1 \begin{problem*}{26.91}
2 As shown in Fig.~26.83, a network of resistors of resistances $R_1$
3 and $R_2$ extends to infinity toward the right. Prove that the total
4 resistance $R_T$ of the infinite network is equal to
6 R_T = R_1 + \sqrt{R_1^2 + 2R_1R_2}
8 (\emph{Hint:} Since the network is infinite, the resistance of the
9 network to the right of points $c$ and $d$ is also equal to $R_T$.)
13 a-/\/\/-c-/\/\/-+-/\/\/-+-...
19 b-/\/\/-d-/\/\/-+-/\/\/-+-...
25 Following the hint, we note that
27 R_T &= R_1 + \p({\frac{1}{R_2} + \frac{1}{R_T}})^{-1} + R_1 \\
28 R_T - 2R_1 &= \p({\frac{1}{R_2} + \frac{1}{R_T}})^{-1} \\
29 (R_T - 2R_1) \cdot \p({\frac{1}{R_2} + \frac{1}{R_T}}) &= 1 \\
30 (R_T - 2R_1) \cdot (R_T + R_2) &= R_2R_T \\
31 R_T^2 - 2R_1R_T + R_2R_T - 2R_1R_2 &= R_2R_T \\
32 0 &= R_T^2 - 2R_1R_T - 2R_1R_2 \;.
34 Plugging this into the quadratic formula
36 R_T = \frac{2R_1 \pm \sqrt{4R_1^2 - 4\cdot1\cdot(-2R_1R_2)}}{2}
37 = R_1 \pm \sqrt{R_1^2 + 2R_1R_2}
38 = \ans{R_1 + \sqrt{R_1^2 + 2R_1R_2}} \;,
40 which is what we set out to show. Note that we chose the $+$ case
43 R_1 < \sqrt{R_1^2 + 2R_1R_2} \;,
45 and $R_T$ must be greater than zero.