question_answer

In the network shown in the Fig, each resistance is \[1\Omega \]. The effective resistance between A and B is

A) \[\frac{4}{3}\Omega \]

B) \[\frac{3}{2}\Omega \]

C) \[7\,\Omega \]

D)        \[\frac{8}{7}\,\Omega \]

Correct Answer: D

Solution :

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 AO and OB remains same. At O, current \[{{i}_{4}}\]returns back without any change. If we detach O from 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 CD, so, equivalent resistance \[=\frac{2\times 1}{2+1}=\frac{2}{3}\Omega \] This equivalent resistance is in series with AC & DB, so, total resistance     \[=\frac{2}{3}+1+1=\frac{8}{3}\Omega \] Now \[\frac{8}{3}\Omega \] is parallel to AB, that is, \[2\,\Omega \] so total resistance \[=\frac{8/3\times 2}{8/3+2}=\frac{16/3}{14/3}=\frac{16}{14}=\frac{8}{7}\Omega \]