---
title: "G-S load flow, j0.12 lines, losses"
subject: "Power System Analysis II"
chapter: 2
key: "semester_5/power_system_analysis_ii:gs-three-bus-j012-losses"
source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/gs-three-bus-j012-losses
site: BCT Engineering
---

# G-S load flow, j0.12 lines, losses

Power System Analysis II · Chapter 2: Load Flow Analysis

- Asked in: 2079 Bhadra, 10 marks; 2070 Chaitra, 10 marks (asked 2 times)
- 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-three-bus-j012-losses
- Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-2

## Question

For a network given below, series impedance of each line is 0 + j0.12 p.u. Shunt admittance of the line is negligible. Calculate: (i) Voltage at bus-2 and bus-3 by using G-S method (upto 2 iteration). (ii) Slack bus real and reactive power. (iii) Network real and reactive power losses. [Figure: three buses with lines 1-2, 1-3 and 2-3; bus 1 is the slack bus with generator, V1 = 1.0∠0°; load at bus 2 = 1.5 + j0.4 p.u; load at bus 3 = 1.2 + j0.5 p.u]

## Short answer

Using G-S with Ybus elements j8.3333 pu (off-diagonal) and −j16.6667 (diagonal), two iterations gave V2 = 0.9576∠−8.87° pu and V3 = 0.9443∠−8.78° pu. Slack bus power: P1 = 2.432 pu, Q1 = 1.005 pu. Since line resistance is zero, real power loss = 0 pu; reactive power loss = 0.418 pu (not fully converged after two iterations).

## Answer

#### Data and Ybus
Each line: $z = j0.12$, $y = 1/j0.12 = -j8.3333$ pu. Every bus has two lines:

$$
Y_{bus} =
\begin{bmatrix}
-j16.6667 & j8.3333 & j8.3333 \\
j8.3333 & -j16.6667 & j8.3333 \\
j8.3333 & j8.3333 & -j16.6667
\end{bmatrix}
$$

Bus 1: slack, $V_1 = 1.0\angle 0^\circ$. Buses 2 and 3 are PQ buses with net injections $S_2 = -1.5 - j0.4$ and $S_3 = -1.2 - j0.5$ pu. Flat start: $V_2^{(0)} = V_3^{(0)} = 1.0\angle 0^\circ$.

G-S equation (latest values used immediately):

$$
V_i^{(r+1)} = \frac{1}{Y_{ii}}\left[\frac{P_i - jQ_i}{V_i^{*(r)}} - \sum_{k \ne i}Y_{ik}V_k\right]
$$

#### (i) Iteration 1
$$
\begin{aligned}
V_2^{(1)} &= \frac{1}{-j16.6667}\left[\frac{-1.5 + j0.4}{1.0} - j8.3333(1.0) - j8.3333(1.0)\right] \\
&= \frac{-1.5 - j16.2667}{-j16.6667} = 0.9760 - j0.0900 = 0.9801\angle -5.2685^\circ \\
V_3^{(1)} &= \frac{1}{-j16.6667}\left[\frac{-1.2 + j0.5}{1.0} - j8.3333(1.0) - j8.3333(0.9760 - j0.0900)\right] \\
&= \frac{-1.95 - j15.9667}{-j16.6667} = 0.9580 - j0.1170 = 0.9651\angle -6.9630^\circ
\end{aligned}
$$

#### Iteration 2
$$
\begin{aligned}
\frac{-1.5 + j0.4}{V_2^{*(1)}} &= -1.4865 + j0.5469 \\
V_2^{(2)} &= \frac{(-1.4865 + j0.5469) - (0.9750 + j16.3167)}{-j16.6667} \\
&= 0.9462 - j0.1477 = 0.9576\angle -8.8715^\circ \\
\frac{-1.2 + j0.5}{V_3^{*(1)}} &= -1.1714 + j0.6650 \\
V_3^{(2)} &= \frac{(-1.1714 + j0.6650) - (1.2307 + j16.2182)}{-j16.6667} \\
&= 0.9332 - j0.1441 = 0.9443\angle -8.7797^\circ
\end{aligned}
$$

Here $\sum_{k \ne 2}Y_{2k}V_k = j8.3333(V_1 + V_3^{(1)}) = 0.9750 + j16.3167$ and $\sum_{k \ne 3}Y_{3k}V_k = j8.3333(V_1 + V_2^{(2)}) = 1.2307 + j16.2182$.

| Iteration | $V_2$ (pu) | $V_3$ (pu) |
|---|---|---|
| 0 | $1.0\angle 0^\circ$ | $1.0\angle 0^\circ$ |
| 1 | $0.9801\angle -5.27^\circ$ | $0.9651\angle -6.96^\circ$ |
| 2 | $0.9576\angle -8.87^\circ$ | $0.9443\angle -8.78^\circ$ |

#### (ii) Slack bus power (with iteration-2 voltages)
$$
\begin{aligned}
I_1 &= Y_{11}V_1 + Y_{12}V_2 + Y_{13}V_3 = 2.4318 - j1.0052 \\
S_1 &= V_1I_1^* = 1.0(2.4318 + j1.0052)
\end{aligned}
$$

So $P_1 = 2.4318$ pu and $Q_1 = 1.0052$ pu.

