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.

The rate of thermal conduction through the first slab is given as
Q1=K1A(T−T1)/L1
The rate of thermal conduction through the second slab is given as
Q2=K2A(T2−T)/L2
Since they are connected end to end, Q=Q1=Q2
⇒KeqA(T2−T1)/(L1+L2) = K1A(T−T1)/L1 = K2A(T2−T)/L2
From the second and third expressions,
T = (K2T2L2+K1T1L1)/(K1L1+K2L2)
Using this value in first and second expressions gives,
Keq=(L1+L2)/(L1/K1+L2/K2)