A wall has two layers A and B made of different materials. The thickness of both layers is the same. The thermal conductivity of A and B are K_A and K_B such that K_A = 3K_B . The temperature across the wall is 20°C. In thermal equilibrium 

The wall of two layers A and B are connected in series. 

Then, heat flowing per second across both wall layers are same i.e 

         $$\begin{gathered} H_1=H_2 \\ \frac{K_{A}A∆{\theta }_A}{L}=\frac{K_{B}A∆{\theta }_B}{L} \Rightarrow K_A∆{\theta }_A=K_B∆{\theta }_B \\ \Rightarrow 3K_B∆{\theta }_A=K_B∆{\theta }_B \left(as K_A=3K_B \right) \\ \Rightarrow 3∆{\theta }_A=∆{\theta }_{B ................\left(i\right)} \\ Also ∆{\theta }_A+∆{\theta }_B={20}^{0}C .............\left(ii\right) \end{gathered}$$

         From (i) & (ii)

         $$\begin{gathered} \therefore ∆{\theta }_A+3∆{\theta }_A={20}^{0}C \Rightarrow 4∆{\theta }_A={20}^{0}C \\ \Rightarrow ∆{\theta }_A=5^{0}C \end{gathered}$$