Question

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 1

`i=(E)/(R_(2))(1-e^((R_(2)t)/(L)))=(12)/(2)(1-e^(2t)/(0.4))`
`i=6(1-e^(-5t))`
`(di)/(dt)=6[0-e^(-5t)(-5)]=30e^(-5t)`
`V_(L)=L(di)/(dt)=(0.4)(30e^(-5t))=12e^(-5t)V`
`i=i_(0)e^(((R_(1)+R_(2)t))/(L)`
`=6e(-4t)/(0.4)=6e^(-10t)`