---
title: "G-S load flow with given Ybus, losses"
subject: "Power System Analysis II"
chapter: 2
key: "semester_5/power_system_analysis_ii:gs-ybus-j12-two-iterations"
source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/gs-ybus-j12-two-iterations
site: BCT Engineering
---

# G-S load flow with given Ybus, losses

Power System Analysis II · Chapter 2: Load Flow Analysis

- Asked in: 2081 Bhadra, 8+2 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-ybus-j12-two-iterations
- Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-2

## Question

A three-bus power system has all the three diagonal elements of Ybus matrix equal to −j12 pu and all the off-diagonal elements equal to j6 pu. The power and bus voltages are as follows:

| Bus No | Pload | Qload | Pgen | Qgen | V | Bus Type |
|---|---|---|---|---|---|---|
| 1 | 2.0 | 0.6 | ? | ? | 1.05∠0 | Slack |
| 2 | 0.7 | 0.5 | 0 | 0 | ? | PQ |
| 3 | 0.6 | 0.3 | 0 | 0 | ? | PQ |

Carry out the load flow analysis (up to 2nd iteration) to compute the unknown variables in the above table starting from assumptions of unknown voltage magnitude as 1 pu and phase voltage angle zero degree using G-S method. Also compute the network real and reactive power losses.

## Short answer

Using G-S method with flat start, after two iterations: V2 = 0.9798∠-5.63° pu, V3 = 0.9882∠-5.59° pu. Slack bus P_G1 = 3.2121 pu, Q_G1 = 1.4908 pu. Real power loss is zero (network pure reactance); reactive power loss is 0.1717 pu.

## Answer

Bus 1 is the slack bus; buses 2 and 3 are load (PQ) buses. Gauss-Seidel is used with the given $Y_{bus}$, flat start $V_2^{(0)} = V_3^{(0)} = 1\angle 0^\circ$ pu, and no acceleration factor.

#### Data
$Y_{ii} = -j12$ pu and $Y_{ij} = j6$ pu. Net injections ($P_i = P_{Gi} - P_{Li}$):

$$
\begin{aligned}
P_2 - jQ_2 &= -0.7 + j0.5 \\
P_3 - jQ_3 &= -0.6 + j0.3
\end{aligned}
$$

#### Gauss-Seidel iterations
**Iteration 1**

$$
\begin{aligned}
V_2^{(1)} &= \frac{1}{Y_{22}}\left[\frac{P_2 - jQ_2}{V_2^{*}} - Y_{21}V_1 - Y_{23}V_3\right] \\
\frac{P_2 - jQ_2}{V_2^{*}} &= \frac{-0.7000 + j0.5000}{1.0000} = -0.7000 + j0.5000 \\
\textstyle\sum_{k\ne 2} Y_{2k}V_k &= j12.3000 \\
V_2^{(1)} &= \frac{(-0.7000 + j0.5000) - (j12.3000)}{-j12.0000} = 0.9833 - j0.0583 \\
&= 0.9851\angle -3.3949^\circ\ \text{pu}
\end{aligned}
$$

$$
\begin{aligned}
V_3^{(1)} &= \frac{1}{Y_{33}}\left[\frac{P_3 - jQ_3}{V_3^{*}} - Y_{31}V_1 - Y_{32}V_2\right] \\
\frac{P_3 - jQ_3}{V_3^{*}} &= \frac{-0.6000 + j0.3000}{1.0000} = -0.6000 + j0.3000 \\
\textstyle\sum_{k\ne 3} Y_{3k}V_k &= 0.3500 + j12.2000 \\
V_3^{(1)} &= \frac{(-0.6000 + j0.3000) - (0.3500 + j12.2000)}{-j12.0000} = 0.9917 - j0.0792 \\
&= 0.9948\angle -4.5644^\circ\ \text{pu}
\end{aligned}
$$

**Iteration 2**

$$
\begin{aligned}
V_2^{(2)} &= \frac{1}{Y_{22}}\left[\frac{P_2 - jQ_2}{V_2^{*}} - Y_{21}V_1 - Y_{23}V_3\right] \\
\frac{P_2 - jQ_2}{V_2^{*}} &= \frac{-0.7000 + j0.5000}{0.9833 + j0.0583} = -0.6793 + j0.5488 \\
\textstyle\sum_{k\ne 2} Y_{2k}V_k &= 0.4750 + j12.2500 \\
V_2^{(2)} &= \frac{(-0.6793 + j0.5488) - (0.4750 + j12.2500)}{-j12.0000} = 0.9751 - j0.0962 \\
&= 0.9798\angle -5.6339^\circ\ \text{pu}
\end{aligned}
$$

$$
\begin{aligned}
V_3^{(2)} &= \frac{1}{Y_{33}}\left[\frac{P_3 - jQ_3}{V_3^{*}} - Y_{31}V_1 - Y_{32}V_2\right] \\
\frac{P_3 - jQ_3}{V_3^{*}} &= \frac{-0.6000 + j0.3000}{0.9917 + j0.0792} = -0.5772 + j0.3486 \\
\textstyle\sum_{k\ne 3} Y_{3k}V_k &= 0.5772 + j12.1506 \\
V_3^{(2)} &= \frac{(-0.5772 + j0.3486) - (0.5772 + j12.1506)}{-j12.0000} = 0.9835 - j0.0962 \\
&= 0.9882\angle -5.5864^\circ\ \text{pu}
\end{aligned}
$$

