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 α, β, γ and x are strictly positive. The bus injections P₁ and P₂ are in pu. If P₁ = mP₂, 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?

The correct option is:
$$\gamma =m\beta$$

Step-by-Step Explanation:
1. Analyze the Given Network from the Image:
From the schematic of the three-bus lossless power network:
- The line reactance between bus 1 and bus 2 is $X_{12}=\alpha x$.
- The line reactance between bus 1 and bus 3 is $X_{13}=\beta x$.
- The line reactance between bus 2 and bus 3 is $X_{23}=\gamma x$.
- The voltage magnitude at all buses is $\left|V_1\right|=\left|V_2\right|=\left|V_3\right|=1\text{pu}$.
- The phase angles at bus 1, bus 2, and bus 3 are ${\theta }_1$, ${\theta }_2$, and ${\theta }_3$ respectively, and their differences are very small.

2. Apply DC Power Flow Approximation:
For a lossless system with flat voltage profile (1 pu) and small angle differences, the active power flow $P_{ij}$ from bus $i$ to bus $j$ is given by:
$$P_{ij}=\frac{{\theta }_i-{\theta }_j}{X_{ij}}$$

3. Condition for Zero Power Flow between Bus 1 and Bus 2:
We are given that the real power flow from bus 1 to bus 2 is 0 pu. Therefore:
$$P_{12}=\frac{{\theta }_1-{\theta }_2}{\alpha x}=0$$
Since $\alpha$ and $x$ are strictly positive, this directly implies:
$${\theta }_1={\theta }_2$$

4. Formulate Power Injection Equations:
Using nodal equations, the power injected at bus 1 ( $P_1$) is:
$$P_1=P_{12}+P_{13}=0+\frac{{\theta }_1-{\theta }_3}{\beta x}=\frac{{\theta }_1-{\theta }_3}{\beta x}$$
Similarly, the power injected at bus 2 ( $P_2$) is:
$$P_2=P_{21}+P_{23}=-P_{12}+\frac{{\theta }_2-{\theta }_3}{\gamma x}=0+\frac{{\theta }_2-{\theta }_3}{\gamma x}$$
Substituting ${\theta }_2={\theta }_1$ into the equation for $P_2$:
$$P_2=\frac{{\theta }_1-{\theta }_3}{\gamma x}$$

5. Relate the Injections:
We are given that $P_1=m{P}_2$. Taking the ratio of $P_1$ to $P_2$:
$$\frac{P_1}{P_2}=\frac{\left(\frac{{\theta }_1-{\theta }_3}{\beta x}\right)}{\left(\frac{{\theta }_1-{\theta }_3}{\gamma x}\right)}=\frac{\gamma x}{\beta x}=\frac{\gamma }{\beta }$$
Substitute $\frac{P_1}{P_2}=m$ into the equation:
$$m=\frac{\gamma }{\beta }$$
Rearranging terms, we obtain:
$$\gamma =m\beta$$