Skip to main content

Ybus, Zbus and 4-bus Ybus formulation

Short answer

Ybus relates injected bus currents to voltages, I = Ybus V; Zbus is its inverse V = Zbus I. For a 4-bus system, diagonal elements Y_ii = y_i0 + sum of all line admittances connected to bus i, off-diagonals Y_ik = –y_ik (zero if no line). Ybus is symmetric, sparse, and diagonally dominant.

What are bus admittance matrix and bus impedance matrix? Formulate the bus admittance matrix for 4-bus system and state the properties of the matrix including general equations of their elements.

Bus admittance and bus impedance matrices

  • Bus admittance matrix YbusY_{bus}: relates injected bus currents to bus voltages, I=YbusV\mathbf{I} = \mathbf{Y}_{bus}\mathbf{V}. It is sparse and formed by inspection.
  • Bus impedance matrix ZbusZ_{bus}: the inverse, V=ZbusI\mathbf{V} = \mathbf{Z}_{bus}\mathbf{I}, with Zbus=Ybus−1Z_{bus} = Y_{bus}^{-1}. It is a full matrix, formed by the building algorithm or inversion, and used mainly in fault analysis.

Formulation for a 4-bus system

Assume lines 1-2, 1-3, 2-3, 2-4 and 3-4 with series admittances yiky_{ik}, and shunt admittances yi0y_{i0} at each bus (half line charging).

   (1)---------y12---------(2)
    |                    / |
    |                 /    |
   y13            y23     y24
    |          /           |
    |       /              |
   (3)---------y34---------(4)

   No line 1-4. Each bus i also has y_i0 to ground.

KCL at each bus:

I1=y10V1+y12(V1−V2)+y13(V1−V3)I2=y20V2+y12(V2−V1)+y23(V2−V3)+y24(V2−V4)I3=y30V3+y13(V3−V1)+y23(V3−V2)+y34(V3−V4)I4=y40V4+y24(V4−V2)+y34(V4−V3)\begin{aligned} I_1 &= y_{10}V_1 + y_{12}(V_1 - V_2) + y_{13}(V_1 - V_3) \\ I_2 &= y_{20}V_2 + y_{12}(V_2 - V_1) + y_{23}(V_2 - V_3) + y_{24}(V_2 - V_4) \\ I_3 &= y_{30}V_3 + y_{13}(V_3 - V_1) + y_{23}(V_3 - V_2) + y_{34}(V_3 - V_4) \\ I_4 &= y_{40}V_4 + y_{24}(V_4 - V_2) + y_{34}(V_4 - V_3) \end{aligned}

In matrix form:

[I1I2I3I4]=[Y11−y12−y130−y12Y22−y23−y24−y13−y23Y33−y340−y24−y34Y44][V1V2V3V4]\begin{bmatrix} I_1 \\ I_2 \\ I_3 \\ I_4 \end{bmatrix} = \begin{bmatrix} Y_{11} & -y_{12} & -y_{13} & 0 \\ -y_{12} & Y_{22} & -y_{23} & -y_{24} \\ -y_{13} & -y_{23} & Y_{33} & -y_{34} \\ 0 & -y_{24} & -y_{34} & Y_{44} \end{bmatrix} \begin{bmatrix} V_1 \\ V_2 \\ V_3 \\ V_4 \end{bmatrix}

with Y11=y10+y12+y13Y_{11} = y_{10} + y_{12} + y_{13}, Y22=y20+y12+y23+y24Y_{22} = y_{20} + y_{12} + y_{23} + y_{24}, Y33=y30+y13+y23+y34Y_{33} = y_{30} + y_{13} + y_{23} + y_{34}, Y44=y40+y24+y34Y_{44} = y_{40} + y_{24} + y_{34}.

General equations of elements

Yii=yi0+∑k≠iyik,Yik=Yki=−yikY_{ii} = y_{i0} + \sum_{k \ne i} y_{ik},\qquad Y_{ik} = Y_{ki} = -y_{ik}

Properties of Ybus

  1. Symmetric about the leading diagonal (no phase shifters).
  2. Sparse: Yik=0Y_{ik} = 0 when no line joins ii and kk (here Y14=0Y_{14} = 0).
  3. Generally diagonally dominant: ∣Yii∣≥∑k≠i∣Yik∣|Y_{ii}| \ge \sum_{k\ne i} |Y_{ik}| when the shunt admittances are small or absent.
  4. Non-singular only when shunt elements to ground exist.
  5. Easy to form by inspection and to modify for adding or removing a line.

✓ Answer checked by Claude Sonnet 5.5

All Power System Analysis II chapter 1 questions →