In the figure shown, self-impedances of the two transmission lines are 1.5 j pu each and 0.5 pu Z_m = j is the mutual impedance. Bus voltages shown in the figure are in pu. Given that δ > 0 , the maximum steady state real power that can be transferred in pu from bus-1 to bus-2 is

The correct option is |E| |V|.

Step-by-Step Explanation:

From the provided figure, we have two parallel transmission lines connecting Bus-1 and Bus-2. Let us analyze the details visible in the schematic:
- The voltage at Bus-1 is represented as $V_1=\left|E\right|\angle \delta$.
- The voltage at Bus-2 is represented as $V_2=\left|V\right|\angle 0^{\circ }$.
- The self-impedance of each transmission line is $Z_s=1.5j\text{pu}$.
- The mutual impedance between the two lines is labeled as $Z_m=0.5j\text{pu}$.
- The dots indicating magnetic coupling are positioned at the receiving-end (Bus-2 side) of both Transmission Line-1 and Transmission Line-2.

Since the currents in both lines flow in the same direction (from Bus-1 to Bus-2), they either both enter the non-dotted terminals and exit through the dotted terminals, or vice versa. Therefore, the mutual coupling is aiding, and the voltage drops across the two lines can be expressed as:

$$V_1-V_2=I_1{Z}_s+I_2{Z}_m$$

$$V_1-V_2=I_2{Z}_s+I_1{Z}_m$$

Due to the symmetry of the parallel connection, the currents in both lines are equal:

$$I_1=I_2=I$$

Substituting $I$ into the voltage drop equation gives:

$$V_1-V_2=I(Z_s+Z_m)$$

The total current $I_{\text{total}}$ transferred from Bus-1 to Bus-2 is the sum of the currents in both lines:

$$I_{\text{total}}=I_1+I_2=2I$$

We can write this in terms of the voltage difference:

$$I_{\text{total}}=\frac{2(V_1-V_2)}{Z_s+Z_m}$$

From this relation, the equivalent impedance $Z_{\text{eq}}$ of the parallel combination is:

$$Z_{\text{eq}}=\frac{V_1-V_2}{I_{\text{total}}}=\frac{Z_s+Z_m}{2}$$

Substituting the given values of self-impedance and mutual impedance:

$$Z_{\text{eq}}=\frac{1.5j+0.5j}{2}=\frac{2.0j}{2}=1.0j\text{pu}$$

The equivalent series reactance between the two buses is therefore $X_{\text{eq}}=1.0\text{pu}$.

The steady-state real power transfer $P$ from Bus-1 to Bus-2 is given by the power transfer equation:

$$P=\frac{\left|E\right|\left|V\right|}{X_{\text{eq}}}sin\left(\delta \right)$$

Substituting $X_{\text{eq}}=1.0$ into the equation gives:

$$P=\left|E\right|\left|V\right|sin\left(\delta \right)$$

For $\delta >0$, the maximum real power transfer occurs at the stability limit when $\delta ={90}^{\circ }$ (so $sin\left(\delta \right)=1$):

$$P_{\text{max}}=\left|E\right|\left|V\right|pu$$