---
title: "G-S one iteration with π-model lines"
subject: "Power System Analysis II"
chapter: 2
key: "semester_5/power_system_analysis_ii:gs-one-iteration-pi-lines-2069"
source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/gs-one-iteration-pi-lines-2069
site: BCT Engineering
---

# G-S one iteration with π-model lines

Power System Analysis II · Chapter 2: Load Flow Analysis

- Asked in: 2069 Chaitra, 3×2+4 marks
- Video: [LOAD FLOW STUDY(GAUSS SEIDEL METHOD) GATE/IES/ISRO/BARC](https://www.youtube.com/watch?v=5gWgmUWoeH4) — ELECTRICAL ENGINEERING CONCEPTS
- 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-pi-lines-2069
- Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-2

## Question

The single line diagram of a power system network is shown in figure below. The per km line series reactance is 0.001 p.u and shunt susceptance is 0.0016 p.u. [Figure: line 1-2 is 100 km, line 1-3 is 200 km, line 2-3 is 400 km] The power and bus voltages are as follows:

| Bus No. | PLoad | QLoad | Pgen | Qgen | V |
|---|---|---|---|---|---|
| 1 | 0 | 0 | ? | ? | 1.05∠0° |
| 2 | 1.0 | 0.6 | 0 | 0 | ? |
| 3 | 0.8 | 0.3 | 0 | 0 | ? |

Using Ybus from the nominal pi-model of lines, compute the voltage at bus 2 & 3 using G-S method for one iteration and also the slack bus complex power using the computed and specified bus voltages. Start the analysis assuming unknown bus voltage magnitudes as 1 p.u. & phase angles zero degree.

## Short answer

Using nominal-π Ybus with line charging, one G-S iteration gives V2=1.0281∠−4.6106° pu and V3=1.0796∠−7.6322° pu. Slack bus complex power is S1=1.6206−j0.1054 pu.

## Answer

Each line is modelled as a nominal-π: series reactance = 0.001 × length, total shunt susceptance = 0.0016 × length, half at each end. Bus 1 is the slack bus ($1.05\angle 0^\circ$); buses 2 and 3 are load buses with $P_2 - jQ_2 = -1.0 + j0.6$ and $P_3 - jQ_3 = -0.8 + j0.3$ pu.

#### Line parameters
| Line | Length (km) | $X$ (pu) | $y = 1/jX$ (pu) | $B/2$ each end (pu) |
|---|---|---|---|---|
| 1-2 | 100 | 0.1 | $-j10$ | $j0.08$ |
| 1-3 | 200 | 0.2 | $-j5$ | $j0.16$ |
| 2-3 | 400 | 0.4 | $-j2.5$ | $j0.32$ |

#### Y-bus
$$
\begin{aligned}
Y_{11} &= -j10 - j5 + j0.08 + j0.16 = -j14.76 \\
Y_{22} &= -j10 - j2.5 + j0.08 + j0.32 = -j12.10 \\
Y_{33} &= -j5 - j2.5 + j0.16 + j0.32 = -j7.02 \\
Y_{12} &= j10, \quad Y_{13} = j5, \quad Y_{23} = j2.5
\end{aligned}
$$

$$
Y_{bus} = \begin{bmatrix} -j14.76 & j10.00 & j5.00 \\ j10.00 & -j12.10 & j2.50 \\ j5.00 & j2.50 & -j7.02 \end{bmatrix}\ \text{pu}
$$

#### Bus voltages after one G-S iteration
**Iteration 1**

$$
\begin{aligned}
V_2^{(1)} &= \frac{1}{Y_{22}}\left[\frac{P_2 - jQ_2}{V_2^{*}} - Y_{21}V_1 - Y_{23}V_3\right] \\
\frac{P_2 - jQ_2}{V_2^{*}} &= \frac{-1.0000 + j0.6000}{1.0000} = -1.0000 + j0.6000 \\
\textstyle\sum_{k\ne 2} Y_{2k}V_k &= j13.0000 \\
V_2^{(1)} &= \frac{(-1.0000 + j0.6000) - (j13.0000)}{-j12.1000} = 1.0248 - j0.0826 \\
&= 1.0281\angle -4.6106^\circ\ \text{pu}
\end{aligned}
$$

$$
\begin{aligned}
V_3^{(1)} &= \frac{1}{Y_{33}}\left[\frac{P_3 - jQ_3}{V_3^{*}} - Y_{31}V_1 - Y_{32}V_2\right] \\
\frac{P_3 - jQ_3}{V_3^{*}} &= \frac{-0.8000 + j0.3000}{1.0000} = -0.8000 + j0.3000 \\
\textstyle\sum_{k\ne 3} Y_{3k}V_k &= 0.2066 + j7.8120 \\
V_3^{(1)} &= \frac{(-0.8000 + j0.3000) - (0.2066 + j7.8120)}{-j7.0200} = 1.0701 - j0.1434 \\
&= 1.0796\angle -7.6322^\circ\ \text{pu}
\end{aligned}
$$

#### Slack bus complex power
Using the specified $V_1$ and the computed $V_2$, $V_3$:

$$
\begin{aligned}
P_1 - jQ_1 &= V_1^{*}\left(Y_{11}V_1 + Y_{12}V_2 + Y_{13}V_3\right) \\
&= (1.05)(1.5434 + j0.1003) = 1.6206 + j0.1054 \\
S_1 &= P_1 + jQ_1 = 1.6206 - j0.1054\ \text{pu}
\end{aligned}
$$

(Bus 1 has no load, so $S_{G1} = S_1$. The negative $Q_1$ shows that the slack generator absorbs reactive power, because the long lines produce a lot of charging current at this light loading.)

**Answer:** $V_2 = 1.0281\angle -4.6106^\circ$ pu, $V_3 = 1.0796\angle -7.6322^\circ$ pu, $S_1 = 1.6206 - j0.1054$ pu.

## Check yourself

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

1. How does the nominal pi-model affect the Ybus compared with series-only lines?
   A. Half of each line's total shunt susceptance is added to the diagonal at each end
   B. The full shunt susceptance is added to the off-diagonals
   C. Shunt terms are ignored
   D. Shunt terms are subtracted from the series admittance at only one end
   Answer: A. Each line contributes y_series plus B/2 at each end to the diagonal, e.g. Y11 = -j10 - j5 + j0.08 + j0.16 = -j14.76. (Key point: Nominal pi Ybus)

2. For the 200 km line 1-3, what are the series admittance and half shunt susceptance?
   A. -j5 and j0.16
   B. -j10 and j0.08
   C. -j2.5 and j0.32
   D. -j5 and j0.32
   Answer: A. X = 0.001*200 = 0.2 so y = 1/j0.2 = -j5; total B = 0.0016*200 = 0.32 so half is j0.16. (Key point: Line parameters)

3. How is the slack bus complex power found, and what does a negative Q1 imply?
   A. S1 = conj of V1*(sum Y1k Vk); the slack absorbs reactive power because of line charging
   B. S1 = V1 I1 directly; the slack supplies large Q
   C. S1 = loads only; Q1 is always positive
   D. S1 = V1^2 Y11; negative Q means a fault
   Answer: A. P1 - jQ1 = V1* sum Y1k Vk, giving Q1 = -0.1054 pu: long lightly loaded lines generate reactive power that the slack absorbs. (Key point: Slack complex power)

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