GATE Papers >> EEE >> 2017 >> Question No 10

Question No. 10 EEE | GATE 2017

$ A\;3 $-bus power system is shown in the figure below, where the diagonal elements of $ Y $-bus matrix are: $ Y_{11}=-j12\;pu,\;Y_{22}=-j15\;pu $ and $ Y_{33}=-j7\;pu. $

The per unit values of the lines reactances $ p,q $ and $ r $ shown in the figure are


Answer : (B) $ p=0.2,q=0.1,r=0.5 $


Solution of Question No 10 of GATE 2017 EEE Paper

$\begin{array}{l}{\mathrm Y}_{11}=-\frac{\mathrm j}{\mathrm q}-\frac{\mathrm j}{\mathrm r}=-\mathrm j12\Rightarrow\frac1{\mathrm q}+\frac1{\mathrm r}=12.....(\mathrm i)\\{\mathrm Y}_{22}=-\frac{\mathrm j}{\mathrm q}-\frac{\mathrm j}{\mathrm p}=-\mathrm j15\Rightarrow\frac1{\mathrm q}+\frac1{\mathrm p}=15.....(\mathrm{ii})\\{\mathrm Y}_{33}=-\frac{\mathrm j}{\mathrm p}-\frac{\mathrm j}{\mathrm r}=-\mathrm j7\Rightarrow\frac1{\mathrm p}+\frac1{\mathrm r}=7\;.....(\mathrm{iii})\end{array}$

Add (i),(ii) & (iii)

2 $\left(\frac1{\mathrm p}+\frac1{\mathrm q}+\frac1{\mathrm r}\right)$ = 34

$\frac1{\mathrm p}+\frac1{\mathrm q}+\frac1{\mathrm r}$ =17  ......(iv)

Subtract one by one (i), (ii) & (iii) from (iv)

$\frac1{\mathrm p}=5\;\;\frac1{\mathrm q}=10\;\;\frac1{\mathrm r}=2$

Hence p = 0.2, q = 0.1, r = 0.5

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