--- title: "G-S voltages after 2nd iteration with shunt" subject: "Power System Analysis II" chapter: 2 key: "semester_5/power_system_analysis_ii:gs-three-bus-shunt-2nd-iteration" source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/gs-three-bus-shunt-2nd-iteration site: BCT Engineering --- # G-S voltages after 2nd iteration with shunt Power System Analysis II · Chapter 2: Load Flow Analysis - Asked in: 2072 Kartik, 10 marks - Video: [Bus Admittance Matrix | Ybus matrix | Power System Analysis | PSA | Ybus Problem 1 | Numericals](https://www.youtube.com/watch?v=lG74k3jMd4k) — Simple n Easy Learning - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/gs-three-bus-shunt-2nd-iteration - Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-2 ## Question In a 3-bus power system the series impedance and shunt admittance of each line are (0.015+j0.12) p.u. and (j0.02) p.u. respectively. Form Y bus and compute the magnitude and phase angles of voltage at bus 2 and 3 after 2nd iteration using G-S method. The data given below are in p.u. [Figure: buses 1, 2 and 3 connected by lines 1-2, 1-3 and 2-3] | Bus | PG | QG | PL | QL | Bus voltage | |---|---|---|---|---|---| | 1 | - | - | - | - | 1.03∠0° | | 2 | 1.5 | - | 0 | 0 | 1.03 | | 3 | 0 | - | 1.2 | 0.5 | - | ## Short answer After 2nd Gauss-Seidel iteration: bus 2 voltage magnitude 1.0300 pu, angle 4.1446°; bus 3 voltage magnitude 0.9891 pu, angle -1.7834°. ## Answer Bus 1 is the slack bus ($1.03\angle 0^\circ$), bus 2 a PV bus ($P_2 = 1.5$ pu, $|V_2| = 1.03$ pu) and bus 3 a PQ bus ($P_3 = -1.2$, $Q_3 = -0.5$ pu). The shunt admittance $j0.02$ pu is taken as the total line charging of each line, so $j0.01$ pu is placed at each end. Flat start: $V_2^{(0)} = 1.03\angle 0^\circ$, $V_3^{(0)} = 1\angle 0^\circ$; no acceleration factor; no Q-limits given. #### Y-bus $$ \begin{aligned} y_{series} &= \frac{1}{0.015 + j0.120} = 1.0256 - j8.2051\ \text{pu} \\ Y_{ii} &= 2y_{series} + 2(j0.01) = 2.0513 - j16.3903 \\ Y_{ij} &= -y_{series} = -1.0256 + j8.2051 \end{aligned} $$ $$ Y_{bus} = \begin{bmatrix} 2.0513 - j16.3903 & -1.0256 + j8.2051 & -1.0256 + j8.2051 \\ -1.0256 + j8.2051 & 2.0513 - j16.3903 & -1.0256 + j8.2051 \\ -1.0256 + j8.2051 & -1.0256 + j8.2051 & 2.0513 - j16.3903 \end{bmatrix}\ \text{pu} $$ #### Gauss-Seidel iterations At the PV bus, $Q_2$ is found from the latest voltages, $V_2$ is updated and its magnitude reset to 1.03 pu. Bus 3 then uses the newest $V_2$. **Iteration 1** $$ \begin{aligned} \sum_{k=1}^{3} Y_{2k}V_k &= 0.0308 - j0.2256 \\ Q_2^{(1)} &= -\text{Im}\{V_2^{*}\textstyle\sum_k Y_{2k}V_k\} = 0.2323\ \text{pu} \end{aligned} $$ $$ \begin{aligned} V_2^{(1)} &= \frac{1}{Y_{22}}\left[\frac{P_2 - jQ_2}{V_2^{*}} - Y_{21}V_1 - Y_{23}V_3\right] \\ \frac{P_2 - jQ_2}{V_2^{*}} &= \frac{1.5000 - j0.2323}{1.0300} = 1.4563 - j0.2256 \\ \textstyle\sum_{k\ne 2} Y_{2k}V_k &= -2.0821 + j16.6564 \\ V_2^{(1)} &= \frac{(1.4563 - j0.2256) - (-2.0821 + j16.6564)}{2.0513 - j16.3903} = 1.0407 + j0.0856 \\ \delta_2 &= 4.7039^\circ \Rightarrow V_2^{(1)} = 1.0300\angle 4.7039^\circ = 1.0265 + j0.0845\ \text{(magnitude reset to }1.03\text{)} \end{aligned} $$ $$ \begin{aligned} V_3^{(1)} &= \frac{1}{Y_{33}}\left[\frac{P_3 - jQ_3}{V_3^{*}} - Y_{31}V_1 - Y_{32}V_2\right] \\ \frac{P_3 - jQ_3}{V_3^{*}} &= \frac{-1.2000 + j0.5000}{1.0000} = -1.2000 + j0.5000 \\ \textstyle\sum_{k\ne 3} Y_{3k}V_k &= -2.8023 + j16.7875 \\ V_3^{(1)} &= \frac{(-1.2000 + j0.5000) - (-2.8023 + j16.7875)}{2.0513 - j16.3903} = 0.9904 - j0.0262 \\ &= 0.9908\angle -1.5151^\circ\ \text{pu} \end{aligned} $$ **Iteration 2** $$ \begin{aligned} \sum_{k=1}^{3} Y_{2k}V_k &= 1.6328 - j0.0469 \\ Q_2^{(2)} &= -\text{Im}\{V_2^{*}\textstyle\sum_k Y_{2k}V_k\} = 0.1861\ \text{pu} \end{aligned} $$ $$ \begin{aligned} V_2^{(2)} &= \frac{1}{Y_{22}}\left[\frac{P_2 - jQ_2}{V_2^{*}} - Y_{21}V_1 - Y_{23}V_3\right] \\ \frac{P_2 - jQ_2}{V_2^{*}} &= \frac{1.5000 - j0.1861}{1.0265 - j0.0845} = 1.4662 - j0.0606 \\ \textstyle\sum_{k\ne 2} Y_{2k}V_k &= -1.8573 + j16.6049 \\ V_2^{(2)} &= \frac{(1.4662 - j0.0606) - (-1.8573 + j16.6049)}{2.0513 - j16.3903} = 1.0261 + j0.0744 \\ \delta_2 &= 4.1446^\circ \Rightarrow V_2^{(2)} = 1.0300\angle 4.1446^\circ = 1.0273 + j0.0744\ \text{(magnitude reset to }1.03\text{)} \end{aligned} $$ $$ \begin{aligned} V_3^{(2)} &= \frac{1}{Y_{33}}\left[\frac{P_3 - jQ_3}{V_3^{*}} - Y_{31}V_1 - Y_{32}V_2\right] \\ \frac{P_3 - jQ_3}{V_3^{*}} &= \frac{-1.2000 + j0.5000}{0.9904 + j0.0262} = -1.1974 + j0.5365 \\ \textstyle\sum_{k\ne 3} Y_{3k}V_k &= -2.7209 + j16.8041 \\ V_3^{(2)} &= \frac{(-1.1974 + j0.5365) - (-2.7209 + j16.8041)}{2.0513 - j16.3903} = 0.9887 - j0.0308 \\ &= 0.9891\angle -1.7834^\circ\ \text{pu} \end{aligned} $$ #### Result after the 2nd iteration | Bus | $\lvert V\rvert$ (pu) | $\delta$ (deg) | |---|---|---| | 2 | 1.0300 | 4.1446 | | 3 | 0.9891 | -1.7834 | **Answer:** $V_2 = 1.0300\angle 4.1446^\circ$ pu and $V_3 = 0.9891\angle -1.7834^\circ$ pu after the 2nd iteration (with $Q_2 = 0.1861$ pu used in that iteration). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Shunt admittance j0.02 pu is the total line charging. What is the diagonal Yii? A. 2y - j0.02 B. 2y + j0.02 only once per bus pair C. 2y + 2(j0.01) = 2.0513 - j16.3903 D. y + j0.01 Answer: C. Half of each line's charging (j0.01) sits at each end, so each bus gets two halves; y = 1.0256 - j8.2051 gives 2.0513 - j16.3903. (Key point: Y-bus with shunt) 2. What is the term for a PQ bus in the update V3 = (1/Y33)[ ... ]? A. (P3 + jQ3)/V3 - Y31V1 - Y32V2 B. (P3 - jQ3)/V3* - Y31V1 - Y32V2 C. (P3 - jQ3)/V3 + Y31V1 + Y32V2 D. P3 - jQ3 only Answer: B. The Gauss-Seidel formula uses the conjugate of the previous V3 and subtracts the other buses' contributions; P3 = -1.2 and Q3 = -0.5 are net injections. (Key point: Bus 3 update) 3. After computing V2 = 1.0407 + j0.0856 in iteration 1, what is used for further steps? A. 1.03 angle 4.7039 degrees B. 1.0407 + j0.0856 unchanged C. 1.0443 angle 4.7039 degrees D. 1.03 angle 0 degrees Answer: A. The magnitude is reset to the specified 1.03 pu while the calculated angle (4.7039 degrees) is kept. (Key point: PV voltage magnitude) 4. What are the voltages after iteration 2? A. V2 = 1.0 angle 0, V3 = 1.0 angle 0 B. V2 = 1.0261 + j0.0744 kept with magnitude 1.0289 C. V2 = 1.03 angle 4.7039 deg, V3 = 0.9908 angle -1.5151 deg D. V2 = 1.03 angle 4.1446 deg, V3 = 0.9891 angle -1.7834 deg Answer: D. The second iteration gives V2 = 1.0300 angle 4.1446 deg (after reset) and V3 = 0.9891 angle -1.7834 deg. 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