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GATE 2017 EE – Question 50

Power Systems · Per-unit quantities, Bus admittance matrix, Gauss-Seidel and Newton-Raphson load flow methods · 2 marks · Multiple choice

The bus admittance matrix for a power system network is

$$\begin{bmatrix} -j39.9 & j20 & j20 \\ j20 & -j39.9 & j20 \\ j20 & j20 & -j39.9 \end{bmatrix} \text{ pu.}$$

There is a transmission line, connected between buses 1 and 3, which is represented by the circuit shown in figure.

[Figure: A pi model with a series reactance of 0.05 pu and a shunt susceptance of 0.05 pu at each end.]

If this transmission line is removed from service, what is the modified bus admittance matrix?

Diagram for GATE 2017 EE question 50
  1. $\begin{bmatrix} -j19.9 & j20 & 0 \\ j20 & -j39.9 & j20 \\ 0 & j20 & -j19.9 \end{bmatrix}$ pu
  2. $\begin{bmatrix} -j39.95 & j20 & 0 \\ j20 & -j39.9 & j20 \\ 0 & j20 & -j39.95 \end{bmatrix}$ pu
  3. $\begin{bmatrix} -j19.95 & j20 & 0 \\ j20 & -j39.9 & j20 \\ 0 & j20 & -j19.95 \end{bmatrix}$ pu
  4. $\begin{bmatrix} -j19.95 & j20 & j20 \\ j20 & -j39.9 & j20 \\ j20 & j20 & -j19.95 \end{bmatrix}$ pu

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Correct answer: (C) $\begin{bmatrix} -j19.95 & j20 & 0 \\ j20 & -j39.9 & j20 \\ 0 & j20 & -j19.95 \end{bmatrix}$ pu

Explanation

The series reactance $0.05$ pu has admittance $-j20$, and each end also has a shunt susceptance of $j0.05$. The line contributed $-j20 + j0.05 = -j19.95$ to each of the diagonal elements $Y_{11}$ and $Y_{33}$, and $+j20$ to $Y_{13}$ and $Y_{31}$. Removing the line adds $+j19.95$ to the diagonals, giving $-j39.9 + j19.95 = -j19.95$, and takes $Y_{13} = Y_{31}$ to 0.