Correct Answer - A::B::C::D
`V_A=0V` (as it is earthed)
`V_C-V_A=5V`
`:. V_C=5V`
`V_B-V_A=2V`
`:. V_B=2V`
`V_D-V_C=10V`
`:. V_D=10+V_C=15V`
`i_(2) Omega = (V_(C ) - V_(B))/(1) = 3A` from `C` to `B` as `V_(C ) gt V_(B)`
`i_(1Omega)=(V_C-V_B)/2=7.5A` from `C` to `B` is `V_CgtV_B`
`i_(2Omega)=(V_D-V_A)`