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

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