Find the input resistance of the circuit between the points A and B of fig. shown

For finding $R_{AB}$, we shall convert the delta CDE of the given figure into its equivalent star as shown in the next figure.

${\mathrm{R}}_{\mathrm{CS}}=\frac{{\mathrm{R}}_1{\mathrm{R}}_2}{{\mathrm{R}}_1+{\mathrm{R}}_2+{\mathrm{R}}_3}=\frac{4\times 8}{4+8+6}=\frac{32}{18}=\frac{16}{9}\mathrm{\Omega }$

${\mathrm{R}}_{\mathrm{DS}}=\frac{{\mathrm{R}}_1{\mathrm{R}}_3}{{\mathrm{R}}_1+{\mathrm{R}}_2+{\mathrm{R}}_3}=\frac{4\times 6}{4+8+6}=\frac{24}{18}=\frac{12}{9}\mathrm{\Omega }$

$\Rightarrow {\mathrm{R}}_{\mathrm{ES}}=\frac{{\mathrm{R}}_2{\mathrm{R}}_3}{{\mathrm{R}}_1+{\mathrm{R}}_2+{\mathrm{R}}_3}=\frac{8\times 6}{4+8+6}=\frac{48}{18}=\frac{24}{9}\mathrm{\Omega }$

$\frac{12}{9}\mathrm{\Omega }, 8\mathrm{\Omega }\text{ are in series. }$

$\text{ Their effective is }\frac{84}{9}\mathrm{\Omega }\text{. }$

$\frac{24}{9}\mathrm{\Omega }, 4\mathrm{\Omega }\text{ are in series. }$

$\text{ Their efective is }\frac{60}{9}\mathrm{\Omega }$

$\frac{84}{9}\mathrm{\Omega }, \frac{60}{9}\mathrm{\Omega }\text{ are in parallel. }$

$\text{ Their effective is }\frac{35}{9}\mathrm{\Omega }\text{. }$

$\text{ This }\frac{35}{9}\mathrm{\Omega }, \frac{16}{9}\mathrm{\Omega }\text{ and }4\mathrm{\Omega }\text{ are in series. }$

${\mathrm{R}}_{\mathrm{AB}}=\frac{35}{9}+\frac{16}{9}+4=\frac{87}{9}\mathrm{\Omega }.$