A) \[R=\frac{r}{2}\]
B) \[R=r\]
C) \[R=\frac{r}{3}\]
D) \[R=2r\]
Correct Answer: A
Solution :
Equivalent resistance \[{{R}_{eq}}=\frac{r}{2}+R=\frac{r+2R}{2}\] \[\therefore \] \[I=\frac{2E}{r+2R}\] For maximum power consumption, I should be maximum so denominator is minimum. For this \[r+2R={{(\sqrt{r}-\sqrt{2}R)}^{2}}2\sqrt{r}\sqrt{2R}\] \[\Rightarrow \] \[\sqrt{r}-\sqrt{2R}=0\] \[\Rightarrow \] \[R=\frac{r}{2}\]You need to login to perform this action.
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