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 \[\frac{P}{Q}=\frac{R}{S}\] ie.,          \[\frac{4}{4}=\frac{4}{4}\] So, the given circuit is a balanced Wheatstones bridge. Hence, the equivalent resistance                                 \[{{R}_{AB}}=\frac{(4+4)\times (4+4)}{(4+4)+(4+4)}\] \[=\frac{8\times 8}{8+8}\] \[=\frac{64}{16}=4\Omega \]