An inductor of inductance `L=400 mH` and resistors of resistances `R_1=2Omega` and `R_2=2Omega` are connected to a battery of emf `E=12 V` as 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 dro across `L` s a function of time? After the steady state is reached, the switch is opened. What is the direction ad the magnitude of current throough `R_1` as a function of time?

Explanation

a. Given `R_1=R_2=2Omega, E=12V` and `L=400mH =0.4H`.
Two parts of the circuit are in parallel with applied battery.
So, the givne circuit can be broken as

Now refere Figure b
This is a simple `L-R` circuit, whose time constant
`tau_L=L/R_2=0.4/2=0.2s`
and steady state current
`i_0=E/R_2=12/2=6A`
Therefoe if switch `S` is closed at time `t=0`, then current in the circuit at any time `t` will be given
`i(t)=(1-e^(-t//tau)L)`
`i(t) =6(1-e^(-t//0.2))`
`= 6 (1-e^(-5t)) =i`
Therefore, potential drop across `L` at any time `t` is
`V=|L(di)/(dt)|=L(30e^(-5t))=(0.4)(30)e^(-5t) or V=12e^(-5t) "volt"`
b. The steady state current in `L` or `R_2` is
`i_0=6A`
Now, as soon as the switch is opened, current in `R_1` is reduced to zero immediately. But in L and `R_2` it decreases exponentially. the situtation is as follows:

Refer figure
Time constant of this circuit would be
`tau'_L^=L/(R_1+R_2)=0.4/((2+2))=0.1s`
`:.` Current through `R_1` at time t is
`i=i_0e^(-t//tauL')=6e^(-t//0.1)`
`or i=6^(-10t)A`
Direction of current in `R_1` is as shown in figure or clockwise.