GATE-2018 (24) :Wyllie – Time average equation

GATE-2016

43. Two horizontal layers have resistivites and thickness of 10ohm m , 5m and 50 ohm m , 10m , respectively. If the two layers are reduced to a single layer, then the coefficient of electrical anisotropy will be......

Solution:

longitudinal conductance ( one layer) S_L = h/ ρ

transverse resistance ( one layer) T_t = h* ρ

longitudinal resistivity (one layer ) ρ_L = h/ S_L

transverse resistivity ( one layer) ρ_t = T_t/ h

For two layer case,

ρ_L =( (h₁+h₂) / ((h₁/ρ₁)+(h₂/ρ₂)) ) -------------(1)

ρ_t = ( (h₁ρ₁ + h₂ρ₂ )/ (h₁+h₂) ) ----------------------(2)

coefficient of anisotropy (λ) = √(ρ_t / ρ_L)

   = √ ( (h₁ρ₁ + h₂ρ₂ )/ (h₁+h₂) ) / ( (h₁+h₂) / ((h₁/ρ₁)+(h₂/ρ₂)) )

   = (1/ h₁ +h₂) *(√( (h₁ρ₁ + h₂ρ₂ )*(h₁/ρ₁)+(h₂/ρ₂) ) ---------(3)

h₁ = 5 m

ρ₁ = 10 Ωm

h₂ = 10 m

ρ₂ = 50 Ωm

Substitute above values in eq (3)

λ= (1/ (5+10) )* (√( (5*10 + 50*10 )*(5/ 10)+(10/ 50) )

λ= (1/ (15) )* (√( (50 + 500)*(1/ 2)+(1/ 5) )

λ= (1/ (15) )* (√( (550)* (7/10) )

λ= (1/ (15) )* (√(385)

λ= (19.6214/ 15)

λ= 1.308