Question

A three bus network is shown in the figure below indicating the p.u. impedances of each element.

The bus admittance matrix, Y-bus, of the network is

• A. $j\left[\begin{array}{ccc}0.3& -0.2& 0\\ -0.2& 0.12& 0.08\\ 0& 0.08& 0.02\end{array}\right]$
• B. $j\left[\begin{array}{ccc}-15& 5& 0\\ 5& 7.5& -12.5\\ 0& -12.5& 2.5\end{array}\right]$
• C. $j\left[\begin{array}{ccc}0.1& 0.2& 0\\ 0.2& 0.12& -0.08\\ 0& -0.08& 0.10\end{array}\right]$
• D. $j\left[\begin{array}{ccc}-10& 5& 0\\ 5& 7.5& 12.5\\ 0& 12.5& 10\end{array}\right]$

Answer 1

The correct option is B $j\left[\begin{array}{ccc}-15& 5& 0\\ 5& 7.5& -12.5\\ 0& -12.5& 2.5\end{array}\right]$


$Z_{10}=j0.1$

$\Rightarrow y_{10}=\frac{1}{Z_{10}}=\frac{1}{j0.1}=-j10p.u.$

$Z_{12}=j0.2$

$\Rightarrow y_{12}=\frac{1}{Z_{12}}=\frac{1}{j0.2}=-j5p.u.$

$Z_{23}=-j0.8$

$\Rightarrow y_{23}=\frac{1}{Z_{23}}=\frac{1}{-j0.08}=j12.5p.u.$

$Z_{30}=j0.1$

$\Rightarrow y_{30}=\frac{1}{Z_{30}}=\frac{1}{j0.1}=-j10p.u.$

$Y_{11}=y10+y_{12}=-j10-j5=-j15p.u.$

$Y_{12}=Y_{21}=-y_{12}=-(-j5)=j5p.u.$

$Y_{31}=Y_{13}=-y_{13}=0$

$Y_{22}=y_{12}+y_{23}=-j5+j12.5=j7.5p.u$

$Y_{33}=y_{30}+y_{23}=-j10+j12.5=j2.5p.u.$

$Y_{BUS}=j\left[\begin{array}{ccc}-15& 5& 0\\ 5& 7.5& -12.5\\ 0& -12.5& 2.5\end{array}\right]$