Question

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 $R\left(in\Omega \right)$ should be

• A. 4/9
• B. 8/9
• C. 2
• D. 18

Answer 1

The correct option is C 2

The equivalent network is

Clearly, the network of resistance is a balanced Wheatstone bridge .So $R_{AE}$ is given by
$\frac{1}{R_{AE}}=\frac{1}{3R}+=\frac{1}{6R}=\frac{2+1}{6R}=\frac{1}{2R}\Rightarrow R_{AE}=2R$
For maximum power transfer $2R=4\Omega \Rightarrow R=\frac{4}{2}=2\Omega$