N-R first iteration, line reactance 0.1

Bus 1 is the slack bus ($1.03\angle 0^\circ$), bus 2 a PV bus ($P_2 = 1.5$ pu, $|V_2| = 1.04$ pu) and bus 3 a PQ bus ($P_3 = -1.2$, $Q_3 = -0.5$ pu). Lines are pure reactances $j0.1$ pu. Flat start: $V_2^{0} = 1.04\angle 0^\circ$, $V_3^{0} = 1.0\angle 0^\circ$. No Q-limits are given for bus 2.

(i) Y-bus

Each line: $y = 1/(j0.1) = -j10$ pu. Each bus has two lines, so $Y_{ii} = -j20$ and $Y_{ij} = j10$:

$$Y_{bus} = \begin{bmatrix} -j20.0000 & j10.0000 & j10.0000 \\ j10.0000 & -j20.0000 & j10.0000 \\ j10.0000 & j10.0000 & -j20.0000 \end{bmatrix} = \begin{bmatrix} 20\angle -90^\circ & 10\angle 90^\circ & 10\angle 90^\circ \\ 10\angle 90^\circ & 20\angle -90^\circ & 10\angle 90^\circ \\ 10\angle 90^\circ & 10\angle 90^\circ & 20\angle -90^\circ \end{bmatrix}\ \text{pu}$$

(ii) Mismatch matrix and Jacobian

With $Y_{ij} = |Y_{ij}|\angle\theta_{ij}$, $V_i = |V_i|\angle\delta_i$ and $\phi_{ij} = \theta_{ij} - \delta_i + \delta_j$:

$$\begin{aligned} P_i &= \sum_{j=1}^{n} |V_i||V_j||Y_{ij}|\cos\phi_{ij} \\ Q_i &= -\sum_{j=1}^{n} |V_i||V_j||Y_{ij}|\sin\phi_{ij} \end{aligned}$$

Jacobian elements (polar form, $j \ne i$):

$$\begin{aligned} \frac{\partial P_i}{\partial \delta_i} &= \sum_{j\ne i} |V_i||V_j||Y_{ij}|\sin\phi_{ij} \\ \frac{\partial P_i}{\partial \delta_j} &= -|V_i||V_j||Y_{ij}|\sin\phi_{ij} \\ \frac{\partial P_i}{\partial |V_i|} &= 2|V_i||Y_{ii}|\cos\theta_{ii} + \sum_{j\ne i} |V_j||Y_{ij}|\cos\phi_{ij} \\ \frac{\partial P_i}{\partial |V_j|} &= |V_i||Y_{ij}|\cos\phi_{ij} \\ \frac{\partial Q_i}{\partial \delta_i} &= \sum_{j\ne i} |V_i||V_j||Y_{ij}|\cos\phi_{ij} \\ \frac{\partial Q_i}{\partial \delta_j} &= -|V_i||V_j||Y_{ij}|\cos\phi_{ij} \\ \frac{\partial Q_i}{\partial |V_i|} &= -2|V_i||Y_{ii}|\sin\theta_{ii} - \sum_{j\ne i} |V_j||Y_{ij}|\sin\phi_{ij} \\ \frac{\partial Q_i}{\partial |V_j|} &= -|V_i||Y_{ij}|\sin\phi_{ij} \end{aligned}$$

Since all $\theta_{ij} = \pm 90^\circ$ and all angles are zero at the start, the $\cos$ terms vanish, so $P_i^0 = 0$:

$$\begin{aligned} P_2^{0} &= 0.0000, \quad Q_2^{0} = 0.5200 \\ P_3^{0} &= 0.0000, \quad Q_3^{0} = -0.7000 \end{aligned}$$ $$\begin{aligned} \Delta P_2 &= 1.5000 - (0.0000) = 1.5000 \\ \Delta P_3 &= -1.2000 - (0.0000) = -1.2000 \\ \Delta Q_3 &= -0.5000 - (-0.7000) = 0.2000 \end{aligned}$$ $$[M^{0}] = \begin{bmatrix} \Delta P_2 \\ \Delta P_3 \\ \Delta Q_3 \end{bmatrix} = \begin{bmatrix} 1.5000 \\ -1.2000 \\ 0.2000 \end{bmatrix}$$

