Physics - Current Electricity Question with Solution | TestHub

At A current is distributed and at B currents are collected. Between A and B, the distribution is symmetrical, as shown in the figure. It appears that current in AO and OB remains same. At O, currenti₄returns  back without any change. If we detach O from AB  there will not be any change in the distribution.
Now, CO & OD will be in series hence its total resistance = 2$\mathrm{\Omega }$
It is in parallel with CD, so, equivalent resistance =$\frac{2\times 1}{2+1}=\frac{2}{3}\mathrm{\Omega }$
This equivalent resistance is in series with AC & DB, so , total resistance$=\frac{2}{3}+1+1=\frac{8}{3}\mathrm{\Omega }$

.
Now,$\frac{8}{3}\mathrm{\Omega }$is parallel toAB, that is, 2$\mathrm{\Omega , }$so total resistance

$=\frac{8/3\times 2}{8/3+2}=\frac{16/3}{14/3}=\frac{16}{14}=\frac{8}{7}\mathrm{\Omega }$