---
title: "Ybus for 3-bus system (2078)"
subject: "Power System Analysis II"
chapter: 1
key: "semester_5/power_system_analysis_ii:ybus-three-bus-2078"
source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/ybus-three-bus-2078
site: BCT Engineering
---

# Ybus for 3-bus system (2078)

Power System Analysis II · Chapter 1: Interconnected Power System

- Asked in: 2078 Bhadra, 8 marks
- Video: [7 Ybus Matrix Formation](https://www.youtube.com/watch?v=AKvmqbv6E58) — AISSMS INSTITUTE OF INFORMATION TECHNOLOGYY - IOIT
- 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-2078
- Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-1

## Question

For a three phase system as shown in figure below, compute the YBus Matrix. [Figure: bus 1 V1 = 1.05∠0; Z12 = 0.02 + j0.04, Z13 = 0.01 + j0.03, Z23 = 0.0125 + j0.025; bus 2: P2 = −4.0 pu, Q2 = −2.5 pu; bus 3: P3 = 2.0 pu, V3 = 1.04 pu, (1.0 ≤ Q3 ≤ 1.5) pu]

## Short answer

Invert line impedances to get admittances. Diagonal elements sum of line admittances at each bus; off-diagonals are negative of the corresponding line admittance. Ybus = [[20–j50, –10+j20, –10+j30]; [–10+j20, 26–j52, –16+j32]; [–10+j30, –16+j32, 26–j62]] pu. Bus 1 slack, Bus2 PQ, Bus3 PV.

## Answer

$Y_{bus}$ depends only on the network impedances; the bus data ($V_1$, $P_2$, $Q_2$, $P_3$, $V_3$, Q limits) are needed later for load flow, not for $Y_{bus}$. No line charging is given, so it is neglected.

#### Step 1: Line admittances
$$
\begin{aligned}
y_{12} &= \frac{1}{0.02 + j0.04} = \frac{0.02 - j0.04}{0.02^2 + 0.04^2} = \frac{0.02 - j0.04}{0.002} = 10 - j20\ \text{pu} \\
y_{13} &= \frac{1}{0.01 + j0.03} = \frac{0.01 - j0.03}{0.001} = 10 - j30\ \text{pu} \\
y_{23} &= \frac{1}{0.0125 + j0.025} = \frac{0.0125 - j0.025}{0.00078125} = 16 - j32\ \text{pu}
\end{aligned}
$$

#### Step 2: Diagonal elements
$$
\begin{aligned}
Y_{11} &= y_{12} + y_{13} = 20 - j50 \\
Y_{22} &= y_{12} + y_{23} = 26 - j52 \\
Y_{33} &= y_{13} + y_{23} = 26 - j62
\end{aligned}
$$

#### Step 3: Off-diagonal elements
$$
Y_{12} = Y_{21} = -10 + j20,\quad Y_{13} = Y_{31} = -10 + j30,\quad Y_{23} = Y_{32} = -16 + j32
$$

**Answer:**

$$
Y_{bus} =
\begin{bmatrix}
20 - j50 & -10 + j20 & -10 + j30 \\
-10 + j20 & 26 - j52 & -16 + j32 \\
-10 + j30 & -16 + j32 & 26 - j62
\end{bmatrix}\ \text{pu}
$$

For the load flow that follows: bus 1 is the slack bus ($V_1 = 1.05\angle 0^\circ$), bus 2 is a PQ (load) bus, and bus 3 is a PV bus ($P_3 = 2.0$, $|V_3| = 1.04$, with Q limits).

## Check yourself

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

1. Which given data are needed to compute Ybus?
   A. Line impedances only
   B. V1, P2, Q2 and P3
   C. Q limits of bus 3
   D. All the data in the figure
   Answer: A. Ybus depends only on network impedances; bus data is for the later load flow. (Key point: Data not needed)

2. y13 = 1/(0.01 + j0.03) equals:
   A. 10 - j30
   B. 100 - j300
   C. 10 + j30
   D. 0.01 - j0.03
   Answer: A. (0.01 - j0.03)/0.001 = 10 - j30 pu. (Key point: Line admittance)

3. Y22 = y12 + y23 equals:
   A. 26 - j52
   B. 20 - j50
   C. 26 - j62
   D. 6 - j12
   Answer: A. (10 - j20) + (16 - j32) = 26 - j52. (Key point: Y22)

4. In this network, bus 3 (P3 and |V3| specified) is a:
   A. Slack bus
   B. PQ bus
   C. PV bus
   D. Reference bus
   Answer: C. P and |V| specified means a PV bus. (Key point: Bus types)

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