Decay of Current
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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
dt
Resistive drop = iR
Applied voltage = V = 0
Since applied voltage, V = Resistive drop + back e.m.f.
0 = iR + L di
dt
or di = - R dt
i L
Integrating both sides we get
loge i = -R t+ K ' where K ' is a constant of integration
L
From initial conditions i.e. when t = 0, i = Im
K ' = loge Im
Hence loge i = -R t + loge Im
or loge 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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Let at any instant current flowing through the coil be i amperes
Self induced e.m.f. è = - L di
dt
Resistive drop = iR
Applied voltage = V = 0
Since applied voltage, V = Resistive drop + back e.m.f.
0 = iR + L di
dt
or di = - R dt
i L
Integrating both sides we get
loge i = -R t+ K ' where K ' is a constant of integration
L
From initial conditions i.e. when t = 0, i = Im
K ' = loge Im
Hence loge i = -R t + loge Im
or loge 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.
For more help in Decay of Current click the button below to submit your homework assignment