Thevenins Theorem (ac)
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Fig. 1: Thevenin equivalent circuit for ac networks.
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.
Solution:
Steps 1 and 2 (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):
$$ \begin{split}
E_{Th} &= { Z_2 E \over Z_1 + Z_2}\\
&= {(-j 2 Ω)(10 V) \over j 8 Ω - j 2 Ω}\\
&={ -j 20 V \over j6}\\
&=3.33 \angle -180^\circ\\
\end{split}$$
Step 5: The Thevenin equivalent circuit is shown in Fig. 6.
Fig. 2: Example 1.
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.
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.
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.
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.
$$Z_{Th} = R_1 || R_2 - j X_C$$
Fig. 8: Example 2.
Fig. 9: Determining the Thevenin impedance for the
network of Fig. 8.
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