Question

Consider a compound slab consisting of two different materials having equal thicknesses and thermal conductivities $K$ and $2K$, respectively.The equivalent thermal conductivity of the slab is

• A. $2/6K$
• B. $\sqrt{2}$K
• C. $3K$
• D. $4/3K$

Answer 1

The correct option is D $4/3K$

$\frac{2\ell }{K_{eq}\left(A\right)}=\frac{\ell }{2KA}+\frac{\ell }{KA}$
(series connection $R=R_1+R_2$)
$\Rightarrow K_{eq}=\frac{4}{3}K$