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

Options

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

Answer

B

Explanation

`k=(k_(1)A_(1)+k_(2)A_(2))/(A_(1)+A_(2))`
For two slabs ofequal thickness and hence equal area,
`k+(k_(1)+k_(2))/(2)`