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

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Solution:

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$\frac{2 \ell}{ K _{ eq }( A )}=\frac{\ell}{2 KA }+\frac{\ell}{ KA }$
$\left(\right.$ series connection $\left.R = R _{1}+ R _{2}\right)$
$\Rightarrow K _{ eq }= \frac{4}{3}\, K$