G-S voltages after 2nd iteration with shunt

Bus 1 is the slack bus ($1.03\angle 0^\circ$), bus 2 a PV bus ($P_2 = 1.5$ pu, $|V_2| = 1.03$ pu) and bus 3 a PQ bus ($P_3 = -1.2$, $Q_3 = -0.5$ pu). The shunt admittance $j0.02$ pu is taken as the total line charging of each line, so $j0.01$ pu is placed at each end. Flat start: $V_2^{(0)} = 1.03\angle 0^\circ$, $V_3^{(0)} = 1\angle 0^\circ$; no acceleration factor; no Q-limits given.

Y-bus

$$\begin{aligned} y_{series} &= \frac{1}{0.015 + j0.120} = 1.0256 - j8.2051\ \text{pu} \\ Y_{ii} &= 2y_{series} + 2(j0.01) = 2.0513 - j16.3903 \\ Y_{ij} &= -y_{series} = -1.0256 + j8.2051 \end{aligned}$$ $$Y_{bus} = \begin{bmatrix} 2.0513 - j16.3903 & -1.0256 + j8.2051 & -1.0256 + j8.2051 \\ -1.0256 + j8.2051 & 2.0513 - j16.3903 & -1.0256 + j8.2051 \\ -1.0256 + j8.2051 & -1.0256 + j8.2051 & 2.0513 - j16.3903 \end{bmatrix}\ \text{pu}$$

Gauss-Seidel iterations

At the PV bus, $Q_2$ is found from the latest voltages, $V_2$ is updated and its magnitude reset to 1.03 pu. Bus 3 then uses the newest $V_2$.

Iteration 1

$$\begin{aligned} \sum_{k=1}^{3} Y_{2k}V_k &= 0.0308 - j0.2256 \\ Q_2^{(1)} &= -\text{Im}\{V_2^{*}\textstyle\sum_k Y_{2k}V_k\} = 0.2323\ \text{pu} \end{aligned}$$ $$\begin{aligned} V_2^{(1)} &= \frac{1}{Y_{22}}\left[\frac{P_2 - jQ_2}{V_2^{*}} - Y_{21}V_1 - Y_{23}V_3\right] \\ \frac{P_2 - jQ_2}{V_2^{*}} &= \frac{1.5000 - j0.2323}{1.0300} = 1.4563 - j0.2256 \\ \textstyle\sum_{k\ne 2} Y_{2k}V_k &= -2.0821 + j16.6564 \\ V_2^{(1)} &= \frac{(1.4563 - j0.2256) - (-2.0821 + j16.6564)}{2.0513 - j16.3903} = 1.0407 + j0.0856 \\ \delta_2 &= 4.7039^\circ \Rightarrow V_2^{(1)} = 1.0300\angle 4.7039^\circ = 1.0265 + j0.0845\ \text{(magnitude reset to }1.03\text{)} \end{aligned}$$ $$\begin{aligned} V_3^{(1)} &= \frac{1}{Y_{33}}\left[\frac{P_3 - jQ_3}{V_3^{*}} - Y_{31}V_1 - Y_{32}V_2\right] \\ \frac{P_3 - jQ_3}{V_3^{*}} &= \frac{-1.2000 + j0.5000}{1.0000} = -1.2000 + j0.5000 \\ \textstyle\sum_{k\ne 3} Y_{3k}V_k &= -2.8023 + j16.7875 \\ V_3^{(1)} &= \frac{(-1.2000 + j0.5000) - (-2.8023 + j16.7875)}{2.0513 - j16.3903} = 0.9904 - j0.0262 \\ &= 0.9908\angle -1.5151^\circ\ \text{pu} \end{aligned}$$

Iteration 2

$$\begin{aligned} \sum_{k=1}^{3} Y_{2k}V_k &= 1.6328 - j0.0469 \\ Q_2^{(2)} &= -\text{Im}\{V_2^{*}\textstyle\sum_k Y_{2k}V_k\} = 0.1861\ \text{pu} \end{aligned}$$ $$\begin{aligned} V_2^{(2)} &= \frac{1}{Y_{22}}\left[\frac{P_2 - jQ_2}{V_2^{*}} - Y_{21}V_1 - Y_{23}V_3\right] \\ \frac{P_2 - jQ_2}{V_2^{*}} &= \frac{1.5000 - j0.1861}{1.0265 - j0.0845} = 1.4662 - j0.0606 \\ \textstyle\sum_{k\ne 2} Y_{2k}V_k &= -1.8573 + j16.6049 \\ V_2^{(2)} &= \frac{(1.4662 - j0.0606) - (-1.8573 + j16.6049)}{2.0513 - j16.3903} = 1.0261 + j0.0744 \\ \delta_2 &= 4.1446^\circ \Rightarrow V_2^{(2)} = 1.0300\angle 4.1446^\circ = 1.0273 + j0.0744\ \text{(magnitude reset to }1.03\text{)} \end{aligned}$$ $$\begin{aligned} V_3^{(2)} &= \frac{1}{Y_{33}}\left[\frac{P_3 - jQ_3}{V_3^{*}} - Y_{31}V_1 - Y_{32}V_2\right] \\ \frac{P_3 - jQ_3}{V_3^{*}} &= \frac{-1.2000 + j0.5000}{0.9904 + j0.0262} = -1.1974 + j0.5365 \\ \textstyle\sum_{k\ne 3} Y_{3k}V_k &= -2.7209 + j16.8041 \\ V_3^{(2)} &= \frac{(-1.1974 + j0.5365) - (-2.7209 + j16.8041)}{2.0513 - j16.3903} = 0.9887 - j0.0308 \\ &= 0.9891\angle -1.7834^\circ\ \text{pu} \end{aligned}$$

Result after the 2nd iteration

Answer: $V_2 = 1.0300\angle 4.1446^\circ$ pu and $V_3 = 0.9891\angle -1.7834^\circ$ pu after the 2nd iteration (with $Q_2 = 0.1861$ pu used in that iteration).