A) 4
B) 3
C) 2
D) 1
Correct Answer: A
Solution :
Let the two resistances be \[{{R}_{1}}\]and\[{{R}_{2}}\] \[S={{R}_{1}}+{{R}_{2}}\] \[\Rightarrow \frac{1}{P}=\frac{1}{{{R}_{1}}}+\frac{1}{{{R}_{2}}}=\frac{{{R}_{1}}+{{R}_{2}}}{{{R}_{1}}{{R}_{2}}}\Rightarrow P=\frac{{{R}_{1}}{{R}_{2}}}{{{R}_{1}}+{{R}_{2}}}\] Now, \[\text{S}=\text{nP}\] \[\Rightarrow \] \[{{R}_{1}}+{{R}_{2}}=n\left( \frac{{{R}_{1}}{{R}_{2}}}{{{R}_{1}}+{{R}_{2}}} \right)\] Now \[{{R}_{1}}={{R}_{2}}=R,\] \[\therefore \] \[2R=\frac{n{{R}^{2}}}{2R}\Rightarrow n=4\]
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