Question

The bus impedance matrix of a 4 – bus power system is given by

\({Z_{bus}} = \left[ {\begin{array}{*{20}{c}} {j0.3435}&{j0.2860}&{j0.2723}&{j0.2277}\\ {j0.2860}&{j0.3408}&{j0.2586}&{j0.2414}\\ {j0.2723}&{j0.2586}&{j0.2791}&{j0.2209}\\ {j0.2277}&{j0.2414}&{j0.2209}&{j0.2791} \end{array}} \right]\)

A branch having an impedance of j 0.2 Ω is connected between bus 2 and the reference Then the values of Z₂₂,new and Z₂₃,new of the bus impedance matrix of the modified network are respectively.

1. j0.5408 Ω and j0.4586 Ω
2. j0.1260 Ω and j0.0956 Ω
3. j0.5408 Ω and j0.0956 Ω
4. j0.1260 Ω and j0.1630 Ω

Answer 1

Correct Answer - Option 2 : j0.1260 Ω and j0.0956 Ω

\({Z_{B\left( {new} \right)}} = {Z_{B\left( {old} \right)}} - \frac{1}{{{Z_{ii}} + {Z_b}}}\left[ {\begin{array}{*{20}{c}} {{Z_{ij}}}\\ \vdots \\ {{Z_{nj}}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{Z_{ji}}}& \ldots &{{Z_{jn}}} \end{array}} \right]\)

New element Z_b = j0.2Ω is connected in i^th bus and reference bus i = 2, n = 4 so

\({Z_{B\left( {new} \right)}}=\frac{1}{{{Z_{ii}} + {Z_b}}}\left[ {\begin{array}{*{20}{c}} {{Z_{12}}}\\ {{Z_{22}}}\\ {{Z_{32}}}\\ {{Z_{42}}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{Z_{21}}}&{{Z_{22}}}&{{Z_{23}}}&{{Z_{24}}} \end{array}} \right]\)

\({Z_{B\left( {new} \right)}}= \frac{1}{{\left[ {j\left( {0.3408} \right) + j\left( {0.2} \right)} \right]}}\left[ {\begin{array}{*{20}{c}} {j0.2860}\\ {j0.3408}\\ {j0.2586}\\ {j0.2414} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {j.0.2860}&{j0.3408}&{j0.2586}&{j0.2414} \end{array}} \right]\)

Given that we are required to find only Z₂₂, Z₂₃

\(\begin{array}{l} Z_{22}^1 = \frac{{{j^2}{{\left( {0.3408} \right)}^2}}}{{j\left( {0.5408} \right)}} = j0.2147\\ Z_{23}^1 = \frac{{{j^2}\left( {0.3408} \right)\left( {0.2586} \right)}}{{j\left( {0.5408} \right)}} = j0.16296\\ {Z_{22\left( {new} \right)}} = {Z_{22\left( {old} \right)}} - Z_{22}^1 = j0.3409 - j0.2147 \end{array}\)

= j 0.1260

\({Z_{23\left( {new} \right)}} = {Z_{23\left( {old} \right)}} - Z_{23}^1 = j0.2586 - j0.16296\)

= j 0.0956