--- title: "N-R two iterations, y12 = 10∠−73.74°" subject: "Power System Analysis II" chapter: 2 key: "semester_5/power_system_analysis_ii:nr-two-bus-y12-10" source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/nr-two-bus-y12-10 site: BCT Engineering --- # N-R two iterations, y12 = 10∠−73.74° Power System Analysis II · Chapter 2: Load Flow Analysis - Asked in: 2073 Chaitra, 10 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-y12-10 - Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-2 ## Question In the power system network shown in figure below, bus 1 is slack bus with V1 = 1.0∠0° pu and bus 2 is a load bus with S2 = 150 MW + j50 MVAR. The line admittance is y12 = 10∠−73.74° pu on a base of 100 MVA. Perform two iterations of Newton Raphson load flow method to obtain the following: i) Voltage magnitude and phase angle of bus 2 ii) Real and reactive power supplied by slack bus and network losses [Figure: bus 1 generator V1 = 1.0∠0°, line Y12 = 2.8 − j9.6, bus 2 load 150 MW, 50 MVAR] ## Short answer Polar Newton-Raphson yields after two iterations V2 = 0.889∠−8.39° pu. Slack power is 1.5833 + j0.7965 pu. Network real power loss is 0.0833 pu (8.33 MW) and reactive loss is 0.2965 pu (29.65 MVar). ## Answer Newton-Raphson (polar form) is applied to the two-bus system. The line admittance is $y_{12} = 10\angle -73.74^\circ = 2.8 - j9.6$ pu, and the load 150 MW + j50 MVAr is $1.5 + j0.5$ 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 $P_2 = -1.50$ pu and $Q_2 = -0.50$ pu (negative, as power is taken out). Unknowns: $\delta_2$, $|V_2|$. Flat start: $|V_2|^{(0)} = 1$, $\delta_2^{(0)} = 0$. $$ \begin{aligned} y_{12} &= 10\angle -73.74^\circ = 2.8 - j9.6 \\ Y_{11} &= Y_{22} = y_{12} = 2.8000 - j9.6000 = 10.0000\angle -73.74^\circ \\ Y_{12} &= Y_{21} = -y_{12} = -2.8000 + j9.6000 = 10.0000\angle 106.26^\circ \end{aligned} $$ #### 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: $$ \begin{aligned} P_2^{(0)} &= 0.0000, \quad \Delta P_2 = -1.5000 - (0.0000) = -1.5000 \\ Q_2^{(0)} &= 0.0000, \quad \Delta Q_2 = -0.5000 - (0.0000) = -0.5000 \end{aligned} $$ Jacobian elements: $$ \begin{aligned} \frac{\partial P_2}{\partial \delta_2} &= (1.0000)(1.0000)(10.0000)\sin(106.26^\circ) = 9.6000 \\ \frac{\partial P_2}{\partial |V_2|} &= 2(1.0000)(10.0000)\cos(-73.74^\circ) + (1.0000)(10.0000)\cos(106.26^\circ) = 2.8000 \\ \frac{\partial Q_2}{\partial \delta_2} &= (1.0000)(1.0000)(10.0000)\cos(106.26^\circ) = -2.8000 \\ \frac{\partial Q_2}{\partial |V_2|} &= -2(1.0000)(10.0000)\sin(-73.74^\circ) - (1.0000)(10.0000)\sin(106.26^\circ) = 9.6000 \end{aligned} $$ $$ \begin{bmatrix} -1.5000 \\ -0.5000 \end{bmatrix} = \begin{bmatrix} 9.6000 & 2.8000 \\ -2.8000 & 9.6000 \end{bmatrix}\begin{bmatrix} \Delta\delta_2 \\ \Delta|V_2| \end{bmatrix} $$ $$ \begin{bmatrix} \Delta\delta_2 \\ \Delta|V_2| \end{bmatrix} = \begin{bmatrix} -0.1300 \\ -0.0900 \end{bmatrix} $$ $$ \begin{aligned} \delta_2^{(1)} &= 0.0000 + (-0.1300) = -0.1300\ \text{rad} = -7.4485^\circ \\ |V_2|^{(1)} &= 1.0000 + (-0.0900) = 0.9100\ \text{pu} \end{aligned} $$ **Iteration 2** Calculated powers and mismatches with the values from iteration 1: $$ \begin{aligned} P_2^{(1)} &= -1.3403, \quad \Delta P_2 = -1.5000 - (-1.3403) = -0.1597 \\ Q_2^{(1)} &= -0.3822, \quad \Delta Q_2 = -0.5000 - (-0.3822) = -0.1178 \end{aligned} $$ Jacobian