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Consider a compound slab consisting of two different materials having equal thickness and thermal conductivities $K$ and $2 K$, respectively. The equivalent thermal conductivity of the slab is:
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Verified Answer
The correct answer is:
$\frac{4}{3} K$
$$
\therefore \frac{2 l}{K_{e g}(A)}=\frac{l}{2 K A}+\frac{l}{K A}
$$
for series connection $\mathrm{R}=\mathrm{R}_1+\mathrm{P}_2$
$$
\Rightarrow K_{e g}=\frac{4}{3} K
$$
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