Question

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?
\includegraphics[width=0.5\linewidth]{41.png}

• A. \gamma = m\beta
• B. \beta = m\gamma
• C. \alpha = m\gamma
• D. \alpha= m\beta

Hint:

By equating the power flows from buses 1 and 2, we establish a relationship between the reactances $\gamma$ and $\beta$.

Answer 1

The Correct Option is A

Step 1: Given information.
(i) 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.
(ii) $P_1 = mP_2$, where $m>0$ and the real power flow from bus 1 to bus 2 is 0 pu. \begin{align*} \text{From the diagram:}
P_1 &= \frac{|V_1||V_3|}{\beta x} \sin(\delta_1 - \delta_3)
P_t &= \frac{1}{\beta x} \sin(\delta_1 - \delta_2)
P_2 &= \frac{|V_2||V_3|}{\gamma x} \sin(\delta_2 - \delta_3)
\text{Given:}
(\delta_1 - \delta_2) &\approx (\delta_2 - \delta_3)
P_1 &= mP_2
\frac{1}{\beta x} \sin(\delta_1 - \delta_2) &= \frac{m}{\gamma x} \sin(\delta_2 - \delta_3)
\gamma &= m\beta \end{align*} Hence the correct option is (A).