A uniformly wound solenoid coil of self inductance `1.8xx10^(-4)H` and resistance `6Omega` is broken up into two identical coils. These identical coils are then connected in parallel across a `12V` battery of negligible resistance. The time constant and steady state current will be

Explanation

For solenoid
`L=(mu_(0)MN^(2)pir^(2))/(l)`
`R=(pN.2pir)/(A)` (`A` area of `X`-section of wire)
`N'=(N)/(2)` , `l'=(1)/(2)`
`L'=(mu_(0)(N//2)^(2)pir^(2))/(l//2)=(L)/(2)`
`R=(p((N)/(2))2pir)/(A)=(R)/(2)`
`L_(eq)=L//4` , `R_(eq)=R//4`
`tau=(L//4)/(R//4)=(L)/(R)=(1.8xx10^(-4))/(6)`
`=0.3xx10^(-4)s`
`I=(E)/(R//4)=(4E)/(R)=(4xx12)/(6)=8A`