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Chapter 13: Problem 2

Find \(R, L, C\), and \(G\) for a coaxial cable with \(a=0.25 \mathrm{~mm}, b=2.50 \mathrm{~mm}\), \(c=3.30 \mathrm{~mm}, \epsilon_{r}=2.0, \mu_{r}=1, \sigma_{c}=1.0 \times 10^{7} \mathrm{~S} / \mathrm{m}, \sigma=1.0 \times 10^{-5} \mathrm{~S} / \mathrm{m}\) and \(f=300 \mathrm{MHz}\).

Short Answer

Expert verified
Based on the given parameters and formulas, calculate the values of R, L, C, and G for the coaxial cable.

Step by step solution

01

Calculate Resistance (R) per unit length

To find the resistance (R) per unit length, we will use the formula: $$ R = \frac{1}{2 \pi a \sigma_c} \times \left(\sqrt{1 + \frac{4 \pi^2 a^2 f^2}{\sigma_c^2}} - 1\right) $$ Now, plug in the given values: \(a = 0.25 \times 10^{-3} \mathrm{m}, \sigma_c = 1.0 \times 10^7 \mathrm{S/m},\) and \(f = 300 \times 10^6 \mathrm{Hz}\).
02

Calculate Inductance (L) per unit length

To calculate the inductance (L) per unit length, we will use the formula: $$ L = \frac{\mu_{r} \mu_{0}}{2 \pi} \ln{\frac{b}{a}} $$ Using the given values: \(a = 0.25 \times 10^{-3} \mathrm{m}, b = 2.5 \times 10^{-3} \mathrm{m},\) and \(\mu_r = 1.\) Also, note that \(\mu_0 = 4 \pi \times 10^{-7} \mathrm{T}\cdot \mathrm{m/A}\).
03

Calculate Capacitance (C) per unit length

To calculate the capacitance (C) per unit length, we will use the formula: $$ C = \frac{2 \pi \epsilon_r \epsilon_0}{\ln{\frac{b}{a}}} $$ Using the given values: \(a = 0.25 \times 10^{-3} \mathrm{m}, b = 2.5 \times 10^{-3} \mathrm{m},\) and \(\epsilon_r = 2.\) Also, note that \(\epsilon_0 = 8.85 \times 10^{-12} \mathrm{F/m}\).
04

Calculate Conductance (G) per unit length

To calculate the conductance (G) per unit length, we will use the formula: $$ G = \frac{1}{2 \pi b \sigma} \times \left(\sqrt{1 + \frac{4 \pi^2 b^2 f^2}{\sigma^2}} - 1\right) $$ Now, plug in the given values: \(b = 2.5 \times 10^{-3} \mathrm{m}, \sigma = 1.0 \times 10^{-5} \mathrm{S/m},\) and \(f = 300 \times 10^6 \mathrm{Hz}\). After calculating all these values, you will get the results for R, L, C, and G.

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Most popular questions from this chapter

For the guide of Problem \(13.14\), and at the \(32 \mathrm{GHz}\) frequency, determine the difference between the group delays of the highest-order mode (TE or TM) and the TEM mode. Assume a propagation distance of \(10 \mathrm{~cm}\).

A rectangular waveguide has dimensions \(a=6 \mathrm{~cm}\) and \(b=4 \mathrm{~cm} .(a)\) Over what range of frequencies will the guide operate single mode? \((b)\) Over what frequency range will the guide support both \(\mathrm{TE}_{10}\) and \(\mathrm{TE}_{01}\) modes and no others?

Two rectangular waveguides are joined end-to-end. The guides have identical dimensions, where \(a=2 b\). One guide is air-filled; the other is filled with a lossless dielectric characterized by \(\epsilon_{r}^{\prime} .(a)\) Determine the maximum allowable value of \(\epsilon_{r}^{\prime}\) such that single-mode operation can be simultaneously assured in both guides at some frequency. \((b)\) Write an expression for the frequency range over which single-mode operation will occur in both guides; your answer should be in terms of \(\epsilon_{r}^{\prime}\), guide dimensions as needed, and other known constants.

The cutoff frequency of the \(m=1 \mathrm{TE}\) and TM modes in an air-filled parallel-plate guide is known to be \(f_{c 1}=7.5 \mathrm{GHz}\). The guide is used at wavelength \(\lambda=1.5 \mathrm{~cm}\). Find the group velocity of the \(m=2 \mathrm{TE}\) and \(\mathrm{TM}\) modes.

Two aluminum-clad steel conductors are used to construct a two-wire transmission line. Let \(\sigma_{\mathrm{Al}}=3.8 \times 10^{7} \mathrm{~S} / \mathrm{m}, \sigma_{\mathrm{St}}=5 \times 10^{6} \mathrm{~S} / \mathrm{m}\), and \(\mu_{\mathrm{St}}=100 \mu \mathrm{H} / \mathrm{m}\). The radius of the steel wire is \(0.5 \mathrm{in.}\), and the aluminum coating is \(0.05\) in. thick. The dielectric is air, and the center-to-center wire separation is 4 in. Find \(C, L, G\), and \(R\) for the line at \(10 \mathrm{MHz}\).
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