Fundamentals of Electromagnetics with Engineering Applications
Fundamentals of Electromagnetics with Engineering Applications
5th Edition
ISBN: 9780471263555
Author: Stuart M. Wentworth
Publisher: John Wiley & Sons
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Chapter 6, Problem 6.1P
To determine

The distributed parameters of RG-223/U coaxial cable at the given frequency.

Expert Solution & Answer
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Answer to Problem 6.1P

The distributed resistance is 3.32 Ω/m , distributed inductance is 223nH/m , distributed conductance is 560×1018S/m and distributed capacitance is 112pF/m .

Explanation of Solution

Given:

The radius a of inner conductor is 0.47 mm and the outer conductor has inner radius b of 1.435 mm .The dielectric material is polyethylene and the conductor is copper. The given frequency is 800 MHz .

Concept used:

The series distributed resistance parameter R is shown below.

  R=12π(1a+1b)πfμσc   ........ (1)

Here, σc is the conductor’s conductance.

The series distributed inductance parameter L is shown below.

  L=μ2πln(ba)   ........ (2)

The shunt distributed conductance parameter G is shown below.

  G=2πσdln(ba)   ........ (3)

Here, σd is the dielectric’s conductance.

The shunt distributed capacitance parameter C is shown below.

  C=2πεln(ba)

Calculation:

The conductance σc of copper is 5.8×107S/m , conductance σd of polyethylene is 1016S/m and relative permittivity εr of polyethylene is 2.26 .

Substitute 0.47×103 for a , 1.435×103 for b , 8×108 for f , 4π×107 for μ0 and 5.8×107 for σc in equation (1).

  R=12π(1 0.47× 10 3 +1 1.435× 10 3 ) π( 8× 10 8 )( 4π× 10 7 ) 5.8× 10 7 3.32 Ω/m

Therefore, the distributed resistance parameter R is 3.32 Ω/m .

Substitute 4π×107 for μ0 , 0.47×103 for a and 1.435×103 for b in equation (2).

  L=4π× 10 72πln( 1.435× 10 3 0.47× 10 3 )=2.23×107223×109H/m

Therefore, the distributed inductance parameter L is 223nH/m .

Substitute 1016 for σd , 0.47×103 for a and 1.435×103 for b in equation (3).

  G=2π× 10 16ln( 1.435× 10 3 0.47× 10 3 )=5.62×1016560×1018S/m

Therefore, the distributed conductance parameter G is 560×1018S/m .

The distributed conductance in dielectric can be written as,

  C=2πεrε0ln(ba)

Substitute 2.26 for εr , 8.854×1012 for ε0 , 0.47×103 for a and 1.435×103 for b in above equation.

  C=2π( 2.26)( 8.854× 10 12 )ln( 1.435× 10 3 0.47× 10 3 )112×1012F/m

Therefore, the distributed capacitance parameter C is 112pF/m .

Conclusion:

Thus, the distributed resistance is 3.32 Ω/m , distributed inductance is 223nH/m , distributed conductance is 560×1018S/m and distributed capacitance is 112pF/m .

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Chapter 6 Solutions

Fundamentals of Electromagnetics with Engineering Applications

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