A composite slab consists of two plates of thicknesses L₁ and L₂ and thermal conductivities K₁ and K₂. The cross-sectional areas are equal. The equivalent thermal conductivity is:

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$\frac{L_{1} L_{2} (K_{1} + K_{2})}{(K_{2} L_{1} + K_{1} L_{2})}$

$\frac{K_{2} L_{1} + K_{1} L_{2}}{L_{1} L_{2} (K_{1} + K_{2})}$

$\frac{K_{1} K_{2} (L_{1} + L_{2})}{K_{2} L_{1} + K_{1} L_{2}}$

$\frac{K_{2} L_{1} + K_{1} L_{2}}{K_{1} K_{2} (L_{1} + L_{2})}$

answer is A.

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Detailed Solution

In series connection
$K_{eq} = \frac{\sum l_{i}}{\sum \frac{l_{i}}{K_{i}}} = \frac{L_{1} + L_{2}}{\frac{L_{1}}{K_{1}} + \frac{L_{2}}{K_{2}}} = \frac{K_{1} K_{2} (L_{1} + L_{2})}{K_{2} L_{1} + K_{1} L_{2}}$

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