Elements of Electromagnetics
Elements of Electromagnetics
7th Edition
ISBN: 9780190698669
Author: Sadiku
Publisher: Oxford University Press
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Chapter 14, Problem 29P
To determine

Calculate the global coefficient matrix of the two-element mesh.

Expert Solution & Answer
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Answer to Problem 29P

The global coefficient matrix is [0.83330.66700.16670.6671.45830.3750.416700.3750.6250.250.16670.41670.250.833]_.

Explanation of Solution

Calculation:

Refer to Figure 14-62 in the textbook for the two-element mesh.

For element 1, local numbering 1-2-3 corresponds to global numbering 1-2-4.

From Figure 14.62, the given values for element 1 are,

x1=0,x2=1,x3=2,y1=1,y2=0,andy3=2.

Calculate the value of P1.

P1=y2y3=02=2

Calculate the value of P2.

P2=y3y1=21=1

Calculate the value of P3.

P3=y1y2=10=1

Calculate the value of Q1.

Q1=x3x2=21=1

Calculate the value of Q2.

Q2=x1x3=02=2

Calculate the value of Q3.

Q3=x2x1=10=1

Consider the expression for the area A.

A=12(P2Q3P3Q2)        (1)

Substitute 1 for P2, 1 for P3, 2 for Q2, and 1 for Q3 in Equation (1).

A=12[((1)(1)(1)(2))]=32

Write the expression for co-efficient Cij.

Cij=14A[PiPj+QiQj]        (2)

By using Equation (2), the coefficient values C11,C12,C13,C21,C22,C23,C31,C32,andC33 are calculated as 56,46,16,46,56,16,16,16,and26 respectively.

Write the expression of the element coefficient matrix C(1).

C(1)=[C11C12C13C21C22C23C31C32C33]

Substitute 56 for C11, 46 for C12, 16 for C13, 46 for C21, 56 for C22, 16 for C23, 16 for C31, 16 for C32, and 26 for C33.

C(1)=[564616465616161626]

For element 2, local numbering 1-2-3 corresponds to global numbering 4-2-3.

From Figure 14.62, the given values for element 2 are,

x1=2,x2=1,x3=3,y1=2,y2=0,andy3=0.

Calculate the value of P1.

P1=y2y3=00=0

Calculate the value of P2.

P2=y3y1=02=2

Calculate the value of P3.

P3=y1y2=20=2

Calculate the value of Q1.

Q1=x3x2=31=2

Calculate the value of Q2.

Q2=x1x3=23=1

Calculate the value of Q3.

Q3=x2x1=12=1

Substitute 2 for P2, 2 for P3, 1 for Q2, and 1 for Q3 in Equation (1) to find area A.

A=12[((2)(1)(2)(1))]=2

By using Equation (2), the coefficient values C11,C12,C13,C21,C22,C23,C31,C32,andC33 are calculated as 48,28,28,28,58,38,28,38,and58 respectively.

Write the expression of the element coefficient matrix C(2).

C(2)=[C11C12C13C21C22C23C31C32C33]

Substitute 48 for C11, 28 for C12, 28 for C13, 28 for C21, 58 for C22, 38 for C23, 28 for C31, 38 for C32, and 58 for C33.

C(2)=[482828285838283858]

Calculate the global coefficient matrix.

C=[C11(1)C12(1)0C13(1)C12(1)C22(1)+C22(2)C23(2)C23(1)+C21(2)0C23(2)C33(2)C31(2)C13(1)C23(1)+C21(2)C31(2)C33(1)+C11(2)]=[0.83330.66700.16670.6671.45830.3750.416700.3750.6250.250.16670.41670.250.833]

Conclusion:

Thus, the global coefficient matrix is [0.83330.66700.16670.6671.45830.3750.416700.3750.6250.250.16670.41670.250.833]_.

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