Consider a compound slab consisting of two different material having equal thickness and thermal conductivities `K` and `2K` respectively. The equivalent thermal conductivity of the slab is

Options

A. `sqrt(2)`
B. 3k
C. `(4)/(3) k`
D. `(2)/(3) k`

Answer

C

Explanation

Equivalent thermal circuit
`1/(R_(eq))=1/(R_(1))+1/(R_(2))rArr (K_(eq)xx2A)/l=(KA)/l+(2KA)/lrArr K_(eq) =3/2 K`