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Ybus for 3-bus lines with shunt susceptance

Short answer

Series admittance y = 2 - j6 pu; half-line charging j0.15 at each end. Diagonals: 4 - j11.7 pu; off-diagonals: -2 + j6 pu. Ybus is symmetric with these values.

The single line diagram of a power system network is shown in figure below. If each line has series impedance of (0.05+j0.15) pu and shunt susceptance of j0.3 pu., find bus Admittance matrix for the system. [Figure: three buses with lines 1-2, 1-3 and 2-3]

Assumption: j0.3j0.3 pu is the total shunt susceptance of each line, so j0.15j0.15 pu is placed at each end (nominal π model).

Step 1: Series admittance of each line

y=10.05+j0.15=0.05−j0.150.052+0.152=0.05−j0.150.025=2−j6 puy = \frac{1}{0.05 + j0.15} = \frac{0.05 - j0.15}{0.05^2 + 0.15^2} = \frac{0.05 - j0.15}{0.025} = 2 - j6\ \text{pu}

Step 2: Diagonal elements

Each bus is connected to two lines, and receives j0.15j0.15 from each:

Y11=Y22=Y33=2(2−j6)+2(j0.15)=4−j11.7 puY_{11} = Y_{22} = Y_{33} = 2(2 - j6) + 2(j0.15) = 4 - j11.7\ \text{pu}

Step 3: Off-diagonal elements

Y12=Y13=Y23=−y=−2+j6 puY_{12} = Y_{13} = Y_{23} = -y = -2 + j6\ \text{pu}

Answer:

Ybus=[4−j11.7−2+j6−2+j6−2+j64−j11.7−2+j6−2+j6−2+j64−j11.7] puY_{bus} = \begin{bmatrix} 4 - j11.7 & -2 + j6 & -2 + j6 \\ -2 + j6 & 4 - j11.7 & -2 + j6 \\ -2 + j6 & -2 + j6 & 4 - j11.7 \end{bmatrix}\ \text{pu}

(If j0.3j0.3 were the half-line charging at each end, each diagonal element would be 4−j11.44 - j11.4.)

✓ Answer checked by Claude Sonnet 5.5

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