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(B) Let $x$ be the equivalent resistance of the infinite circuit. Since the circuit is infinite,adding one more section of $2 \Omega$ resistors in ser

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$5.5$

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(B) Let $x$ be the equivalent resistance of the infinite circuit. Since the circuit is infinite,adding one more section of $2 \Omega$ resistors in series and parallel does not change the total equivalent resistance $x$.The circuit can be viewed as a $2 \Omega$ resistor in series with the top branch,a $2 \Omega$ resistor in series with the bottom branch,and the equivalent resistance $x$ in parallel with the vertical $2 \Omega$ resistor.Thus,the equivalent resistance $x$ is given by:$x = 2 + 2 + \frac{2x}{2+x}$$x - 4 = \frac{2x}{2+x}$$(x - 4)(x + 2) = 2x$$x^2 + 2x - 4x - 8 = 2x$$x^2 - 4x - 8 = 0$Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$x = \frac{4 \pm \sqrt{(-4)^2 - 4(1)(-8)}}{2(1)}$$x = \frac{4 \pm \sqrt{16 + 32}}{2} = \frac{4 \pm \sqrt{48}}{2}$$x = \frac{4 \pm 4\sqrt{3}}{2} = 2 \pm 2\sqrt{3}$Since resistance must be positive,we take $x = 2 + 2\sqrt{3} \approx 2 + 2(1.732) = 5.464 \Omega$.Rounding to the nearest option,we get $5.5 \Omega$.

The equivalent resistance between the points $A$ and $B$ in the following infinite circuit is (in $\Omega$)

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