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(C) Let the equivalent resistance between $A$ and $B$ be $X$. Since the circuit is infinite,the part of the circuit to the right of the first $R_1$ an

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$\frac{(R_1 - R_2) + \sqrt{R_1^2 + R_2^2 + 6 R_1 R_2}}{2}$

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(C) Let the equivalent resistance between $A$ and $B$ be $X$. Since the circuit is infinite,the part of the circuit to the right of the first $R_1$ and $R_2$ is equivalent to a resistance $kX$ in parallel with $R_2$,all in series with $R_1$. Thus,the equivalent resistance $X$ is given by: $X = R_1 + \frac{R_2 (kX)}{R_2 + kX}$ $X(R_2 + kX) = R_1(R_2 + kX) + R_2 kX$ $kX^2 + R_2 X = R_1 R_2 + R_1 kX + R_2 kX$ $kX^2 + X(R_2 - R_1 k - R_2 k) - R_1 R_2 = 0$ Given $k = 1/2$: $\frac{1}{2} X^2 + X(R_2 - \frac{R_1}{2} - \frac{R_2}{2}) - R_1 R_2 = 0$ $\frac{1}{2} X^2 + X(\frac{R_2 - R_1}{2}) - R_1 R_2 = 0$ $X^2 + X(R_2 - R_1) - 2 R_1 R_2 = 0$ Using the quadratic formula $X = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $X = \frac{-(R_2 - R_1) + \sqrt{(R_2 - R_1)^2 - 4(1)(-2 R_1 R_2)}}{2}$ $X = \frac{(R_1 - R_2) + \sqrt{R_1^2 + R_2^2 - 2 R_1 R_2 + 8 R_1 R_2}}{2}$ $X = \frac{(R_1 - R_2) + \sqrt{R_1^2 + R_2^2 + 6 R_1 R_2}}{2}$

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