elements: $$ \begin{aligned} \frac{\partial P_2}{\partial \delta_2} &= (0.9100)(1.0000)(10.0000)\sin(113.71^\circ) = 8.3320 \\ \frac{\partial P_2}{\partial |V_2|} &= 2(0.9100)(10.0000)\cos(-73.74^\circ) + (1.0000)(10.0000)\cos(113.71^\circ) = 1.0751 \\ \frac{\partial Q_2}{\partial \delta_2} &= (0.9100)(1.0000)(10.0000)\cos(113.71^\circ) = -3.6590 \\ \frac{\partial Q_2}{\partial |V_2|} &= -2(0.9100)(10.0000)\sin(-73.74^\circ) - (1.0000)(10.0000)\sin(113.71^\circ) = 8.3160 \end{aligned} $$ $$ \begin{bmatrix} -0.1597 \\ -0.1178 \end{bmatrix} = \begin{bmatrix} 8.3320 & 1.0751 \\ -3.6590 & 8.3160 \end{bmatrix}\begin{bmatrix} \Delta\delta_2 \\ \Delta|V_2| \end{bmatrix} $$ $$ \begin{bmatrix} \Delta\delta_2 \\ \Delta|V_2| \end{bmatrix} = \begin{bmatrix} -0.0164 \\ -0.0214 \end{bmatrix} $$ $$ \begin{aligned} \delta_2^{(2)} &= -0.1300 + (-0.0164) = -0.1464\ \text{rad} = -8.3885^\circ \\ |V_2|^{(2)} &= 0.9100 + (-0.0214) = 0.8886\ \text{pu} \end{aligned} $$ #### i) Bus 2 voltage after the 2nd iteration $|V_2| = 0.8886$ pu, $\delta_2 = -8.389^\circ$, i.e. $V_2 = 0.8886\angle -8.389^\circ$ pu. #### ii) Slack bus power and network losses $$ \begin{aligned} S_1 = P_1 + jQ_1 &= V_1\left[Y_{11}V_1 + Y_{12}V_2\right]^{*} = 1.5830 + j0.7976\ \text{pu} \\ &= 158.30\ \text{MW} + j79.76\ \text{MVAr} \end{aligned} $$ $$ \begin{aligned} I_{12} &= (V_1 - V_2)y_{12} = 1.5830 - j0.7976 = 1.7726\angle -26.7403^\circ\ \text{pu} \\ S_{loss} &= S_{12} + S_{21} = V_1I_{12}^{*} - V_2I_{12}^{*} = 0.0880 + j0.3016\ \text{pu} \\ P_{loss} &= 8.80\ \text{MW}, \quad Q_{loss} = 30.16\ \text{MVAr} \end{aligned} $$ (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 $0.0830 + j0.2976$ 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 $V_2 = 0.8879\angle -8.420^\circ$ pu.) **Answer:** $V_2 = 0.8886\angle -8.389^\circ$ pu; slack bus supplies $P_1 = 158.30$ MW and $Q_1 = 79.76$ MVAr; losses $\approx 8.80$ MW and $30.16$ MVAr. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the load 150 MW + j50 MVAr used in the N-R equations on a 100 MVA base? A. P2 = +1.5, Q2 = +0.5 B. P2 = 150, Q2 = 50 C. P2 = -1.5, Q2 = -0.5 D. P2 = -0.15, Q2 = -0.05 Answer: C. Divide by 100 MVA and take the negative sign because the power is taken out of the bus. (Key point: Per-unit load) 2. Which Jacobian is correct at the first iteration (Y12 = 10 at 106.26 deg, Y22 = 10 at -73.74 deg)? A. [9.6 -2.8; 2.8 9.6] B. [2.8 9.6; 9.6 -2.8] C. [10 0; 0 10] D. [9.6 2.8; -2.8 9.6] Answer: D. dP/ddelta = 10 sin(106.26) = 9.6, dP/dV = 20cos(-73.74) + 10cos(106.26) = 2.8, dQ/ddelta = -2.8, dQ/dV = 9.6. (Key point: Jacobian at flat start) 3. What are the bus 2 values after iteration 1? A. |V2| = 0.91 pu, delta = -7.45 deg B. |V2| = 0.89 pu, delta = -8.39 deg C. |V2| = 1.09 pu, delta = +7.45 deg D. |V2| = 0.91 pu, delta = -0.13 deg Answer: A. The corrections are -0.13 rad and -0.09 pu, applied to the flat start 1 pu and 0 rad. (Key point: First update) 4. How are the slack bus power and network loss obtained after the 2nd iteration? A. S1 = V1[Y11V1 + Y12V2]*; loss from line current as S12 + S21 B. S1 = specified load; loss = 0 C. S1 = V2 I*; loss = P2 only D. S1 = 1.5 + j0.5 always Answer: A. S1 = 1.583 + j0.798 pu from the Y-bus row, and the line current gives loss 0.088 + j0.302 pu (8.8 MW, 30.16 MVAr). 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