Physics - Current Electricity Question with Solution | TestHub

At$\mathrm{A}$current is distributed and at$\mathrm{B}$currents are collected. Between$\mathrm{A}$ and$\mathrm{B}$, the distribution is symmetrical, as shown in the figure. It appears that current in$\mathrm{AO}$and$\mathrm{OB}$remains the same. At$\mathrm{O}$, current$i_4$returns back without any change. If we detach$\mathrm{O}$from$\mathrm{AB}$there will not be any change in the distribution.
Now,$\mathrm{CO}$&$\mathrm{OD}$will be in series hence it's 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, the total resistance$=\frac{2}{3}+1+1=\frac{8}{3} \mathrm{\Omega }$

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

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