Question

The network shown in the given figure has impedances in p.u. as indicated. The diagonal element $Y_{22}$ of the bus admittance matrix $Y_{BUS}$ of the network is

• A. -j 19
• B. +j 20
• C. + j 0.2
• D. -j 19.95

Answer 1

The correct option is D -j 19.95

• Impedance between bus(1) and bus (2) $=Z_{12}=j0.1$

admittance$=Y_{12}=\frac{1}{Z_{12}}=\frac{1}{j0.1}=-j10p.u.$

• Impedance between bus(2) and bus (3) $=Z_{23}=j0.1$

admittance$=Y_{23}=\frac{1}{Z_{23}}=\frac{1}{j0.1}=-j10p.u.$

• Impedance between bus(2) and ground

$=Z_{20}=-j20p.u.$
admittance$Y_{20}=\frac{1}{Z_{20}}=\frac{1}{-j20}=j0.05p.u.$
$Y_{22}=Y_{20}+Y_{12}+Y_{23}$
$=j0.05-j10-j10$
$=-j19.95p.u.$