---
title: "G-S one iteration, generator at bus 3"
subject: "Power System Analysis II"
chapter: 2
key: "semester_5/power_system_analysis_ii:gs-one-iteration-pv-bus3"
source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/gs-one-iteration-pv-bus3
site: BCT Engineering
---

# G-S one iteration, generator at bus 3

Power System Analysis II · Chapter 2: Load Flow Analysis

- Asked in: 2076 Chaitra, 8 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-one-iteration-pv-bus3
- Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-2

## Question

For the network given below, compute the bus voltage magnitude and phase angles for 1 iterations by Gauss-Seidal method taking initial voltage magnitude estimates of 1.0 pu and phase angle estimates of 0° for the required buses. [Figure: line 1-2 j0.06, line 1-3 j0.05, line 2-3 j0.08; V1 = 1.05∠0° pu; |V3| = 1.03 pu, PG3 = 2 pu; load at bus 2 SL2 = 3 + j1.5 pu]

## Short answer

After one Gauss-Seidel iteration: V2 = 0.9953∠-5.93° pu, V3 = 1.03∠1.12° pu, with Q3 = 0.103 pu.

## Answer

Bus 1 is the slack bus ($1.05\angle 0^\circ$), bus 2 a PQ (load) bus with $P_2 = -3$ pu, $Q_2 = -1.5$ pu, and bus 3 a PV bus with $P_3 = 2$ pu, $|V_3| = 1.03$ pu. Initial estimates: $V_2^{(0)} = 1.0\angle 0^\circ$; for the PV bus the specified magnitude is kept, $V_3^{(0)} = 1.03\angle 0^\circ$. Bus 2 is solved first, then bus 3 with the new $V_2$.

#### Y-bus
$$
\begin{aligned}
y_{12} &= 1/j0.06 = -j16.6667 \\
y_{13} &= 1/j0.05 = -j20.0000 \\
y_{23} &= 1/j0.08 = -j12.5000
\end{aligned}
$$

$$
Y_{bus} = \begin{bmatrix} -j36.6667 & j16.6667 & j20.0000 \\ j16.6667 & -j29.1667 & j12.5000 \\ j20.0000 & j12.5000 & -j32.5000 \end{bmatrix}\ \text{pu}
$$

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

$$
\begin{aligned}
V_2^{(1)} &= \frac{1}{Y_{22}}\left[\frac{P_2 - jQ_2}{V_2^{*}} - Y_{21}V_1 - Y_{23}V_3\right] \\
\frac{P_2 - jQ_2}{V_2^{*}} &= \frac{-3.0000 + j1.5000}{1.0000} = -3.0000 + j1.5000 \\
\textstyle\sum_{k\ne 2} Y_{2k}V_k &= j30.3750 \\
V_2^{(1)} &= \frac{(-3.0000 + j1.5000) - (j30.3750)}{-j29.1667} = 0.9900 - j0.1029 \\
&= 0.9953\angle -5.9315^\circ\ \text{pu}
\end{aligned}
$$

$$
\begin{aligned}
\sum_{k=1}^{3} Y_{3k}V_k &= 1.2857 - j0.1000 \\
Q_3^{(1)} &= -\text{Im}\{V_3^{*}\textstyle\sum_k Y_{3k}V_k\} = 0.1030\ \text{pu}
\end{aligned}
$$

$$
\begin{aligned}
V_3^{(1)} &= \frac{1}{Y_{33}}\left[\frac{P_3 - jQ_3}{V_3^{*}} - Y_{31}V_1 - Y_{32}V_2\right] \\
\frac{P_3 - jQ_3}{V_3^{*}} &= \frac{2.0000 - j0.1030}{1.0300} = 1.9417 - j0.1000 \\
\textstyle\sum_{k\ne 3} Y_{3k}V_k &= 1.2857 + j33.3750 \\
V_3^{(1)} &= \frac{(1.9417 - j0.1000) - (1.2857 + j33.3750)}{-j32.5000} = 1.0300 + j0.0202 \\
\delta_3 &= 1.1227^\circ \Rightarrow V_3^{(1)} = 1.0300\angle 1.1227^\circ = 1.0298 + j0.0202\ \text{(magnitude reset to }1.03\text{)}
\end{aligned}
$$

#### Result after one iteration
| Bus | $\lvert V\rvert$ (pu) | $\delta$ (deg) |
|---|---|---|
| 1 | 1.05 | 0 |
| 2 | 0.9953 | -5.9315 |
| 3 | 1.0300 | 1.1227 |

**Answer:** $V_2 = 0.9953\angle -5.9315^\circ$ pu, $V_3 = 1.0300\angle 1.1227^\circ$ pu (with $Q_3 = 0.1030$ pu).

## Check yourself

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

1. Which bus is solved first and with which values for bus 3?
   A. Slack bus first
   B. Bus 2 first using V3 = 1.03 angle 0; then bus 3 using new V2
   C. Both simultaneously
   D. Bus 3 first, then bus 2
   Answer: B. Buses are updated in order; the PV bus 3 starts at its specified magnitude 1.03 with 0 degrees, and bus 3's update uses the updated V2. (Key point: Order of solution)

2. For load bus 2 with SL2 = 3 + j1.5, the term (P2 - jQ2)/V2* at V2 = 1 is
   A. 3 - j1.5
   B. -3 - j1.5
   C. 3 + j1.5
   D. -3 + j1.5
   Answer: D. P2 = -3 and Q2 = -1.5, so P2 - jQ2 = -3 + j1.5. (Key point: Load bus formula)

3. Why is Q3 = 0.1030 computed before updating V3?
   A. Q is not specified at a PV bus; it is needed in (P3 - jQ3)/V3*
   B. It is a given load
   C. To adjust the slack bus
   D. To check the voltage limits
   Answer: A. At the PV bus only P and |V| are specified, so Q3 is estimated from the latest voltages. (Key point: PV bus Q3)

4. After V3 is computed as 1.0300 + j0.0202 (magnitude 1.0302), what is V3?
   A. 1.03 angle 1.1227 degrees
   B. 1.0 angle 1.1227 degrees
   C. 1.0302 angle 1.1227 degrees
   D. 1.03 angle 0 degrees
   Answer: A. Magnitude is reset to 1.03 pu and the computed angle 1.1227 degrees is retained. (Key point: Result)

## Related in Power System Analysis II

- [G-S two iterations, lines 0.026+j0.11](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Ags-two-iterations-2075) — Power System Analysis II, chapter 2
- [G-S two iterations, lines 0.02+j0.11](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Ags-two-iterations-2080) — Power System Analysis II, chapter 2
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- [Line flows by G-S after one iteration](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Ags-line-flows-2071) — Power System Analysis II, chapter 2
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- [G-S first iteration, 3-bus with PV bus](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Ags-first-iteration-2076-asoj) — Power System Analysis II, chapter 2