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example: 3.2-4
The voltage waveform applied to an inductor of 2H is shown in Fig. 3.2-13. (i) Find the magnitude of 
current in the inductor at t 
=
9 s if the initial current is zero. (ii) What must have been the initial current 
if the current at 9 s is found to be 0A? (iii) Plot current in the inductor in the first case.
v
(
t
)
2 H
+
–
i
L
t
(s)
2
5
4
3
5 6 7
1
v
(
t
)
(V)
Fig. 3.2-13 
Circuit and waveform for Example 3.2-4 
Solution
(i) Inductor current at any instant is given by its initial current plus the V-s product divided by 
inductance value where V-s is the area under the voltage curve from initial instant to the instant at 
which the current is calculated. The initial instant is 0 s here. Final instant is 9 s. The V-s available 
in this range is 25V-s. Inductance value is 2H. Initial current is 0. Therefore, inductor current at
t 
=
9 s is 12.5A. In fact it is 12.5A from t 
= 
6 s onwards.
(ii) 25Wb-T of additional flux linkage will change the current of a 2H inductor by 12.5A. Hence if 
current at t 
=
9s is observed to be 0A, then the initial current must have been –12.5A.
(iii) The equations for v(t) in various time intervals are shown in the following:
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3.24
Single Element Circuits
v t
t
t
t
t
t
t
( )
(
)
=
≤ <
≤ <
−
−
≤ <
≤ ≤




5
0
1
5
1
5
5 5
5
5
6
0
6
9
for 
for 
for 
for 


The inductor current can be found by integrating v(t) and dividing the integral by L.
i t
t
t
t
t
t
( )
.
.
. (
)
.
. (
)
=
+
≤ <
+
−
≤ <
+
−
0 1 25
0
1
1 25 2 5
1
1
5
11 25 2 5
5
2
for 
for 
−−
−
≤ <
≤ ≤







1 25
5
5
6
12 5
6
9
2
. (
)
.
t
t
t
for 
for 
This is plotted in Fig. 3.2-14.
14
12
10
8
8
6
8
9
4
4
2
2
1
3
5
7
i
L
(A)
t
(s)
Fig. 3.2-14 
Plot of current in 2H inductor in Example 3.2-4