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An inductor of inductance L = 400 mH and...

An inductor of inductance L = 400 mH and resistors of resistances `R_(1) = 2 Omega and R_(2) = 2 Omega` are connected to a battery of emf 12V as shown in the figure. The internal resisrance of the battery is negligible. The switch S is closed at t = 0. The potential drop across L as a function of time is

A

`12/te^(-31) V`

B

`6(1-e^(-t//0.2))V`

C

`12 e^(-5t)V`

D

`5 e^(-5t)V`

Text Solution

Verified by Experts

The correct Answer is:
C
`V_(L)=epsilone-(R_(2)t)/L=12.e-(2t)/(400xx10^(-3))=12 e^(-5t)`
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