Ybus modified by phase-shifting transformer

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.)