For the balanced 3-phase transmission line shown, consider the following cases:

Case-1: $\left|V_1\right|=1.1\text{ p.u., }\left|V_2\right|=0.9\text{ p.u., }Z=0.75\angle 0°\text{ p.u., }{\delta }_{12}=0°$

Case-2: $\left|V_1\right|=1.1\text{ p.u., }\left|V_2\right|=0.9\text{ p.u., }Z=0.75\angle 90°\text{ p.u., }{\delta }_{12}=90°$

Which of the following statements is/are correct about real and reactive power loss in the line?

The correct options/statements are:
1. Real power loss in Case-1 is more than that in Case-2
2. Reactive power loss in Case-2 is more than that in Case-1

Analysis:
As shown in the single-line diagram, a balanced 3-phase transmission line connects generator G1 (at Bus 1 with voltage phase $V_1\angle {\theta }_1$) and generator G2 (at Bus 2 with voltage phase $V_2\angle {\theta }_2$). The transmission line has a series impedance of $Z=Z_L\angle {\theta }_L=R+jX$, where $R$ is the resistance and $X$ is the reactance of the line.

Using Bus 2 voltage phase as the reference angle (0°), the voltage at Bus 1 is represented as $V_1=\left|V_1\right|\angle {\delta }_{12}$ and the voltage at Bus 2 is $V_2=\left|V_2\right|\angle 0°$.
The current $I$ flowing through the transmission line is given by Ohm's Law:

$$I=\frac{V_1-V_2}{Z}$$

The real power loss ($P_{\text{loss}}$) and reactive power loss ($Q_{\text{loss}}$) in the line are calculated using:

$$P_{\text{loss}}={\left|I\right|}^{2}R$$

$$Q_{\text{loss}}={\left|I\right|}^{2}X$$

Case-1 Evaluation:
Given parameters for Case-1:
- $\left|V_1\right|=1.1$ p.u., $\left|V_2\right|=0.9$ p.u.
- Line impedance $Z=0.75\angle 0°$ p.u. (purely resistive, so $R=0.75$ p.u. and reactance $X=0$ p.u.)
- Voltage angle difference ${\delta }_{12}=0°$
Substituting these values, the line current $I_1$ is:

$$I_1=\frac{1.1\angle 0°-0.9\angle 0°}{0.75\angle 0°}=\frac{0.2}{0.75}=0.2667\text{p.u.}$$

Using the current value to calculate the losses:
- Real power loss in Case-1:

$$P_{\text{loss},1}={\left|I_1\right|}^{2}R={\left(0.2667\right)}^2\times 0.75\approx 0.0533\text{p.u.}$$

- Reactive power loss in Case-1:

$$Q_{\text{loss},1}={\left|I_1\right|}^{2}X={\left(0.2667\right)}^2\times 0=0\text{p.u.}$$

Case-2 Evaluation:
Given parameters for Case-2:
- $\left|V_1\right|=1.1$ p.u., $\left|V_2\right|=0.9$ p.u.
- Line impedance $Z=0.75\angle 90°$ p.u. (purely inductive, so $R=0$ p.u. and reactance $X=0.75$ p.u.)
- Voltage angle difference ${\delta }_{12}=90°$
Substituting these complex phasor representations:
$V_1=1.1\angle 90°=j1.1$ and $V_2=0.9\angle 0°=0.9$
The line current $I_2$ is:

$$I_2=\frac{1.1\angle 90°-0.9\angle 0°}{0.75\angle 90°}=\frac{j1.1-0.9}{j0.75}=\frac{1.1}{0.75}+j\frac{0.9}{0.75}=1.4667+j1.2\text{p.u.}$$

The squared magnitude of this current is:

$${\left|I_2\right|}^2={\left(1.4667\right)}^2+{\left(1.2\right)}^2\approx 2.1511+1.44=3.5911\text{p.u.}$$

Using the current value to calculate the losses:
- Real power loss in Case-2:

$$P_{\text{loss},2}={\left|I_2\right|}^{2}R=3.5911\times 0=0\text{p.u.}$$

- Reactive power loss in Case-2:

$$Q_{\text{loss},2}={\left|I_2\right|}^{2}X=3.5911\times 0.75\approx 2.6933\text{p.u.}$$

Comparison of Losses:
1. Real Power Loss:
In Case-1, $P_{\text{loss},1}\approx 0.0533$ p.u., while in Case-2, $P_{\text{loss},2}=0$ p.u.
Thus, the real power loss in Case-1 is greater than that in Case-2.

2. Reactive Power Loss:
In Case-1, $Q_{\text{loss},1}=0$ p.u., while in Case-2, $Q_{\text{loss},2}\approx 2.6933$ p.u.
Thus, the reactive power loss in Case-2 is greater than that in Case-1.