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