A battery of internal resistance $4$ $\Omega$ is connected to the network of resistances as shown. In order to give the maximum power to the network, the value of $R$ (in $\Omega $) should be
IIT 1995, Diffcult
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(c) Clearly, the network of resistances is a balanced Wheatstone bridge. So ${R_{AB}}$ is given by
$\frac{1}{{{R_{AB}}}} = \frac{1}{{3R}} + \frac{1}{{6R}} = \frac{{2 + 1}}{{6R}} = \frac{1}{{2R}}$$ ==>$ ${R_{AB}} = 2R$
For maximum power transfer $2R = 4\,\Omega $$==>$ $R = \frac{4}{2} = 2\,\Omega $
$\frac{1}{{{R_{AB}}}} = \frac{1}{{3R}} + \frac{1}{{6R}} = \frac{{2 + 1}}{{6R}} = \frac{1}{{2R}}$$ ==>$ ${R_{AB}} = 2R$
For maximum power transfer $2R = 4\,\Omega $$==>$ $R = \frac{4}{2} = 2\,\Omega $
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