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# Consider a compound slab consisting of two different materials having equal thickness and thermal conductivities K and 2K respectively. The equivalent thermal conductivity of the slabs taken in series is

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a
25K
b
3K
c
2K
d
43K
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detailed solution

Correct option is D

resistance in series is Rseries=R1+R2 where R = lKA, here l is thickness or length, A is area of cross section,  K is coefficient of thermal conductivity when 2 slabs of different coefficient of thermal conductivity K,2K ,  having same thickness and area of cross section are taken in series effective thermal conductivity is Kseries, given by, l+lKseries=lK+l2K ⇒2Kseries=32K ⇒Kseries=43K

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