Thevenin's theorem, as stated for sinusoidal ac circuits, is changed
only to include the term
impedance instead of
resistance; that is,
Any two-terminal linear ac network can be replaced with an
equivalent circuit consisting of a voltage source and an impedance in
series, as shown in Fig. 1.
Fig. 1: Thevenin equivalent circuit for ac networks.
Since the reactances of a circuit are frequency dependent, the Thevenin
circuit found for a particular network is applicable only at one frequency.
The steps required to apply this method to dc circuits are repeated
here with changes for sinusoidal ac circuits. As before, the only change
is the replacement of the term resistance with impedance. Again,
dependent and independent sources will be treated separately.
Independent Sources
- Remove that portion of the network across which the Thevenin
equivalent circuit is to be found.
- Mark the terminals of the remaining two-terminal
network.
- Calculate $Z_{Th}$ by first setting all voltage and current sources to
zero (short circuit and open circuit, respectively) and then finding
the resulting impedance between the two marked terminals.
- Calculate $E_{Th}$ by first replacing the voltage and current sources
and then finding the open-circuit voltage between the marked
terminals.
- Draw the Thevenin equivalent circuit with the portion of the
circuit previously removed replaced between the terminals of the
Thevenin equivalent circuit.
Example 1: Find the Thevenin equivalent circuit for the network
external to resistor R in Fig. 2.
Fig. 2: Example 1.
Solution:
Steps 1 and 2 (Fig. 2):
Fig. 3: Assigning the subscripted impedances to the network of Fig. 2.
Fig. 4: Determining the Thevenin impedance for the
network of Fig. 2.
$Z_1 = j X_L = j 8 Ω$ $Z_2 = -j X_C =- j 2 Ω$
Step 3 (Fig. 4):
$$ \begin{split}
Z_{Th} &= { Z_1Z_2 \over Z_1 + Z_2} = { (j 8 Ω)(-j 2 Ω) \over j 8 Ω - j 2 Ω} \\
&= { - j^2 16 Ω \over j 6} = {16 Ω \over 6 \angle 90^\circ}\\
&=2.67 \angle -90^\circ\\
\end{split}$$
Step 4 (Fig. 5):
Fig. 5: Determining the open-circuit Thevenin
voltage for the network of Fig. 1.
$$ \begin{split}
E_{Th} &= { Z_2 E \over Z_1 + Z_2}\\
&= {(-j 2 Ω)(10 V) \over j 8 Ω - j 2 Ω}\\
&={ -j 20 V \over j6}\\
&=3.33 \angle -180^\circ\\
\end{split}$$
Step 5: The Thevenin equivalent circuit is shown in Fig. 6.

Fig. 6: The Thevenin equivalent circuit for the network of Fig. 1.
Dependent Sources
For dependent sources with a controlling variable not in the network
under investigation, the procedure indicated above can be applied. However, for dependent sources of the other type, where the controlling variable is part of the network to which the theorem is to be applied, another
approach must be employed. The necessity for a different approach will be
demonstrated in an example to follow. The method is not limited to dependent sources of the latter type. It can also be applied to any dc or sinusoidal
ac network. However, for networks of independent sources, the method of
application employed in Chapter 8 and presented in the first portion of this
section is generally more direct, with the usual savings in time and errors.
The new approach to Thevenins theorem can best be introduced at
this stage in the development by considering the Thevenin equivalent
circuit of Fig. 7(a). As indicated in Fig. 7(b), the open-circuit
terminal voltage (Eoc) of the Thevenin equivalent circuit is the Thevenin
equivalent voltage; that is,
$$ \bbox[10px,border:1px solid grey]{E_{oc} = E_{Th}} \tag{1}$$
If the external terminals are short circuited as in Fig. 7(c), the
resulting short-circuit current is determined by
$$\bbox[10px,border:1px solid grey]{ I_{sc} = { E_{Th} \over Z_{Th}}} \tag{2}$$
or, rearranged,
$$\bbox[10px,border:1px solid grey]{ Z_{Th} = { E_{Th} \over I_{sc} }} \tag{3}$$
Equations (1) and (3) indicate that for any linear bilateral dc or
ac network with or without dependent sources of any type, if the open-circuit terminal voltage of a portion of a network can be determined
along with the short-circuit current between the same two terminals, the
Thevenin equivalent circuit is effectively known. A few examples will
make the method quite clear. The advantage of the method, which was
stressed earlier in this section for independent sources, should now be
more obvious. The current $I_{sc}$, which is necessary to find $Z_{Th}$, is in general more difficult to obtain since all of the sources are present.
Example 2:
Determine the Thevenin equivalent circuit for the network of
Fig. 8.
Fig. 8: Example 2.
Solution: Since for each approach the Thevenin voltage is found in
exactly the same manner, it will be determined first. From Fig. 8,
where $I_{X_{C}} = 0$,
$$ V_{R_{2}} = - { R_2 (\mu V) \over R_1 + R_2} = - { \mu R_2 V \over R_1 + R_2} $$
See Fig. 9.
Fig. 9: Determining the Thevenin impedance for the
network of Fig. 8.
$$Z_{Th} = R_1 || R_2 - j X_C$$
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