---
title: "Ybus for network with three sources"
subject: "Power System Analysis II"
chapter: 1
key: "semester_5/power_system_analysis_ii:ybus-three-sources-2080"
source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/ybus-three-sources-2080
site: BCT Engineering
---

# Ybus for network with three sources

Power System Analysis II · Chapter 1: Interconnected Power System

- Asked in: 2080 Bhadra, 4 marks
- Video: [Bus Admittance Matrix](https://www.youtube.com/watch?v=1gK_ot7Plqs) — Dr.Nagaraja Kumari
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/ybus-three-sources-2080
- Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-1

## Question

For the network shown below obtain a bus admittance matrix. [Figure: three buses; line 1-2 = 0.02 + j0.05, line 1-3 = 0.02 + j0.06, line 2-3 = 0.01 + j0.03; source E1 with x = j0.02 connected to bus 1 through transformer T1 of j0.08; source E2 with x = j0.13 connected to bus 2 through transformer T2 of j0.12; source E3 with x = j0.1 connected to bus 3 through transformer T3 of j0.1]

## Short answer

Combine each generator and its transformer series reactance into a shunt admittance to ground. Compute line admittances from impedances. Diagonal entries: sum of all admittances connected to the bus; off-diagonals: negative of line admittance. Ybus = [[11.8966–j42.2414, –6.8966+j17.2414, –5+j15]; [–6.8966+j17.2414, 16.8966–j51.2414, –10+j30]; [–5+j15, –10+j30, 15–j50]] pu.

## Answer

Each source reactance is in series with its transformer, so each source branch is a shunt admittance from its bus to the reference (generator emfs become current injections and do not enter $Y_{bus}$).

#### Admittances
| Branch | Impedance (pu) | Admittance (pu) |
|---|---|---|
| Bus 1 to ground | $j0.02 + j0.08 = j0.10$ | $-j10$ |
| Bus 2 to ground | $j0.13 + j0.12 = j0.25$ | $-j4$ |
| Bus 3 to ground | $j0.10 + j0.10 = j0.20$ | $-j5$ |
| Line 1-2 | $0.02 + j0.05$ | $6.8966 - j17.2414$ |
| Line 1-3 | $0.02 + j0.06$ | $5.0000 - j15.0000$ |
| Line 2-3 | $0.01 + j0.03$ | $10.0000 - j30.0000$ |

#### Elements
$$
\begin{aligned}
Y_{11} &= -j10 + (6.8966 - j17.2414) + (5 - j15) = 11.8966 - j42.2414 \\
Y_{22} &= -j4 + (6.8966 - j17.2414) + (10 - j30) = 16.8966 - j51.2414 \\
Y_{33} &= -j5 + (5 - j15) + (10 - j30) = 15.0000 - j50.0000 \\
Y_{12} &= -y_{12} = -6.8966 + j17.2414 \\
Y_{13} &= -y_{13} = -5.0000 + j15.0000 \\
Y_{23} &= -y_{23} = -10.0000 + j30.0000
\end{aligned}
$$

**Answer:**

$$
Y_{bus} =
\begin{bmatrix}
11.8966 - j42.2414 & -6.8966 + j17.2414 & -5 + j15 \\
-6.8966 + j17.2414 & 16.8966 - j51.2414 & -10 + j30 \\
-5 + j15 & -10 + j30 & 15 - j50
\end{bmatrix}\ \text{pu}
$$

## Check yourself

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

1. How is a generator with series reactance and transformer treated in Ybus?
   A. Ignored completely
   B. As a shunt admittance from its bus to reference using the series sum of reactances
   C. As a series line between two buses
   D. As an off-diagonal term
   Answer: B. The emf is a current injection; the series reactance gives a shunt admittance to ground. (Key point: Source branches)

2. What is the shunt admittance at bus 1 (j0.02 + j0.08)?
   A. j10
   B. -j10
   C. -j12.5
   D. -j50
   Answer: B. 1/j0.10 = -j10 pu. (Key point: Bus 1 shunt)

3. Y12 equals:
   A. +(6.8966 - j17.2414)
   B. -6.8966 + j17.2414
   C. 11.8966 - j42.2414
   D. -5 + j15
   Answer: B. Mutual admittance is the negative of the 1-2 line admittance. (Key point: Y12)

4. Y11 is the sum of:
   A. Only the line admittances at bus 1
   B. Source branch -j10 plus lines 1-2 and 1-3 admittances
   C. Lines 2-3 and 1-3
   D. Source branches of all buses
   Answer: B. -j10 + (6.8966 - j17.2414) + (5 - j15) = 11.8966 - j42.2414. (Key point: Y11)

## Related in Power System Analysis II

- [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×
- [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
- [Ybus with generator and transformer impedances](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Aybus-generators-transformers-table) — Power System Analysis II, chapter 1
- [Ybus modified by mid-line tap changer](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Aybus-tap-changer-midpoint) — Power System Analysis II, chapter 1
- [Ybus for 3-bus system (2078)](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Aybus-three-bus-2078) — Power System Analysis II, chapter 1
- [Ybus and bus types (parallel lines)](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Aybus-bus-types-2074) — Power System Analysis II, chapter 1