Sometimes we can make effective use of source transformations to derive a Thévenin or Norton equivalent circuit. For example, we can derive the Thévenin and Norton equivalents of the circuit shown in Fig. 4.45 by making the series of source transformations shown in Fig. 4.48. This technique is most useful when the network contains only independent sources. The presence of dependent sources requires retaining the identity of the controlling voltages and/or currents, and this constraint usually prohibits continued reduction of the circuit by source transformations. We discuss the problem of finding the Thévenin equivalent when a circuit contains dependent sources in Example 4.10.
24 Æ RTh =
VTh
isc
= 32
4 = 8 Æ. isc =
16
4 = 4A.
v2 = 16V.
v2
20
Step 1:
Source transformation
5 4
25 V 3 A
a
b
20 5
Step 2:
Parallel sources and parallel resistors combined
4 3 A 5 A
a
b
Step 3:
Source transformation; series resistors combined, producing the Thévenin equivalent circuit
4 a
b 4 8 A
a
b 8 4 A
Step 4:
Source transformation, producing the Norton equivalent circuit 8
a
b 32 V
Figure 4.48 Step-by-step derivation of the Thévenin and Norton equivalents of the circuit shown in Fig. 4.45.
8
32 V
a
b
Figure 4.47 The Thévenin equivalent of the circuit shown in Fig. 4.45.
Example 4.10 Finding the Thévenin Equivalent of a Circuit with a Dependent Source
Find the Thévenin equivalent for the circuit con- taining dependent sources shown in Fig. 4.49.
Figure 4.49A circuit used to illustrate a Thévenin equivalent when the circuit contains dependent sources.
Solution
The first step in analyzing the circuit in Fig. 4.49 is to recognize that the current labeled must be zero. (Note the absence of a return path for to enter the left-hand portion of the circuit.) The open-circuit, or Thévenin, voltage will be the volt- age across the resistor. With
The current iis
In writing the equation for i, we recognize that the Thévenin voltage is identical to the control voltage.
When we combine these two equations, we obtain
To calculate the short-circuit current, we place a short circuit across a,b. When the terminals a,b are shorted together, the control voltage is reduced to zero. Therefore, with the short in place, the circuit shown in Fig. 4.49 becomes the one shown in Fig. 4.50. With the short circuit shunting the resistor, all the current from the dependent current source appears in the short, so
isc = -20i.
25 Æ v
VTh = -5V.
i =
5 - 3v 2000 =
5 - 3VTh 2000 . VTh = vab = (-20i)(25) = -500i.
ix = 0, 25 Æ
ix
ix
25 2 k
5 V 3 v 20 i
vab
v
a
b i
ix
Figure 4.50The circuit shown in Fig. 4.49 with terminals a and b short-circuited.
As the voltage controlling the dependent volt- age source has been reduced to zero, the current controlling the dependent current source is
Combining these two equations yields a short-circuit current of
From and we get
Figure 4.51 illustrates the Thévenin equivalent for the circuit shown in Fig. 4.49. Note that the ref- erence polarity marks on the Thévenin voltage source in Fig. 4.51 agree with the preceding equa- tion for
Figure 4.51The Thévenin equivalent for the circuit shown in Fig. 4.49.
100
5 V
a
b VTh.
RTh = VTh
isc = -5
-50 * 103 = 100 Æ. VTh
isc
isc = -20(2.5) = -50mA.
i = 5
2000 = 2.5mA.
25 2 k
5 V 20 i
isc
a
b i
4.11 More on Deriving a Thévenin Equivalent 117
4.11 More on Deriving a Thévenin Equivalent
The technique for determining that we discussed and illustrated in Section 4.10 is not always the easiest method available. Two other meth- ods generally are simpler to use. The first is useful if the network contains only independent sources. To calculate for such a network, we first deactivate all independent sources and then calculate the resistance seen looking into the network at the designated terminal pair. A voltage source is deactivated by replacing it with a short circuit. A current source is deac- tivated by replacing it with an open circuit. For example, consider the cir- cuit shown in Fig. 4.52. Deactivating the independent sources simplifies the circuit to the one shown in Fig. 4.53. The resistance seen looking into the terminals a,b is denoted which consists of the resistor in series with the parallel combinations of the 5 and resistors. Thus,
(4.63) Note that the derivation of with Eq. 4.63 is much simpler than the same derivation with Eqs. 4.57–4.62.
RTh Rab = RTh = 4 +
5 * 20
25 = 8 Æ. 20 Æ
4 Æ Rab,
RTh RTh
20
5 4
25 V 3 A vab
a
b
Figure 4.52 A circuit used to illustrate a Thévenin equivalent.
a
b 5
20
4
Rab
Figure 4.53 The circuit shown in Fig. 4.52 after deac- tivation of the independent sources.
Objective 5—Understand Thévenin and Norton equivalents 4.16 Find the Thévenin equivalent circuit with respect
to the terminals a,b for the circuit shown.
Answer:
4.17 Find the Norton equivalent circuit with respect to the terminals a,b for the circuit shown.
Answer: (directed toward a), .
4.18 A voltmeter with an internal resistance of is used to measure the voltage in the circuit shown. What is the voltmeter reading?
Answer: 120 V.
vAB
100 kÆ
RN = 7.5 Æ IN = 6 A
Vab = VTh = 64.8 V, RTh = 6 Æ.
18 mA 60 k
12 k 15 k
36 V vAB
A
B
8 12
2
10 15 A
a
b
12
5 8
20 72 V
a
b
NOTE: Also try Chapter Problems 4.64, 4.68, and 4.72.
A S S E S S M E N T P R O B L E M S
If the circuit or network contains dependent sources, an alternative procedure for finding the Thévenin resistance is as follows. We first deactivate all independent sources, and we then apply either a test voltage source or a test current source to the Thévenin terminals a,b. The Thévenin resistance equals the ratio of the voltage across the test source to the cur- rent delivered by the test source. Example 4.11 illustrates this alternative procedure for finding RTh,using the same circuit as Example 4.10.
RTh
Example 4.11 Finding the Thévenin Equivalent Using a Test Source
Find the Thévenin resistance for the circuit in Fig. 4.49, using the alternative method described.
Solution
We first deactivate the independent voltage source from the circuit and then excite the circuit from the terminals a,b with either a test voltage source or a test current source. If we apply a test voltage source, we will know the voltage of the dependent voltage source and hence the controlling current i. Therefore we opt for the test voltage source. Figure 4.54 shows the circuit for computing the Thévenin resistance.
Figure 4.54An alternative method for computing the Thévenin resistance.
i 2 k
25 3 vT
20 i
iT
vT
RTh The externally applied test voltage source is
denoted and the current that it delivers to the circuit is labeled To find the Thévenin resistance, we simply solve the circuit shown in Fig. 4.54 for the ratio of the voltage to the current at the test source;
that is, From Fig. 4.54,
(4.64)
(4.65) We then substitute Eq. 4.65 into Eq. 4.64 and solve the resulting equation for the ratio :
(4.66)
(4.67)
From Eqs. 4.66 and 4.67,
(4.68) RTh =
vT
iT = 100 Æ. iT
vT
= 1 25 -
6 200 =
50 5000 =
1 100. iT =
vT
25 - 60vT
2000, vT>iT
i = -3vT
2 mA.
iT = vT
25 + 20i, RTh = vT>iT.
iT. vT,
In general, these computations are easier than those involved in com- puting the short-circuit current. Moreover, in a network containing only resistors and dependent sources, you must use the alternative method, because the ratio of the Thévenin voltage to the short-circuit current is indeterminate. That is, it is the ratio 0>0.