Find current passing through the battery and each resistor.

Explanation

Method (I) :
It is to see that potential difference across
each resistor is 30 V.
`:.` current is each resistors are 30/2 = 15 A,
`30/3=10` A and `30/6 5`A.
Method (II) :
By ohm's law `i=V/R_(eq)rArr1/R_(eq)=1/2+1/3+1/6`
`=1Omega`
`R_(eq)=1OmegarArr i=30/1=30 A`
Now distribute this current in the resistors in their inverse ratio.

Current total in 3 `Omega` and 6 `Omega` is 15 A it will be
divided as 10 A and 5 A.
The method (I) is better. But you will not find
such an case every where.