The bus impedance matrix of a 4-bus power system is given by Zbus- j0-3435 j0-2860 j0-2723 j0-2277- j0-2860 j0-3408 0-2586 j0-2414- j0-2723 j0-2586 j0-2791 j0-2209- j0-2277 j0-2414 j0-2209 j0-2791 -A branch having an impedance of j0-2 - is connected between bus 2 and the reference- Then the values of Z22- new and Z23- new of the bus impedance matrix of the modified network are respectively

Z_B(new)=Z_B(old) 1Z_jj+Z_b[ Z_1_j; ⋮; Z_n_j ][ Z_j_1 amp;.. amp;Z_n_j ] New elements (Z_b) is connected between j^th and reference bus. ∴1Z_ij+Z_b[ ...

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Question

The bus impedance matrix of a 4-bus power system is given by
$Z_{b u s} = ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} j 0.3435 & j 0.2860 & j 0.2723 & j 0.2277 j 0.2860 & j 0.3408 & 0.2586 & j 0.2414 j 0.2723 & j 0.2586 & j 0.2791 & j 0.2209 j 0.2277 & j 0.2414 & j 0.2209 & j 0.2791 \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦$
A branch having an impedance of $j 0.2 Ω$ is connected between bus 2 and the reference. Then the values of $Z_{22, n e w}$ and $Z_{23, n e w}$ of the bus impedance matrix of the modified network are respectively

j 0.5408 Ω

and

j 0.4586 Ω

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j 0.1260 Ω

and

j 0.0956 Ω

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j 0.5408 Ω

and

0.0956 Ω

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j 0.1260 Ω

and

j 0.1630 Ω

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Solution

The correct option is B $j 0.1260 Ω$ and $j 0.0956 Ω$

$Z_{B} (n e w) = Z_{B} (o l d) - \frac{1}{Z_{j j} + Z_{b}} ⎡ ⎢ ⎢ ⎢ ⎣ \begin{matrix} Z_{1_{j}} ⋮ Z_{n_{j}} \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎦ [\begin{matrix} Z_{j_{1}} & . . & Z_{n_{j}} \end{matrix}]$
New elements $(Z_{b})$ is connected between $j^{t h}$ and reference bus.
$∴ \frac{1}{Z_{i j} + Z_{b}} ⎡ ⎢ ⎢ ⎢ ⎣ \begin{matrix} Z_{12} Z_{22} Z_{32} Z_{42} \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎦ [\begin{matrix} Z_{21} & Z_{22} & Z_{23} & Z_{24} \end{matrix}]$
$= \frac{1}{(j 0.3408 + j 0.2)} ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} j 0.2860 j 0.3408 j 0.2586 j 0.2414 \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ [\begin{matrix} j 0.2860 & j 0.3408 & j 0.2586 & j 0.2414 \end{matrix}]$
We require only changes in $Z_{22} Z_{23}$
$⎡ ⎢ ⎢ ⎢ ⎣ \begin{matrix} - & - & - & - - j 0.2147 & - & j 0.16296 & - - & - & - & - - & - & - & - \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎦$
$Z_{22} (n e w) = Z_{22} (o l d) - j 0.2147 = j 0.1260$
$Z_{23} (n e w) = Z_{23} (o l d) - j 0.16296$
$= j 0.2586 - j 0.16296 = j 0.0956 Ω$

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