An inductor of inductance L=400 mH and resistor of resistance `R_(1) = 2(Omega) and R_(2) = 2 (Omega)` are connected to a battery of emf E = 12 Vas shown in the figure. The internal resistance of the battery is negligible. The switch S is closed at time t =0. What is the potential drop across L as a function of time? After the steady state is reached, the switch is opened. What is the direction and the magnitude of current through `R_(1)` as a function of time?

Answer

A, B, C

Explanation

This is a question on growth and rise of current.
Growth of current: Let at any instant of time t the current be as shown in the figure.
Applying Kirochoff's law in the loop ABCDFGA
we get, starting from G moving clockwise
`E-L(dl_(2))/(dt)-(I_2)(R_2)=0`
or `(I_2)=(E)/(R^2)[1-e^(-(R/L)t)]`

Also, we know that the emf (V) produced across hte inductor
`V=-(d phi)/(dt)=-(d)/(dt)[LI_(2)]=-L(dI_(2))/(dt)`
=-L(d)/(dt)[(E)/(R_2)( 1-e^((-R_(2))/(L)t)]`
`V=-E^((R_2)/(-L)t)`. Here the negative sign shows the opposite to the growth of current.
`:. V+12e(2)/(400xx10^(-3))t=12 e^(-5t)volt`.
DECAY OF CURRENT:When the switch is opened, the branch AG is out of hte circuit CBFDC(in clockwise direction).
Applying Kirchhoff's law
`I(R_(1)+R_(2))-(-(LdI)/(dt)=0`
`:. (dI)/(I)=-(R_(1)+R_(2))/(L) int_(0)^(t) dt`
`:. I=(I_0)e^(-(R_(1)+R_(2)t)/(L)`
here, (R_(1)+R_(2))/(L)=(2+2)/(400xx10^(-3))=10`
and `(I_0)=(E)/(R_(1)+R_(2))=12/4=3A`
`:. I=3e^(-10t)A, clockwise
Alternatively, you may directly find the time constant
`(tau)=(l)/(R_(1)+R_(2))` and use the equation `i=(i_0)e^(-t//tau)` where `(I-0)=6A`.