#### (iii) Network losses
Line flows $S_{ik} = V_i[y_{ik}(V_i - V_k)]^*$:

| Line | $S_{ik}$ (pu) | $S_{ki}$ (pu) | Loss $S_{ik} + S_{ki}$ |
|---|---|---|---|
| 1-2 | $1.2307 + j0.4485$ | $-1.2307 - j0.2426$ | $j0.2059$ |
| 1-3 | $1.2011 + j0.5567$ | $-1.2011 - j0.3464$ | $j0.2103$ |
| 2-3 | $-0.0121 + j0.1068$ | $0.0121 - j0.1053$ | $j0.0015$ |

**Answer:**
- $V_2 = 0.9576\angle -8.87^\circ$ pu, $V_3 = 0.9443\angle -8.78^\circ$ pu after 2 iterations.
- Slack bus: $P_1 = 2.432$ pu, $Q_1 = 1.005$ pu.
- Losses: real power loss $= 0$ (lines have no resistance); reactive power loss $= 0.418$ pu.

Note: after only two iterations the solution has not fully converged, so $S_1$ minus the loads ($2.432 - 2.7 = -0.27$) does not yet equal the losses; with more iterations $P_1$ tends to 2.7 pu.

## Check yourself

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

1. Each line has z = j0.12 pu in a 3-bus triangle. What is the diagonal element Y11?
   A. -j8.3333
   B. -j16.6667
   C. j16.6667
   D. -j4.1667
   Answer: B. Each line has y = -j8.3333 and bus 1 has two lines, so Y11 = -j16.6667, and the off-diagonals are +j8.3333. (Key point: Ybus formation)

2. In the G-S equation, what is used for the power at bus 2 (load 1.5 + j0.4)?
   A. P2 - jQ2 = 1.5 - j0.4
   B. P2 - jQ2 = -1.5 + j0.4
   C. P2 - jQ2 = -1.5 - j0.4
   D. P2 - jQ2 = 1.5 + j0.4
   Answer: B. A load is a negative injection, S2 = -1.5 - j0.4, so P2 - jQ2 = -1.5 + j0.4. (Key point: Net injection of load buses)

3. When computing V3 in iteration 1, which value of V2 is used?
   A. The flat-start value 1.0∠0°
   B. The newly computed V2 of the same iteration
   C. The value from the previous run
   D. The slack bus value V1
   Answer: B. G-S uses the latest available values immediately, so the new V2 = 0.9760 - j0.0900 is used. (Key point: Gauss-Seidel update rule)

4. Why is the network real power loss zero, even though P1 = 2.432 pu is not exactly 2.7 pu after two iterations?
   A. Because the loads draw no real power
   B. Because lines have no resistance; the gap is only due to non-convergence after two iterations
   C. Because the slack bus absorbs the loss
   D. Because the shunt admittance is negligible
   Answer: B. Lines are purely reactive so P loss is 0; P1 tends toward 2.7 pu as the iterations converge. (Key point: Real power loss and slack power)

## Related in Power System Analysis II

- [G-S first iteration with line charging](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Ags-first-iteration-line-charging) — Power System Analysis II, chapter 2
- [N-R first iteration, line reactance 0.1](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Anr-first-iteration-x0-1) — Power System Analysis II, chapter 2
- [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
- [Ybus for 3-bus lines with shunt susceptance](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Aybus-three-bus-shunt-2071) — Power System Analysis II, chapter 1
- [G-S load flow with given Ybus, losses](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Ags-ybus-j12-two-iterations) — Power System Analysis II, chapter 2
- [G-S voltages after 2nd iteration with shunt](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Ags-three-bus-shunt-2nd-iteration) — Power System Analysis II, chapter 2

## Related in other subjects

- [Double-stub shunt tuner for admittance load](https://www.bctengineer.com/question.md?key=semester_7%2Frf_and_microwave_engineering%3Adouble-stub-admittance-load) — RF and Microwave Engineering, chapter 2
- [Charging reactive power of line at no load](https://www.bctengineer.com/question.md?key=semester_8%2Fhigh_voltage_engineering%3Acharging-reactive-power-no-load) — High Voltage Engineering, chapter 3, asked 2×
- [Series R, X and shunt admittance from ABCD](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Aline-constants-from-abcd) — Power System Analysis I, chapter 5
- [12 km 33 kV line: losses, zero-regulation capacitor](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3A12km-33kv-losses-zero-regulation) — Power System Analysis I, chapter 6