The equivalent resistance across $A$ and $B$ is

• A

  $2 \Omega$
• B

  $3 \Omega$
• Correct answer
  C

  $4 \Omega$
• D

  $5 \Omega$

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 Wheatstone's bridge. Hence, the equivalent resistance $R_{A B}=\frac{(4+4) \times(4+4)}{(4+4)+(4+4)}=\frac{8 \times 8}{8+8}=\frac{64}{16}=4 \Omega$

Hence, the correct answer is option: 'C'.