Jacobian elements:

$$\begin{aligned} \frac{\partial P_2}{\partial \delta_2} &= (1.0400)(1.0300)(10.0000)\sin(90.00^\circ) + (1.0400)(1.0000)(10.0000)\sin(90.00^\circ) = 21.1120 \\ \frac{\partial P_2}{\partial \delta_3} &= -(1.0400)(1.0000)(10.0000)\sin(90.00^\circ) = -10.4000 \\ \frac{\partial P_2}{\partial |V_3|} &= (1.0400)(10.0000)\cos(90.00^\circ) = 0.0000 \\ \frac{\partial P_3}{\partial \delta_2} &= -(1.0000)(1.0400)(10.0000)\sin(90.00^\circ) = -10.4000 \\ \frac{\partial P_3}{\partial \delta_3} &= (1.0000)(1.0300)(10.0000)\sin(90.00^\circ) + (1.0000)(1.0400)(10.0000)\sin(90.00^\circ) = 20.7000 \\ \frac{\partial P_3}{\partial |V_3|} &= 2(1.0000)(20.0000)\cos(-90.00^\circ) + (1.0300)(10.0000)\cos(90.00^\circ) + (1.0400)(10.0000)\cos(90.00^\circ) = 0.0000 \\ \frac{\partial Q_3}{\partial \delta_2} &= -(1.0000)(1.0400)(10.0000)\cos(90.00^\circ) = 0.0000 \\ \frac{\partial Q_3}{\partial \delta_3} &= (1.0000)(1.0300)(10.0000)\cos(90.00^\circ) + (1.0000)(1.0400)(10.0000)\cos(90.00^\circ) = 0.0000 \\ \frac{\partial Q_3}{\partial |V_3|} &= -2(1.0000)(20.0000)\sin(-90.00^\circ) - (1.0300)(10.0000)\sin(90.00^\circ) - (1.0400)(10.0000)\sin(90.00^\circ) = 19.3000 \end{aligned}$$ $$[J^{0}] = \begin{bmatrix} 21.1120 & -10.4000 & 0.0000 \\ -10.4000 & 20.7000 & 0.0000 \\ 0.0000 & 0.0000 & 19.3000 \end{bmatrix}$$

(The off-diagonal blocks are zero at the flat start because the lines have no resistance.)

(iii) First iteration

$$\begin{bmatrix} 1.5000 \\ -1.2000 \\ 0.2000 \end{bmatrix} = \begin{bmatrix} 21.1120 & -10.4000 & 0.0000 \\ -10.4000 & 20.7000 & 0.0000 \\ 0.0000 & 0.0000 & 19.3000 \end{bmatrix}\begin{bmatrix} \Delta\delta_2 \\ \Delta\delta_3 \\ \Delta|V_3| \end{bmatrix}$$ $$\begin{bmatrix} \Delta\delta_2 \\ \Delta\delta_3 \\ \Delta|V_3| \end{bmatrix} = [J^{0}]^{-1}\begin{bmatrix} 1.5000 \\ -1.2000 \\ 0.2000 \end{bmatrix} = \begin{bmatrix} 0.0565 \\ -0.0296 \\ 0.0104 \end{bmatrix}$$

Updated values:

$$\begin{aligned} \delta_2^{1} &= 0 + (0.0565) = 0.0565\ \text{rad} = 3.235^\circ \\ \delta_3^{1} &= 0 + (-0.0296) = -0.0296\ \text{rad} = -1.696^\circ \\ |V_3|^{1} &= 1.0000 + (0.0104) = 1.0104\ \text{pu} \end{aligned}$$

So $V_2^{1} = 1.0400\angle 3.235^\circ$ pu and $V_3^{1} = 1.0104\angle -1.696^\circ$ pu.

Reactive power at bus 2 (with the new voltages):

$$\begin{aligned} Q_2^{1} &= -\text{Im}\{V_2^{*}\textstyle\sum_k Y_{2k}V_k\} = 0.4682\ \text{pu} \\ Q_{G2} &= Q_2 + Q_{L2} = 0.4682\ \text{pu} \end{aligned}$$

Slack bus power:

$$\begin{aligned} P_1 - jQ_1 &= V_1^{*}\textstyle\sum_k Y_{1k}V_k = (1.0300)(-0.2879 - j0.1174) \\ &= -0.2966 - j0.1209 \\ \Rightarrow P_1 &= -0.2966\ \text{pu},\quad Q_1 = 0.1209\ \text{pu} \\ P_{G1} &= P_1 + P_{L1} = -0.2966 + 1.00 = 0.7034\ \text{pu} \\ Q_{G1} &= Q_1 + Q_{L1} = 0.1209 + 0.50 = 0.6209\ \text{pu} \end{aligned}$$

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

$$\begin{aligned} P_{loss} &= \textstyle\sum P_i = 0.0000\ \text{pu} \\ Q_{loss} &= \textstyle\sum Q_i = 0.1347\ \text{pu} \end{aligned}$$

Answer (after the 1st N-R iteration): $\delta_2 = 3.235^\circ$, $V_3 = 1.0104\angle -1.696^\circ$ pu, $Q_{G2} = 0.4682$ pu, $P_{G1} = 0.7034$ pu, $Q_{G1} = 0.6209$ pu.