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

The correct option is:
$\left[\begin{array}{ccc}-\frac{3}{4}j& \frac{1}{4}j& \frac{1}{4}j\\ \frac{1}{4}j& -\frac{3}{4}j& \frac{1}{4}j\\ \frac{1}{4}j& \frac{1}{4}j& -\frac{3}{4}j\end{array}\right]$

Step-by-Step Explanation:

1. Network Impedance and Admittance Values
Based on the provided network diagram:
- The three main buses are labeled as Bus-1, Bus-2, and Bus-3, each connected to ideal voltage source generators.
- These three buses connect to a central junction point (let's define this junction as Node 4) through branch impedances:
$Z_1=Z_2=Z_3=j\Omega$
- The central junction (Node 4) is connected to ground through an impedance:
$Z_4=j\Omega$
To formulate the admittance matrix, we convert these impedances into admittances ($y=\frac{1}{Z}$):
$y_1=y_2=y_3=y_4=\frac{1}{j}=-j\text{S}$

2. Formulating the Complete 4x4 Admittance Matrix
Let us construct the admittance matrix for all 4 nodes (Bus 1, Bus 2, Bus 3, and the central Node 4):
- Diagonal elements:
$Y_{11}=y_1=-j$
$Y_{22}=y_2=-j$
$Y_{33}=y_3=-j$
$Y_{44}=y_1+y_2+y_3+y_4=(-j)+(-j)+(-j)+(-j)=-4j$
- Off-diagonal elements:
Since there are no direct line connections between Bus 1, Bus 2, and Bus 3:
$Y_{12}=Y_{21}=0$
$Y_{13}=Y_{31}=0$
$Y_{23}=Y_{32}=0$
For connections to the central Node 4:
$Y_{14}=Y_{41}=-y_1=j$
$Y_{24}=Y_{42}=-y_2=j$
$Y_{34}=Y_{43}=-y_3=j$
Thus, the 4x4 admittance matrix $Y$ is:
$Y=\left[\begin{array}{cccc}-j& 0& 0& j\\ 0& -j& 0& j\\ 0& 0& -j& j\\ j& j& j& -4j\end{array}\right]$

3. Eliminating Node 4 using Kron Reduction
Since the central node (Node 4) does not have a generator connected to it, we can eliminate Node 4 using the Kron reduction formula:
$Y_{bus,ik}=Y_{ik}-\frac{Y_{i4}{Y}_{4k}}{Y_{44}}$
Let us calculate the terms for our 3x3 reduced admittance matrix:
- Diagonal Elements ($i=k$):
$Y_{bus,11}=Y_{11}-\frac{Y_{14}{Y}_{41}}{Y_{44}}=-j-\frac{j\cdot j}{-4j}=-j-\frac{-1}{-4j}=-j-\frac{1}{4j}=-j+\frac{1}{4}j=-\frac{3}{4}j$
By symmetry:
$Y_{bus,22}=Y_{bus,33}=-\frac{3}{4}j$
- Off-Diagonal Elements ($i\ne k$):
$Y_{bus,12}=Y_{12}-\frac{Y_{14}{Y}_{42}}{Y_{44}}=0-\frac{j\cdot j}{-4j}=-\frac{-1}{-4j}=-\frac{1}{4j}=\frac{1}{4}j$
By symmetry:
$Y_{bus,12}=Y_{bus,21}=Y_{bus,13}=Y_{bus,31}=Y_{bus,23}=Y_{bus,32}=\frac{1}{4}j$

Thus, putting these values back into the 3x3 matrix yields the final Y_bus:
$Y_{bus}=\left[\begin{array}{ccc}-\frac{3}{4}j& \frac{1}{4}j& \frac{1}{4}j\\ \frac{1}{4}j& -\frac{3}{4}j& \frac{1}{4}j\\ \frac{1}{4}j& \frac{1}{4}j& -\frac{3}{4}j\end{array}\right]$