An electric network is shown in Fig. 4. 71. The resistance across AB will be : 

Resistances 3, 4 and 5 are in series. 

and 6ohm resistance is parallel to the other 3 resistances. 

${\mathrm{R}}_{\mathrm{AB}}=\frac{\left(6\mathrm{\Omega }\right)\times \left[\left(3\mathrm{\Omega }\right) + \left(4\mathrm{\Omega }\right) + \left(5\mathrm{\Omega }\right)\right]}{\left(6\mathrm{\Omega }\right) +\left(3\mathrm{\Omega }\right) + \left(4\mathrm{\Omega }\right) + \left(5\mathrm{\Omega }\right)}=\frac{\left(6\mathrm{\Omega }\right) \times \left(12\mathrm{\Omega }\right)}{\left(6\mathrm{\Omega }\right) + \left(12\mathrm{\Omega }\right)}=4\mathrm{\Omega }$D