G-S first iteration, 3-bus with PV bus

Bus 1 is the slack bus ($1.03\angle 0^\circ$), bus 2 a PV bus ($P_2 = 0.3$ pu, $|V_2| = 1.01$ pu) and bus 3 a PQ bus ($P_3 = -0.4$, $Q_3 = -0.2$ pu). Initial values: $V_2^{(0)} = 1.01\angle 0^\circ$, $V_3^{(0)} = 1\angle 0^\circ$. No Q-limits are given.

Y-bus

$$\begin{aligned} y_{12} &= 1/j0.2 = -j5.0000 \\ y_{13} &= 1/j0.1 = -j10.0000 \\ y_{23} &= 1/j0.1 = -j10.0000 \end{aligned}$$ $$Y_{bus} = \begin{bmatrix} -j15.0000 & j5.0000 & j10.0000 \\ j5.0000 & -j15.0000 & j10.0000 \\ j10.0000 & j10.0000 & -j20.0000 \end{bmatrix}\ \text{pu}$$

First Gauss-Seidel iteration

Iteration 1

$$\begin{aligned} \sum_{k=1}^{3} Y_{2k}V_k &= 0.0000 \\ Q_2^{(1)} &= -\text{Im}\{V_2^{*}\textstyle\sum_k Y_{2k}V_k\} = 0.0000\ \text{pu} \end{aligned}$$ $$\begin{aligned} V_2^{(1)} &= \frac{1}{Y_{22}}\left[\frac{P_2 - jQ_2}{V_2^{*}} - Y_{21}V_1 - Y_{23}V_3\right] \\ \frac{P_2 - jQ_2}{V_2^{*}} &= \frac{0.3000}{1.0100} = 0.2970 \\ \textstyle\sum_{k\ne 2} Y_{2k}V_k &= j15.1500 \\ V_2^{(1)} &= \frac{(0.2970) - (j15.1500)}{-j15.0000} = 1.0100 + j0.0198 \\ \delta_2 &= 1.1232^\circ \Rightarrow V_2^{(1)} = 1.0100\angle 1.1232^\circ = 1.0098 + j0.0198\ \text{(magnitude reset to }1.01\text{)} \end{aligned}$$ $$\begin{aligned} V_3^{(1)} &= \frac{1}{Y_{33}}\left[\frac{P_3 - jQ_3}{V_3^{*}} - Y_{31}V_1 - Y_{32}V_2\right] \\ \frac{P_3 - jQ_3}{V_3^{*}} &= \frac{-0.4000 + j0.2000}{1.0000} = -0.4000 + j0.2000 \\ \textstyle\sum_{k\ne 3} Y_{3k}V_k &= -0.1980 + j20.3981 \\ V_3^{(1)} &= \frac{(-0.4000 + j0.2000) - (-0.1980 + j20.3981)}{-j20.0000} = 1.0099 - j0.0101 \\ &= 1.0100\angle -0.5730^\circ\ \text{pu} \end{aligned}$$

At the start the voltages are such that $\sum_k Y_{2k}V_k = 0$, so $Q_2^{(1)} = 0$.

Slack bus power after the first iteration

$$\begin{aligned} P_1 - jQ_1 &= V_1^{*}\textstyle\sum_k Y_{1k}V_k = (1.0300)(0.0020 - j0.3019) \\ &= 0.0021 - j0.3110 \\ \Rightarrow P_1 &= 0.0021\ \text{pu},\quad Q_1 = 0.3110\ \text{pu} \\ P_{G1} &= P_1 + P_{L1} = 0.0021 + 0.00 = 0.0021\ \text{pu} \\ Q_{G1} &= Q_1 + Q_{L1} = 0.3110 + 0.00 = 0.3110\ \text{pu} \end{aligned}$$

Total loss = sum of the calculated net injections at all buses (equal to the sum of all line losses):

$$\begin{aligned} P_{loss} &= \textstyle\sum P_i = 0.0000\ \text{pu} \\ Q_{loss} &= \textstyle\sum Q_i = 0.0180\ \text{pu} \end{aligned}$$

(The lines are pure reactances, so the real loss is zero; the small value of $P_1$ is the mismatch of an unconverged first iteration.)

Answer: $Q_2 = 0.0000$ pu, $V_2 = 1.0100\angle 1.1232^\circ$ pu, $V_3 = 1.0100\angle -0.5730^\circ$ pu, $P_1 = 0.0021$ pu, $Q_1 = 0.3110$ pu after the first iteration.