Two slabs A and B of different materials but of the same thickness are joined as shown in fig. (2). The thermal conductivities of A and B are K1 and K2 respectively. The thermal conductivity of the composite slab is

The thermal resistance of a slab of length I, area A and thermal conductivity K is given by R=l/(KA)For slab A, R=(l/2)/K1AFor slab B, R2=(l/2)/K2AThe two rods are joined in series. Hence, their thermal resistance R=R1+R2or    lKA=(l/2)K1A+(l/2)K2A  or 1K=121K1+1K2or    K=2K1K2K1+K2