When an emf is applied to an inductor, the current cannot rise instantly due to the back emf generated by the coil, which opposes the flow of current. As the current rises in an inductance, the back emf causes the rate of rise in current to reduce over time. Similarly, when the source is removed, the energy stored in the magnetic field cannot cause the current to fall immediately to zero due to the self-induced back emf.

1. When an emf is applied to an inductor the current cannot rise instantaneously
due to the emf generated by the coil (back emf, ‘e’) which opposes the flow
of current.
Transients - Inductive Circuits
At the instant the voltage is
applied, i = 0
At any instant:
e
iR
-
S
V
+
R
L
I
or
V = e + iR
also induced emf ‘e’
di
dt
e = V – iR = L
=
di
dt
V
L
A/s
initial rate of change in current
dt
di
V = L
V = iR + L
di
dt
final current
Io =
V
R
final current
rate of change of current
=
L
R
seconds=
V/L
V/R
The time constant, T =

2. As the current rises in an inductance, the back emf opposes the supply voltage and
causes the rate of rise in current to reduce as time goes on. After approximately 5 time
constants the current is within 1% of its maximum value.
Transients - Inductive Circuits
If the current continued to rise at its initial rate, it would reach its maximum value in
one time constant (T) seconds.
Initial rate of rise of current, di/dt = VS / L A/s
Maximum current possible, I = VS / R A
Current rise at any instant, i = I(1 – e – Rt /L
) V
I=VS /R
T
( L/R)
0 20 40 60 80 t (msec)
2.0
1.5
1.0
0.5
0
-
S
VS 100V
+
R = 50Ω
L 800 mH
I

3. If the source emf is removed and the inductor is short circuited the energy stored in the
magnetic field will collapse. The current cannot fall immediately to zero as the collapse
is opposed by the self induced emf which tries to maintain the current at its original
value.
Transients - Inductive Circuits
After 5 time constants the current is said to have reached its final value although
the equation allows for a continuously decaying state.
Initial rate of decay of current, di/dt = -VL / L A/s
Initial current possible, I = VL / R
Current decay at any instant, i = Ie – Rt /L
I=VS /R
T
( L/R)
0 20 40 60 80 t (msec)
2.0
1.5
1.0
0.5
0
-
+
R = 50Ω
L 800 mH
I

4. Transients – Inductive Circuits
Activity
1. A 200mH inductor is connected to a 20V dc supply. If the coil
resistance is 10Ω, determine a) the time constant, b) the initial rate of
increase in current, c) the time to reach a fully charged state and d) the
current at t = L/R s.
2. A coil of inductance 0.5H and resistance 25Ω is suddenly connected to
a 100V supply. Find a) the current after 5 seconds, b) the time at which
vL = vR and c) the final steady state current.
3. A coil of resistance 20Ω and inductance 250mH is connected to a 40V
dc supply and the current is allowed to reach its final steady state after
which the supply is removed and the coil instantly short circuited.
Calculate a) the initial rate of decrease of current and b) the current
after L/R seconds and c) the time for the current to be reduced by 20%.

5. What Have We Learnt
The time constant for a LR network is T = LR seconds and after 5LR seconds
the capacitor is fully charged or discharged.
The rate of change of current is continuously opposed by the induced emf
(back emf) in the coil.
0
10
20
30
40
50
60
70
80
90
100
110
120
0
29
59
88
118
147
176
206
235
265
294
324
353
382
412
441
471
500
t
i
I
0
1
2
3
4
5
6
7
8
9
10
0
59
118
176
235
294
353
412
471
I
t
i
The decaying current (i) at any instant in time can be found using the formula
i = Io e – Rt /L
The rising current (i) at any instant in time can be found using the formula
i = Io (1 – e -R t /L
)

6. What Have We Learnt
The time constant for a LR network is T = LR seconds and after 5LR seconds
the capacitor is fully charged or discharged.
The rate of change of current is continuously opposed by the induced emf
(back emf) in the coil.
0
10
20
30
40
50
60
70
80
90
100
110
120
0
29
59
88
118
147
176
206
235
265
294
324
353
382
412
441
471
500
t
i
I
0
1
2
3
4
5
6
7
8
9
10
0
59
118
176
235
294
353
412
471
I
t
i
The decaying current (i) at any instant in time can be found using the formula
i = Io e – Rt /L
The rising current (i) at any instant in time can be found using the formula
i = Io (1 – e -R t /L
)