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

Options

A. `0.1xx10^(-4)s`
B. `0.2xx10^(-1)s`
C. `0.3xx10^(-4)s`
D. `0.4xx10^(-4)s`

Answer

C

Explanation

`L.=L/2=(1.8xx10^(-4))/2=0.9xx10^(-4)H`
`L_(eq)=(L.)/2=0.45xx10^(-4)HandR.=R/2=3Omega`
`R_(eq)=1.5Omega`
`tau=L_(eq)/R_(eq)=(0.45xx10^(-4))/1.5=0.3xx10^(-4)sec`.