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Decay of Current

Consider an inductive coil of resistance R ohms. Inductance L henrys and having a current of I amperes flowing through it. If the source of supply is removed from its circuit, the current stars decreasing and due to decrease in current, self induced e.m.f. will be set up, which will oppose the current.

Let at any instant current flowing through the coil be i amperes

Self induced e.m.f.  è = - L di

Resistive drop = iR
Applied voltage = V = 0
Since applied voltage, V = Resistive drop + back e.m.f.

0 = iR + L di

or  di - R dt
      i      L   

Integrating both sides we get

loge i =  -R t+ K ' where K ' is a constant of integration

From initial conditions i.e. when t = 0, i = Im

K ' = loge  Im

Hence loge i = -R t + logIm

or  log  i  =     -R  t
           Im        L

or i = Im e -R  t  = Imet/T   where T = L the time constant.
                L                                  R '

The rise and decay of current is shown in.

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