---
title: "Load flow equations from Ybus"
subject: "Power System Analysis II"
chapter: 2
key: "semester_5/power_system_analysis_ii:load-flow-equations-from-ybus"
source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/load-flow-equations-from-ybus
site: BCT Engineering
---

# Load flow equations from Ybus

Power System Analysis II · Chapter 2: Load Flow Analysis

- Asked in: 2073 Shrawan, 4 marks
- Video: [Load flow equation (prove) \ Y Bus \ type of buses](https://www.youtube.com/watch?v=017AmuXRmUQ) — ENG mahmoud hamdy
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/load-flow-equations-from-ybus
- Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-2

## Question

Starting from Y-Bus, Bus voltage and injected current relationship deduce the basic load flow equations.

## Short answer

From I=YV, Ii = sum Yik Vk. Complex power: Si = Pi+jQi = Vi Ii*. In polar form, Pi = sum|Vi||Vk||Yik|cos(θik-δi+δk), Qi = -sum|Vi||Vk||Yik|sin(θik-δi+δk). These are the non-linear load flow equations solved iteratively.

## Answer

Load flow equations relate the power injected at each bus to the bus voltages through the bus admittance matrix. They follow from the nodal equation $I_{bus} = Y_{bus}V_{bus}$.

#### Step 1: Current injection at bus i
For an $n$-bus network,

$$
\begin{bmatrix} I_1 \\ I_2 \\ \vdots \\ I_n \end{bmatrix} = \begin{bmatrix} Y_{11} & Y_{12} & \cdots & Y_{1n} \\ Y_{21} & Y_{22} & \cdots & Y_{2n} \\ \vdots & & & \vdots \\ Y_{n1} & Y_{n2} & \cdots & Y_{nn} \end{bmatrix} \begin{bmatrix} V_1 \\ V_2 \\ \vdots \\ V_n \end{bmatrix}
$$

so the current injected at bus $i$ is

$$
I_i = \sum_{k=1}^{n} Y_{ik}V_k
$$

#### Step 2: Complex power injection
The net complex power injected at bus $i$ is $S_i = P_i + jQ_i = V_iI_i^{*}$. Taking the conjugate,

$$
\begin{aligned}
P_i - jQ_i &= V_i^{*}I_i \\
&= V_i^{*}\sum_{k=1}^{n} Y_{ik}V_k
\end{aligned}
$$

Here $P_i = P_{Gi} - P_{Li}$ and $Q_i = Q_{Gi} - Q_{Li}$.

#### Step 3: Polar form (used in N-R)
Put $V_i = |V_i|\angle\delta_i$ and $Y_{ik} = |Y_{ik}|\angle\theta_{ik}$:

$$
P_i - jQ_i = \sum_{k=1}^{n}|V_i||V_k||Y_{ik}|\angle(\theta_{ik} - \delta_i + \delta_k)
$$

Equating real and imaginary parts:

$$
\begin{aligned}
P_i &= \sum_{k=1}^{n}|V_i||V_k||Y_{ik}|\cos(\theta_{ik} - \delta_i + \delta_k) \\
Q_i &= -\sum_{k=1}^{n}|V_i||V_k||Y_{ik}|\sin(\theta_{ik} - \delta_i + \delta_k)
\end{aligned}
$$

These are the **static load flow equations**: $2n$ real, non-linear equations in $|V|$ and $\delta$, solved by iteration.

#### Step 4: Form used in Gauss-Seidel
Separating the $k = i$ term in Step 2:

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

At each bus two of the four quantities ($P$, $Q$, $|V|$, $\delta$) are given and the other two are found by solving these equations.

## Check yourself

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

1. What is the current injected at bus i in terms of Ybus?
   A. Ii = sum Yik / Vk
   B. Ii = sum over k of Yik Vk
   C. Ii = Vi / Yii
   D. Ii = sum Vk - Vi
   Answer: B. From I_bus = Y_bus V_bus, Ii = sum_k Yik Vk. (Key point: Current injection)

2. Which expression follows from S_i = Vi Ii*?
   A. Pi + jQi = Vi* sum Yik Vk
   B. Pi - jQi = Vi* sum Yik Vk
   C. Pi - jQi = Vi sum Yik Vk
   D. Pi - jQi = sum Yik Vk / Vi*
   Answer: B. Taking the conjugate of S_i = Vi Ii* gives Pi - jQi = Vi* Ii = Vi* sum Yik Vk. (Key point: Complex power conjugate)

3. In polar form, which gives the reactive power Qi?
   A. +sum |Vi||Vk||Yik| cos(theta_ik - delta_i + delta_k)
   B. -sum |Vi||Vk||Yik| sin(theta_ik - delta_i + delta_k)
   C. +sum |Vi||Vk||Yik| sin(theta_ik - delta_i + delta_k)
   D. -sum |Vi||Vk||Yik| cos(theta_ik - delta_i + delta_k)
   Answer: B. Pi - jQi equals the complex sum, so Qi is the negative of the imaginary part: -sum ... sin(...). (Key point: Polar form and real/imag parts)

## Related in Power System Analysis II

- [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
- [G-S voltage and power flow, 2-bus](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Ags-two-bus-power-flow-new) — Power System Analysis II, chapter 2
- [Node equations and Ybus, three-source network](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Aybus-node-equations-three-sources) — Power System Analysis II, chapter 1, asked 2×
- [Need for load flow; NR and decoupled equations](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Aload-flow-need-nr-decoupled) — Power System Analysis II, chapter 2
- [N-R load flow corrections and algorithm](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Anr-load-flow-derivation) — Power System Analysis II, chapter 2, asked 2×

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