question_answer

The equivalent resistance across A and B is

A)  \[2\,\Omega \]   

B)                                         \[3\,\Omega \]               

C)  \[4\,\Omega \]               

D)         \[5\,\Omega \]

Correct Answer: C

Solution :

The equivalent circuit can be redrawn as We have, \[\frac{P}{Q}=\frac{R}{S}\] i.e.,        \[\frac{4}{4}=\frac{4}{4}\] So, the given circuit is a balanced Wheatstones bridge. Hence, the equivalent resistance \[{{R}_{AB}}=\frac{(4+4)\times (4+4)}{(4+4)+(4+4)}=\frac{8\times 8}{8+8}=\frac{64}{16}=4\,\Omega \]