--- title: "Initial mismatch and Jacobian, 2-bus system" subject: "Power System Analysis II" chapter: 2 key: "semester_5/power_system_analysis_ii:nr-two-bus-mismatch-jacobian" source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/nr-two-bus-mismatch-jacobian site: BCT Engineering --- # Initial mismatch and Jacobian, 2-bus system Power System Analysis II · Chapter 2: Load Flow Analysis - Asked in: 2076 Chaitra, 8 marks - Video: [Electric Power Systems | Solved Problems | Newton-Raphson Load Flow Analysis](https://www.youtube.com/watch?v=XkrVoVtf2Zk) — Engineering Devotion - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/nr-two-bus-mismatch-jacobian - Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-2 ## Question For the system given below, find the initial power mismatch matrix and initial Jacobian matrix. Take a flat start of V2(0) = 1.0 pu and δ2(0) = 0°. [Figure: bus 1 generator with 1.0∠0°, line j0.1 to bus 2, load 1 + j0.5 at bus 2] ## Short answer Initial mismatches: ΔP2 = -1.0 pu, ΔQ2 = -0.5 pu. Initial Jacobian: ∂P2/∂δ2 = 10, ∂P2/∂|V2| = 0, ∂Q2/∂δ2 = 0, ∂Q2/∂|V2| = 10. ## Answer 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 $[J^{0}] = \begin{bmatrix} 10 & 0 \\ 0 & 10 \end{bmatrix}$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For this two-bus system, P2 reduces to A. -10|V2| sin(delta2) B. 10|V2|^2 C. 10|V2| cos(delta2) D. 10|V2| sin(delta2) Answer: D. theta21 = 90 degrees, so cos(90 - delta2) = sin(delta2), and the self-term has cos(-90) = 0, giving P2 = 10|V2| sin delta2. (Key point: Power equations) 2. What are DeltaP2 and DeltaQ2 at the flat start? A. 0 and 0 B. -1.0 and -0.5 C. -1.0 and 0.5 D. 1.0 and 0.5 Answer: B. Calculated P2 and Q2 are 0 at the flat start, and specified values are -1.0 and -0.5 (load), so mismatches are -1.0 and -0.5. (Key point: Initial mismatch) 3. What is dQ2/d|V2| at the flat start? A. 20 B. 10 C. -10 D. 0 Answer: B. dQ2/d|V2| = -10cos(delta2) + 20|V2| = -10 + 20 = 10. (Key point: Jacobian diagonal) 4. Why are dP2/d|V2| and dQ2/d delta2 equal to zero at the start? A. They are proportional to sin(delta2) and delta2 = 0 B. The line is lossless so there is no coupling at all C. The bus is a PV bus D. |V2| = 1 makes them zero Answer: A. dP2/d|V2| = 10 sin delta2 and dQ2/d delta2 = 10|V2| sin delta2, both zero at delta2 = 0. 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