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`

Answer

D

Explanation

As voltage drop across `8Omega=sqrt(2xx8)`
`= 4V ( :. P = V^2/R)`
Therefore voltage drop across `3Omega = 3V`
[ `:.` 4V is divided in ratio of resistances between 12 and `3Omega` ]
Hence power dissipated in `3Omega= ((3)^2)/3 = 3` watt