Ybus of five-bus system

Series admittance of each line

$$y = \frac{1}{0.05 + j0.15} = \frac{0.05 - j0.15}{0.025} = 2 - j6\ \text{pu}$$

Shunt admittances are neglected, so $Y_{ii}$ = (number of lines at bus $i$) × $y$ and $Y_{ik} = -y$ for each connected pair.

Lines at each bus

Off-diagonal: $Y_{12} = Y_{15} = Y_{23} = Y_{25} = Y_{34} = Y_{45} = -2 + j6$; $Y_{13} = Y_{14} = Y_{24} = Y_{35} = 0$.

Answer:

$$Y_{bus} = \begin{bmatrix} 4 - j12 & -2 + j6 & 0 & 0 & -2 + j6 \\ -2 + j6 & 6 - j18 & -2 + j6 & 0 & -2 + j6 \\ 0 & -2 + j6 & 4 - j12 & -2 + j6 & 0 \\ 0 & 0 & -2 + j6 & 4 - j12 & -2 + j6 \\ -2 + j6 & -2 + j6 & 0 & -2 + j6 & 6 - j18 \end{bmatrix}\ \text{pu}$$

Driving-point and transfer admittances

• Driving-point (self) admittances are the diagonal elements $Y_{ii}$: $Y_{11} = Y_{33} = Y_{44} = 4 - j12$ pu and $Y_{22} = Y_{55} = 6 - j18$ pu. $Y_{ii}$ is the current injected at bus $i$ per unit voltage at bus $i$ with all other buses shorted to ground.
• Transfer (mutual) admittances are the off-diagonal elements $Y_{ik}$: $-2 + j6$ pu for directly connected buses and 0 otherwise. $Y_{ik}$ is the current at bus $i$ per unit voltage at bus $k$ with all other buses shorted.