Correct Answer - C
As the supply is form d.c. battery,
`X_(L) = omega L = 2 pi n L = 0`, and
`X_(C ) = (1)/(omega C) = (1)/(2 pi n C) = prop`
`:. C` draws no current, and both inductors L have no role. Three resistances of `9 Omega` each are in parallel.
`(1)/(R_(P)) + (1)/(R ) + (1)/(R) + (1)/(R) = (1)/(9) + (1)/(9) + (1)/(9) = (3)/(9)`
`R_(P) = 9//3 Omega = 3 Omega`.
Total current the battery `= (E)/(R ) = (18)/(3) = 6 A`
Its divides equally among three resistances.
Current through each resistance `= (6)/(3) A = 2 A`