Question

The bus admittance matrix of the network shown in the given figure, for which the marked parameters are per unit impedance, is _______.

 

• A. \( \begin{bmatrix} 0.3 & -0.2 \\ -0.2 & 0.2 \end{bmatrix}^{-1} \)
• B. \( \begin{bmatrix} 0.3 & 0.2 \\ 0.2 & 0.2 \end{bmatrix} \)
• C. \( \begin{bmatrix} 0.3 & -0.2 \\ -0.2 & 0.2 \end{bmatrix} \)
• D. \( \begin{bmatrix} 15 & -5 \\ -5 & 5 \end{bmatrix} \)

Hint:

Convert Z to Y: \( Y = 1/Z \). Diagonal = sum of connections. Off-diagonal = –line admittance.

Answer 1

The Correct Option is D

Step 1: Convert Impedance to Admittance
Given:
- Line 1–0: \( Z = 0.1 \Rightarrow Y = \frac{1}{0.1} = 10 \)
- Line 1–2: \( Z = 0.2 \Rightarrow Y = \frac{1}{0.2} = 5 \)

Step 2: Construct Bus Admittance Matrix (Y_bus)
Using admittance matrix formula for a 2-bus system: \[ Y_{\text{bus}} = \begin{bmatrix} Y_{11} & Y_{12} \\ Y_{21} & Y_{22} \end{bmatrix} \] Where:
- \( Y_{11} = Y_{10} + Y_{12} = 10 + 5 = 15 \)
- \( Y_{12} = Y_{21} = -5 \)
- \( Y_{22} = 5 \)

Thus, \[ Y_{\text{bus}} = \begin{bmatrix} 15 & -5 \\ -5 & 5 \end{bmatrix} \]
Conclusion:
Option (4) correctly represents the bus admittance matrix in per unit.