Question

Calculate the distributed inductance of a two-wire (ribbon) cable transmission line when the frequency is 86.5 MHz. The radius of the copper conductor is 1.42 mm and the distance between them is 6.10 mm. Following are the properties of the dielectric material: Relative permittivity $\epsilon_r$; 2.40 Relative permeability $\mu_r$; 1 Conductivity $\sigma_d$ 2.51 x 10$^{-6}$ S/m. And the properties of the conductor are: Conductivity $\sigma_c$ 5.80 x 10$^7$ S/m. Relative permeability $\mu_r$; 1 The absolute permittivity $\epsilon_o$ = 8.854 x 10$^{-12}$ F/m and the absolute permeability $\mu_o$ = 4$\pi$ x 10$^{-7}$ H/m Enter the answer in m$\Omega$/m rounded to three significant digits.

          Calculate the distributed inductance of a two-wire (ribbon) cable transmission line when the frequency is
86.5 MHz. The radius of the copper conductor is 1.42 mm and the distance between them is 6.10 mm.
Following are the properties of the dielectric material:
Relative permittivity $\epsilon_r$; 2.40
Relative permeability $\mu_r$; 1
Conductivity $\sigma_d$ 2.51 x 10$^{-6}$ S/m.
And the properties of the conductor are:
Conductivity $\sigma_c$ 5.80 x 10$^7$ S/m.
Relative permeability $\mu_r$; 1
The absolute permittivity $\epsilon_o$ = 8.854 x 10$^{-12}$ F/m and the absolute permeability $\mu_o$ = 4$\pi$ x 10$^{-7}$ H/m
Enter the answer in m$\Omega$/m rounded to three significant digits.

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Calculate the distributed inductance of a two-wire (ribbon) cable transmission line when the frequency is
86.5 MHz. The radius of the copper conductor is 1.42 mm and the distance between them is 6.10 mm.
Following are the properties of the dielectric material:
Relative permittivity ; 2.40
Relative permeability ; 1
Conductivity 2.51 x 10^-6 S/m.
And the properties of the conductor are:
Conductivity 5.80 x 10^7 S/m.
Relative permeability ; 1
The absolute permittivity = 8.854 x 10^-12 F/m and the absolute permeability = 4π x 10^-7 H/m
Enter the answer in mΩ/m rounded to three significant digits.

Added by Alexandra S.

University Physics with Modern Physics

Hugh D. Young 14th Edition

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Solved by Expert Patha Sharma

02/27/2022

Step 1

The inductance per unit length of a single wire transmission line can be calculated using the formula: L = \frac{{\mu_0 \mu_r}}{{2\pi}} \ln\left(\frac{{D}}{{d}}\right) where L is the inductance per unit length, \mu_0 is the absolute permeability, \mu_r is the Show more…

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Calculate the distributed inductance of a two-wire (ribbon) cable transmission line when the frequency is 80.5MHz. The radius of the copper conductor is 1.42mm and the distance between them is 6.10mm. Following are the properties of the dielectric material: Relative permittivity epsi lon_(r);2.40 Relative permeability mu _(r);1 Conductivity sigma _(d)2.51 imes 10^(-6)(S)/(m). And the properties of the conductor are: Conductivity sigma _(c)5.80 imes 10^(7)(S)/(m). Relative permeability mu _(r);1 The absolute permittivity epsi lon_(o)=8.854 imes 10^(-12)(F)/(m) and the absolute permeability mu _(0)=4pi imes 10^(-7)(H)/(m) Enter the answer in m(Omega )/(m) rounded to three significant digits. Calculate the distributed inductance of a two-wire (ribbon)cable transmission line when the frequency is 8G.5 MHz. The radius of the copper conductor is 1.42 mm and the distance between them is 6.10 mm. Following are the properties of the dielectric material: Relative permittivity cr2.40 Relative permeability r1 Conductivity O 2.51 10-6 S/m. And the properties of the conductor are Conductivity 5.80 x 107 S/m. Relative permeability r;1 The absolute permittivity Co = 8.854 x 10-12 F/m and the absolute permeability o = 4rt x 10-7 H/m Enter the answer in mQ/m rounded to three significant digits

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