In the circuit shown in figure, power developed across `1Omega,2Omegaand3Omega` resistances are in the ratio.

Options

A. `1:2:3`
B. `4:2:27`
C. `6:4:9`
D. `2:1:27`

Answer

B

Explanation

Current distribution is as shown in figure

`therefore` Powers are
`P_(1)=I_(1)^(2)R_(1)=4/9i^(2),P_(2)=I_(2)^(2)R_(2)=2/9i^(2),andP_(3)=I_(3)^(2)R_(3)=3i^(2)=27/9i^(2)`.
`rArr" "P_(1):P_(2):P_(3): :4:2:27`