---
title: "G-S two iterations, lines 0.026+j0.11"
subject: "Power System Analysis II"
chapter: 2
key: "semester_5/power_system_analysis_ii:gs-two-iterations-2075"
source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/gs-two-iterations-2075
site: BCT Engineering
---

# G-S two iterations, lines 0.026+j0.11

Power System Analysis II · Chapter 2: Load Flow Analysis

- Asked in: 2075 Asoj, 12 marks
- Video: [Gauss Seidal Method Solved Problem -2 |  GS Method | Power flow Analysis | Power System Analysis](https://www.youtube.com/watch?v=MQBtptz5VaY) — 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-two-iterations-2075
- Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-2

## Question

Perform the two iterations of gauss-seidal load flow analysis for the data table given below to find the unknown variables of the busses after first iteration. Line impedance of each line is (0.026 + j0.11) pu.

| Bus no | PG | QG | PL | QL | Bus voltage | Type | Configuration |
|---|---|---|---|---|---|---|---|
| 1 | ? | ? | 1.0 | 0.5 | 1.03∠0 | Slack | 1-2 |
| 2 | 1.5 | ? | 0.5 | 0 | 1.03 | PV | 2-3 |
| 3 | 0.0 | 0 | 1.2 | 0.5 | ? | PQ | 3-1 |

## Short answer

Using flat start, after first Gauss‑Seidel iteration: Q2 = 0.2660 pu; V2 = 1.0300∠2.75° pu (magnitude reset to 1.03); V3 = 0.9869∠−2.02° pu.

## Answer

Bus 1 is the slack bus, bus 2 a PV (generator) bus and bus 3 a PQ (load) bus. Lines 1-2, 2-3 and 3-1 each have $z = 0.026 + j0.110$ pu (no shunt admittance given). Gauss-Seidel is used with no acceleration factor and no Q-limits (none are given).

#### Bus data
| Bus | Type | Specified | Unknown |
|---|---|---|---|
| 1 | Slack | $V_1 = 1.03\angle 0^\circ$ | $P_1$, $Q_1$ |
| 2 | PV | $P_2 = 1.5 - 0.5 = 1.0$, $\lvert V_2\rvert = 1.03$ | $Q_2$, $\delta_2$ |
| 3 | PQ | $P_3 = -1.2$, $Q_3 = -0.5$ | $\lvert V_3\rvert$, $\delta_3$ |

Initial values: $V_2^{(0)} = 1.03\angle 0^\circ$, $V_3^{(0)} = 1.0\angle 0^\circ$.

#### Y-bus
$$
\begin{aligned}
y_{12} &= \frac{1}{0.0260 + j0.1100} = 2.0351 - j8.6099\ \text{pu} \\
y_{23} &= \frac{1}{0.0260 + j0.1100} = 2.0351 - j8.6099\ \text{pu} \\
y_{13} &= \frac{1}{0.0260 + j0.1100} = 2.0351 - j8.6099\ \text{pu}
\end{aligned}
$$

$$
Y_{bus} = \begin{bmatrix} 4.0701 - j17.2198 & -2.0351 + j8.6099 & -2.0351 + j8.6099 \\ -2.0351 + j8.6099 & 4.0701 - j17.2198 & -2.0351 + j8.6099 \\ -2.0351 + j8.6099 & -2.0351 + j8.6099 & 4.0701 - j17.2198 \end{bmatrix}\ \text{pu}
$$

#### Gauss-Seidel iterations
For the PV bus, $Q_2$ is calculated first, then $V_2$ is found and its magnitude reset to 1.03 pu, keeping the new angle. Bus 3 uses the newest $V_2$.