#### Results after the 2nd iteration
| Bus | Voltage (pu) |
|---|---|
| 1 | $1.05\angle 0^\circ$ |
| 2 | $0.9798\angle -5.6339^\circ$ |
| 3 | $0.9882\angle -5.5864^\circ$ |

**Slack bus power**

$$
\begin{aligned}
P_1 - jQ_1 &= V_1^{*}\textstyle\sum_k Y_{1k}V_k = (1.0500)(1.1543 - j0.8484) \\
&= 1.2121 - j0.8908 \\
\Rightarrow P_1 &= 1.2121\ \text{pu},\quad Q_1 = 0.8908\ \text{pu} \\
P_{G1} &= P_1 + P_{L1} = 1.2121 + 2.00 = 3.2121\ \text{pu} \\
Q_{G1} &= Q_1 + Q_{L1} = 0.8908 + 0.60 = 1.4908\ \text{pu}
\end{aligned}
$$

#### Network losses
Line flows, using $y_{ij} = -Y_{ij} = -j6$ pu and $S_{ij} = V_i[(V_i - V_j)y_{ij}]^{*}$:

| Line | $S_{ij}$ (pu) | $S_{ji}$ (pu) | Loss (pu) |
|---|---|---|---|
| 1-2 | 0.6060 + j0.4719 | -0.6060 - j0.3827 | j0.0892 |
| 1-3 | 0.6060 + j0.4189 | -0.6060 - j0.3369 | j0.0821 |
| 2-3 | -0.0048 - j0.0491 | 0.0048 + j0.0496 | j0.0004 |

The network is purely reactive (no resistance), so the real power loss is zero:

$$
\begin{aligned}
P_{loss} &= P_1 + P_2 + P_3 = 1.2121 + (-0.6108) + (-0.6012) = 0.0000\ \text{pu} \\
Q_{loss} &= Q_1 + Q_2 + Q_3 = 0.8908 + (-0.4318) + (-0.2873) = 0.1717\ \text{pu}
\end{aligned}
$$

(Here $P_2$, $P_3$, $Q_2$, $Q_3$ are recalculated with the 2nd-iteration voltages.)

**Answer:** $V_2 = 0.9798\angle -5.6339^\circ$ pu, $V_3 = 0.9882\angle -5.5864^\circ$ pu, $P_{G1} = 3.2121$ pu, $Q_{G1} = 1.4908$ pu, real power loss $= 0.0000$ pu, reactive power loss $= 0.1717$ pu.

## Check yourself

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

1. For bus 2 (load 0.7 + j0.5, no generation), what is P2 - jQ2 in the G-S formula?
   A. 0.7 - j0.5
   B. -0.7 + j0.5
   C. -0.7 - j0.5
   D. 0.7 + j0.5
   Answer: B. Net injection is S2 = -0.7 - j0.5, so P2 - jQ2 = -0.7 + j0.5. (Key point: Net injection setup)

2. In iteration 1 for bus 2, why is the sum Y21V1 + Y23V3 = j12.30?
   A. j6(1.05) + j6(1.0) = j6.3 + j6 = j12.3
   B. j12(1.05)
   C. j6(1.0) + j6(1.0) = j12
   D. -j6(1.05) - j6(1.0)
   Answer: A. Y21 = Y23 = j6, with V1 = 1.05 and V3 = 1.0, giving j6.3 + j6 = j12.3. (Key point: Iteration 1 sum term)

3. How is the network loss computed from bus powers?
   A. Ploss = P1 + P2 + P3 and Qloss = Q1 + Q2 + Q3 (net injections)
   B. Ploss = P1 - P2 - P3
   C. Only P1
   D. Loss = load sum
   Answer: A. Total net injection equals loss; here P loss = 0 and Q loss = 0.1717 pu. (Key point: Reactive power loss sum)

4. Slack bus P1 = 1.2121 pu, with load 2.0 pu at bus 1. What is PG1?
   A. 1.2121 - 2.0 = -0.7879
   B. 1.2121 + 2.0 = 3.2121
   C. 2.0 - 1.2121
   D. 1.2121
   Answer: B. P1 is the net injection PG1 - PL1, so PG1 = P1 + PL1 = 3.2121 pu. (Key point: Slack generation)

## Related in Power System Analysis II

- [Network losses from given bus voltages](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Anetwork-losses-given-voltages) — Power System Analysis II, chapter 2
- [G-S with PV bus Q limits and losses](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Ags-pv-bus-q-limits-2071) — Power System Analysis II, chapter 2
- [3-bus load flow with Q limits (cut off)](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Aload-flow-reference-load-generator-bus) — Power System Analysis II, chapter 2
- [Node equations and Ybus, six-bus network](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Aybus-six-bus-node-equations) — Power System Analysis II, chapter 1
- [Ybus for 3-bus system (2078)](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Aybus-three-bus-2078) — Power System Analysis II, chapter 1
- [Ybus derivation for 3-bus π-model network](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Aybus-three-bus-derivation) — Power System Analysis II, chapter 1, asked 2×

## Related in other subjects

- [Interconnected system SLD with bus voltages](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Ainterconnected-sld-bus-voltages) — Power System Analysis I, chapter 1