Figure shows a circuit consisting of an ideal cell, an inductor `L`, and a resistor `R`, connected in series. Let switch `S` be closed at `t = 0`. Suppose at `t = 0`, the current in the inductor is `i_(o)`, then find out the equation of current as a function of time.

Explanation

Let an instant `t` current in the circuit is `i` which is increasing at the rate `di//dt`
Writing `KVL` along the circuit, we have `epsilon-L(di)/(dt)-i R=0`
`rArr L(di)/(dt)=epsilon-iR rArr underset(i_(0))overset(i)int(di)/(epsilon-iR)=underset(0)overset(t)int(dt)/L`
`rArr "ln"((epsilon-iR)/(epsilon-i_(0)R))=-(Rt)/L`
`rArr epsilon-iR=(epsilon-i_(0)R)e^(-Rt//L) rArr i=(epsilon-i(epsilon-i_(0)R)e^(Rt//L))/R`
Equivalent self inductance:
Series combination:
from (1) and (2) `L=L_(1)+L_(2)`(neglecting mutual inductance)
Parallel Combination:
From figure `V_(A)-V_(B)=L_(1)(dt_(1))/(dt)=L_(2)(dt_(2))/(dt)` ...(3)
also `i=i_(1)+i_(2)`
or `(di)/(dt)=(di_(1))/(dt)+(di_(2))/(dt)`
or `(V_(A)-V_(B))/L=(V_(A)-V_(B))/L_(1)+(V_(A)-V_(B))/L_(2)` (neglecting mutual inductance)