---
title: "Branch impedances from given Ybus"
subject: "Power System Analysis II"
chapter: 1
key: "semester_5/power_system_analysis_ii:branch-impedances-from-ybus"
source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/branch-impedances-from-ybus
site: BCT Engineering
---

# Branch impedances from given Ybus

Power System Analysis II · Chapter 1: Interconnected Power System

- Asked in: 2071 Chaitra, 5 marks
- Video: [Power Systems | Tutorial - 13 | Solved Problems on Admittance Bus Matrix (Y-bus)](https://www.youtube.com/watch?v=iYIo-so0i8k) — Engineering Devotion
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/branch-impedances-from-ybus
- Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-1

## Question

For a 3-bus network, the bus admittance matrix is given as follows: Ybus = [−j30 j12 j18; j12 −j25 j13; j18 j13 −j31] pu, determine the respective branch impedances.

## Short answer

Series branch admittances: y12 = -j12 ⇒ z12 = j0.0833 pu; y13 = -j18 ⇒ z13 = j0.0556 pu; y23 = -j13 ⇒ z23 = j0.0769 pu. No shunt branches.

## Answer

For a network with no mutual coupling, $Y_{ik} = -y_{ik}$ and $Y_{ii} = y_{i0} + \sum_{k \ne i} y_{ik}$. So the series branch admittances come from the off-diagonal terms and the shunt branches from the row sums.

#### Series branches
$$
\begin{aligned}
y_{12} &= -Y_{12} = -j12 &\Rightarrow\ z_{12} &= \frac{1}{-j12} = j0.0833\ \text{pu} \\
y_{13} &= -Y_{13} = -j18 &\Rightarrow\ z_{13} &= \frac{1}{-j18} = j0.0556\ \text{pu} \\
y_{23} &= -Y_{23} = -j13 &\Rightarrow\ z_{23} &= \frac{1}{-j13} = j0.0769\ \text{pu}
\end{aligned}
$$

#### Shunt branches
$$
\begin{aligned}
y_{10} &= Y_{11} + Y_{12} + Y_{13} = -j30 + j12 + j18 = 0 \\
y_{20} &= Y_{21} + Y_{22} + Y_{23} = j12 - j25 + j13 = 0 \\
y_{30} &= Y_{31} + Y_{32} + Y_{33} = j18 + j13 - j31 = 0
\end{aligned}
$$

So there are no shunt elements to ground.

```text
            j0.0833
     (1)-------------(2)
       \            /
        \          /
  j0.0556\        / j0.0769
          \      /
           \    /
            (3)
```

**Answer:** $z_{12} = j0.0833$ pu, $z_{13} = j0.0556$ pu, $z_{23} = j0.0769$ pu (purely inductive, i.e. $X_{12} = 1/12$, $X_{13} = 1/18$, $X_{23} = 1/13$ pu); no shunt branches.

## Check yourself

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

1. Which element gives the series admittance y12?
   A. Y11
   B. Y12 with sign change, y12 = -Y12
   C. Y11 + Y22
   D. Y12 as is
   Answer: B. Off-diagonal Y_ik = -y_ik. (Key point: Series branch from Ybus)

2. With Y12 = j12, z12 is:
   A. j0.0833
   B. -j0.0833
   C. j12
   D. -j12
   Answer: A. y12 = -j12, so z12 = 1/(-j12) = j0.0833. (Key point: Impedance z12)

3. The shunt admittance y_i0 is found from:
   A. The diagonal alone
   B. Sum of the row (Y_ii plus off-diagonals)
   C. The transpose
   D. The determinant
   Answer: B. Y_ii = y_i0 + sum y_ik so y_i0 = row sum. (Key point: Shunt branch)

4. Here the row sums are all zero, which means:
   A. No shunt branches to ground
   B. Large shunts
   C. The matrix is unsymmetric
   D. Bus 1 is slack
   Answer: A. Zero row sums mean y_i0 = 0. (Key point: Result)

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