---
title: "Line flows by G-S after one iteration"
subject: "Power System Analysis II"
chapter: 2
key: "semester_5/power_system_analysis_ii:gs-line-flows-2071"
source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/gs-line-flows-2071
site: BCT Engineering
---

# Line flows by G-S after one iteration

Power System Analysis II · Chapter 2: Load Flow Analysis

- Asked in: 2071 Chaitra, 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-line-flows-2071
- Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-2

## Question

The following bus and line data are available for a power system, determine the line flows by Gauss Seidel method using voltages obtained after 1 iteration. Consider flat voltage start for all the load buses.

Bus data:

| Bus No | PG, pu | QG, pu | PL, pu | QL, pu | |V|, pu | δ, degree |
|---|---|---|---|---|---|---|
| 1 | - | - | - | - | 1.02 | 0 |
| 2 | - | - | 0.8 | 0.4 | - | - |
| 3 | - | - | 1.0 | 0.6 | - | - |

Line data:

| Line | Series admittance, pu |
|---|---|
| 1-2 | 0.01+j0.05 |
| 2-3 | 0.007+j0.037 |
| 3-1 | 0.012+j0.064 |

## Short answer

Treat line data as series impedances. After one G-S iteration: V2=0.9968∠−0.8802°, V3=0.9872∠−1.7706° pu. Line flows: S12=0.3918+j0.3971, S23=0.4468+j0.1770, S31=−0.5599−j0.3931 pu; total loss=0.0104+j0.0543 pu.

## Answer

The line values $0.01 + j0.05$ pu etc. have the size of series **impedances** (as admittances they would be far too small), so they are treated as series impedances. Bus 1 is the slack bus ($1.02\angle 0^\circ$); buses 2 and 3 are load buses: $P_2 = -0.8$, $Q_2 = -0.4$, $P_3 = -1.0$, $Q_3 = -0.6$ pu. Flat start $V_2^{(0)} = V_3^{(0)} = 1\angle 0^\circ$.

#### Y-bus
$$
\begin{aligned}
y_{12} &= \frac{1}{0.010 + j0.050} = 3.8462 - j19.2308 \\
y_{23} &= \frac{1}{0.007 + j0.037} = 4.9365 - j26.0931 \\
y_{31} &= \frac{1}{0.012 + j0.064} = 2.8302 - j15.0943
\end{aligned}
$$

$$
Y_{bus} = \begin{bmatrix} 6.6763 - j34.3251 & -3.8462 + j19.2308 & -2.8302 + j15.0943 \\ -3.8462 + j19.2308 & 8.7827 - j45.3239 & -4.9365 + j26.0931 \\ -2.8302 + j15.0943 & -4.9365 + j26.0931 & 7.7667 - j41.1874 \end{bmatrix}\ \text{pu}
$$

#### Gauss-Seidel: first iteration
**Iteration 1**

$$
\begin{aligned}
V_2^{(1)} &= \frac{1}{Y_{22}}\left[\frac{P_2 - jQ_2}{V_2^{*}} - Y_{21}V_1 - Y_{23}V_3\right] \\
\frac{P_2 - jQ_2}{V_2^{*}} &= \frac{-0.8000 + j0.4000}{1.0000} = -0.8000 + j0.4000 \\
\textstyle\sum_{k\ne 2} Y_{2k}V_k &= -8.8596 + j45.7085 \\
V_2^{(1)} &= \frac{(-0.8000 + j0.4000) - (-8.8596 + j45.7085)}{8.7827 - j45.3239} = 0.9967 - j0.0153 \\
&= 0.9968\angle -0.8802^\circ\ \text{pu}
\end{aligned}
$$

$$
\begin{aligned}
V_3^{(1)} &= \frac{1}{Y_{33}}\left[\frac{P_3 - jQ_3}{V_3^{*}} - Y_{31}V_1 - Y_{32}V_2\right] \\
\frac{P_3 - jQ_3}{V_3^{*}} &= \frac{-1.0000 + j0.6000}{1.0000} = -1.0000 + j0.6000 \\
\textstyle\sum_{k\ne 3} Y_{3k}V_k &= -7.4074 + j41.4786 \\
V_3^{(1)} &= \frac{(-1.0000 + j0.6000) - (-7.4074 + j41.4786)}{7.7667 - j41.1874} = 0.9868 - j0.0305 \\
&= 0.9872\angle -1.7706^\circ\ \text{pu}
\end{aligned}
$$