**Iteration 1**

$$
\begin{aligned}
\sum_{k=1}^{3} Y_{2k}V_k &= 0.0611 - j0.2583 \\
Q_2^{(1)} &= -\text{Im}\{V_2^{*}\textstyle\sum_k Y_{2k}V_k\} = 0.2660\ \text{pu}
\end{aligned}
$$

$$
\begin{aligned}
V_2^{(1)} &= \frac{1}{Y_{22}}\left[\frac{P_2 - jQ_2}{V_2^{*}} - Y_{21}V_1 - Y_{23}V_3\right] \\
\frac{P_2 - jQ_2}{V_2^{*}} &= \frac{1.0000 - j0.2660}{1.0300} = 0.9709 - j0.2583 \\
\textstyle\sum_{k\ne 2} Y_{2k}V_k &= -4.1312 + j17.4781 \\
V_2^{(1)} &= \frac{(0.9709 - j0.2583) - (-4.1312 + j17.4781)}{4.0701 - j17.2198} = 1.0418 + j0.0500 \\
\delta_2 &= 2.7499^\circ \Rightarrow V_2^{(1)} = 1.0300\angle 2.7499^\circ = 1.0288 + j0.0494\ \text{(magnitude reset to }1.03\text{)}
\end{aligned}
$$

$$
\begin{aligned}
V_3^{(1)} &= \frac{1}{Y_{33}}\left[\frac{P_3 - jQ_3}{V_3^{*}} - Y_{31}V_1 - Y_{32}V_2\right] \\
\frac{P_3 - jQ_3}{V_3^{*}} &= \frac{-1.2000 + j0.5000}{1.0000} = -1.2000 + j0.5000 \\
\textstyle\sum_{k\ne 3} Y_{3k}V_k &= -4.6153 + j17.6256 \\
V_3^{(1)} &= \frac{(-1.2000 + j0.5000) - (-4.6153 + j17.6256)}{4.0701 - j17.2198} = 0.9863 - j0.0348 \\
&= 0.9869\angle -2.0203^\circ\ \text{pu}
\end{aligned}
$$

**Iteration 2**

$$
\begin{aligned}
\sum_{k=1}^{3} Y_{2k}V_k &= 1.2346 - j0.0838 \\
Q_2^{(2)} &= -\text{Im}\{V_2^{*}\textstyle\sum_k Y_{2k}V_k\} = 0.1473\ \text{pu}
\end{aligned}
$$

$$
\begin{aligned}
V_2^{(2)} &= \frac{1}{Y_{22}}\left[\frac{P_2 - jQ_2}{V_2^{*}} - Y_{21}V_1 - Y_{23}V_3\right] \\
\frac{P_2 - jQ_2}{V_2^{*}} &= \frac{1.0000 - j0.1473}{1.0288 - j0.0494} = 0.9766 - j0.0962 \\
\textstyle\sum_{k\ne 2} Y_{2k}V_k &= -3.8038 + j17.4310 \\
V_2^{(2)} &= \frac{(0.9766 - j0.0962) - (-3.8038 + j17.4310)}{4.0701 - j17.2198} = 1.0261 + j0.0351 \\
\delta_2 &= 1.9572^\circ \Rightarrow V_2^{(2)} = 1.0300\angle 1.9572^\circ = 1.0294 + j0.0352\ \text{(magnitude reset to }1.03\text{)}
\end{aligned}
$$

$$
\begin{aligned}
V_3^{(2)} &= \frac{1}{Y_{33}}\left[\frac{P_3 - jQ_3}{V_3^{*}} - Y_{31}V_1 - Y_{32}V_2\right] \\
\frac{P_3 - jQ_3}{V_3^{*}} &= \frac{-1.2000 + j0.5000}{0.9863 + j0.0348} = -1.1973 + j0.5492 \\
\textstyle\sum_{k\ne 3} Y_{3k}V_k &= -4.4939 + j17.6596 \\
V_3^{(2)} &= \frac{(-1.1973 + j0.5492) - (-4.4939 + j17.6596)}{4.0701 - j17.2198} = 0.9839 - j0.0411 \\
&= 0.9848\angle -2.3932^\circ\ \text{pu}
\end{aligned}
$$

