---
title: "G-S with PV bus Q limits and losses"
subject: "Power System Analysis II"
chapter: 2
key: "semester_5/power_system_analysis_ii:gs-pv-bus-q-limits-2071"
source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/gs-pv-bus-q-limits-2071
site: BCT Engineering
---

# G-S with PV bus Q limits and losses

Power System Analysis II · Chapter 2: Load Flow Analysis

- Asked in: 2071 Shrawan, 8 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-pv-bus-q-limits-2071
- 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 Ybus of the system is Ybus = [4−j2 −2+j6 −2+j6; −2+j6 4−j12 −2+j6; −2+j6 −2+j6 4−j12] (element (1,1) printed as 4−j2). [Figure: line 1-2 is 100 km, line 2-3 is 200 km, line 1-3 is 400 km] If the power and bus voltages with different buses 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 | 1.2 | ? | 1.0 | PV |
| 3 | 0.6 | 0.3 | 0 | 0 | ? | PQ |

Assume limit on Qgen of 0.1 ≤ Qgen ≤ 0.4. Carry out the load flow analysis (only two iterations starting from assumptions of unknown voltage magnitudes as 1 p.u. and phase angles zero degree) using G-S to compute the unknown parameters in above table. Also compute the network real and reactive power losses.

## Short answer

Correct Y11=4−j12 pu. G-S: after iteration 1 QG2=0.2 pu (within limits); after iteration 2 QG2=0.0829 pu violates lower limit, so QG2 fixed at 0.1 pu and bus 2 treated as PQ. Voltages after two iterations: V2=0.9997∠2.2359°, V3=0.9860∠−1.0398° pu.

## Answer

The (1,1) element printed as $4 - j2$ is a misprint: each row of this $Y_{bus}$ must sum to zero (no shunt elements), so $Y_{11} = 4 - j12$ pu is used. Line lengths are not needed because $Y_{bus}$ is given.

#### Bus data
| Bus | Type | Specified (net) | Unknown |
|---|---|---|---|
| 1 | Slack | $V_1 = 1.05\angle 0^\circ$ | $P_{G1}$, $Q_{G1}$ |
| 2 | PV | $P_2 = 1.2 - 0.7 = 0.5$, $\lvert V_2\rvert = 1.0$ | $Q_{G2}$, $\delta_2$ |
| 3 | PQ | $P_3 = -0.6$, $Q_3 = -0.3$ | $\lvert V_3\rvert$, $\delta_3$ |

Limits: $0.1 \le Q_{G2} \le 0.4$ pu, and $Q_2 = Q_{G2} - Q_{L2} = Q_{G2} - 0.5$. Start: $V_2^{(0)} = 1\angle 0^\circ$, $V_3^{(0)} = 1\angle 0^\circ$.

$$
Y_{bus} = \begin{bmatrix} 4.0000 - j12.0000 & -2.0000 + j6.0000 & -2.0000 + j6.0000 \\ -2.0000 + j6.0000 & 4.0000 - j12.0000 & -2.0000 + j6.0000 \\ -2.0000 + j6.0000 & -2.0000 + j6.0000 & 4.0000 - j12.0000 \end{bmatrix}\ \text{pu}
$$

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

$$
\begin{aligned}
\sum_{k=1}^{3} Y_{2k}V_k &= -0.1000 + j0.3000 \\
Q_2^{(1)} &= -\text{Im}\{V_2^{*}\textstyle\sum_k Y_{2k}V_k\} = -0.3000\ \text{pu}
\end{aligned}
$$

$Q_{G2} = Q_2 + Q_{L2} = 0.2000$ pu lies within $0.10 \le Q_{G2} \le 0.40$, so bus 2 remains a PV bus.

