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Q. 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

Odisha JEEOdisha JEE 2009Electromagnetic Induction

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 {2R})^2+2 \sqrt r \sqrt {2R} $
$\Rightarrow \sqrt r- \sqrt {2R}=0 $
$\Rightarrow R=\frac {r}{2} $