Thevenin Equivalent Circuit of Inductor

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. Solution:
a. Applying Thevenin's theorem to the $80mH$ inductor (Fig. 3) yields $$ \bbox[10px,border:1px solid grey]{R_{TH} = {R \over N} = {20kΩ \over 2} = 10kΩ}$$ Applying the voltage divider rule (Fig. 4), $$E_{TH} = {(R_2 + R_3) E \over R_1 + R_2 + R_3}$$ $$ = {(4kΩ + 16kΩ) 12 \over 20kΩ+4kΩ + 16kΩ}$$ $$ \bbox[10px,border:1px solid grey]{E_{TH}= 6V}$$ The thevenin equivalent circuit is shown in Fig. 5. $$\begin{split} i_L &= I_m(1-e^{-t/\tau}) = {E_{Th} \over R_{TH}}(1-e^{-t/\tau})\\ \tau &={ L \over R_{TH}} = {80mH \over 10kΩ} = 8\mu s\\ i_L &= {6 \over 10kΩ}(1-e^{-t/8\mu s})\\ \end{split} $$ $$\bbox[10px,border:1px solid grey]{i_L = 0.6 \times 10^{-3}(1-e^{-t/8\mu s})}$$ and $$ \bbox[10px,border:1px solid grey]{v_L = E_{TH} e^{-t/\tau} = 6e^{-t/8\mu s}}$$ b.