Ybus and bus types (parallel lines)

Assumptions: line 1-3 is a reactance $j0.2$ pu (the "0.2" is taken as a misprint for $j0.2$). G1 is an ideal source at bus 1 (no reactance given). The G2 transformer reactance $j0.2$ is treated as a source branch at bus 2, i.e. a shunt admittance from bus 2 to reference (Norton form), as for the other generator branches in this course.

Step 1: Branch admittances

• Two parallel lines 1-2: $j0.4 \parallel j0.4 = j0.2$, so $y_{12} = 1/j0.2 = -j5$
• Line 2-3: $y_{23} = 1/j0.2 = -j5$
• Line 1-3: $y_{13} = 1/j0.2 = -j5$
• Shunt at bus 2: $Y = +j2$ (given) and transformer branch $1/j0.2 = -j5$, so $y_{20} = j2 - j5 = -j3$

Step 2: Elements

$$\begin{aligned} Y_{11} &= y_{12} + y_{13} = -j5 - j5 = -j10 \\ Y_{22} &= y_{12} + y_{23} + y_{20} = -j5 - j5 - j3 = -j13 \\ Y_{33} &= y_{13} + y_{23} = -j10 \\ Y_{12} &= Y_{13} = Y_{23} = +j5 \end{aligned}$$

Answer:

$$Y_{bus} = \begin{bmatrix} -j10 & j5 & j5 \\ j5 & -j13 & j5 \\ j5 & j5 & -j10 \end{bmatrix}\ \text{pu}$$

(If the G2 transformer is left out and only the network shunt $j2$ is kept at bus 2, $Y_{22} = -j8$; all other elements are unchanged.)

Types of buses