# 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.
Solution:
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.
$$\begin{split} i_L &= I_m(1-e^{-t/\tau}) = {E_{Th} \over R_{TH}}(1-e^{-t/\tau})\\ \tau &={ L \over R_{TH}} = {80mH \over 10kΩ} = 8\mu s\\ i_L &= {6 \over 10kΩ}(1-e^{-t/8\mu s})\\ \end{split}$$ $$\bbox[10px,border:1px solid grey]{i_L = 0.6 \times 10^{-3}(1-e^{-t/8\mu s})}$$ and $$\bbox[10px,border:1px solid grey]{v_L = E_{TH} e^{-t/\tau} = 6e^{-t/8\mu s}}$$ b. Fig. 6: The resulting waveforms for $i_L$ and $v_L$ for the network of Fig. 2.