 delivers all of $P_1$ to bus 3 through the $\beta x$ line, $P_1=\dfrac{\theta_1-\theta_3}{\beta x}$, and bus 2 delivers $P_2=\dfrac{\theta_2-\theta_3}{\gamma x}$. Since $\theta_1=\theta_2$, $\dfrac{P_1}{P_2}=\dfrac{\gamma}{\beta}=m$, i.e. $\gamma=m\beta$.