In the figure,the value of the resistor to be | vedclass

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(B) Let the equivalent resistance of the entire infinite ladder network between $A$ and $B$ be $R_0$. If we add one more elementary set to the network

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$R(\sqrt{3} - 1)$

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(B) Let the equivalent resistance of the entire infinite ladder network between $A$ and $B$ be $R_0$. If we add one more elementary set to the network,the total resistance remains $R_0$. Consider the first elementary set consisting of two resistors $R$ in series with the upper and lower arms,and one resistor $R$ in parallel. The remaining infinite network to the right of this set also has an equivalent resistance $R_0$. The total resistance $R_{AB}$ is given by the parallel combination of the vertical resistor $R$ and the series combination of the upper arm resistor $R$,the right-side equivalent resistance $R_0$,and the lower arm resistor $R$. Thus,$R_{AB} = \frac{R(R + R_0 + R)}{R + (R + R_0 + R)} = R_0$. $R(2R + R_0) = R_0(3R + R_0)$ $2R^2 + RR_0 = 3RR_0 + R_0^2$ $R_0^2 + 2RR_0 - 2R^2 = 0$. Solving this quadratic equation for $R_0$ using the formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $R_0 = \frac{-2R \pm \sqrt{(2R)^2 - 4(1)(-2R^2)}}{2} = \frac{-2R \pm \sqrt{4R^2 + 8R^2}}{2} = \frac{-2R \pm \sqrt{12R^2}}{2} = \frac{-2R \pm 2R\sqrt{3}}{2} = R(\sqrt{3} - 1)$ (taking the positive value).

In the figure,the value of the resistor to be connected between $C$ and $D$ so that the equivalent resistance of the entire circuit between $A$ and $B$ does not change with the number of elementary sets used is

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