A) Infinite
B) \[2\,\Omega \]
C) \[\frac{1+\sqrt{5}}{2}\Omega \]
D) Zero
Correct Answer: C
Solution :
Similar to Q. No. 30. By formula \[R={{R}_{1}}+\frac{{{R}_{2}}\times R}{{{R}_{2}}+R}\] \ \[R=1+\frac{1\times R}{1+R}\] Þ \[{{R}^{2}}+R=1+R+R\] Þ \[{{R}^{2}}-R-1=0\] or \[R=\frac{1\pm \sqrt{1+4}}{2}\] \[=\frac{1\pm \sqrt{5}}{2}\] Since R cannot be negative, hence \[R=\frac{1+\sqrt{5}}{2}\Omega \]
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