$$
\begin{aligned}
V_2^{(1)} &= \frac{1}{Y_{22}}\left[\frac{P_2 - jQ_2}{V_2^{*}} - Y_{21}V_1 - Y_{23}V_3\right] \\
\frac{P_2 - jQ_2}{V_2^{*}} &= \frac{0.5000 + j0.3000}{1.0000} = 0.5000 + j0.3000 \\
\textstyle\sum_{k\ne 2} Y_{2k}V_k &= -4.1000 + j12.3000 \\
V_2^{(1)} &= \frac{(0.5000 + j0.3000) - (-4.1000 + j12.3000)}{4.0000 - j12.0000} = 1.0150 + j0.0450 \\
\delta_2 &= 2.5385^\circ \Rightarrow V_2^{(1)} = 1.0000\angle 2.5385^\circ = 0.9990 + j0.0443\ \text{(magnitude reset to }1.00\text{)}
\end{aligned}
$$

$$
\begin{aligned}
V_3^{(1)} &= \frac{1}{Y_{33}}\left[\frac{P_3 - jQ_3}{V_3^{*}} - Y_{31}V_1 - Y_{32}V_2\right] \\
\frac{P_3 - jQ_3}{V_3^{*}} &= \frac{-0.6000 + j0.3000}{1.0000} = -0.6000 + j0.3000 \\
\textstyle\sum_{k\ne 3} Y_{3k}V_k &= -4.3638 + j12.2055 \\
V_3^{(1)} &= \frac{(-0.6000 + j0.3000) - (-4.3638 + j12.2055)}{4.0000 - j12.0000} = 0.9870 - j0.0154 \\
&= 0.9871\angle -0.8912^\circ\ \text{pu}
\end{aligned}
$$

**Iteration 2**

$$
\begin{aligned}
\sum_{k=1}^{3} Y_{2k}V_k &= 0.5457 + j0.4417 \\
Q_2^{(2)} &= -\text{Im}\{V_2^{*}\textstyle\sum_k Y_{2k}V_k\} = -0.4171\ \text{pu}
\end{aligned}
$$

$Q_{G2} = Q_2 + Q_{L2} = 0.0829$ pu violates the limit $0.10 \le Q_{G2} \le 0.40$. So $Q_{G2}$ is fixed at 0.10 pu, i.e. $Q_2 = 0.10 - 0.50 = -0.4000$ pu, and bus 2 is treated as a **PQ bus** in this iteration (its voltage magnitude is not reset).

$$
\begin{aligned}
V_2^{(2)} &= \frac{1}{Y_{22}}\left[\frac{P_2 - jQ_2}{V_2^{*}} - Y_{21}V_1 - Y_{23}V_3\right] \\
\frac{P_2 - jQ_2}{V_2^{*}} &= \frac{0.5000 + j0.4000}{0.9990 - j0.0443} = 0.4818 + j0.4218 \\
\textstyle\sum_{k\ne 2} Y_{2k}V_k &= -3.9819 + j12.2528 \\
V_2^{(2)} &= \frac{(0.4818 + j0.4218) - (-3.9819 + j12.2528)}{4.0000 - j12.0000} = 0.9989 + j0.0390 \\
&= 0.9997\angle 2.2359^\circ\ \text{pu}
\end{aligned}
$$

$$
\begin{aligned}
V_3^{(2)} &= \frac{1}{Y_{33}}\left[\frac{P_3 - jQ_3}{V_3^{*}} - Y_{31}V_1 - Y_{32}V_2\right] \\
\frac{P_3 - jQ_3}{V_3^{*}} &= \frac{-0.6000 + j0.3000}{0.9870 + j0.0154} = -0.6030 + j0.3133 \\
\textstyle\sum_{k\ne 3} Y_{3k}V_k &= -4.3318 + j12.2155 \\
V_3^{(2)} &= \frac{(-0.6030 + j0.3133) - (-4.3318 + j12.2155)}{4.0000 - j12.0000} = 0.9859 - j0.0179 \\
&= 0.9860\angle -1.0398^\circ\ \text{pu}
\end{aligned}
$$

#### Unknowns after the 2nd iteration
In the 2nd iteration bus 2 hit its lower Q-limit, so $Q_{G2} = 0.1$ pu and $|V_2|$ drifted slightly from 1.0 to 0.9997 pu.

