---
title: "Ybus modified by phase-shifting transformer"
subject: "Power System Analysis II"
chapter: 1
key: "semester_5/power_system_analysis_ii:ybus-phase-shifter-midpoint"
source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/ybus-phase-shifter-midpoint
site: BCT Engineering
---

# Ybus modified by phase-shifting transformer

Power System Analysis II · Chapter 1: Interconnected Power System

- Asked in: 2082 Bhadra (new course), 3+4 marks
- Video: [BUS ADMITTANCE MATRIX - PART - 20 - ONE PROBLEM - PHASE SHIFTING TRANSFORMER](https://www.youtube.com/watch?v=Ax4YzlIDat0) — 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-phase-shifter-midpoint
- Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-1

## Question

Consider a 3-bus power system network. The series impedance and shunt admittance of each line are 0.026 + j0.11 pu and j0.04 pu respectively. Find: (i) the bus admittance matrix. (ii) the modified bus admittance matrix if a tap changing transformer inserted in mid-point between bus 2 and 3 having turn ratio t = e^(j30°). [Figure: Gen 1 at bus 1, Gen 2 at bus 2, load at bus 3; lines 1-2, 1-3 and 2-3]

## Short answer

Original Ybus: diagonals 4.0701 - j17.1798 pu, off-diagonals -2.0351 + j8.6099 pu. With phase shifter t=e^(j30°) at line 2-3 midpoint, Y23=2.5425+j8.4739, Y32=-6.0674+j6.4389; other elements unchanged, making Ybus unsymmetric.

## Answer

Assumption: $j0.04$ pu is the total shunt admittance of each line, so $j0.02$ pu is placed at each end.

#### (i) Bus admittance matrix
$$
y = \frac{1}{0.026 + j0.11} = 2.0351 - j8.6099\ \text{pu}\ (= 8.8471\angle -76.70^\circ)
$$

$$
\begin{aligned}
Y_{ii} &= 2y + 2(j0.02) = 4.0701 - j17.1798 \\
Y_{ik} &= -y = -2.0351 + j8.6099
\end{aligned}
$$

$$
Y_{bus} =
\begin{bmatrix}
4.0701 - j17.1798 & -2.0351 + j8.6099 & -2.0351 + j8.6099 \\
-2.0351 + j8.6099 & 4.0701 - j17.1798 & -2.0351 + j8.6099 \\
-2.0351 + j8.6099 & -2.0351 + j8.6099 & 4.0701 - j17.1798
\end{bmatrix}
$$

#### (ii) Phase-shifting transformer at the mid-point of line 2-3
The transformer has $t = e^{j30^\circ}$ ($|t| = 1$), with the "1" side towards bus 2. For an ideal shifter, $V_b = tV_a$ and, being lossless, $V_aI_a^* = V_bI_b^*$, so $I_a = t^*I_b$.

Splitting the line into two halves $z/2$ and referring the bus-2 half to the bus-3 side ($\times |t|^2 = 1$) gives total series admittance $y$ on the bus-3 side. The two-port equations are:

$$
\begin{bmatrix} I_2 \\ I_3 \end{bmatrix} =
\begin{bmatrix} |t|^2y & -t^*y \\ -t\,y & y \end{bmatrix}
\begin{bmatrix} V_2 \\ V_3 \end{bmatrix}
$$

Since $|t|^2 = 1$, $Y_{22}$ and $Y_{33}$ do not change. Only the mutual terms change:

$$
\begin{aligned}
Y_{23} &= -t^*y = -(1\angle -30^\circ)(8.8471\angle -76.70^\circ) = 8.8471\angle 73.30^\circ = 2.5425 + j8.4739 \\
Y_{32} &= -t\,y = -(1\angle 30^\circ)(8.8471\angle -76.70^\circ) = 8.8471\angle 133.30^\circ = -6.0674 + j6.4389
\end{aligned}
$$

**Answer:**

$$
Y_{bus,new} =
\begin{bmatrix}
4.0701 - j17.1798 & -2.0351 + j8.6099 & -2.0351 + j8.6099 \\
-2.0351 + j8.6099 & 4.0701 - j17.1798 & 2.5425 + j8.4739 \\
-2.0351 + j8.6099 & -6.0674 + j6.4389 & 4.0701 - j17.1798
\end{bmatrix}\ \text{pu}
$$

With a phase shifter, $Y_{bus}$ is **no longer symmetric** ($Y_{23} \ne Y_{32}$), because $t$ is complex. (If the shifter is taken the other way round, $Y_{23}$ and $Y_{32}$ swap.)

## Check yourself

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

1. Which statement about Ybus with t = e^(j30°) is correct?
   A. It remains symmetric
   B. It becomes non-symmetric, Y23 differs from Y32
   C. All elements are unchanged
   D. Diagonals change
   Answer: B. Mutual terms are -t* y and -t y which differ for complex t. (Key point: Phase shifter effect)

2. Why do Y22 and Y33 not change?
   A. |t|^2 = 1
   B. Shunts cancel
   C. t is real
   D. They are zero
   Answer: A. The diagonal factor is |t|^2 = 1. (Key point: Diagonals)

3. j0.04 total shunt per line is placed as:
   A. j0.04 at each end
   B. j0.02 at each end
   C. j0.08 at each end
   D. j0.01 at each end
   Answer: B. Nominal pi: half at each end. (Key point: Shunt placement)

4. What is |Y23| and |Y32| after the shift?
   A. 8.8471
   B. |y|*2
   C. 17.69
   D. 0
   Answer: A. |t| = 1 so the magnitudes equal |y| = 8.8471. (Key point: Magnitude)

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