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

Options

A. `6e^(-5t)V`
B. `(12)/(t)e^(-3t)V`
C. `6(1-e^((-1)/(0.2)))V`
D. `12e^(-5t)V`

Answer

D

Explanation

E(across BC)=`L(dI_(2))/(dt)+R_(2)I_(2)` [using Kirchhoff's law]…(i)
`I_(2)=I_(0)(I-e^(-t//t_(0)))`

At t = 0, `I_(0)=(E)/(R_(2))=(12)/(2)=6A`
`tau_(L)=t_(0)=(L)/(R)=(400xx10^(-3))/(2Omega)=0.2s`
`therefore" "I_(2)=6(1-e^(-t//0.2))` [using Eq. (i)]
Potential drop area L = `E-R_(2)I_(2)`
`=12-2xx6(1-e^(-t//0.2))=12e^(-t//0.2)=12e^(-5t)V`