#### Unknown variables after the 2nd iteration
**Slack bus power** (with the 2nd-iteration voltages)

$$
\begin{aligned}
P_1 - jQ_1 &= V_1^{*}\textstyle\sum_k Y_{1k}V_k = (1.0300)(0.1462 - j0.3897) \\
&= 0.1505 - j0.4014 \\
\Rightarrow P_1 &= 0.1505\ \text{pu},\quad Q_1 = 0.4014\ \text{pu} \\
P_{G1} &= P_1 + P_{L1} = 0.1505 + 1.00 = 1.1505\ \text{pu} \\
Q_{G1} &= Q_1 + Q_{L1} = 0.4014 + 0.50 = 0.9014\ \text{pu}
\end{aligned}
$$

Total loss = sum of the calculated net injections at all buses (equal to the sum of all line losses):

| Bus | $P_i$ (pu) | $Q_i$ (pu) |
|---|---|---|
| 1 | 0.1505 | 0.4014 |
| 2 | 1.0764 | 0.2011 |
| 3 | -1.2006 | -0.4911 |
| Sum | 0.0263 | 0.1114 |

$$
\begin{aligned}
P_{loss} &= \textstyle\sum P_i = 0.0263\ \text{pu} \\
Q_{loss} &= \textstyle\sum Q_i = 0.1114\ \text{pu}
\end{aligned}
$$

| Unknown | Value |
|---|---|
| $V_2$ | $1.0300\angle 1.9572^\circ$ pu |
| $Q_{G2} = Q_2 + Q_{L2}$ | $0.1473$ pu |
| $V_3$ | $0.9848\angle -2.3932^\circ$ pu |
| $P_{G1}$ | $1.1505$ pu |
| $Q_{G1}$ | $0.9014$ pu |

Note: $Q_{G2} = 0.1473$ pu is the value used in the 2nd iteration (found from the 1st-iteration voltages). Recomputing $Q_2$ from the final voltages gives 0.2011 pu (see the table above). The two differ because Gauss-Seidel has not yet converged.

**Answer:** after two iterations $V_2 = 1.0300\angle 1.9572^\circ$ pu, $V_3 = 0.9848\angle -2.3932^\circ$ pu, $Q_{G2} = 0.1473$ pu, $P_{G1} = 1.1505$ pu, $Q_{G1} = 0.9014$ pu, and real power loss $\approx 0.0263$ pu.

## Check yourself

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

1. Which set of bus types and unknowns is correct for this system?
   A. All buses PQ with unknown P and Q
   B. Bus 1 slack (V1 unknown), bus 2 PQ, bus 3 PV
   C. Bus 1 PV, bus 2 slack, bus 3 PQ
   D. Bus 1 slack (P1, Q1 unknown), bus 2 PV (Q2, delta2 unknown), bus 3 PQ (|V3|, delta3 unknown)
   Answer: D. Bus 1 has fixed V and angle (slack), bus 2 has fixed P and |V| (PV), bus 3 has fixed P and Q (PQ). (Key point: Bus types)

2. With z = 0.026 + j0.11 pu, the series admittance of each line is 2.0351 - j8.6099 pu. What is Y11?
   A. 1.0175 - j4.3050
   B. 4.0701 - j17.2198
   C. -4.0701 + j17.2198
   D. 2.0351 - j8.6099
   Answer: B. Bus 1 is connected to two lines, so Y11 = 2y = 4.0701 - j17.2198. (Key point: Line admittance)

3. Why is Q2 computed at each iteration of the Gauss-Seidel method?
   A. Because Q2 is specified and only checked
   B. Because it is the unknown of the PV bus needed in the update formula (P2 - jQ2)/V2*
   C. To compute the line charging
   D. To find the slack bus angle
   Answer: B. At a PV bus Q is not specified, so it is estimated from the latest voltages (Q2 = -Im{V2* sum Y2k Vk}) and used in the voltage update. (Key point: Why the Q2 is computed)

4. In computing V3 in an iteration, which value of V2 is used?
   A. The average of old and new V2
   B. V2 = 1.03 angle 0 always
   C. The newly updated V2 (after magnitude reset) from the same iteration
   D. The initial V2(0)
   Answer: C. Gauss-Seidel uses the most recently updated voltages immediately, so bus 3 uses the new V2. (Key point: Using newest values)

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