A) 4
B) 3
C) 2
D) 1
Correct Answer: A
Solution :
If two resistances are \[{{R}_{1}}\] and \[{{R}_{2}}\] then \[S={{R}_{1}}+{{R}_{2}}\] and \[P=\frac{{{R}_{1}}{{R}_{2}}}{({{R}_{1}}+{{R}_{2}})}\] From given condition S = nP i.e. \[({{R}_{1}}+{{R}_{2}})=n\,\left( \frac{{{R}_{1}}{{R}_{2}}}{{{R}_{1}}+{{R}_{2}}} \right)\] Þ\[{{({{R}_{1}}+{{R}_{2}})}^{2}}=n\,\,{{R}_{1}}{{R}_{2}}\]Þ\[{{({{R}_{1}}-{{R}_{2}})}^{2}}+4{{R}_{1}}{{R}_{2}}=n{{R}_{1}}{{R}_{2}}\] So \[n=4+\frac{{{({{R}_{1}}-{{R}_{2}})}^{2}}}{{{R}_{1}}{{R}_{2}}}.\] Hence minimum value of n is 4.
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