Find current passing through the battery and each resistor.

Explanation

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

Current total in `3 Omega" and "6 Omega`