Seven identical rods of material of thermal conductivity k are connected as shown in the figure. All the rods are of identical length l and cross-sectional area A.
If the Junction A is kept at 100° C  and the Junction B is kept at 0°C , what would be the temperatures of the junction C, D and E (${\mathrm{\theta }}_{\mathrm{C}},{\mathrm{\theta }}_{\mathrm{D}}$ and ${\mathrm{\theta }}_{\mathrm{E}}$) in the steady state?

This problem can be solved like electric current problem.
Let ${\mathrm{R}}_1,{\mathrm{R}}_2,{\mathrm{R}}_3,{\mathrm{R}}_4,{\mathrm{R}}_5,{\mathrm{R}}_6\mathrm{and}{\mathrm{R}}_7$be the rates of heat flow through AE, EB, AC, CD, CE, ED and DB, respectively.

Since ${\mathrm{R}}_1={\mathrm{R}}_2,{\mathrm{\theta }}_{\mathrm{E}}={50}^0\mathrm{C}$
${\mathrm{R}}_5={\mathrm{R}}_6,{\mathrm{R}}_3={\mathrm{R}}_4+{\mathrm{R}}_5={\mathrm{R}}_7$    ....(i)
${\mathrm{R}}_4+{\mathrm{R}}_6={\mathrm{R}}_7 ....\left(\mathrm{ii}\right)$
${\mathrm{R}}_4+{\mathrm{R}}_6={\mathrm{R}}_7$
$\frac{\mathrm{kA}\left({\mathrm{\theta }}_{\mathrm{c}}-50\right)}{\mathrm{l}}=\frac{\mathrm{kA}}{\mathrm{l}}\left(50-{\mathrm{\theta }}_{\mathrm{D}}\right)$
$\begin{array}{l}\frac{\mathrm{kA}}{\mathrm{l}}\left(100-{\mathrm{\theta }}_{\mathrm{c}}\right)=\frac{\mathrm{kA}}{\mathrm{l}}\left({\mathrm{\theta }}_{\mathrm{C}}-50\right)+\frac{\mathrm{kA}}{\mathrm{l}}\left({\mathrm{\theta }}_{\mathrm{C}}-{\mathrm{\theta }}_{\mathrm{D}}\right)=\frac{\mathrm{kA}}{\mathrm{l}}{\mathrm{\theta }}_{\mathrm{D}}\\ {\mathrm{\theta }}_{\mathrm{C}}+{\mathrm{\theta }}_{\mathrm{D}}=100\\ 2{\mathrm{\theta }}_{\mathrm{C}}-2{\mathrm{\theta }}_{\mathrm{D}}=50\Rightarrow {\mathrm{\theta }}_{\mathrm{C}}=62.5^{\circ }\mathrm{C}\\ {\mathrm{\theta }}_{\mathrm{D}}=37.5^0\mathrm{C}\therefore {\mathrm{\theta }}_{\mathrm{C}}>{\mathrm{\theta }}_{\mathrm{E}}>{\mathrm{\theta }}_{\mathrm{D}}\end{array}$