A 3-bus network is shown. Consider generators as ideal voltage sources. If rows 1, 2 and 3 of the Y_bus matrix correspond to bus 1, 2 and 3 respectively, then Y_bus of the network is

Correct Answer:
$\left[\begin{array}{ccc}-\frac{3}{4}j& \frac{1}{4}j& \frac{1}{4}j\\ \frac{1}{4}j& -\frac{3}{4}j& \frac{1}{4}j\\ \frac{1}{4}j& \frac{1}{4}j& -\frac{3}{4}j\end{array}\right]$

Step-by-Step Explanation:

1. Analyze the Network Diagram:
From the given 3-bus network diagram:

We identify:
- Three main buses: Bus-1, Bus-2, and Bus-3, each connected to ideal generator sources (which represent the connection to the reference/ground).
- A central star-point junction node (let us designate it as Node 4).
- Three branch impedances connecting each bus to the central node (Node 4):
$Z_1=j\mathrm{\Omega }$ (connecting Bus-1 to Node 4)
$Z_2=j\mathrm{\Omega }$ (connecting Bus-2 to Node 4)
$Z_3=j\mathrm{\Omega }$ (connecting Bus-3 to Node 4)
- One branch impedance connecting the central node (Node 4) to the reference ground:
$Z_4=j\mathrm{\Omega }$

2. Calculate the Admittances:
Convert the branch impedances to admittances using the relation $y=\frac{1}{Z}$:
$y_1=\frac{1}{j}=-j\text{S}$
$y_2=\frac{1}{j}=-j\text{S}$
$y_3=\frac{1}{j}=-j\text{S}$
$y_4=\frac{1}{j}=-j\text{S}$

3. Formulate the Augmented Admittance Matrix (including the central node):
We set up a $4\times 4$ admittance matrix including Bus-1, Bus-2, Bus-3, and the central Node 4:
- The self-admittances are:
$Y_{11}=y_1=-j$
$Y_{22}=y_2=-j$
$Y_{33}=y_3=-j$
$Y_{44}=y_1+y_2+y_3+y_4=-j-j-j-j=-4j$
- The mutual admittances between the buses and the central Node 4 are:
$Y_{14}=Y_{41}=-y_1=j$
$Y_{24}=Y_{42}=-y_2=j$
$Y_{34}=Y_{43}=-y_3=j$
All other mutual admittances among Bus-1, Bus-2, and Bus-3 are zero since there are no direct lines connecting them.

Thus, the augmented admittance matrix is:
$Y_{\text{augmented}}=\left[\begin{array}{cccc}-j& 0& 0& j\\ 0& -j& 0& j\\ 0& 0& -j& j\\ j& j& j& -4j\end{array}\right]$

4. Eliminate Node 4 (Kron Reduction):
To find the equivalent $3\times 3$ bus admittance matrix ($Y_{\text{bus}}$), we eliminate Node 4 using Kron reduction:
$Y_{jk,\text{new}}=Y_{jk}-\frac{Y_{j4}{Y}_{4k}}{Y_{44}}$
for $j,k\in \{1,2,3\}$.

- Diagonal Elements (j = k):
$Y_{11,\text{new}}=Y_{11}-\frac{Y_{14}{Y}_{41}}{Y_{44}}=-j-\frac{\left(j\right)\left(j\right)}{-4j}=-j-\frac{-1}{-4j}=-j-\frac{1}{4j}=-j+\frac{1}{4}j=-\frac{3}{4}j$
Due to the symmetry of the network:
$Y_{22,\text{new}}=Y_{33,\text{new}}=-\frac{3}{4}j$

- Off-Diagonal Elements (j ≠ k):
$Y_{12,\text{new}}=Y_{12}-\frac{Y_{14}{Y}_{42}}{Y_{44}}=0-\frac{\left(j\right)\left(j\right)}{-4j}=-\frac{-1}{-4j}=-\frac{1}{4j}=\frac{1}{4}j$
By symmetry, all off-diagonal terms are identical:
$Y_{12,\text{new}}=Y_{13,\text{new}}=Y_{21,\text{new}}=Y_{23,\text{new}}=Y_{31,\text{new}}=Y_{32,\text{new}}=\frac{1}{4}j$

5. Construct the Final Y_bus Matrix:
Putting the computed terms together, the bus admittance matrix of the network is:
$Y_{\text{bus}}=\left[\begin{array}{ccc}-\frac{3}{4}j& \frac{1}{4}j& \frac{1}{4}j\\ \frac{1}{4}j& -\frac{3}{4}j& \frac{1}{4}j\\ \frac{1}{4}j& \frac{1}{4}j& -\frac{3}{4}j\end{array}\right]$