5S Graphic World: Network Theorems

Network Theorems

9.1 INTRODUCTION
This chapter will introduce the important fundamental theorems of network analysis. Included are the superposition, Thévenin’s, Norton’s, maximum power transfer, substitution, Millman’s, and reciprocity theorems. We will consider a number of areas of application for each. Athorough understanding of each theorem is important because a number of the theorems will be applied repeatedly in the material to follow.
9.2 SUPERPOSITION THEOREM
The superposition theorem, like the methods of the last chapter, can be used to ﬁnd the solution to networks with two or more sources that are not in series or parallel. The most obvious advantage of this method is that it does not require the use of a mathematical technique such as determinants to ﬁnd the required voltages or currents. Instead, each source is treated independently, and the algebraic sum is found to determine a particular unknown quantity of the network. The superposition theorem states the following:
The current through, or voltage across, an element in a linear bilateral network is equal to the algebraic sum of the currents or voltages produced independently by each source.
When one is applying the theorem, it is possible to consider the effects of two sources at the same time and reduce the number of networks that have to be analyzed, but, in general,
(9.1)
To consider the effects of each source independently requires that sources be removed and replaced without affecting the ﬁnal result. To
Number of networks Number of to be analyzed independent sources
Th
Network Theorems
322  NETWORK THEOREMS Th
EI
FIG. 9.1 Removing the effects of ideal sources. Rint RintI Rint
E
Rint
FIG. 9.2 Removing the effects of practical sources.
remove a voltage source when applying this theorem, the difference in potential between the terminals of the voltage source must be set to zero (short circuit); removing a current source requires that its terminals be opened (open circuit). Any internal resistance or conductance associated with the displaced sources is not eliminated but must still be considered. Figure 9.1 reviews the various substitutions required when removing an ideal source, and Figure 9.2 reviews the substitutions with practical sources that have an internal resistance.
(b) P2 = I22R
R
I2
(c) PT = IT2R
R
IT
(a) P1 = I12R
R
I1
FIG. 9.3 Demonstration of the fact that superposition is not applicable to power effects.
The total current through any portion of the network is equal to the algebraic sum of the currents produced independently by each source. That is, for a two-source network, if the current produced by one source is in one direction, while that produced by the other is in the opposite direction through the same resistor, the resulting current is the difference of the two and has the direction of the larger. If the individual currents are in the same direction, the resulting current is the sum of two in the direction of either current. This rule holds true for the voltage across a portion of a network as determined by polarities, and it can be extended to networks with any number of sources. The superposition principle is not applicable to power effects since the power loss in a resistor varies as the square (nonlinear) of the current or voltage. For instance, the current through the resistor R of Fig. 9.3(a) is I1 due to one source of a two-source network. The current through the same resistor due to the other source is I2 as shown in Fig. 9.3(b). Applying the superposition theorem, the total current through the resistor due to both sources is IT,asshown in Fig. 9.3(c) with
IT I1 I2
The power delivered to the resistor in Fig. 9.3(a) is P1 I2 1R
while the power delivered to the same resistor in Fig. 9.3(b) is P2 I2 2R
If we assume that the total power delivered in Fig. 9.3(c) can be obtained by simply adding the power delivered due to each source, we ﬁnd that PT P1 P2 I2 1R I2 2R I2 T R or I2 T I2 1 I2 2
SUPERPOSITION THEOREM  323Th
R1 6 E + – 30 V I 3 A
I1
FIG. 9.4 Example 9.1.
R1 6
I
I
I
3 A
(a) (b)
E 30 V + –
I′ 1 I″1
R1 6
FIG. 9.5 (a) The contribution of I to I1; (b) the contribution of E to I1.
As shown in Fig. 9.5(a), the source current will choose the shortcircuit path, and I′1 0 A. If we applied the current divider rule,
I′1 0 A
Setting I to zero amperes will result in the network of Fig. 9.5(b), with the current source replaced by an open circuit. Applying Ohm’s law,
I″1 5 A
Since I′1 and I″1 have the same deﬁned direction in Fig. 9.5(a) and (b), the current I1 is the sum of the two, and I1 I′1 I″1 0 A 5 A 5 A
Note in this case that the current source has no effect on the current through the 6- resistor. The voltage across the resistor must be ﬁxed at 30 V because they are parallel elements.
