Correct Answer - D
As voltage drop across
`8Omega=sqrt(2xx8)=4V(thereforep=V^(2)/(R))`
Therefore voltage drop across`3Omega=3v`
[`therefore`4V is divided in ration of resistances between `1Omega and 3Omega`]
Hence power dissipated in `3Omega=(3)^(2)/(3)=3"watt"`