**Slack bus:**

$$
\begin{aligned}
P_1 - jQ_1 &= V_1^{*}\textstyle\sum_k Y_{1k}V_k = (1.0500)(0.1037 - j0.7334) \\
&= 0.1089 - j0.7701 \\
\Rightarrow P_1 &= 0.1089\ \text{pu},\quad Q_1 = 0.7701\ \text{pu} \\
P_{G1} &= P_1 + P_{L1} = 0.1089 + 2.00 = 2.1089\ \text{pu} \\
Q_{G1} &= Q_1 + Q_{L1} = 0.7701 + 0.60 = 1.3701\ \text{pu}
\end{aligned}
$$

#### Network losses
| Bus | $P_i$ (pu) | $Q_i$ (pu) |
|---|---|---|
| 1 | 0.1089 | 0.7701 |
| 2 | 0.5151 | -0.4002 |
| 3 | -0.6001 | -0.2981 |
| Sum | 0.0239 | 0.0718 |

$$
\begin{aligned}
P_{loss} &= \textstyle\sum P_i = 0.0239\ \text{pu} \\
Q_{loss} &= \textstyle\sum Q_i = 0.0718\ \text{pu}
\end{aligned}
$$

**Answer:** $V_2 = 0.9997\angle 2.2359^\circ$ pu, $Q_{G2} = 0.1$ pu (at its lower limit), $V_3 = 0.9860\angle -1.0398^\circ$ pu, $P_{G1} = 2.1089$ pu, $Q_{G1} = 1.3701$ pu, $P_{loss} = 0.0239$ pu, $Q_{loss} = 0.0718$ pu.

## Check yourself

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

1. Why is Y11 taken as 4 - j12 instead of the printed 4 - j2?
   A. Because the line lengths change it
   B. Because each row of a Ybus with no shunt elements must sum to zero
   C. Because bus 1 is the slack bus
   D. Because the PV bus needs a larger diagonal
   Answer: B. With no shunt elements, the diagonal equals minus the sum of the off-diagonals in that row, so Y11 = 4 - j12. (Key point: Ybus misprint check)

2. In iteration 2, Q_G2 computed is 0.0829 pu with limits 0.1 to 0.4. What is done?
   A. Keep bus 2 as PV with |V2| = 1.0
   B. Fix Q_G2 = 0.1 pu and treat bus 2 as PQ, not resetting |V2|
   C. Fix Q_G2 = 0.4 pu and treat bus 2 as slack
   D. Set Q_G2 = 0.0829 and ignore the limit
   Answer: B. The limit is violated, so Q is held at the violated limit (0.1) and the bus acts as PQ, letting its voltage magnitude float. (Key point: Q-limit check at PV bus)

3. How is Q2 (net injection) related to Q_G2 for bus 2?
   A. Q2 = Q_G2 + Q_L2
   B. Q2 = Q_G2 - Q_L2 = Q_G2 - 0.5
   C. Q2 = Q_L2 - Q_G2 only if Q_G2 > 0.4
   D. Q2 = Q_G2
   Answer: B. Net injection is generation minus load, so Q2 = Q_G2 - 0.5, e.g. 0.1 - 0.5 = -0.4 pu. (Key point: Net Q at PV bus)

4. How are the slack bus power and the network losses obtained after the iterations?
   A. P1 - jQ1 = V1* sum(Y1k Vk); losses = sum of all bus net injections P and Q
   B. Losses = P_G1 only
   C. P1 is specified, so only Q1 is computed
   D. Losses = difference of the load voltages
   Answer: A. The slack injection comes from V1* times the Ybus-voltage sum, and the total of all net bus injections equals the network loss (0.0239 + j0.0718 pu). (Key point: Slack power and losses)

## 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
- [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
- [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
- [Ybus with nominal π model, line lengths](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Aybus-km-lines-pi-model) — Power System Analysis II, chapter 1, asked 2×
- [N-R Jacobian and first iteration (new course)](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Anr-jacobian-ybus-j12-new) — Power System Analysis II, chapter 2
- [SLG fault at bus 3, currents and voltages](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Aslg-fault-bus3-three-generators) — Power System Analysis II, chapter 5

## Related in other subjects

- [Initial line design for 200 MW over 120 km](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Aline-design-200mw-120km) — Transmission and Distribution System Design, chapter 2
- [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
- [Insulator discs per string, 180 MW over 200 km](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Ainsulator-discs-180mw-200km) — Transmission and Distribution System Design, chapter 3
- [Nominal-T 100 km line, 20 MW at 110 kV](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Anominal-t-100km-20mw) — Power System Analysis I, chapter 5