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492 ⏐⏐⏐ INDUCTORS
E
E Th :
12 V
20 k�
R 1
20 k�
R 1
+
v L
–
E 12 V E Th
4 k�
R 2
R Th :
R 3
20 k�
R 1
16 k�
FIG. 12.33
Determining E Th for the network of Fig. 12.31.
i L
4 k�
R 2
L = 80 mH
FIG. 12.31
Example 12.7.
R 3
16 k�
For the other form, the equation v C � Ee �t/t is a close match with
v L � Ee �t/t , permitting a derivation similar to that employed for Eq.
(10.25):
E
t � t loge�� (12.22)
v
The similarities between the above and the equations in Chapter 10
should make the equation for t fairly easy to obtain.
12.11 THÉVENIN EQUIVALENT: t � L/R Th
In Chapter 10 (“Capacitors”), we found that there are occasions when
the circuit does not have the basic form of Fig. 12.14. The same is true
for inductive networks. Again, it is necessary to find the Thévenin
equivalent circuit before proceeding in the manner described in this
chapter. Consider the following example.
EXAMPLE 12.7 For the network of Fig. 12.31:
a. Find the mathematical expression for the transient behavior of the
current iL and the voltage vL after the closing of the switch (Ii �
0 mA).
b. Draw the resultant waveform for each.
Solutions:
a. Applying Thévenin’s theorem to the 80-mH inductor (Fig. 12.32)
yields
R 20 k�
RTh � � ���10 k�
N 2
R Th
4 k�
R 2
R 3
16 k�
R Th
R 1
L
20 k�
FIG. 12.32
Determining R Th for the network of Fig. 12.31.
R2 + R3 =
4 k� + 16 k�
= 20 k�
Applying the voltage divider rule (Fig. 12.33),
(R2 � R3)E ETh ���
R1 � R2 � R3 (4 k��16 k�)(12 V) (20 k�)(12 V)
���� ��� �6 V
20 k��4 k��16 k� 40 k�
The Thévenin equivalent circuit is shown in Fig. 12.34. Using Eq.
(12.8),