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Node equations and Ybus, six-bus network

Short answer

Node equations by KCL using branch admittances yield Ybus with diagonals: -j20, -j24.98, -j27.49, -j20, -j27.49, -j20 pu for buses 1–6; off-diagonals: j20 (1-2), j2.5 (2-3,2-5), j20 (3-4), j5 (3-5), j20 (5-6) and symmetric counterparts.

Obtain node equations and compute bus admittance matrix (YBus) of the network shown in figure below. All voltages and impedances are marked in pu. [Figure: G1 at bus 1 connected through a transformer j0.05 to bus 2; Line-1 from bus 2 to bus 3; Line-3 from bus 2 to bus 5; Line-2 from bus 3 to bus 5; bus 3 connected through a transformer j0.05 to bus 4, which has a load; bus 5 connected through a transformer j0.05 to bus 6, where G2 is connected]

Line data are as follows:

LineSeries reactance, puShunt susceptance, pu
Line-10.400.02
Line-20.200.00
Line-30.400.02

Assumptions: the given shunt susceptance of each line is the total line charging, split as B/2B/2 at each end (nominal π). Generators and load are injections at buses 1, 6 and 4.

Step 1: Branch admittances

BranchBusesXX (pu)y=1/jXy = 1/jXB/2B/2 at each end
Transformer1-20.05−j20-j200
Line-12-30.40−j2.5-j2.5j0.01j0.01
Line-32-50.40−j2.5-j2.5j0.01j0.01
Line-23-50.20−j5-j50
Transformer3-40.05−j20-j200
Transformer5-60.05−j20-j200

Step 2: Node equations (KCL, IiI_i = injected current)

I1=y12(V1−V2)I2=y12(V2−V1)+y23(V2−V3)+y25(V2−V5)+j0.01V2+j0.01V2I3=y23(V3−V2)+y35(V3−V5)+y34(V3−V4)+j0.01V3I4=y34(V4−V3)I5=y25(V5−V2)+y35(V5−V3)+y56(V5−V6)+j0.01V5I6=y56(V6−V5)\begin{aligned} I_1 &= y_{12}(V_1 - V_2) \\ I_2 &= y_{12}(V_2 - V_1) + y_{23}(V_2 - V_3) + y_{25}(V_2 - V_5) + j0.01V_2 + j0.01V_2 \\ I_3 &= y_{23}(V_3 - V_2) + y_{35}(V_3 - V_5) + y_{34}(V_3 - V_4) + j0.01V_3 \\ I_4 &= y_{34}(V_4 - V_3) \\ I_5 &= y_{25}(V_5 - V_2) + y_{35}(V_5 - V_3) + y_{56}(V_5 - V_6) + j0.01V_5 \\ I_6 &= y_{56}(V_6 - V_5) \end{aligned}

Step 3: Elements

Y11=−j20Y22=−j20−j2.5−j2.5+j0.02=−j24.98Y33=−j2.5−j5−j20+j0.01=−j27.49Y44=−j20Y55=−j2.5−j5−j20+j0.01=−j27.49Y66=−j20\begin{aligned} Y_{11} &= -j20 \\ Y_{22} &= -j20 - j2.5 - j2.5 + j0.02 = -j24.98 \\ Y_{33} &= -j2.5 - j5 - j20 + j0.01 = -j27.49 \\ Y_{44} &= -j20 \\ Y_{55} &= -j2.5 - j5 - j20 + j0.01 = -j27.49 \\ Y_{66} &= -j20 \end{aligned}

Off-diagonal: Y12=j20Y_{12} = j20, Y23=j2.5Y_{23} = j2.5, Y25=j2.5Y_{25} = j2.5, Y34=j20Y_{34} = j20, Y35=j5Y_{35} = j5, Y56=j20Y_{56} = j20; all others zero.

Answer:

Ybus=[−j20j200000j20−j24.98j2.50j2.500j2.5−j27.49j20j5000j20−j20000j2.5j50−j27.49j200000j20−j20] puY_{bus} = \begin{bmatrix} -j20 & j20 & 0 & 0 & 0 & 0 \\ j20 & -j24.98 & j2.5 & 0 & j2.5 & 0 \\ 0 & j2.5 & -j27.49 & j20 & j5 & 0 \\ 0 & 0 & j20 & -j20 & 0 & 0 \\ 0 & j2.5 & j5 & 0 & -j27.49 & j20 \\ 0 & 0 & 0 & 0 & j20 & -j20 \end{bmatrix}\ \text{pu}

✓ Answer checked by Claude Sonnet 5.5

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