A composite slab is prepared by pasting two plates of thicknesses $L_{1}$ and $L_{2}$ and thermal conductivities $K_{1}$ and $K_{2}$ . The slabs have equal cross-sectional area. Find the equivalent conductivity of the slab.

K_{e q} = \frac{L_{1} + L_{2}}{\frac{L_{1}}{K_{1}} + \frac{L_{2}}{K_{2}}}

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K_{e q} = \frac{L_{1} + L_{2}}{K_{1} K_{2}}

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K_{e q} = \frac{L_{1}}{K_{1} + K_{2}} + \frac{L_{2}}{K_{1} + K_{2}}

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K_{e q} = \frac{L_{1} L_{2}}{L_{1} + L_{2}}

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Solution

The correct option is A $K_{e q} = \frac{L_{1} + L_{2}}{\frac{L_{1}}{K_{1}} + \frac{L_{2}}{K_{2}}}$
The rate of thermal conduction through the first slab is given as
$Q_{1} = K_{1} A \frac{T - T_{1}}{L_{1}}$
The rate of thermal conduction through the second slab is given as
$Q_{2} = K_{2} A \frac{T_{2} - T}{L_{2}}$
Since they are connected end to end,
$Q = Q_{1} = Q_{2}$

$⟹ K_{e q} A \frac{T_{2} - T_{1}}{L_{1} + L_{2}} = K_{1} A \frac{T - T_{1}}{L_{1}} = K_{2} A \frac{T_{2} - T}{L_{2}}$

From the second and third expressions,
$T = \frac{\frac{K_{2} T_{2}}{L_{2}} + \frac{K_{1} T_{1}}{L_{1}}}{\frac{K_{1}}{L_{1}} + \frac{K_{2}}{L_{2}}}$

Using this value in first and second expressions gives,
$K_{e q} = \frac{L_{1} + L_{2}}{\frac{L_{1}}{K_{1}} + \frac{L_{2}}{K_{2}}}$

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A composite slab is prepared by pasting two plates of thicknesses L1 and L2 and thermal conductivities K1 and K2. The slabs have equal cross-sectional area. Find the equivalent conductivity of the slab.

A composite slab is prepared by pasting two plates of thicknesses $L_{1}$ and $L_{2}$ and thermal conductivities $K_{1}$ and $K_{2}$ . The slabs have equal cross-sectional area. Find the equivalent conductivity of the slab.