--- title: "N-R one iteration, P2 = 0.4 pu" subject: "Power System Analysis II" chapter: 2 key: "semester_5/power_system_analysis_ii:nr-one-iteration-y40-p0-4" source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/nr-one-iteration-y40-p0-4 site: BCT Engineering --- # N-R one iteration, P2 = 0.4 pu Power System Analysis II · Chapter 2: Load Flow Analysis - Asked in: 2074 Asoj, 8 marks - Video: [Newton Raphson Power Flow Example Part 1](https://www.youtube.com/watch?v=-XeG8WAjS7E) — GeneralPAC: Power System Tutorials - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/nr-one-iteration-y40-p0-4 - Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-2 ## Question Figure below shows the single line diagram of 3 bus power system network. Determine the jacobian matrix, and perform load flow by N-R method up to one iteration. [Figure: bus 1 slack bus 1∠0° pu; bus 2 generator bus |V2| = 1.05 pu with P2 = 0.4 pu; bus 3 load bus (5 + j4) pu; line admittances y12 = −j40 pu, y13 = −j20 pu, y23 = −j20 pu] ## Short answer Jacobian matrix: J = [63, −21, 0; −21, 41, 0; 0, 0, 39]. Flat-start mismatches: ΔP2=0.4, ΔP3=−5, ΔQ3=−3. After one iteration: V2 = 1.05∠−2.37° pu, V3 = 0.9231∠−8.20° pu. ## Answer Bus 1 is the slack bus ($1\angle 0^\circ$), bus 2 a generator (PV) bus ($P_2 = 0.4$ pu, $|V_2| = 1.05$ pu) and bus 3 a load bus ($P_3 = -5$ pu, $Q_3 = -4$ pu). Flat start: $V_2^0 = 1.05\angle 0^\circ$, $V_3^0 = 1\angle 0^\circ$. No Q-limits are given. #### Y-bus With $y_{12} = -j40$, $y_{13} = -j20$, $y_{23} = -j20$ pu: $$ \begin{aligned} Y_{11} &= y_{12} + y_{13} = -j60, \quad Y_{22} = y_{12} + y_{23} = -j60, \quad Y_{33} = y_{13} + y_{23} = -j40 \\ Y_{12} &= -y_{12} = j40, \quad Y_{13} = -y_{13} = j20, \quad Y_{23} = -y_{23} = j20 \end{aligned} $$ $$ Y_{bus} = \begin{bmatrix} -j60.0000 & j40.0000 & j20.0000 \\ j40.0000 & -j60.0000 & j20.0000 \\ j20.0000 & j20.0000 & -j40.0000 \end{bmatrix}\ \text{pu} $$ #### Calculated powers and mismatches 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, $j \ne i$): $$ \begin{aligned} \frac{\partial P_i}{\partial \delta_i} &= \sum_{j\ne i} |V_i||V_j||Y_{ij}|\sin\phi_{ij} \\ \frac{\partial P_i}{\partial \delta_j} &= -|V_i||V_j||Y_{ij}|\sin\phi_{ij} \\ \frac{\partial P_i}{\partial |V_i|} &= 2|V_i||Y_{ii}|\cos\theta_{ii} + \sum_{j\ne i} |V_j||Y_{ij}|\cos\phi_{ij} \\ \frac{\partial P_i}{\partial |V_j|} &= |V_i||Y_{ij}|\cos\phi_{ij} \\ \frac{\partial Q_i}{\partial \delta_i} &= \sum_{j\ne i} |V_i||V_j||Y_{ij}|\cos\phi_{ij} \\ \frac{\partial Q_i}{\partial \delta_j} &= -|V_i||V_j||Y_{ij}|\cos\phi_{ij} \\ \frac{\partial Q_i}{\partial |V_i|} &= -2|V_i||Y_{ii}|\sin\theta_{ii} - \sum_{j\ne i} |V_j||Y_{ij}|\sin\phi_{ij} \\ \frac{\partial Q_i}{\partial |V_j|} &= -|V_i||Y_{ij}|\sin\phi_{ij} \end{aligned} $$ At the flat start all $\cos$ terms are zero ($\theta = \pm 90^\circ$, angles zero), so $P_2^0 = P_3^0 = 0$: $$ \begin{aligned} P_2^{0} &= 0.0000, \quad Q_2^{0} = 3.1500 \\ P_3^{0} &= 0.0000, \quad Q_3^{0} = -1.0000 \end{aligned} $$ $$ \begin{aligned} \Delta P_2 &= 0.4000 - (0.0000) = 0.4000 \\ \Delta P_3 &= -5.0000 - (0.0000) = -5.0000 \\ \Delta Q_3 &= -4.0000 - (-1.0000) = -3.0000 \end{aligned} $$ #### Jacobian matrix $$ \begin{aligned} \frac{\partial P_2}{\partial \delta_2} &= (1.0500)(1.0000)(40.0000)\sin(90.00^\circ) + (1.0500)(1.0000)(20.0000)\sin(90.00^\circ) = 63.0000 \\ \frac{\partial P_2}{\partial \delta_3} &= -(1.0500)(1.0000)(20.0000)\sin(90.00^\circ) = -21.0000 \\ \frac{\partial P_2}{\partial |V_3|} &= (1.0500)(20.0000)\cos(90.00^\circ) = 0.0000 \\ \frac{\partial P_3}{\partial \delta_2} &= -(1.0000)(1.0500)(20.0000)\sin(90.00^\circ) = -21.0000 \\ \frac{\partial P_3}{\partial \delta_3} &= (1.0000)(1.0000)(20.0000)\sin(90.00^\circ) + (1.0000)(1.0500)(20.0000)\sin(90.00^\circ) = 41.0000 \\ \frac{\partial P_3}{\partial |V_3|} &= 2(1.0000)(40.0000)\cos(-90.00^\circ) + (1.0000)(20.0000)\cos(90.00^\circ) + (1.0500)(20.0000)\cos(90.00^\circ) = 0.0000 \\ \frac{\partial Q_3}{\partial \delta_2} &= -(1.0000)(1.0500)(20.0000)\cos(90.00^\circ) = 0.0000 \\ \frac{\partial Q_3}{\partial \delta_3} &= (1.0000)(1.0000)(20.0000)\cos(90.00^\circ) + (1.0000)(1.0500)(20.0000)\cos(90.00^\circ) = 0.0000 \\ \frac{\partial Q_3}{\partial |V_3|} &= -2(1.0000)(40.0000)\sin(-90.00^\circ) - (1.0000)(20.0000)\sin(90.00^\circ) - (1.0500)(20.0000)\sin(90.00^\circ) = 39.0000 \end{aligned} $$ $$ [J^{0}] = \begin{bmatrix} 63.0000 & -21.0000 & 0.0000 \\ -21.0000 & 41.0000 & 0.0000 \\ 0.0000 & 0.0000 & 39.0000 \end{bmatrix} $$ #### First iteration $$ \begin{bmatrix} 0.4000 \\ -5.0000 \\ -3.0000 \end{bmatrix} = \begin{bmatrix} 63.0000 & -21.0000 & 0.0000 \\ -21.0000 & 41.0000 & 0.0000 \\ 0.0000 & 0.0000 & 39.0000 \end{bmatrix}\begin{bmatrix} \Delta\delta_2 \\ \Delta\delta_3 \\ \Delta|V_3| \end{bmatrix} $$ $$ \begin{bmatrix} \Delta\delta_2 \\ \Delta\delta_3 \\ \Delta|V_3| \end{bmatrix} = [J^{0}]^{-1}\begin{bmatrix} 0.4000 \\ -5.0000 \\ -3.0000 \end{bmatrix} = \begin{bmatrix} -0.0414 \\ -0.1431 \\ -0.0769 \end{bmatrix} $$ $$ \begin{aligned} \delta_2^{1} &= 0 + (-0.0414) = -0.0414\ \text{rad} = -2.370^\circ \\ \delta_3^{1} &= 0 + (-0.1431) = -0.1431\ \text{rad} = -8.201^\circ \\ |V_3|^{1} &= 1.0000 + (-0.0769) = 0.9231\ \text{pu} \end{aligned} $$ So $V_2^{1} = 1.0500\angle -2.370^\circ$ pu and $V_3^{1} = 0.9231\angle -8.201^\circ$ pu. **Reactive power at bus 2 and slack bus power** (with the new voltages): $$ \begin{aligned} Q_2^{1} &= -\text{Im}\{V_2^{*}\textstyle\sum_k Y_{2k}V_k\} = 4.9016\ \text{pu} \\ Q_{G2} &= Q_2 + Q_{L2} = 4.9016\ \text{pu} \end{aligned} $$ $$ \begin{aligned} P_1 - jQ_1 &= V_1^{*}\textstyle\sum_k Y_{1k}V_k = (1.0000)(4.3703 + j0.2368) \\ &= 4.3703 + j0.2368 \\ \Rightarrow P_1 &= 4.3703\ \text{pu},\quad Q_1 = -0.2368\ \text{pu} \\ P_{G1} &= P_1 + P_{L1} = 4.3703 + 0.00 = 4.3703\ \text{pu} \\ Q_{G1} &= Q_1 + Q_{L1} = -0.2368 + 0.00 = -0.2368\ \text{pu} \end{aligned} $$ **Answer:** $J^0 = \begin{bmatrix} 63 & -21 & 0 \\ -21 & 41 & 0 \\ 0 & 0 & 39 \end{bmatrix}$; after one iteration $\delta_2 = -2.370^\circ$, $V_3 = 0.9231\angle -8.201^\circ$ pu, $Q_{G2} = 4.9016$ pu, $P_1 = 4.3703$ pu, $Q_1 = -0.2368$ pu. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What size is the Jacobian for this system with bus 2 PV and bus 3 PQ? A. 2 x 2 B. 3 x 3 C. 4 x 4 D. 6 x 6 Answer: B. There are three unknowns (delta2, delta3, |V3|) and three equations (P2, P3, Q3), giving a 3 x 3 Jacobian. (Key point: Jacobian size) 2. How is dP2/ddelta2 obtained, and what is its value? A. -(1.05)(20) = -21 B. (1.05)(40) + (1.05)(20) = 63 C. 2(1.05)(60) = 126 D. (1.05)(40) = 42 Answer: B. dP2/ddelta2 = sum over j of |V2||Vj||Y2j| sin(phi) = 1.05(40) + 1.05(20) = 63. (Key point: dP2/ddelta2) 3. What is dQ3/d|V3| at the flat start? A. 39 B. 41 C. 40 D. -39 Answer: A. -2(1)(40)sin(-90) - (1)(20) - (1.05)(20) = 80 - 20 - 21 = 39. (Key point: dQ3/d|V3|) 4. With Q3^0 = -1.0 and load Q = 4 pu, what are DeltaP3 and DeltaQ3? A. -5 and +3 B. +5 and -3 C. -5 and -4 D. -5 and -3 Answer: D. DeltaP3 = -5 - 0 = -5 and DeltaQ3 = -4 - (-1) = -3. 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