A composite slab consists of two slabs  A and  B of different materials but of the same thickness placed in contact as shown in figure. The thermal conductivity of  A and B are  $k_1$ and $k_2$ respectively. A steady temperature difference of ${12}^{\circ }\text{C}$ is maintained across the composite slab. If  $k_1=\frac{k_2}{2}$, the temperature difference across slab  A will be: 

Let temperature of junction be  ${T_{\text{o}}}^{\circ }\text{C}$
Temperature difference across slab  $A={\left(T_1-T_{\text{o}}\right)}^{\circ }\text{C}$
Temperature difference across slab  $B={\left(T_o-T_2\right)}^{\circ }\text{C}$
Since both slabs are connected in series, heat current through both will be equal 

$\begin{array}{l}\frac{k_{1}A\left(T_1-T_0\right)}{\frac{l}{2}}=\frac{k_{2}A\left(T_{\mathrm{o}}-T_2\right)}{\frac{l}{2}}\\ \text{ As }{k}_1=\frac{k_2}{2}\\ \Rightarrow \frac{k_2}{2}\left(T_1-T_{\mathrm{o}}\right)=\left(T_{\mathrm{o}}-T_2\right)k_2\\ \Rightarrow T_1-T_{\mathrm{o}}=2T_{\mathrm{o}}-2T_2\end{array}$

$\begin{array}{l}\Rightarrow T_1+2T_2=3T_0\dots .\left(1\right)\\ \text{ As }{T}_1-T_2={12}^{\circ }\mathrm{C}\\ \Rightarrow T_2=T_1-12\dots ..\left(2\right)\\ \text{ From (1) and (2), we get }\\ \Rightarrow T_1+2\left(T_1-12\right)=3T_{\mathrm{o}}\\ \Rightarrow 3T_1-24=3T_0\\ \Rightarrow T_1-T_0=8^{\circ }\mathrm{C}\end{array}$