1. **Problem Statement:** We have a 3-bus power system with bus 1 as slack bus and bus 2 as PQ bus with given power demands $P_2 = -0.8$ pu and $Q_2 = -0.3$ pu. The goal is to perform one iteration of the Newton-Raphson load flow method starting from flat start ($|V_2|=1.0$ pu, $\delta_2=0$) to find updated voltage magnitude and angle at bus 2. 2. **Y-bus Matrix:** Given line impedance $j0.1$ pu between bus 1 and 2, admittance is $$Y_{12} = \frac{1}{j0.1} = -j10 = j10 \quad \text{(since admittance is imaginary)}$$ Thus, $$Y = \begin{bmatrix} -j10 & j10 \\ j10 & -j10 \end{bmatrix}$$ 3. **Power Flow Equations:** Power injection at bus 2 is $$S_2 = P_2 + jQ_2 = V_2 I_2^* = V_2 \left( \sum_{k=1}^2 Y_{2k} V_k \right)^*$$ Current at bus 2: $$I_2 = Y_{21} V_1 + Y_{22} V_2 = j10 \times 1.05 - j10 V_2$$ 4. **Mismatch Equations:** Define mismatches as $$\Delta P = P_2^{spec} - P_2^{calc}$$ $$\Delta Q = Q_2^{spec} - Q_2^{calc}$$ Using power flow formulas, $$P_2 = |V_2||V_1| B_{21} \sin(\delta_2 - \delta_1) = 10.5 |V_2| \sin \delta_2$$ $$Q_2 = |V_2||V_1| B_{21} \cos(\delta_2 - \delta_1) - |V_2|^2 B_{22} = 10.5 |V_2| \cos \delta_2 - 10.5 |V_2|^2$$ At flat start $|V_2|=1$, $\delta_2=0$, calculate: $$P_2 = 10.5 \times 1 \times \sin 0 = 0$$ $$Q_2 = 10.5 \times 1 \times \cos 0 - 10.5 \times 1^2 = 10.5 - 10.5 = 0$$ Mismatch: $$\Delta P = -0.8 - 0 = -0.8$$ $$\Delta Q = -0.3 - 0 = -0.3$$ 5. **Jacobian Matrix:** Partial derivatives at flat start: $$\frac{\partial P_2}{\partial \delta_2} = 10.5 |V_2| \cos \delta_2 = 10.5$$ $$\frac{\partial P_2}{\partial |V_2|} = 10.5 \sin \delta_2 = 0$$ $$\frac{\partial Q_2}{\partial \delta_2} = 10.5 |V_2| \sin \delta_2 = 0$$ $$\frac{\partial Q_2}{\partial |V_2|} = -10.5 \cos \delta_2 - 2 \times 10.5 |V_2| = -10.5 - 21 = -31.5$$ Jacobian: $$J = \begin{bmatrix} 10.5 & 0 \\ 0 & -31.5 \end{bmatrix}$$ 6. **Calculate Voltage Corrections:** $$\begin{bmatrix} \Delta \delta_2 \\ \Delta |V_2| \end{bmatrix} = J^{-1} \begin{bmatrix} \Delta P \\ \Delta Q \end{bmatrix} = \begin{bmatrix} \frac{1}{10.5} & 0 \\ 0 & \frac{1}{-31.5} \end{bmatrix} \begin{bmatrix} -0.8 \\ -0.3 \end{bmatrix} = \begin{bmatrix} -0.0762 \\ 0.0095 \end{bmatrix}$$ 7. **Update Voltage Magnitude and Angle:** $$|V_2|^{new} = 1.0 + 0.0095 = 1.0095$$ $$\delta_2^{new} = 0 - 0.0762 = -0.0762 \text{ radians}$$ **Interpretation:** The voltage magnitude at bus 2 slightly increases from 1.0 to approximately 1.0095 pu, and the voltage angle shifts slightly negative, indicating a small phase lag relative to the slack bus. This is consistent with the load demands and system admittance.