G-S voltage and power flow, 2-bus

Bus 1 is the slack bus ($V_1 = 1\angle 0^\circ$). At bus 2 the net injected power is the capacitor injection minus the load:

$$S_2 = P_2 + jQ_2 = j1.0 - (0.5 + j1.0) = -0.5 + j0\ \text{pu}$$

So bus 2 is a PQ bus with $P_2 = -0.5$, $Q_2 = 0$. The load $S_{d1}$ at bus 1 does not affect $V_2$.

Y-bus

$y_{12} = 1/j0.75 = -j1.3333$ pu:

$$Y_{bus} = \begin{bmatrix} -j1.3333 & j1.3333 \\ j1.3333 & -j1.3333 \end{bmatrix}$$

Gauss-Seidel equation for bus 2

$$\begin{aligned} V_2^{(k+1)} &= \frac{1}{Y_{22}}\left[\frac{P_2 - jQ_2}{V_2^{(k)*}} - Y_{21}V_1\right] \\ &= \frac{1}{-j1.3333}\left[\frac{-0.5}{V_2^{(k)*}} - j1.3333\right] \\ &= 1 - \frac{j0.375}{V_2^{(k)*}} \end{aligned}$$

Iteration 1 (flat start $V_2^{(0)} = 1\angle 0^\circ$):

$$V_2^{(1)} = 1 - \frac{j0.375}{1} = 1.0000 - j0.3750 = 1.0680\angle -20.556^\circ$$

Iteration 2:

$$V_2^{(2)} = 1 - \frac{j0.375}{1.0000 + j0.3750} = 0.8767 - j0.3288 = 0.9363\angle -20.556^\circ$$

Continuing until the change is below $10^{-4}$ pu:

$$V_2 = 0.9115\angle -24.29^\circ\ \text{pu}$$

Check (exact solution): with $Q_2 = 0$ and a lossless line, $|V_2| = \cos\delta_2$ and $P_2 = 1.3333|V_2|\sin\delta_2 = -0.5$, so $\sin 2\delta_2 = -0.75$, giving $\delta_2 = -24.30^\circ$ and $|V_2| = 0.9114$ pu, which matches.

Power flow between buses 1 and 2

$$\begin{aligned} I_{12} &= \frac{V_1 - V_2}{j0.75} = 0.5000 - j0.2256 = 0.5486\angle -24.29^\circ\ \text{pu} \\ S_{12} &= V_1I_{12}^{*} = 0.5000 + j0.2256\ \text{pu} \\ S_{21} &= V_2(-I_{12})^{*} = -0.5000\ \text{pu} \end{aligned}$$

Line loss $= S_{12} + S_{21} = j0.2257$ pu (reactive only, as the line has no resistance).

Answer: $V_2 \approx 0.911\angle -24.29^\circ$ pu; power sent from bus 1 to bus 2 is $S_{12} = 0.500 + j0.226$ pu, and bus 2 receives $0.5$ pu real power with zero reactive power.