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A battery of internal resistance 4Ω is connected to the network of resistances as shown. In order to give the maximum power to the network, the value of R (in Ω) should be

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Detailed Solution

The equivalent network is

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}\Rightarrow\,R_{AB}=2R

For maximum power transfer

2R=4\Omega\,\Rightarrow\,R=\frac{4}{2}=2\Omega

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