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A coil of self inductance 10 mH and resi...

A coil of self inductance 10 mH and resistance `0.1Omega` is connected through a switch to a battery of internal resistance `0.9Omega` . After the switch is closed, the time taken for the current to attain 80% of the saturation value is: [take ln 5 = 1.6]

A

`0.016 `

B

`0.002s`

C

`0.324s `

D

`0.10 s`

Text Solution

Verified by Experts

The correct Answer is:
A
`I_s=V/R=V/(0.9+0.1)=V/1`
`I=I_s(1-e^(-(Rt)/L))`
`0.8 V =V(1-e^((1xxt)/10xx10^(-3)))`
`0.8 =1-e^(-100t)implies 0.2 =e^(-100t)`
`e^(-100t) =5`
`t=(ln5)/100 =0.016s`
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