question_answer

In the given figure, equivalent resistance between A and B will be                                           [CBSE PMT 2000]

A)         \[\frac{14}{3}\,\,\Omega \]

B)         \[\frac{3}{14}\,\,\Omega \]

C)         \[\frac{9}{14}\,\,\Omega \]

D)         \[\frac{14}{9}\,\,\Omega \]

Correct Answer: A

Solution :

                                         The given circuit is a balanced Wheatstone bridge, hence it can be redrawn as follows \[\Rightarrow \,\,\,\,\,\,{{R}_{eq}}=\frac{7\times 14}{(7+14)}=\frac{14}{3}\Omega \].