A battery of internal resistance 4 Ω is connected to the network of resistances, as shown in the figure. In order that the maximum power can be delivered to the network, the value of R in Ω should be

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The given figure is a balanced form of Wheatstone bridge therefore, 6R resistance can be removed.

$∴ \frac{1}{R_{P}} = \frac{1}{R_{1}} + \frac{1}{R_{2}} = \frac{1}{3 R} + \frac{1}{6 R} = \frac{1}{2 R}$

$\Rightarrow R_{P} = 2 R .$

The power delivered to the network is maximum when internal resistance = External resistance.

$4 = 2 R$

$\Rightarrow R = 2 Ω$

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