Voltages after one iteration: $V_1 = 1.02\angle 0^\circ$, $V_2 = 0.9968\angle -0.8802^\circ$, $V_3 = 0.9872\angle -1.7706^\circ$ pu.

#### Line flows
$S_{ij} = V_iI_{ij}^{*}$ with $I_{ij} = (V_i - V_j)y_{ij}$ (no shunt admittance). For line 1-2:

$$
\begin{aligned}
I_{12} &= (V_1 - V_2)y_{12} = (1.0200 - (0.9967 - j0.0153))(3.8462 - j19.2308) = 0.3841 - j0.3893 \\
S_{12} &= V_1I_{12}^{*} = 0.3918 + j0.3971\ \text{pu} \\
S_{21} &= V_2(-I_{12})^{*} = -0.3888 - j0.3821\ \text{pu}
\end{aligned}
$$

All lines:

| Line | $S_{ij}$ (pu) | $S_{ji}$ (pu) | Loss (pu) |
|---|---|---|---|
| 1-2 | 0.3918 + j0.3971 | -0.3888 - j0.3821 | 0.0030 + j0.0150 |
| 2-3 | 0.4468 + j0.1770 | -0.4452 - j0.1684 | 0.0016 + j0.0086 |
| 3-1 | -0.5599 - j0.3931 | 0.5656 + j0.4239 | 0.0058 + j0.0307 |

Total loss $= 0.0104 + j0.0543$ pu. Slack bus injection $S_1 = S_{12} + S_{13} = 0.9574 + j0.8210$ pu.

**Answer:** $S_{12} = 0.3918 + j0.3971$, $S_{23}$ and $S_{31}$ as in the table; total line loss $= 0.0104$ pu real and 0.0543 pu reactive (with first-iteration voltages).

## Check yourself

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

1. Why are 0.01+j0.05 etc. treated as series impedances and not admittances?
   A. Because they are given in ohms
   B. Because as admittances they would be unrealistically small; their size matches impedances
   C. Because the lines are short
   D. Because bus 1 is the slack
   Answer: B. Values like 0.01+j0.05 are typical per-unit impedances, so each line admittance is y = 1/z. (Key point: Interpreting line data)

2. How is the diagonal element Y22 formed here (no shunts)?
   A. Sum of the admittances of lines connected to bus 2
   B. Sum of impedances of lines at bus 2
   C. Negative of y12 only
   D. Average of y12 and y23
   Answer: A. Yii is the sum of admittances of all lines meeting at bus i; off-diagonals are minus the line admittance. (Key point: Ybus formation)

3. Which expression gives the Gauss-Seidel update of a load-bus voltage?
   A. V = (1/Yii)[(P - jQ)/V* - sum Yik Vk (k not i)]
   B. V = Yii[(P + jQ)/V - sum Yik Vk]
   C. V = (P - jQ)/Yii
   D. V = V* sum Yik
   Answer: A. The scheduled (P - jQ)/V* term minus the effect of the other buses, divided by Yii, gives the new voltage; P and Q are negative for loads. (Key point: G-S update for load bus)

4. How are line flow and line loss computed once voltages are known?
   A. S_ij = Vi (Iij)* with Iij = (Vi - Vj) yij; loss = S_ij + S_ji
   B. S_ij = Vi Vj yij; loss = S_ij - S_ji
   C. S_ij = Iij^2 yij; loss = S_ij
   D. S_ij = Vi/yij; loss = 2 S_ij
   Answer: A. Current from the voltage difference times yij, then S = V I*; adding the flows at both ends gives the line loss. (Key point: Line flow computation)

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