A battery of internal resistance $4 \mathrm{\Omega }$ is connected to the network of resistances as shown in the figure. In order that the maximum power be delivered to the network, the value of $R$ (in $\mathrm{\Omega }$) should be

The given circuit is a balanced Wheatstone's bridge.



Thus, no current will flow across $6R$. The given circuit will now be equivalent to:-




For maximum power, net external resistance = Total internal resistance
or                     $2R=4 \mathrm{\Omega }$
or                      $R=2 \mathrm{\Omega }$