#### Question

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

#### Solution

It is equivalent to the series combination of 2 resistors.

∴ *R*_{S} = *R*_{1} +*R*_{2}

Resistance of a conducting slab, `R = l/(KA)`

`(L_1 + L_2)/(K_SA)=(L_1)/(K_1A)+ (L_2)/(K_2A)`

`(L_1 +L_2)/(K_s) = L_1/K_1 + L_2/K_2`

`(L_1 + L_2 )/K_s = (L_1K_2 + L_2K_1)/(K_1 xx K_2)`

`K_s = ((L_1 + L_2) (K_1 K_2))/(L_1k_2 + L_2K_1 `

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Solution A Composite Slab is Prepared by Pasting Two Plates of Thickness L1 and L2 and Thermal Conductivites K1 and K2. the Slabs Have Equal Cross-sectional Area. Find the Equivalent Conductivity of the Concept: Thermal Expansion of Solids.