Ybus modified by mid-line tap changer

(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.

  (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.