detailed solution

Correct option is B

The given network of resistances is a balanced Wheatstone’s bridge. Therefore, no current flows through the $6\mathrm{\Omega }$ resistor and it can be omitted. The equivalent resistance of the network is that of a parallel $R_1=R+R+R=3R$combination of  and $R_2=R+R+4R=6R$ which is

$R^{\mathrm{\prime }}=\frac{R_1{R}_2}{R_1+R_2}=\frac{3R\times 6R}{3R+6R}=2R$

The power delivered will be maximum if the resistance of the $4\mathrm{\Omega }=R^{\mathrm{\prime }}=2R$ or $R=2\mathrm{\Omega }$

detailed solution

Correct answer is 2

The given network of resistances is a balanced Wheatstone’s bridge. Therefore, no current flows through the $6\mathrm{\Omega }$ resistor and it can be omitted. The equivalent resistance of the network is that of a parallel $R_1=R+R+R=3R$combination of  and $R_2=R+R+4R=6R$ which is

$R^{\mathrm{\prime }}=\frac{R_1{R}_2}{R_1+R_2}=\frac{3R\times 6R}{3R+6R}=2R$

The power delivered will be maximum if the resistance of the $4\mathrm{\Omega }=R^{\mathrm{\prime }}=2R$ or $R=2\mathrm{\Omega }$