c
(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 \(2R = 4\,\Omega \)\(==>\) \(R = \frac{4}{2} = 2\,\Omega \)
