In the circuit shown in figure heat developed across `2Omega, 4Omega and 3Omega` resistances are in the ratio

Options

A. `2:4:3`
B. `1:4:27`
C. `4:8:27`
D. `8:4:27`

Answer

D

Explanation

Current through `2Omega, I_(1)=(2I)/(3)`
Hence produced per second, `H_(1)=I_(2)^(2)xx=((2I)/(3))^(2)xx2=(8I^(2))/(9)`
Current through, `4Omega, I_(2)=(I)/(3)`
Hence produced per second `H_(2)=I_(2)^(2)xx4=((I)/(3))^(2)xx4=(4I^(2))/(9)`
Current through, `3Omega, I`
Heat produced, `H_(3)=I^(2)xx3=3I^(2)=(27I^(2))/(9)`
`therefore H_(1):H_(2):H_(3)=8:4:27`