Power dissipated across the `8 Omega` in the circuit shown here is `2 W`. The power dissiated in watt units across the `3 Omega` is

Options

A. `2.0`
B. `1.0`
C. `0.5`
D. `3.0`

Explanation

As voltage drop across `8Omega=sqrt(2xx8)`
`=4V(because P=(V^(2))/(R))`
Therefore voltage drop across `3Omega=3V`
[`because 4V` is divided in ratio of resistances between `1Omega` and `3Omega`]
Hence power dissipated in `3Omega=((3)^(2))/(3)=3` watt