A) 4
B) 3
C) 2
D) 1
Correct Answer: A
Solution :
| [a] Let resistances be\[{{R}_{1}}\]and\[{{R}_{2}}\], then equivalent of these two is given by, |
| \[S={{R}_{1}}+{{R}_{2}}\] |
| and \[P=\frac{{{R}_{1}}{{R}_{2}}}{{{R}_{1}}+{{R}_{2}}}\] |
| \[\therefore \]\[({{R}_{1}}+{{R}_{2}})=\frac{n\times {{R}_{1}}{{R}_{2}}}{{{R}_{1}}+{{R}_{2}}}\] (from\[S=nP\]) |
| \[\Rightarrow \] \[{{({{R}_{1}}+{{R}_{2}})}^{2}}=n{{R}_{1}}{{R}_{2}}\] |
| \[\Rightarrow \] \[n=\left[ \frac{R_{1}^{2}+R_{2}^{2}+2{{R}_{1}}{{R}_{2}}}{{{R}_{1}}{{R}_{2}}} \right]\] |
| \[=\left[ \frac{{{R}_{1}}}{{{R}_{2}}}+\frac{{{R}_{2}}}{{{R}_{1}}}+2 \right]\] (i) |
| 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,\[n\](minimum value)\[=2+2=4\] [from Eq. (i)] |
You need to login to perform this action.
You will be redirected in
3 sec
