# Thevenin Equivalent Circuit

Some capacitor's networks does not have simple series form of Fig. 1. It will then be necessary first to find the Thevenin equivalent circuit for the network external to the capacitive element.
Fig. 1: Basic charging network
$E_{Th}$ will then be the source voltage $E$ of Eqs. (1) and $R_{Th}$ will be the resistance R. The time constant is then $\tau =R_{Th}C$. $$\bbox[10px,border:1px solid grey]{v_C = V_f + (V_i - V_f)e^{-t/\tau}} \tag{1}$$
The following example will clarify the Thevenin equivalent circuit derivation for a network which is not possible to be solved with the series circuit formulas.
Example 1: For the network of Fig. 2:
Fig. 2: For example 1.
a. Find the mathematical expression for the transient behavior of the voltage $v_C$ and the current $i_C$ following the closing of the switch (position 1 at t = 0 s).
b. Find the mathematical expression for the voltage $v_C$ and current $i_C$ as a function of time if the switch is thrown into position 2 at t = 9 ms.
c. Draw the resultant waveforms of parts (a) and (b) on the same time axis
Solution:
a. Applying Thevenin's theorem to the $0.2\mu F$ capacitor, we obtain Fig. 3: $$\begin{split} R_{Th} &= R_1 || R_2 + R_3\\ &={(60 kΩ \times 30 kΩ) \over 90 kΩ} + 10 kΩ\\ &= 20 kΩ+ 10kΩ= 30kΩ\\ E_{Th} &= {R_2 E \over R_2 + R_1}\\ &= {(30kΩ 21 v) \over 30kΩ+60 kΩ} = 7 v \end{split}$$
Fig. 3: Thevenin's equivalent for the network of Fig. 2.
The resultant Thevenin equivalent circuit with the capacitor replaced is shown in Fig. 4. Using Eq. (1) with $V_f = E_{Th}$ and $V_i = 0 v$, we find that $$v_C = V_f + (V_i - V_f)e^{-t/\tau}$$ becomes $$v_C = E_{Th} + (0 v - E_{Th})e^{-t/\tau}$$ $$v_C = E_{Th} (1 - e^{-t/\tau})$$ with $$\tau = RC = (30 kΩ)(0.2 \mu F) = 6 ms$$ so that $$v_C = 7 v (1 - e^{-t/6ms})$$ For the current: $$\begin{split} i_C &= {E \over R_{Th}}e^{-t/RC} \\ &= {7 v \over 30 kΩ}e^{-t/6ms} \\ i_C &= (0.233 \times 10^{-3})e^{-t/6ms} \end{split}$$
Fig. 4: Substituting the Thevenin equivalent for the network of Fig. 2.
b. At t = 9 ms, charging a capacitor is given by $$\begin{split} v_C & = E_{Th}(1 - e^{-t/\tau} ) \\ & = 7 (1 - e^{-9ms/6ms} ) = 7 (1 - e^{-1.5}) \\ & = 7 (1 - 0.223 ) = 5.44 v \end{split}$$ and current $$\begin{split} i_C &= {E \over R_{Th}}e^{-t/RC} \\ &= {7 V \over 30 kΩ}e^{-9ms/6ms} \\ i_C &= (0.233 \times 10^{-3})e^{-1.5} = (0.233 \times 10^{-3}) 0.223\\ i_C &= 0.052 mA \end{split}$$ Now that we have found the initial values before discharging the capacitor by switching towards point (2) Using Eq. (1) with $V_f = 0 v$ and $V_i = 5.44 v$, we find that $$v_C = V_f + (V_i - V_f)e^{-t/\tau}$$ becomes $$\begin{split} v_C &= 0 v + (5.44 v - 0 V)e^{-t/\tau}\\ &=5.44e^{-t/\tau}\\ \end{split}$$ with $$\tau = R_4C = (10 kΩ)(0.2 \mu F) = 2 ms$$ and $$v_C = 5.44e^{-t/2ms}$$ while $$\begin{split} i_C &= {V_i \over R} e^{-t/\tau}\\ &= -{5.44 \over 10kΩ} e^{-t/2ms}\\ &= -(0.054 \times 10^{-3} e^{-t/2ms} \end{split}$$ c. See Fig. 5.
Fig. 5: The resulting waveforms for the network of Fig. 2.