Initial mismatch and Jacobian, 2-bus system

Bus 1 is the slack bus ($1.0\angle 0^\circ$) and bus 2 is a load (PQ) bus with $P_2^{sp} = -1.0$ pu and $Q_2^{sp} = -0.5$ pu. Unknowns: $\delta_2$ and $|V_2|$. Flat start: $|V_2|^{0} = 1.0$, $\delta_2^{0} = 0$.

Y-bus

$y_{12} = 1/j0.1 = -j10$ pu, so

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

Power equations at bus 2

$$\begin{aligned} P_2 &= |V_2||V_1||Y_{21}|\cos(\theta_{21} - \delta_2 + \delta_1) + |V_2|^2|Y_{22}|\cos\theta_{22} \\ &= 10|V_2|\cos(90^\circ - \delta_2) = 10|V_2|\sin\delta_2 \\ Q_2 &= -|V_2||V_1||Y_{21}|\sin(\theta_{21} - \delta_2 + \delta_1) - |V_2|^2|Y_{22}|\sin\theta_{22} \\ &= -10|V_2|\cos\delta_2 + 10|V_2|^2 \end{aligned}$$

Initial power mismatch

At $|V_2| = 1$, $\delta_2 = 0$: $P_2^{0} = 10(1)\sin 0 = 0.0000$ and $Q_2^{0} = -10 + 10 = 0.0000$.

$$\begin{aligned} \Delta P_2 &= P_2^{sp} - P_2^{0} = -1.0 - 0 = -1.0000 \\ \Delta Q_2 &= Q_2^{sp} - Q_2^{0} = -0.5 - 0 = -0.5000 \end{aligned}$$ $$[\Delta M^{0}] = \begin{bmatrix} \Delta P_2 \\ \Delta Q_2 \end{bmatrix} = \begin{bmatrix} -1.0000 \\ -0.5000 \end{bmatrix}$$

Initial Jacobian

$$\begin{aligned} \frac{\partial P_2}{\partial \delta_2} &= 10|V_2|\cos\delta_2 = 10 \\ \frac{\partial P_2}{\partial |V_2|} &= 10\sin\delta_2 = 0 \\ \frac{\partial Q_2}{\partial \delta_2} &= 10|V_2|\sin\delta_2 = 0 \\ \frac{\partial Q_2}{\partial |V_2|} &= -10\cos\delta_2 + 20|V_2| = -10 + 20 = 10 \end{aligned}$$ $$[J^{0}] = \begin{bmatrix} 10.0000 & 0.0000 \\ 0.0000 & 10.0000 \end{bmatrix}$$

(For use in the first iteration: $\Delta\delta_2 = -1/10 = -0.1000$ rad $= -5.730^\circ$ and $\Delta|V_2| = -0.5/10 = -0.0500$, giving $V_2^{1} = 0.9500\angle -5.730^\circ$ pu.)

Answer: $[\Delta M^{0}] = [-1.0000,\ -0.5000]^T$ and $[J^{0}] = \begin{bmatrix} 10 & 0 \\ 0 & 10 \end{bmatrix}$.