Thevenin Equivalent Circuit of Inductor

In Chapter 9 ("Capacitors"), we found that there are occasions when the circuit does not have the basic form of [Fig. 1]. The same is true for inductive networks. Again, it is necessary to find the Thevenin equivalent circuit before proceeding in the manner described in this chapter.
Fig. 1: Standard inductive circuit.
Consider the following example.
Example 1: For the network of [Fig. 2]:
a. Find the mathematical expression for the transient behavior of the current iL and the voltage $v_L$ after the closing of the switch ($I_i = 0 mA$).
b. Draw the resultant waveform for each.
Fig. 2: For Example 1.
a. Applying Thevenin's theorem to the $80mH$ inductor ([Fig. 3]) yields
$$ \bbox[10px,border:1px solid grey]{R_{TH} = {R \over N} = {20kΩ \over 2} = 10kΩ}$$
Fig. 3: Determining $R_{Th}$ for the network of Fig. 2
Applying the voltage divider rule ([Fig. 4]),
$$E_{TH} = {(R_2 + R_3) E \over R_1 + R_2 + R_3}$$ $$ = {(4kΩ + 16kΩ) 12 \over 20kΩ+4kΩ + 16kΩ}$$
$$ \bbox[10px,border:1px solid grey]{E_{TH}= 6V}$$
Fig. 4: Determining $E_{Th}$ for the network of Fig. 2.
The thevenin equivalent circuit is shown in [Fig. 5].
Fig. 5: The resulting thevenin equivalent circuit for the network of [Fig. 2].
$$\begin{split} i_L &= I_m(1-e^{-t/\tau}) = {E_{Th} \over R_{TH}}(1-e^{-t/\tau})\\ \tau &={ L \over R_{TH}} = {80mH \over 10kΩ} = 8\mu s\\ i_L &= {6 \over 10kΩ}(1-e^{-t/8\mu s})\\ \end{split} $$
$$\bbox[10px,border:1px solid grey]{i_L = 0.6 \times 10^{-3}(1-e^{-t/8\mu s})}$$
$$ \bbox[10px,border:1px solid grey]{v_L = E_{TH} e^{-t/\tau} = 6e^{-t/8\mu s}}$$
Fig. 6: The resulting waveforms for $i_L$ and $v_L$ for the network of [Fig. 2].

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