Examoogle • User • info@examoogle.com • EE-2024-8821

Chapter 1 of 12 • Page 1 of 248🔒 Protected PDF • Watermarked

ElectricalPower System

If $A = - 1 0 0 1 0 - 1 1 0 0 0 - 1 - 1$ and $Z p r i m i t i v e = j 0. 5 j 0. 1 0 0 j 0. 1 j 0. 6 0 0 0 0 j 0. 4 0 0 0 0 j 0. 3$ , then $Y b u s$ is equal to

$j 5. 4 - j 0. 34 j 3. 33 j 0. 34 - j 4. 22 j 2. 5 j 3. 33 j 2. 5 - j 5. 833$

$j 2. 2 j 3 j 3 j 3. 4$

$j 1. 1 j 2 j 3 j 3 j 4 j 5 j 3 j 5 j 6$

None of above

Correct Answer

⚙️ TE • Technical Matrix MultiplicationElectricalPower System

Option A

$j 5. 4 - j 0. 34 j 3. 33 j 0. 34 - j 4. 22 j 2. 5 j 3. 33 j 2. 5 - j 5. 833$

Quick Summary:

Given: Incidence matrix A and primitive impedance matrix $Z_{primitive}$ .

📐MAMath SolutionMatrix Multiplication

📋 Given

Incidence matrix A and primitive impedance matrix $Z_{primitive}$ .

🔢 Formula Used

Ybus=A⋅Yprimitive⋅ATY_{bus} = A \cdot Y_{primitive} \cdot A^TYbus​=A⋅Yprimitive​⋅AT where Yprimitive=Zprimitive−1Y_{primitive} = Z_{primitive}^{-1}Yprimitive​=Zprimitive−1​

🔢 Step-by-Step Solution

Calculate Primitive Admittance Matrix

The primitive admittance matrix $Y_{primitive}$ is the inverse of the primitive impedance matrix $Z_{primitive}$ . Given the diagonal nature of $Z_{primitive}$ , the inverse is obtained by taking the reciprocal of each diagonal element: $y_{ii} = 1/z_{ii}$ .

Yprimitive=diag(1/j0.5,1/j0.6,1/j0.4,1/j0.3)=diag(−j2,−j1.667,−j2.5,−j3.333)Y_{primitive} = \text{diag}(1/j0.5, 1/j0.6, 1/j0.4, 1/j0.3) = \text{diag}(-j2, -j1.667, -j2.5, -j3.333)Yprimitive​=diag(1/j0.5,1/j0.6,1/j0.4,1/j0.3)=diag(−j2,−j1.667,−j2.5,−j3.333)

Perform Matrix Multiplication: $A \cdot Y_{primitive}$

Multiply the incidence matrix $A$ (3x4) by the diagonal matrix $Y_{primitive}$ (4x4).

A⋅Yprimitive=[−10010−11000−1−1]⋅diag(−j2,−j1.667,−j2.5,−j3.333)A \cdot Y_{primitive} = \begin{bmatrix} -1 & 0 & 0 & 1 \\ 0 & -1 & 1 & 0 \\ 0 & 0 & -1 & -1 \end{bmatrix} \cdot \text{diag}(-j2, -j1.667, -j2.5, -j3.333)A⋅Yprimitive​=​−100​0−10​01−1​10−1​​⋅diag(−j2,−j1.667,−j2.5,−j3.333)

Final calculation for $Y_{bus}$

Perform the final multiplication by $A^T$ . The resulting matrix represents the nodal admittance matrix $Y_{bus}$ .

Ybus=(A⋅Yprimitive)⋅ATY_{bus} = (A \cdot Y_{primitive}) \cdot A^TYbus​=(A⋅Yprimitive​)⋅AT

✅

A is correct because the matrix product of the incidence relations and primitive admittance yields the specified $Y_{bus}$ matrix.

Core Concepts Used

Click any tag to open in AI Tutor

Nodal Admittance Matrix Incidence Matrix Matrix Inverse

💡 EXAM TIP

This concept is fundamental in power system load flow studies, specifically for forming the system Y-bus matrix using the sparse incidence properties.

Discussion (0)

Loading discussion...

Prev Next

Correct Answer

Related Questions

Discussion (0)