---
title: "G-S first iteration with PV bus"
subject: "Power System Analysis II"
chapter: 2
key: "semester_5/power_system_analysis_ii:gs-pv-bus-first-iteration"
source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/gs-pv-bus-first-iteration
site: BCT Engineering
---

# G-S first iteration with PV bus

Power System Analysis II · Chapter 2: Load Flow Analysis

- Asked in: 2078 Kartik, 10 marks
- Video: [Gauss Seidal Method Solved Problem -1 |  GS Method | Power flow Analysis | Power System Analysis](https://www.youtube.com/watch?v=2NRI6gZd_7o) — 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-pv-bus-first-iteration
- Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-2

## Question

For the system shown in figure, determine the voltage at the end of the first iteration by G-S method. Neglect limits on reactive power generation. Bus 1: Slack bus, V = 1∠0°; Bus 2: P-V bus, V = 1.05 pu, PG = 3 pu; Bus 3: P-Q bus, PL = 4 pu, QL = 2 pu. [Figure: line 1-2 j0.3, line 1-3 j0.2, line 2-3 j0.2; generators at buses 1 and 2, load at bus 3]

## Short answer

At end of first iteration: Q2 = 0.4375 pu, V2 = 1.05∠18.08° pu, V3 = 0.8335∠-16.52° pu. Computed by forming Ybus, finding Q2, updating V2 resetting magnitude, then V3.

## Answer

Bus 1 is the slack bus ($1\angle 0^\circ$), bus 2 a PV bus ($P_2 = 3$ pu, $|V_2| = 1.05$ pu) and bus 3 a PQ bus ($P_3 = -4$, $Q_3 = -2$ pu). Initial values: $V_2^{(0)} = 1.05\angle 0^\circ$, $V_3^{(0)} = 1\angle 0^\circ$. Q-limits are neglected.

#### Y-bus
$$
\begin{aligned}
y_{12} &= 1/j0.3 = -j3.3333 \\
y_{13} &= 1/j0.2 = -j5.0000 \\
y_{23} &= 1/j0.2 = -j5.0000
\end{aligned}
$$

$$
Y_{bus} = \begin{bmatrix} -j8.3333 & j3.3333 & j5.0000 \\ j3.3333 & -j8.3333 & j5.0000 \\ j5.0000 & j5.0000 & -j10.0000 \end{bmatrix}\ \text{pu}
$$

#### First Gauss-Seidel iteration
**Iteration 1**

$$
\begin{aligned}
\sum_{k=1}^{3} Y_{2k}V_k &= -j0.4167 \\
Q_2^{(1)} &= -\text{Im}\{V_2^{*}\textstyle\sum_k Y_{2k}V_k\} = 0.4375\ \text{pu}
\end{aligned}
$$

$$
\begin{aligned}
V_2^{(1)} &= \frac{1}{Y_{22}}\left[\frac{P_2 - jQ_2}{V_2^{*}} - Y_{21}V_1 - Y_{23}V_3\right] \\
\frac{P_2 - jQ_2}{V_2^{*}} &= \frac{3.0000 - j0.4375}{1.0500} = 2.8571 - j0.4167 \\
\textstyle\sum_{k\ne 2} Y_{2k}V_k &= j8.3333 \\
V_2^{(1)} &= \frac{(2.8571 - j0.4167) - (j8.3333)}{-j8.3333} = 1.0500 + j0.3429 \\
\delta_2 &= 18.0834^\circ \Rightarrow V_2^{(1)} = 1.0500\angle 18.0834^\circ = 0.9981 + j0.3259\ \text{(magnitude reset to }1.05\text{)}
\end{aligned}
$$

$$
\begin{aligned}
V_3^{(1)} &= \frac{1}{Y_{33}}\left[\frac{P_3 - jQ_3}{V_3^{*}} - Y_{31}V_1 - Y_{32}V_2\right] \\
\frac{P_3 - jQ_3}{V_3^{*}} &= \frac{-4.0000 + j2.0000}{1.0000} = -4.0000 + j2.0000 \\
\textstyle\sum_{k\ne 3} Y_{3k}V_k &= -1.6296 + j9.9907 \\
V_3^{(1)} &= \frac{(-4.0000 + j2.0000) - (-1.6296 + j9.9907)}{-j10.0000} = 0.7991 - j0.2370 \\
&= 0.8335\angle -16.5227^\circ\ \text{pu}
\end{aligned}
$$

**Answer (end of first iteration):** $Q_2 = 0.4375$ pu, $V_2 = 1.0500\angle 18.0834^\circ$ pu, $V_3 = 0.8335\angle -16.5227^\circ$ pu.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. With y12 = -j3.333, y13 = -j5 and y23 = -j5, what is Y33?
   A. -j10
   B. +j10
   C. -j5
   D. -j8.333
   Answer: A. Y33 = y13 + y23 = -j5 - j5 = -j10. (Key point: Y-bus from reactances)

2. In the first iteration, Q2 comes out as 0.4375 pu. How is it computed?
   A. It is given in the data
   B. Q2 = -Im{V2* (Y21V1 + Y22V2 + Y23V3)} using initial voltages
   C. Q2 = QL2
   D. Q2 = P2 tan(18 deg)
   Answer: B. With V1 = 1, V2 = 1.05, V3 = 1 the sum is -j0.4167, and -Im{1.05 x (-j0.4167)} = 0.4375. (Key point: Q2 at PV bus)

3. V2 first comes out as 1.0500 + j0.3429 (magnitude 1.1045). What is done next?
   A. Set the angle to zero
   B. Keep 1.0500 + j0.3429
   C. Set magnitude to 1.05 keeping angle 18.08 degrees
   D. Set magnitude to 1.0 pu
   Answer: C. At a PV bus the voltage magnitude is fixed, so only the angle (18.0834 degrees) is retained. (Key point: Magnitude reset)

4. For the PQ bus 3 the load enters as P3 - jQ3 equal to
   A. 4 - j2
   B. -4 + j2
   C. -4 - j2
   D. 4 + j2
   Answer: B. P3 = -4 and Q3 = -2 (load), so P3 - jQ3 = -4 + j2. (Key point: Load bus update)

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