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A uniformly wound solenoid coil of self ...

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

A

`0.1xx10^(-4)s`

B

`0.2xx10^(-1)s`

C

`0.3xx10^(-4)s`

D

`0.4xx10^(-4)s`

Text Solution

Verified by Experts

The correct Answer is:
C
`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`.
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