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Power Factor

A two bus sys...

Question

A two bus system is as shown below,

Corresponding Z-bus matrix is $Z = [\begin{matrix} j 0.5 & j 0.5 j 0.5 & j 4.5 \end{matrix}]$

If the line between bus 1 and bus 2 is replaced with another line having reactance $(j 2) Ω$ , then modified $Z_{b u s}$ is

A

[\begin{matrix} j 0.5 & j 4.5 j 2.5 & j 4.5 \end{matrix}]

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B

[\begin{matrix} j 2.5 & j 0.5 j 0.5 & j 2.5 \end{matrix}]

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C

[\begin{matrix} j 0.5 & j 4.5 j 2.5 & j 4.5 \end{matrix}]

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D

[\begin{matrix} j 0.5 & j 0.5 j 0.5 & j 2.5 \end{matrix}]

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Solution

The correct option is D $[\begin{matrix} j 0.5 & j 0.5 j 0.5 & j 2.5 \end{matrix}]$
Existing system and bus matrix is

$Z_{B u s} = [\begin{matrix} j 0.5 & j 0.5 j 0.5 & j 4.5 \end{matrix}]$

Modifying line with reactance j2 is equivalent to adding a line in parallel with impedance j4.
Thus it is type 4 modification.

$[Z_{n e w}] = [\begin{matrix} Z_{o l d} \end{matrix}] - \frac{1}{Z_{11} + Z_{22} - 2 Z_{12} + Z_{s}} ⎡ ⎢ ⎣ \begin{matrix} s u b s t r a c t 2^{n d} c o l u m n f r o m f i r s t c o l u m n \end{matrix} ⎤ ⎥ ⎦ [T r a n s p o s e$ ]

$[Z_{B u s}]_{n e w} = [\begin{matrix} j 0.5 & j 0.5 j 0.5 & j 4.5 \end{matrix}] - \frac{1}{j 0.5 + j 4.5 - 2 (j 0.5) + j 4} [\begin{matrix} j 0 - j 4 \end{matrix}] [j 0 - j 4]$

$= [\begin{matrix} j 0.5 & j 0.5 j 0.5 & j 4.5 \end{matrix}] - \frac{1}{j 8} [\begin{matrix} 0 & 0 0 & - 16 \end{matrix}]$

$= [\begin{matrix} j 0.5 & j 0.5 j 0.5 & j 4.5 \end{matrix}] - ⎡ ⎢ ⎣ \begin{matrix} 0 & 0 0 & - \frac{16}{j 8} \end{matrix} ⎤ ⎥ ⎦$

$= ⎡ ⎢ ⎣ \begin{matrix} j 0.5 & j 0.5 j 0.5 & j 4.5 - \frac{j 16}{8} \end{matrix} ⎤ ⎥ ⎦ = [\begin{matrix} j 0.5 & j 0.5 j 0.5 & j 2.5 \end{matrix}]$

Alternative:
New system will

First branch:

Type - 2 modification

$Z = [\begin{matrix} j 0.5 & j 0.5 j 0.5 & j 0.5 + j 2 \end{matrix}] = [\begin{matrix} j 0.5 & j 0.5 j 0.5 & j 2.5 \end{matrix}]$

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