Question

A battery of internal resistance $4\Omega$ 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 $\Omega$ should be

Answer 1

Answer: 2

The given figure is a balanced form of Wheatstone bridge therefore, 6R resistance can be removed.
$\therefore \frac{1}{R_P}=\frac{1}{R_1}+\frac{1}{R_2}=\frac{1}{3R}+\frac{1}{6R}=\frac{1}{2R}$
$\Rightarrow R_P=2R.$
The power delivered to the network is maximum when internal resistance = External resistance.
$4=2R$
$\Rightarrow R=2\Omega$