Question

A power system network consists of three elements 0 to 1, 1 to 2, 2 to 0 of per unit impedances 0.3, 0.5 and 0.5 respectively.
The value of $z_{12}$ elements in $Z_{bus}$ matrix $(2\times 2)$ will be ________pu.
(upto 3 decimal place )

• A. 0.115

Answer 1

The correct option is A 0.115
For the power system finding admittance value,

$y_{10}=\frac{1}{Z_{10}}=\frac{1}{0.3}=\frac{10}{3}$

$y_{12}=\frac{1}{Z_{12}}=\frac{1}{0.5}=2$

$y_{20}=\frac{1}{Z_{20}}=\frac{1}{0.5}=2$

$\therefore Admittancematrix,Y_{bus}=\left[\begin{array}{cc}10/3+2& -2\\ -2& 4\end{array}\right]=\left[\begin{array}{cc}5.33& -2\\ -2& 4\end{array}\right]$

$\therefore Impedancematrix,$
$Z_{bus}=[Y_{bus}{]}^{-1}=\frac{1}{(5.33\times 4)-(2\times 2)}\times \left[\begin{array}{cc}4& 2\\ 2& 5.33\end{array}\right]$

$=0.0577\times \left[\begin{array}{cc}4& 2\\ 2& 5.33\end{array}\right]=\left[\begin{array}{cc}0.230& 0.115\\ 0.115& 0.308\end{array}\right]$