---
title: "Ybus for 3-bus lines with shunt susceptance"
subject: "Power System Analysis II"
chapter: 1
key: "semester_5/power_system_analysis_ii:ybus-three-bus-shunt-2071"
source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/ybus-three-bus-shunt-2071
site: BCT Engineering
---

# Ybus for 3-bus lines with shunt susceptance

Power System Analysis II · Chapter 1: Interconnected Power System

- Asked in: 2071 Shrawan, 5 marks
- Video: [BUS ADMITTANCE MATRIX - PART - 04 - FORMATION OF Y-BUS - TWO PROBLEMS](https://www.youtube.com/watch?v=eWJvRTL-aAw) — zenmurali Sachithanandam
- 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-bus-shunt-2071
- Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-1

## Question

The single line diagram of a power system network is shown in figure below. If each line has series impedance of (0.05+j0.15) pu and shunt susceptance of j0.3 pu., find bus Admittance matrix for the system. [Figure: three buses with lines 1-2, 1-3 and 2-3]

## Short answer

Series admittance y = 2 - j6 pu; half-line charging j0.15 at each end. Diagonals: 4 - j11.7 pu; off-diagonals: -2 + j6 pu. Ybus is symmetric with these values.

## Answer

Assumption: $j0.3$ pu is the total shunt susceptance of each line, so $j0.15$ pu is placed at each end (nominal π model).

#### Step 1: Series admittance of each line
$$
y = \frac{1}{0.05 + j0.15} = \frac{0.05 - j0.15}{0.05^2 + 0.15^2} = \frac{0.05 - j0.15}{0.025} = 2 - j6\ \text{pu}
$$

#### Step 2: Diagonal elements
Each bus is connected to two lines, and receives $j0.15$ from each:

$$
Y_{11} = Y_{22} = Y_{33} = 2(2 - j6) + 2(j0.15) = 4 - j11.7\ \text{pu}
$$

#### Step 3: Off-diagonal elements
$$
Y_{12} = Y_{13} = Y_{23} = -y = -2 + j6\ \text{pu}
$$

**Answer:**

$$
Y_{bus} =
\begin{bmatrix}
4 - j11.7 & -2 + j6 & -2 + j6 \\
-2 + j6 & 4 - j11.7 & -2 + j6 \\
-2 + j6 & -2 + j6 & 4 - j11.7
\end{bmatrix}\ \text{pu}
$$

(If $j0.3$ were the half-line charging at each end, each diagonal element would be $4 - j11.4$.)

## Check yourself

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

1. 1/(0.05 + j0.15) equals:
   A. 2 - j6
   B. 0.05 - j0.15
   C. 2 + j6
   D. 20 - j60
   Answer: A. (0.05 - j0.15)/0.025 = 2 - j6. (Key point: Line admittance)

2. With total line susceptance j0.3 pu, the charging at each line end is:
   A. j0.3
   B. j0.15
   C. j0.6
   D. j0.075
   Answer: B. Half of the total is placed at each end. (Key point: Charging split)

3. Each bus has two lines, so Y11 is:
   A. 4 - j11.7
   B. 4 - j12.3
   C. 2 - j5.85
   D. 4 - j6
   Answer: A. 2(2 - j6) + 2(j0.15) = 4 - j11.7. (Key point: Y11)

4. Shunt charging affects the off-diagonals?
   A. Yes, increases them
   B. Yes, decreases them
   C. No, they remain -2 + j6
   D. Only for line 1-2
   Answer: C. Shunts connect to ground and affect only diagonals. (Key point: Off-diagonal)

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