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 slab have equal cross-sectional area. Find the equivalent conductivity of the composite slab.

Explanation

`R_(eq) = R_(1) + R_(2)`
`(L_(1) + L_(2))/(K_(eq)A) = L_(1)/(K_(1)A) +L_(2)/(K_(2)A)`
`K_(eq) = (L_(1)+L_(2))/(L_(1)/K_(1)+L_(2)/K_(2))`