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Branch impedances from given Ybus

Short answer

Series branch admittances: y12 = -j12 ⇒ z12 = j0.0833 pu; y13 = -j18 ⇒ z13 = j0.0556 pu; y23 = -j13 ⇒ z23 = j0.0769 pu. No shunt branches.

For a 3-bus network, the bus admittance matrix is given as follows: Ybus = [−j30 j12 j18; j12 −j25 j13; j18 j13 −j31] pu, determine the respective branch impedances.

For a network with no mutual coupling, Yik=−yikY_{ik} = -y_{ik} and Yii=yi0+∑k≠iyikY_{ii} = y_{i0} + \sum_{k \ne i} y_{ik}. So the series branch admittances come from the off-diagonal terms and the shunt branches from the row sums.

Series branches

y12=−Y12=−j12⇒ z12=1−j12=j0.0833 puy13=−Y13=−j18⇒ z13=1−j18=j0.0556 puy23=−Y23=−j13⇒ z23=1−j13=j0.0769 pu\begin{aligned} y_{12} &= -Y_{12} = -j12 &\Rightarrow\ z_{12} &= \frac{1}{-j12} = j0.0833\ \text{pu} \\ y_{13} &= -Y_{13} = -j18 &\Rightarrow\ z_{13} &= \frac{1}{-j18} = j0.0556\ \text{pu} \\ y_{23} &= -Y_{23} = -j13 &\Rightarrow\ z_{23} &= \frac{1}{-j13} = j0.0769\ \text{pu} \end{aligned}

Shunt branches

y10=Y11+Y12+Y13=−j30+j12+j18=0y20=Y21+Y22+Y23=j12−j25+j13=0y30=Y31+Y32+Y33=j18+j13−j31=0\begin{aligned} y_{10} &= Y_{11} + Y_{12} + Y_{13} = -j30 + j12 + j18 = 0 \\ y_{20} &= Y_{21} + Y_{22} + Y_{23} = j12 - j25 + j13 = 0 \\ y_{30} &= Y_{31} + Y_{32} + Y_{33} = j18 + j13 - j31 = 0 \end{aligned}

So there are no shunt elements to ground.

            j0.0833
     (1)-------------(2)
       \            /
        \          /
  j0.0556\        / j0.0769
          \      /
           \    /
            (3)

Answer: z12=j0.0833z_{12} = j0.0833 pu, z13=j0.0556z_{13} = j0.0556 pu, z23=j0.0769z_{23} = j0.0769 pu (purely inductive, i.e. X12=1/12X_{12} = 1/12, X13=1/18X_{13} = 1/18, X23=1/13X_{23} = 1/13 pu); no shunt branches.

✓ Answer checked by Claude Sonnet 5.5

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