At t = 0, switch S is closed, calculate
(i) initial rate of increase of current, i.e., `(di)/(dt)"at t"=0`.
(ii) `(di)/(dt)` at time when current in the circuit is `0.5A`.
(iii) Current at t = `0.6` s.
(iv) rate at which energy of magnetic field is increasing, rate of heat produced in resistance and rate at which energy is supplied by battery when i = 0.5 A.
(v) energy stored in inductor in steady state.

Explanation

We know that
`E=iR+"L dt/dt"`
`(dt)/(dt)=(E-iR)/(L)`
Current in the circuit at any time t,
`i=i_(0)(1-e^(-t//tau_(L)))`
where, `i_(0)=E//R,tau_(L)=L//R`
(i) Rate of increase of current `(di)/(dt)=(E-iR)/(L)`, when t = 0, `i` = 0
`|(di)/(dt)|_(t=0)=(E)/(L)=(12)/(2)=6A//s`
(ii) At `t=0.5s,(di)/(dt)=(E-iR)/(L)=(12-0.5xx4)/(2)=(12-2)/(2)=5A//s`
(iii) At t = `0.6` s, current i = `(E)/(R)(1-e^(-(Rt)/(L)))`
`=(12)/(4)(1-e^((4xx0.6)/(0.5)))=3(1-e^(2.4))A`
(iv) Energy stored in inductor in steady state is given by
`U=(1)/(2)Li^(2)rArr(dU)/(dt)=Li(di)/(dt)`
`(di)/(dt)=(E-iR)/(L)=(12-0.5xx4)/(2)=5A//s`
`(dU)/(dt)=2xx0.5xx5=5J//s`
Power produced per second, `P=i^(2)R=(0.5)^(2)xx4=1J//s`
Power supplied by battery `=Ei=0.5xx12=6J//s`
(v) We have `i_(0)=E//R=12//4=3A`
Energy stored in inductor in steady state,
`U=(1)/(2)Li_(0)^(2)=(1)/(2)xx2xx(3)^(2)=9J//s`