A) Less than \[4\,\Omega \]
B) \[4\,\Omega \]
C) More than \[4\,\Omega \] but less than \[12\,\Omega \]
D) \[12\,\Omega \]
Correct Answer: C
Solution :
Similar to Q. No. 30 \[R=2+2+\frac{2\times R}{2+R}\] Þ \[2R+{{R}^{2}}=8+4R+2R\] Þ \[{{R}^{2}}-4R-8=0\] Þ \[R=\frac{4\pm \sqrt{16+32}}{2}=2\pm 2\sqrt{3}\] R cannot be negative, hence \[R=2+2\sqrt{3}=5.46\Omega \]
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