The line reactances of a power network are as follows:

Line No

From Bus

To Bus

Reactance

1

0

1

0.2 pu

2

1

2

0.4 pu

The bus impedance matrix with ‘0’ as reference bus is

Question

Question

This question was previously asked in

ESE Electrical 2018 Official Paper

\(\left[ {\begin{array}{*{20}{c}} {0.2}&{0.4}\\ {0.4}&{0.6} \end{array}} \right]\)
\(\left[ {\begin{array}{*{20}{c}} {0.4}&{0.2}\\ {0.2}&{0.6} \end{array}} \right]\)
\(\left[ {\begin{array}{*{20}{c}} {0.2}&{0.2}\\ {0.2}&{0.6} \end{array}} \right]\)
\(\left[ {\begin{array}{*{20}{c}} {0.6}&{0.2}\\ {0.2}&{0.4} \end{array}} \right]\)

Answer 1

By considering line - 1,

F1 U.B. N.J 06.09.2019 D11

Here Z_BUSis formed between new bus & reference bus, so type - 1 formation

∴ Z_Bus = [j 0.2]

By adding line - 2 to the above Bus

F1 U.B. N.J 06.09.2019 D12

Here Z_BUSis formed between new bus & old bus, so type - 2 formation

\({Z_{Bus}} = \left[ {\begin{array}{*{20}{c}} {j\;0.2}&{j\;0.2}\\ {j\;0.2}&{j\;0.2 + j\;0.4} \end{array}} \right]\)

\(= \left[ {\begin{array}{*{20}{c}} {j\;0.2}&{j\;0.2}\\ {j\;0.2}&{j\;0.6} \end{array}} \right]\)