A slab consists of two layers of different materials of the same thickness and having thermal conductivities `K_(1)` and `K_(2)`. The equivalent thermal conductivity of the slab is

Options

A. `K_(1)+K_(2)`
B. `(2K_(1)K_(2))/(K_(1)+K_(2))`
C. `(K_(1)+K_(2))/(K_(1)K_(2))`
D. `K_(1)K_(2)`

Explanation

in parallel
`R_(eq)=(R_(1)R_(2))/(R_(1)+R_(2))`
`R_(1)=(l)/(k_(1)A)R_(2)=(l)/(k_(2)A)`
`implies R_(eq)=(((l)/(A))*((l)/(A))((1)/(k_(1)k_(2))))/((l)/(A)((1)/(k_(1)+k_(2))))`
`impliesR_(eq)=(l)/(A)((1)/(k_(1)+k_(2)))`
`k=(k_(1)+k_(2))/(2)`