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Ybus with shunt admittances

Short answer

Parallel lines 2-3 combine to –j10. Shunt admittances j5 at buses 1 and 2 add to diagonal. Y11 = j5–j10 = –j5, Y22 = j5–j10–j10 = –j15, Y33 = –j10; off-diagonals Y12=Y21=j10, Y23=Y32=j10, Y13=Y31=0. Ybus = [[–j5, j10, 0]; [j10, –j15, j10]; [0, j10, –j10]] pu.

Compute Y-bus matrix for the following power system network shown in figure below. [Figure: G1 at bus 1 and G2 at bus 3; line 1-2 x1 = j0.1 pu; two parallel lines between buses 2 and 3, x2 = j0.2 pu and x3 = j0.2 pu; shunt admittances y1 = j5 pu at bus 1 and y2 = j5 pu at bus 2 to ground]

Step 1: Branch admittances

  • Line 1-2: y12=1/j0.1=−j10y_{12} = 1/j0.1 = -j10
  • Two parallel lines 2-3: j0.2∥j0.2=j0.1j0.2 \parallel j0.2 = j0.1, so y23=1/j0.1=−j10y_{23} = 1/j0.1 = -j10 (or −j5−j5-j5 - j5)
  • No line 1-3: y13=0y_{13} = 0
  • Shunt admittances (given as admittances): y10=j5y_{10} = j5, y20=j5y_{20} = j5, y30=0y_{30} = 0

Generators are sources at buses 1 and 3; no internal reactance is given, so they do not add to YbusY_{bus}.

Step 2: Diagonal elements

Y11=y10+y12=j5−j10=−j5Y22=y20+y12+y23=j5−j10−j10=−j15Y33=y23=−j10\begin{aligned} Y_{11} &= y_{10} + y_{12} = j5 - j10 = -j5 \\ Y_{22} &= y_{20} + y_{12} + y_{23} = j5 - j10 - j10 = -j15 \\ Y_{33} &= y_{23} = -j10 \end{aligned}

Step 3: Off-diagonal elements

Y12=Y21=j10,Y23=Y32=j10,Y13=Y31=0Y_{12} = Y_{21} = j10,\qquad Y_{23} = Y_{32} = j10,\qquad Y_{13} = Y_{31} = 0

Answer:

Ybus=[−j5j100j10−j15j100j10−j10] puY_{bus} = \begin{bmatrix} -j5 & j10 & 0 \\ j10 & -j15 & j10 \\ 0 & j10 & -j10 \end{bmatrix}\ \text{pu}

Note the zero at Y13Y_{13} (no direct line), which shows the sparsity of YbusY_{bus}, and that the capacitive shunts reduce ∣Y11∣|Y_{11}| and ∣Y22∣|Y_{22}|.

✓ Answer checked by Claude Sonnet 5.5

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