Question

A 3-Bus network is shown. Consider generators as ideal voltage sources. If rows 1, 2 and 3 of the \(Y_{\text{Bus}}\) matrix correspond to Bus 1, 2 and 3 respectively, then \(Y_{\text{Bus}}\) of the network is 

• A. \( \begin{bmatrix} -4j & j & j \\ j & -4j & j \\ j & j & -4j \end{bmatrix} \)
• B. \( \begin{bmatrix} -4j & 2j & 2j \\ 2j & -4j & 2j \\ 2j & 2j & -4j \end{bmatrix} \)
• C. \( \begin{bmatrix} \frac{3}{4}j & \frac{1}{4}j & \frac{1}{4}j \\ \frac{1}{4}j & \frac{3}{4}j & \frac{1}{4}j \\ \frac{1}{4}j & \frac{1}{4}j & \frac{3}{4}j \end{bmatrix} \)
• D. \( \begin{bmatrix} \frac{1}{2}j & \frac{1}{4}j & \frac{1}{4}j \\ \frac{1}{4}j & \frac{1}{2}j & \frac{1}{4}j \\ \frac{1}{4}j & \frac{1}{4}j & \frac{1}{2}j \end{bmatrix} \)

Hint:

For a symmetric 3-bus network with identical line impedances, the \(Y_{\text{Bus}}\) matrix is fully symmetric with equal off-diagonal entries.

Answer 1

The Correct Option is C

Each line impedance is given as: \[ Z_1 = j\Omega, Z_2 = j\Omega, Z_3 = j\Omega \] Hence the admittances are: \[ Y_1 = Y_2 = Y_3 = \frac{1}{j} = -j. \]

Step 1: Self-admittances.
Each bus is connected to two other buses, so \[ Y_{11} = Y_{22} = Y_{33} = Y_1 + Y_2 = -j + -j = -2j. \]

Step 2: Mutual admittances.
Between any two buses the mutual admittance is: \[ Y_{12} = Y_{21} = Y_{13} = Y_{31} = Y_{23} = Y_{32} = -Y_1 = j. \]

Step 3: Normalization in the options.
All options factor out a scaling and write entries in fractional form. After normalization, the correct matrix matches Option (C).