--- title: "N-R two iterations, 2-bus system" subject: "Power System Analysis II" chapter: 2 key: "semester_5/power_system_analysis_ii:nr-two-bus-two-iterations" source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/nr-two-bus-two-iterations site: BCT Engineering --- # N-R two iterations, 2-bus system Power System Analysis II · Chapter 2: Load Flow Analysis - Asked in: 2073 Shrawan, 10 marks - Video: [LOAD FLOW ANALYSIS – PART - 37 – N-R METHOD - PROBLEM – 01 – TWO-BUS SYSTEM](https://www.youtube.com/watch?v=CrrfwNpIMvE) — zenmurali Sachithanandam - 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-two-iterations - Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-2 ## Question For the 2-bus power system, bus-1 is slack bus with V1 = 1.0∠0°. A load of 100 MW and 50 MVar is taken from bus-2. The line impedance is 0.12+j0.16 p.u on a base of 100 MVA. Using Newton Raphson load flow technique, determine the following after 2nd iteration. i) Voltage magnitude and phase angle in degree of bus-2. ii) Real and reactive power supplied by slack bus and network losses. ## Short answer Using polar Newton-Raphson: after two iterations V2 = 0.723∠−7.73° pu. Slack bus supplies P1 = 1.2387 pu, Q1 = 0.8416 pu. Real power loss is 0.2387 pu (23.87 MW) and reactive loss is 0.3416 pu (34.16 MVar). ## Answer Newton-Raphson (polar form) is applied to the two-bus system; the line has $z = 0.12 + j0.16$ pu, and the load of 100 MW + j50 MVAr becomes $1 + j0.5$ pu on the 100 MVA base. #### Bus data and Y-bus Base 100 MVA. Bus 1: slack, $V_1 = 1\angle 0^\circ$. Bus 2: load bus with $P_2 = -1.00$ pu and $Q_2 = -0.50$ pu (negative, as power is taken out). Unknowns: $\delta_2$, $|V_2|$. Flat start: $|V_2|^{(0)} = 1$, $\delta_2^{(0)} = 0$. $$ \begin{aligned} y_{12} &= \frac{1}{0.12 + j0.16} = \frac{0.12 - j0.16}{0.04} = 3 - j4 \\ Y_{11} &= Y_{22} = y_{12} = 3.0000 - j4.0000 = 5.0000\angle -53.13^\circ \\ Y_{12} &= Y_{21} = -y_{12} = -3.0000 + j4.0000 = 5.0000\angle 126.87^\circ \end{aligned} $$ #### Equations used 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, $j \ne i$): $$ \begin{aligned} \frac{\partial P_i}{\partial \delta_i} &= \sum_{j\ne i} |V_i||V_j||Y_{ij}|\sin\phi_{ij} \\ \frac{\partial P_i}{\partial \delta_j} &= -|V_i||V_j||Y_{ij}|\sin\phi_{ij} \\ \frac{\partial P_i}{\partial |V_i|} &= 2|V_i||Y_{ii}|\cos\theta_{ii} + \sum_{j\ne i} |V_j||Y_{ij}|\cos\phi_{ij} \\ \frac{\partial P_i}{\partial |V_j|} &= |V_i||Y_{ij}|\cos\phi_{ij} \\ \frac{\partial Q_i}{\partial \delta_i} &= \sum_{j\ne i} |V_i||V_j||Y_{ij}|\cos\phi_{ij} \\ \frac{\partial Q_i}{\partial \delta_j} &= -|V_i||V_j||Y_{ij}|\cos\phi_{ij} \\ \frac{\partial Q_i}{\partial |V_i|} &= -2|V_i||Y_{ii}|\sin\theta_{ii} - \sum_{j\ne i} |V_j||Y_{ij}|\sin\phi_{ij} \\ \frac{\partial Q_i}{\partial |V_j|} &= -|V_i||Y_{ij}|\sin\phi_{ij} \end{aligned} $$ For bus 2, $[\Delta P_2;\ \Delta Q_2] = J\,[\Delta\delta_2;\ \Delta|V_2|]$. #### Newton-Raphson iterations **Iteration 1** Calculated powers and mismatches with the values from iteration 0: $$ \begin{aligned} P_2^{(0)} &= 0.0000, \quad \Delta P_2 = -1.0000 - (0.0000) = -1.0000 \\ Q_2^{(0)} &= 0.0000, \quad \Delta Q_2 = -0.5000 - (0.0000) = -0.5000 \end{aligned} $$ Jacobian elements: $$ \begin{aligned} \frac{\partial P_2}{\partial \delta_2} &= (1.0000)(1.0000)(5.0000)\sin(126.87^\circ) = 4.0000 \\ \frac{\partial P_2}{\partial |V_2|} &= 2(1.0000)(5.0000)\cos(-53.13^\circ) + (1.0000)(5.0000)\cos(126.87^\circ) = 3.0000 \\ \frac{\partial Q_2}{\partial \delta_2} &= (1.0000)(1.0000)(5.0000)\cos(126.87^\circ) = -3.0000 \\ \frac{\partial Q_2}{\partial |V_2|} &= -2(1.0000)(5.0000)\sin(-53.13^\circ) - (1.0000)(5.0000)\sin(126.87^\circ) = 4.0000 \end{aligned} $$ $$ \begin{bmatrix} -1.0000 \\ -0.5000 \end{bmatrix} = \begin{bmatrix} 4.0000 & 3.0000 \\ -3.0000 & 4.0000 \end{bmatrix}\begin{bmatrix} \Delta\delta_2 \\ \Delta|V_2| \end{bmatrix} $$ $$ \begin{bmatrix} \Delta\delta_2 \\ \Delta|V_2| \end{bmatrix} = \begin{bmatrix} -0.1000 \\ -0.2000 \end{bmatrix} $$ $$ \begin{aligned} \delta_2^{(1)} &= 0.0000 + (-0.1000) = -0.1000\ \text{rad} = -5.7296^\circ \\ |V_2|^{(1)} &= 1.0000 + (-0.2000) = 0.8000\ \text{pu} \end{aligned} $$ **Iteration 2** Calculated powers and mismatches with the values from iteration 1: $$ \begin{aligned} P_2^{(1)} &= -0.7875, \quad \Delta P_2 = -1.0000 - (-0.7875) = -0.2125 \\ Q_2^{(1)} &= -0.3844, \quad \Delta Q_2 = -0.5000 - (-0.3844) = -0.1156 \end{aligned} $$ Jacobian elements: $$ \begin{aligned} \frac{\partial P_2}{\partial \delta_2} &= (0.8000)(1.0000)(5.0000)\sin(132.60^\circ) = 2.9444 \\ \frac{\partial P_2}{\partial |V_2|} &= 2(0.8000)(5.0000)\cos(-53.13^\circ) + (1.0000)(5.0000)\cos(132.60^\circ) = 1.4157 \\ \frac{\partial Q_2}{\partial \delta_2} &= (0.8000)(1.0000)(5.0000)\cos(132.60^\circ) = -2.7075 \\ \frac{\partial Q_2}{\partial |V_2|} &= -2(0.8000)(5.0000)\sin(-53.13^\circ) - (1.0000)(5.0000)\sin(132.60^\circ) = 2.7195 \end{aligned} $$ $$ \begin{bmatrix} -0.2125 \\ -0.1156 \end{bmatrix} = \begin{bmatrix} 2.9444 & 1.4157 \\ -2.7075 & 2.7195 \end{bmatrix}\begin{bmatrix} \Delta\delta_2 \\ \Delta|V_2| \end{bmatrix} $$ $$ \begin{bmatrix} \Delta\delta_2 \\ \Delta|V_2| \end{bmatrix} = \begin{bmatrix} -0.0350 \\ -0.0773 \end{bmatrix} $$ $$ \begin{aligned} \delta_2^{(2)} &= -0.1000 + (-0.0350) = -0.1350\ \text{rad} = -7.7345^\circ \\ |V_2|^{(2)} &= 0.8000 + (-0.0773) = 0.7227\ \text{pu} \end{aligned} $$ #### i) Bus 2 voltage after the 2nd iteration $|V_2| = 0.7227$ pu, $\delta_2 = -7.735^\circ$, i.e. $V_2 = 0.7227\angle -7.735^\circ$ pu. #### ii) Slack bus power and network losses $$ \begin{aligned} S_1 = P_1 + jQ_1 &= V_1\left[Y_{11}V_1 + Y_{12}V_2\right]^{*} = 1.2408 + j0.8439\ \text{pu} \\ &= 124.08\ \text{MW} + j84.39\ \text{MVAr} \end{aligned} $$ $$ \begin{aligned} I_{12} &= (V_1 - V_2)y_{12} = 1.2408 - j0.8439 = 1.5006\angle -34.2208^\circ\ \text{pu} \\ S_{loss} &= S_{12} + S_{21} = V_1I_{12}^{*} - V_2I_{12}^{*} = 0.2702 + j0.3603\ \text{pu} \\ P_{loss} &= 27.02\ \text{MW}, \quad Q_{loss} = 36.03\ \text{MVAr} \end{aligned} $$ (The loss is found from the line current with the 2nd-iteration voltages. If it is instead taken as $S_1 - S_{load}$, it gives $0.2408 + j0.3439$ pu; the two differ because after two iterations the calculated bus 2 power has not yet reached the specified load (the solution is not fully converged). The converged answer is $V_2 = 0.7071\angle -8.130^\circ$ pu.) **Answer:** $V_2 = 0.7227\angle -7.735^\circ$ pu; slack bus supplies $P_1 = 124.08$ MW and $Q_1 = 84.39$ MVAr; losses $\approx 27.02$ MW and $36.03$ MVAr. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is y12 for z = 0.12 + j0.16 pu? A. 3 - j4 B. 0.12 - j0.16 C. 8.33 - j6.25 D. 3 + j4 Answer: A. y = 1/z = (0.12 - j0.16)/0.04 = 3 - j4, with magnitude 5 and angle -53.13 deg. (Key point: Line admittance) 2. At the flat start, calculated P2 and Q2 are zero. What are DeltaP2 and DeltaQ2? A. +1.0 and +0.5 B. -1.0 and -0.5 C. -0.5 and -1.0 D. 0 and 0 Answer: B. The load 1 + j0.5 pu is a negative injection, so mismatch = -1.0 - 0 and -0.5 - 0. (Key point: Initial mismatch) 3. Solving [-1; -0.5] = [4 3; -3 4][d_delta; d_V] gives what? A. d_delta = -0.2, d_V = -0.1 B. d_delta = +0.1, d_V = +0.2 C. d_delta = -0.1, d_V = -0.2 D. d_delta = -1.0, d_V = -0.5 Answer: C. The determinant is 25; the solution is -0.1 rad and -0.2 pu, giving V2 = 0.8 pu at -5.73 deg. (Key point: First N-R correction) 4. Why do the line-current loss and the S1 - Sload loss differ after two iterations? A. Line charging is ignored B. The slack bus has no losses C. The solution is not yet converged, so calculated bus 2 power differs from the load D. The base MVA is wrong Answer: C. Only at full convergence does S1 - Sload equal the line loss; the converged V2 is 0.7071 at -8.13 deg. 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