Question

At `t=0` , switch `S` is closed, calculate
(a) initial rate of increase of current i.e. `(di)/(dt)` at `t=0`
(b) `(di)/(dt)` at time when current in the circiut is `1A`
(c) current at `t=0.5sec`
(d) rate at which energy of magnetic field is incresing, rate of heat produced in resistance and rate at which energy is supplied by battery when `i=1A`
(e) energy stored in inductor in steady state

Answer 1

`E=iR+Ldi//dt`
`(di)/(dt)=(E-iR)/(L)`
Current in the circuit at any time `t`
`i=i_(0)(1-e^(-t//tau))`
where `i_(0)0E//R` . `tau=L//R`
(a) `(di)/(dt)=(E-iR)/(L)` , when `t=0` , `i=0`
`(di)/(dt)|_(t=0)=(E)/(L)=(8)/(1)=88A//sec`
(b) `i=1A` , `(di)/(dt)=(E-iR)/(L)=(8-1xx2)/(1)=6Asec`
(c) `i=(E)/(R)(1-e^(-(Rt)/(L)))`
`=(8)/(2)(1-e^(-(2xx0.5)/(L)))=4(1-1//e)`
(d) `U=(1)/(2)Li^(2)` , `(dU)/(dt)=Li(di)/(dt)`
`(di)/(dt)=(E-iR)/(L)=(8-1xx2)/(1)=6A//sec`
`(dU)/(dt)=1xx1xx6=xJ//sec`
heat produced per sec, `p=I^(2)r=(1)^(2)XX2=2l//sec`
power supplied by battery `=E_(i)=8xx1=8J//sec`
(e) `i_(0)=E//R=8//2=4A`
`U=(1)/(2)Li_(0)^(2)=(1)/(2)xx1xx(4)^(2)=8J//sec`