---
title: "Node equations and Ybus, six-bus network"
subject: "Power System Analysis II"
chapter: 1
key: "semester_5/power_system_analysis_ii:ybus-six-bus-node-equations"
source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/ybus-six-bus-node-equations
site: BCT Engineering
---

# Node equations and Ybus, six-bus network

Power System Analysis II · Chapter 1: Interconnected Power System

- Asked in: 2073 Shrawan, 6 marks
- Video: [BUS ADMITTANCE MATRIX - PART - 01 - FORMATION OF Y-BUS - ADVANTAGES](https://www.youtube.com/watch?v=rWjoK28YMkk) — 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-six-bus-node-equations
- Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-1

## Question

Obtain node equations and compute bus admittance matrix (YBus) of the network shown in figure below. All voltages and impedances are marked in pu. [Figure: G1 at bus 1 connected through a transformer j0.05 to bus 2; Line-1 from bus 2 to bus 3; Line-3 from bus 2 to bus 5; Line-2 from bus 3 to bus 5; bus 3 connected through a transformer j0.05 to bus 4, which has a load; bus 5 connected through a transformer j0.05 to bus 6, where G2 is connected]

Line data are as follows:

| Line | Series reactance, pu | Shunt susceptance, pu |
|---|---|---|
| Line-1 | 0.40 | 0.02 |
| Line-2 | 0.20 | 0.00 |
| Line-3 | 0.40 | 0.02 |

## Short answer

Node equations by KCL using branch admittances yield Ybus with diagonals: -j20, -j24.98, -j27.49, -j20, -j27.49, -j20 pu for buses 1–6; off-diagonals: j20 (1-2), j2.5 (2-3,2-5), j20 (3-4), j5 (3-5), j20 (5-6) and symmetric counterparts.

## Answer

Assumptions: the given shunt susceptance of each line is the total line charging, split as $B/2$ at each end (nominal π). Generators and load are injections at buses 1, 6 and 4.

#### Step 1: Branch admittances
| Branch | Buses | $X$ (pu) | $y = 1/jX$ | $B/2$ at each end |
|---|---|---|---|---|
| Transformer | 1-2 | 0.05 | $-j20$ | 0 |
| Line-1 | 2-3 | 0.40 | $-j2.5$ | $j0.01$ |
| Line-3 | 2-5 | 0.40 | $-j2.5$ | $j0.01$ |
| Line-2 | 3-5 | 0.20 | $-j5$ | 0 |
| Transformer | 3-4 | 0.05 | $-j20$ | 0 |
| Transformer | 5-6 | 0.05 | $-j20$ | 0 |

#### Step 2: Node equations (KCL, $I_i$ = injected current)
$$
\begin{aligned}
I_1 &= y_{12}(V_1 - V_2) \\
I_2 &= y_{12}(V_2 - V_1) + y_{23}(V_2 - V_3) + y_{25}(V_2 - V_5) + j0.01V_2 + j0.01V_2 \\
I_3 &= y_{23}(V_3 - V_2) + y_{35}(V_3 - V_5) + y_{34}(V_3 - V_4) + j0.01V_3 \\
I_4 &= y_{34}(V_4 - V_3) \\
I_5 &= y_{25}(V_5 - V_2) + y_{35}(V_5 - V_3) + y_{56}(V_5 - V_6) + j0.01V_5 \\
I_6 &= y_{56}(V_6 - V_5)
\end{aligned}
$$

#### Step 3: Elements
$$
\begin{aligned}
Y_{11} &= -j20 \\
Y_{22} &= -j20 - j2.5 - j2.5 + j0.02 = -j24.98 \\
Y_{33} &= -j2.5 - j5 - j20 + j0.01 = -j27.49 \\
Y_{44} &= -j20 \\
Y_{55} &= -j2.5 - j5 - j20 + j0.01 = -j27.49 \\
Y_{66} &= -j20
\end{aligned}
$$

Off-diagonal: $Y_{12} = j20$, $Y_{23} = j2.5$, $Y_{25} = j2.5$, $Y_{34} = j20$, $Y_{35} = j5$, $Y_{56} = j20$; all others zero.

**Answer:**

$$
Y_{bus} =
\begin{bmatrix}
-j20 & j20 & 0 & 0 & 0 & 0 \\
j20 & -j24.98 & j2.5 & 0 & j2.5 & 0 \\
0 & j2.5 & -j27.49 & j20 & j5 & 0 \\
0 & 0 & j20 & -j20 & 0 & 0 \\
0 & j2.5 & j5 & 0 & -j27.49 & j20 \\
0 & 0 & 0 & 0 & j20 & -j20
\end{bmatrix}\ \text{pu}
$$

## Check yourself

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

1. A line with total shunt susceptance 0.02 pu is split in the nominal pi as:
   A. j0.02 at one end
   B. j0.01 at each end
   C. j0.04 at each end
   D. j0.02 at each end
   Answer: B. Total charging is divided equally at the two ends. (Key point: Line charging)

2. Bus 2 connects to bus 1 via transformer, and lines 2-3 and 2-5. Y22 is:
   A. -j24.98
   B. -j25
   C. -j27.49
   D. -j20
   Answer: A. -j20 - j2.5 - j2.5 + j0.02 = -j24.98. (Key point: Y22)

3. Why is Y11 = -j20?
   A. Only the transformer 1-2 is connected to bus 1, with no charging
   B. Bus 1 has line charging
   C. Bus 1 is the load bus
   D. Bus 1 connects to three lines
   Answer: A. Bus 1 has only the j0.05 transformer: 1/j0.05 = -j20. (Key point: Y11)

4. Y35 equals:
   A. -j5
   B. j5
   C. j20
   D. j2.5
   Answer: B. Line 3-5 has y = -j5, so Y35 = +j5. (Key point: Off-diagonal)

## Related in Power System Analysis II

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