A coaxial cable has shielding with inner diameter 8.5 mm and conductorwith outer diameter 2.5 mm. Assume that the permittivity of the insulatoris 2.10.(a) (1 marks) If that capacitance of the cable must not be greater than120 pF, then what is the longest that we can make the cable?(b) (1 marks) Calculate the characteristic impedance of this cable.
Question
A coaxial cable has shielding with inner diameter 8.5 mm and conductorwith outer diameter 2.5 mm. Assume that the permittivity of the insulatoris 2.10.(a) (1 marks) If that capacitance of the cable must not be greater than120 pF, then what is the longest that we can make the cable?(b) (1 marks) Calculate the characteristic impedance of this cable.
Solution
(a) The capacitance per unit length for a coaxial cable is given by the formula:
C = (2πε) / ln(b/a)
where: ε is the permittivity of the insulator, a is the radius of the inner conductor, b is the radius of the outer conductor, ln is the natural logarithm.
Given that ε = 2.1ε0 = 2.1 * 8.85 * 10^-12 F/m, a = 2.5 mm / 2 = 1.25 mm = 1.25 * 10^-3 m, b = 8.5 mm / 2 = 4.25 mm = 4.25 * 10^-3 m,
we can substitute these values into the formula to find the capacitance per unit length:
C = (2π * 2.1 * 8.85 * 10^-12 F/m) / ln(4.25 * 10^-3 m / 1.25 * 10^-3 m) C ≈ 50.9 pF/m
If the total capacitance of the cable must not exceed 120 pF, then the maximum length of the cable is given by:
L = C_total / C_per_unit_length L = 120 pF / 50.9 pF/m L ≈ 2.36 m
So, the cable can be at most approximately 2.36 meters long.
(b) The characteristic impedance of a coaxial cable is given by the formula:
Z = (1 / (2π * sqrt(με))) * ln(b/a)
where μ is the permeability of the insulator. For most insulators, μ is approximately equal to μ0, the permeability of free space, which is 4π * 10^-7 H/m.
Substituting the given values into the formula, we get:
Z = (1 / (2π * sqrt(4π * 10^-7 H/m * 2.1 * 8.85 * 10^-12 F/m))) * ln(4.25 * 10^-3 m / 1.25 * 10^-3 m) Z ≈ 50 Ohms
So, the characteristic impedance of the cable is approximately 50 Ohms.
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