The equivalent resistance between the points $A$ and $B$ in the following circuit is

• A

  $0.5 \Omega$
• Correct answer
  B

  $5.5 \Omega$
• C

  $0.05 \Omega$
• D

  $5 \Omega$

Solution

Let $x$ be the equivalent resistance of the circuit, then
$\begin{array}{ll}\therefore \quad & R_{A B}=x \\ \Rightarrow & \frac{x \times 2}{2+x}+2+2=x \\ \Rightarrow & \frac{2 x}{x+2}=x-4 \\ \Rightarrow \quad & 2 x=(x-4)(x+2) \\ \Rightarrow \quad & 2 x=x^2+2 x-4 x-8 \\ \Rightarrow \quad & x^2-4 x-8=0 \\ \Rightarrow \quad & \frac{-(-4) \pm \sqrt{(-4)^2-4 \times 1 \times(-8)}}{2 \times 1} \\ & =\frac{4 \pm \sqrt{48}}{2}=\frac{4 \pm 4 \sqrt{3}}{2} \\ & =2+2 \sqrt{3}=5.46 \Omega \approx 5.5 \Omega\end{array}$

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