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A coaxial cable with an inner diameter of 1 mm and outer diameter of 2.4 mm is filled with a dielectric of relative permittivity 10.89. Given \(\mu_0=4\pi\times10^{-7}\,\text{H/m}\) and \(\varepsilon_0=\dfrac{10^{-9}}{36\pi}\,\text{F/m}\), the characteristic impedance of the cable is
- 33Ω
- 100Ω
- 143.3Ω
- 43.4Ω
Correct answer: 33Ω
Solution
For a coaxial cable, \(Z_0=\dfrac{60}{\sqrt{\varepsilon_r}}\ln(b/a)\), where \(a\) and \(b\) are the inner and outer radii. Here \(a=0.5\) mm and \(b=1.2\) mm, so \(b/a=2.4\); substituting \(\varepsilon_r=10.89\) gives approximately \(33\,\Omega\).
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