A) 4
B) 3
C) 2
D) 1
Correct Answer: A
Solution :
Let resistances are\[{{R}_{1}}\]and\[{{R}_{2}}\], then \[S={{R}_{1}}+{{R}_{2}}\] and \[P=\frac{{{R}_{1}}{{R}_{2}}}{{{R}_{1}}+{{R}_{2}}}\] \[\therefore \] \[({{R}_{1}}+{{R}_{2}})=\frac{n\times {{R}_{1}}{{R}_{2}}}{{{R}_{1}}+{{R}_{2}}}\] [form\[S=np\]] \[\Rightarrow \] \[{{({{R}_{1}}+{{R}_{2}})}^{2}}=n{{R}_{1}}{{R}_{2}}\] \[\Rightarrow \]\[n=\left[ \frac{R_{1}^{2}+R_{2}^{2}+2{{R}_{1}}{{R}_{2}}}{{{R}_{1}}{{R}_{2}}} \right]=\left[ \frac{{{R}_{1}}}{{{R}_{2}}}+\frac{{{R}_{2}}}{{{R}_{1}}}+2 \right]\] We know, Arithmetic Mean\[\ge \]Geometric Mean \[\frac{\frac{{{R}_{1}}}{{{R}_{2}}}+\frac{{{R}_{2}}}{{{R}_{1}}}}{2}\ge \sqrt{\frac{{{R}_{1}}}{{{R}_{2}}}\times \frac{{{R}_{2}}}{{{R}_{1}}}}\] \[\Rightarrow \] \[\frac{{{R}_{1}}}{{{R}_{2}}}+\frac{{{R}_{2}}}{{{R}_{1}}}\ge 2\] So, n (min. value)\[=2+2=4\]
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