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494 ⏐⏐⏐ INDUCTORS
E Th
26.4 V
R Th
10.4 k�
FIG. 12.37
Thévenin equivalent circuit for the network of
Fig. 12.36 for t ≥ 0 s.
L
i L
680 mH
6 mA
a. Find the initial current through the coil. Pay particular attention to its
direction.
b. Find the mathematical expression for the current i L following the
closing of the switch S 2.
c. Sketch the waveform for i L.
Solutions:
a. Using Ohm’s law, the initial current through the coil is determined
by
E 6 V
Ii ��� ��� ��6 mA
R3 1 k�
b. Applying Thévenin’s theorem:
RTh � R1 � R2 � 2.2 k��8.2 k��10.4 k�
ETh � IR1 � (12 mA)(2.2 k�) � 26.4 V
The Thévenin equivalent network appears in Fig. 12.37.
The steady-state current can then be determined by substituting
the short-circuit equivalent for the inductor:
E 26.4 V
If �����2.54 mA
RTh 10.4 k�
The time constant:
t � � �65.39 ms
Applying Eq. (12.12):
iL � If � (Ii � If)e �t/t
L 680 mH
� �
RTh 10.4 k�
� 2.54 mA � (�6 mA � 2.54 mA)e
�t/(65.39 ms)
� 2.54 mA � 8.54 mAe
c. Note Fig. 12.38.
3
2
1
0
–1
–2
–3
–4
–5
–6 mA
i L (mA)
1τ τ
τ τ = 65.39 �s �
�t/65.39 ms
2τ τ 3τ τ 4τ τ 5τ τ
FIG. 12.38
The current i L for the network of Fig. 12.37.
2.54 mA
t