Question

2. Leakage Resistance of a Coax Cable The leakage resistivity of a coax cable's insulation is measured as follows. A known potential difference V is applied across the two conductors in a length L of the cable; the leakage current I that flows between them (through the insulation) is measured; and the resistivity is calculated from the voltage and current. The inner conductor has radius a and the outer conductor has inner radius b. The insulator fills the space between them. The dielectric constant of the insulator is K (a) Find the resistivity ρ of the insulating material in terms of the given quantities (b) What is the capacitance C of a length L of the cable? (c) Calculate the time constant τ RC in terms of K,p,L,b,a, as needed, and universal (d) In a particular case, a -0.8 mm, b-3 mm, and k 3.2. When a 30 meter length of Hint: Use an expression you derived in Coop Hint: Use an expression you derived in Coop, but modified to take κ into account constants. How does it depend on the length L? cable is tested, the resistance measured is V/1-16 G12. Find ρ,C, and τ for this cable plastic jacket dielectric insulator outer conductor inner conductor

Answer 1

a) Measured resistance = R dielectric constant = K Consider a cylindrical layer of radius r thickness dr and length L. Resistance of this layer is dR = rho dr/2 pi rL R = integral dR R = integral^b_a rho dr/2 pi rL R = rho/2 pi L integral^b_a dr/r R = rho/2 pi L [Log(r)]^b_a \r\n R = rho/2 pi L [log[b] - log[a]] R = rho/2 pi L [log(b/a)] b) what is the capacitance of a length of the cable Capacitance of cable is 'C' length of the cable is 'L' V_0 - V_i = -integral^0_i E^bar middot dr^bar where E = lambda/2 pi epsilon_0 r V_0 - V_i = -integral^b_a lambda/2 pi epsilon_0 r dr V_0 - V_i = -lambda/2 pi epsilon_0 integral^b_R 1/r dr \r\n V_0 - V_i = -lambda/2 pi epsilon_0 [log r]^b_a V_0 - V_i = -lambda/2 pi epsilon_0 [log b - log a] V_0 - V_i = -lambda/2 pi epsilon_0 log(b/a) using Q = CV Q = C[Q/2