Question

P3: Determine the resistance of a hemispherical grounding electrode of radius a if the ground is not homogeneous, but has a conductivity σ1 for a < r < b, and σ2 for r b, where b > a, and r is the distance from the grounding electrode center. (Ch.10 7 points)

Answer 1

"Conductivity, sigma = { sigma_1, a < r < b sigma_2, b < r < infinity Radius = a If distance is assumed 'r"" , then Surface area i.e, cross sectional area, S = 2 pi r^2 Let Minute thickness in hemisphere surface = delta r Resistance for minute thickness delta r, delta R = delta r/sigma 5 = delta r/sigma 2 pi r^2 (1) For a < r < b, sigma = sigma_1 delta R = delta r/sigma_1 2 pi r^2 integral dR = 1/sigma_1 2 pi integral^b_a 1/r^2 dr R_1 = 1/2 pi sigma_1 [-1/r]^b_a R_1 = 1/2 pi sigma_1 [1/a - 1/b] ... (1) (2) For r > b, sigma = sigma_2 delta R = delta r/sigma_2 2 pi r^2 integral dR = 1/sigma_2 2 pi integral&infinity_b 1/r^2 dr R-2 = 1/sigma_2 2 pi [-1/r]^infinity_b R_2 = 1/2 pi sigma_2 b ... (2) These two resistance R_1 & R_2 will be in parallel, R_eq = R_1 Vertical-bar Vertical-bar R_2 1/R_eq = 1/R_1 + 1/R_2"