Ybus for 3-bus system (2078)

$Y_{bus}$ depends only on the network impedances; the bus data ($V_1$, $P_2$, $Q_2$, $P_3$, $V_3$, Q limits) are needed later for load flow, not for $Y_{bus}$. No line charging is given, so it is neglected.

Step 1: Line admittances

$$\begin{aligned} y_{12} &= \frac{1}{0.02 + j0.04} = \frac{0.02 - j0.04}{0.02^2 + 0.04^2} = \frac{0.02 - j0.04}{0.002} = 10 - j20\ \text{pu} \\ y_{13} &= \frac{1}{0.01 + j0.03} = \frac{0.01 - j0.03}{0.001} = 10 - j30\ \text{pu} \\ y_{23} &= \frac{1}{0.0125 + j0.025} = \frac{0.0125 - j0.025}{0.00078125} = 16 - j32\ \text{pu} \end{aligned}$$

Step 2: Diagonal elements

$$\begin{aligned} Y_{11} &= y_{12} + y_{13} = 20 - j50 \\ Y_{22} &= y_{12} + y_{23} = 26 - j52 \\ Y_{33} &= y_{13} + y_{23} = 26 - j62 \end{aligned}$$

Step 3: Off-diagonal elements

$$Y_{12} = Y_{21} = -10 + j20,\quad Y_{13} = Y_{31} = -10 + j30,\quad Y_{23} = Y_{32} = -16 + j32$$

Answer:

$$Y_{bus} = \begin{bmatrix} 20 - j50 & -10 + j20 & -10 + j30 \\ -10 + j20 & 26 - j52 & -16 + j32 \\ -10 + j30 & -16 + j32 & 26 - j62 \end{bmatrix}\ \text{pu}$$

For the load flow that follows: bus 1 is the slack bus ($V_1 = 1.05\angle 0^\circ$), bus 2 is a PQ (load) bus, and bus 3 is a PV bus ($P_3 = 2.0$, $|V_3| = 1.04$, with Q limits).