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. It has been shown in the figure. It appears that current in $\mathrm{AO}$ and $\mathrm{OB}$ remains same. At $\mathrm{O}$, current ${\mathrm{i}}_4$ returns back without any change. If we detach $\mathrm{O}$ from $\mathrm{AB}$ there will not be any change in distribution. Now, CO & OD will be in series hence its total resistance $=2\Omega$ It is in parallel with $\mathrm{CD}$, so, equivalent resistance $=\frac{2\times 1}{2+1}=\frac{2}{3}\Omega$ This equivalent resistance is in series with $\mathrm{AC}$ & $\mathrm{DB}$, so, total resistance $=\frac{2}{3}+1+1=\frac{8}{3}\Omega$
Now $\frac{8}{3}\Omega$ is parallel to $\mathrm{AB}$, that is, $2\Omega$, so total resistance $=\frac{8/3\times 2}{8/3+2}=\frac{16/3}{14/3}=\frac{16}{14}=\frac{8}{7}\Omega$ [Alt:
Between $\mathrm{C}$ & D, the equivalent resistance is given by $1/\mathrm{r}=\frac{1}{{\mathrm{r}}_3}+\frac{1}{\left({\mathrm{r}}_4+{\mathrm{r}}_5\right)}=1+\frac{1}{2}=\frac{3}{2}$ Equivalent resistance along $\mathrm{ACDB}=1+\frac{2}{3}+1=\frac{8}{3}$ $\therefore$ Effective resistance between $\mathrm{A}$ and $\mathrm{B}$ is $\frac{1}{\mathrm{R}}=\frac{3}{8}+\frac{1}{2}=\frac{7}{8}\text{ or }\mathrm{R}=\frac{8}{7}]$