---
title: "Ybus modified by mid-line tap changer"
subject: "Power System Analysis II"
chapter: 1
key: "semester_5/power_system_analysis_ii:ybus-tap-changer-midpoint"
source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/ybus-tap-changer-midpoint
site: BCT Engineering
---

# Ybus modified by mid-line tap changer

Power System Analysis II · Chapter 1: Interconnected Power System

- Asked in: 2083 Baishakh (new course), 3+4 marks
- Video: [PS105 Effect of Tap Changing Transformer](https://www.youtube.com/watch?v=6s76GXKO3Rc) — Lectures in Electrical Engineering
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/ybus-tap-changer-midpoint
- Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-1

## Question

For the power system network shown below, obtain the bus admittance matrix. The series impedance and shunt half-charging admittance of each line are 0.05 + j0.15 p.u. and j0.03 p.u. respectively. If a tap-changing transformer having turn ratio (t) = 1:1.05 is inserted mid-point between buses 1 and 2, find the modified bus admittance matrix. [Figure: three buses, each with a generator, lines 1-2, 1-3 and 2-3]

## Short answer

Original Ybus: diagonals 4 - j11.94 pu, off-diagonals -2 + j6 pu. With tap t=1.05 midpoint in line 1-2, Y11=4.0975 - j12.2325, Y22=3.9025 - j11.6475, Y12=Y21= -1.9976 + j5.9929; other elements unchanged.

## Answer

#### (a) Original Ybus
Series admittance of each line:

$$
y = \frac{1}{0.05 + j0.15} = 2 - j6\ \text{pu}
$$

Half-line charging $j0.03$ at each end; each bus has two lines:

$$
\begin{aligned}
Y_{ii} &= 2(2 - j6) + 2(j0.03) = 4 - j11.94 \\
Y_{ik} &= -y = -2 + j6
\end{aligned}
$$

$$
Y_{bus} =
\begin{bmatrix}
4 - j11.94 & -2 + j6 & -2 + j6 \\
-2 + j6 & 4 - j11.94 & -2 + j6 \\
-2 + j6 & -2 + j6 & 4 - j11.94
\end{bmatrix}
$$

#### (b) With the tap-changing transformer at the mid-point of line 1-2
Assumptions: the ideal transformer (ratio $1 : t$, $t = 1.05$, "1" on the bus-1 side) has negligible own impedance; the line's series impedance is split into two halves $z/2$; the line-charging terms stay at the line ends.

```text
  (1)--[ z/2 ]--)|( 1:t --[ z/2 ]--(2)
```

Referring the bus-1-side half to the t-side multiplies its impedance by $t^2$, so the transformer branch is an ideal $1:t$ in series with

$$
z_{eq} = t^2\frac{z}{2} + \frac{z}{2} = \frac{z(1 + t^2)}{2},\qquad
y_{eq} = \frac{2y}{1 + t^2} = \frac{4 - j12}{2.1025} = 1.9025 - j5.7075
$$

Using the off-nominal model ($t^2y_{eq}$ at bus 1, $y_{eq}$ at bus 2, $-t\,y_{eq}$ mutual):

$$
\begin{aligned}
t^2y_{eq} &= 1.1025(1.9025 - j5.7075) = 2.0975 - j6.2925 \\
-t\,y_{eq} &= -1.05(1.9025 - j5.7075) = -1.9976 + j5.9929
\end{aligned}
$$

Modified elements (only those involving line 1-2 change):

$$
\begin{aligned}
Y_{11} &= t^2y_{eq} + y_{13} + 2(j0.03) = 4.0975 - j12.2325 \\
Y_{22} &= y_{eq} + y_{23} + 2(j0.03) = 3.9025 - j11.6475 \\
Y_{12} &= Y_{21} = -t\,y_{eq} = -1.9976 + j5.9929 \\
Y_{33}, Y_{13}, Y_{23} &\ \text{unchanged}
\end{aligned}
$$

**Answer:**

$$
Y_{bus,new} =
\begin{bmatrix}
4.0975 - j12.2325 & -1.9976 + j5.9929 & -2 + j6 \\
-1.9976 + j5.9929 & 3.9025 - j11.6475 & -2 + j6 \\
-2 + j6 & -2 + j6 & 4 - j11.94
\end{bmatrix}\ \text{pu}
$$

The matrix stays symmetric because $t$ is real.

## Check yourself

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

1. With series y = 2 - j6 and half-charging j0.03 per end, two lines per bus give Y_ii =
   A. 4 - j11.94
   B. 4 - j12.06
   C. 4 - j11.97
   D. 4 - j12
   Answer: A. 2(2 - j6) + 2(j0.03) = 4 - j11.94. (Key point: Original diagonal)

2. Splitting z into two halves and referring one half through 1:t gives y_eq =
   A. y
   B. 2y/(1+t^2)
   C. 2y/(1+t)
   D. y/t^2
   Answer: B. z_eq = z(1+t^2)/2, so y_eq = 2y/(1+t^2). (Key point: Equivalent admittance)

3. In the modified matrix, the line 1-2 contribution to Y11 is:
   A. y_eq
   B. t^2 y_eq
   C. t y_eq
   D. y
   Answer: B. Bus-1 side gets t^2 y_eq. (Key point: Y11 change)

4. Why does the matrix remain symmetric?
   A. t is real
   B. Shunts are zero
   C. Lines are equal
   D. Bus 3 unchanged
   Answer: A. A real tap gives Y12 = Y21 = -t y_eq. (Key point: Symmetry)

## Related in Power System Analysis II

- [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 for 3-bus lines with shunt susceptance](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Aybus-three-bus-shunt-2071) — Power System Analysis II, chapter 1
- [Ybus modified by phase-shifting transformer](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Aybus-phase-shifter-midpoint) — Power System Analysis II, chapter 1
- [Ybus of five-bus system](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Aybus-five-bus) — 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 for network with three sources](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Aybus-three-sources-2080) — Power System Analysis II, chapter 1

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