Zbus vs Ybus; off-nominal transformer model

Zbus versus Ybus

Ybus for a regulating (off-nominal tap) transformer

Model the transformer as an ideal transformer of ratio $1 : t$ in series with its per-unit leakage admittance $y = 1/z$ placed on the tap (bus q) side. Bus p is on the "1" side, bus q on the "t" side.

  p   Ip      1 : t     x    y     Iq   q
  o----->----)|(------o--[===]----<-----o
  Vp              Vx = t Vp

Ideal transformer: $V_x = tV_p$. Let $I_{xq} = y(V_x - V_q)$ be the current leaving node x towards q, so $I_q = -I_{xq}$. Being lossless, $V_pI_p^* = V_xI_{xq}^*$, which with real $t$ gives $I_p = tI_{xq}$:

$$\begin{aligned} I_q &= y(V_q - tV_p) = -tyV_p + yV_q \\ I_p &= t\,y(tV_p - V_q) = t^2yV_p - tyV_q \end{aligned}$$

In matrix form:

$$\begin{bmatrix} I_p \\ I_q \end{bmatrix} = \begin{bmatrix} t^2y & -ty \\ -ty & y \end{bmatrix} \begin{bmatrix} V_p \\ V_q \end{bmatrix}$$

So the transformer adds $t^2y$ to $Y_{pp}$, $y$ to $Y_{qq}$ and $-ty$ to $Y_{pq}$, $Y_{qp}$.

Equivalent π-model

Match a π circuit (series $A$, shunts $B$ at p and $C$ at q) to the matrix: $Y_{pq} = -A$, $Y_{pp} = A + B$, $Y_{qq} = A + C$:

$$A = ty,\qquad B = t^2y - ty = t(t - 1)y,\qquad C = y - ty = (1 - t)y$$

  p          t*y            q
  o--------[=====]----------o
  |                         |
 [ t(t-1)y ]          [ (1-t)y ]
  |                         |
 ===                       ===

If $t = 1$, the shunts vanish and the model becomes the simple series admittance $y$. For $t \ne 1$ the two shunt branches have opposite signs (one acts inductive, the other capacitive), which is how the tap changes reactive power flow and bus voltage.