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WB JEE Medical WB JEE Medical Solved Paper-2009

  • question_answer
    Two identical cells each of emf E and internal resistance r are connected in parallel with an-external resistance R. To get maximum power developed across R, the value of R is

    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 \[r+2R={{(\sqrt{r}-\sqrt{2}R)}^{2}}2\sqrt{r}\sqrt{2R}\] \[\Rightarrow \]               \[\sqrt{r}-\sqrt{2R}=0\] \[\Rightarrow \]               \[R=\frac{r}{2}\]


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