question_answer

In the network shown in figure 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 :

  By symmetry, currents \[{{i}_{1}}\,\,and\,\,{{i}_{2}}\] form A is same as \[{{i}_{1}}\,\,and\,\,{{i}_{2}}\] reaching B. As the same current is flowing from A to B and O to B. O can be treated as detached from AB. Now, CO and 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 and DB, so \[\operatorname{total} resistance = \frac{2}{3} +1 +1 = \frac{8}{3}\,\Omega \] Now, \[\frac{8}{3}\,\Omega \] is parallel to AB that is \[2\,\Omega \]. So, total resistance \[=\,\,\frac{(8/3)\times 2}{(8/3)+2}=\frac{\frac{16}{3}}{\frac{14}{3}}=\frac{16}{14}=\frac{8}{7}\,\Omega \]