In the circuit shown in figure potential differences across `6Omega` resistance is 4 volt. Find `V` and `i` values across each resistance. Also find emf `E` of the applied battery.

Explanation

`8Omega, 2Omega` (Resultant of `6Omega` and `3Omega`) and `3Omega` (resultant of `12 Omega` and `4 Omega)` are in series. Therefore, potential drop across them should be in direct ratio of resistance. So, using this concept we can find the potential difference across other resistors. For examples, potential across `2Omega` was `4 V`. So, potential difference across `8Omega`(which is four times of `2Omega`) should be `16 V`, Similarly, potential difference across `3Omega`( which is 1.5 times of `2 Omega`) should be `16 V`. Similarly, potential difference across `3Omega`(which is 1.5 times of `2Omega`) should be 1.5 times or `6 V`.
Once` V` and `R` are knwon we can find i across that resistane. for example ltbr `i_(12Omega)=6/12=1/2A`
`i_(8Omega)=16/8=2A` etc.