30 V 6
E R1
(0 )I 0 6
RscI Rsc R1
This ﬁnal relationship between current levels is incorrect, however, as can be demonstrated by taking the total current determined by the superposition theorem and squaring it as follows: I2 T (I1 I2)2 I2 1 I2 2 2I1I2
which is certainly different from the expression obtained from the addition of power levels. In general, therefore,
the total power delivered to a resistive element must be determined using the total current through or the total voltage across the element and cannot be determined by a simple sum of the power levels established by each source.
EXAMPLE 9.1 Determine I1 for the network of Fig. 9.4.
Solution: Setting E 0 V for the network of Fig. 9.4 results in the network of Fig. 9.5(a), where a short-circuit equivalent has replaced the 30-V source.
FIG. 9.8 The effect of E2 on the current I3.
R1
24
R2 12
+
– E2 48 V
4
I″ 3
R3
54-V battery replaced by short circuit
R2 12
+
– E2 48 V
4
I″ 3
R3
R1 24
8
RT
324  NETWORK THEOREMS
Solution: Considering the effects of a 54-V source (Fig. 9.7): RT R1 R2 R3 24 12 4 24 3 27
I 2 A 54 V 27 E1 RT
Th
E1 – +
54 V R2 12
R3 = 4
48-V battery replaced by short circuit
R1
24
R1
24
R3 4 R 2 12
3
RT
E1
+
–
I 3
I
I 3
54 V
FIG. 9.7 The effect of E1 on the current I3.
EXAMPLE 9.2 Using superposition, determine the current through the 4- resistor of Fig. 9.6. Note that this is a two-source network of the type considered in Chapter 8.
R1
24
+
– E1 54 V R2 12
+
– E2 48 V
R3
4
I3
FIG. 9.6 Example 9.2.
Using the current divider rule,
I′3 1.5 A
Considering the effects of the 48-V source (Fig. 9.8): RT R3 R1 R2 4 24 12 4 8 12
I″3 4 A 48 V 12 E2 RT
24 A 16
(12 )(2 A) 12 4
R2I R2 R3
SUPERPOSITION THEOREM  325Th
4
I"3 = 4 A
I'3 = 1.5 A
FIG. 9.9 The resultant current for I3.
R2 6
R1
12
E 36 V + –
I
I2
9 A
FIG. 9.10 Example 9.3.
Current source replaced by open circuit
36 VE
I 2
R1
12
+
–
R2 6
FIG. 9.11 The contribution of E to I2.
R2 6
R1
12
I = 9 A
I″2
FIG. 9.12 The contribution of I to I2.
I′ 2 = 2 A I″ 2 = 6 A I2 = 8 AR 2 6 R2 6
Same direction
FIG. 9.13 The resultant current for I2.
The total current through the 4- resistor (Fig. 9.9) is
I3 I″3 I′3 4 A 1.5 A 2.5 A (direction of I″3)
EXAMPLE 9.3 a. Using superposition, ﬁnd the current through the 6- resistor of the network of Fig. 9.10.
b. Demonstrate that superposition is not applicable to power levels.
Solutions:
a. Considering the effect of the 36-V source (Fig. 9.11):
I′2 2 A
Considering the effect of the 9-A source (Fig. 9.12): Applying the current divider rule,
I″2 6 A
The total current through the 6- resistor (Fig. 9.13) is I2 I′2 I″2 2 A 6 A 8 A
108 A 18
(12 )(9 A) 12 6
R1I R1 R2
36 V 12 6
E R1 R2
E RT
b. The power to the 6- resistor is Power I2R (8 A)2(6 ) 384 W
The calculated power to the 6- resistor due to each source, misusing the principle of superposition, is P1 (I′2)2R (2 A)2(6 ) 24 W P2 (I″2)2R (6 A)2(6 ) 216 W P1 P2 240 W 384 W
326  NETWORK THEOREMS
Obviously, x y z, or 24 W 216 W 384 W, and superposition does not hold. However, for a linear relationship, such as that between the voltage and current of the ﬁxed-type 6- resistor, superposition can be applied, as demonstrated by the graph of Fig. 9.15, where a b c, or 2 A 6 A 8 A.
Th
4
3 2
1 0
12 24 36 48 V6Ω (V)
I (A)
a
c Linear curveb
8
7
6
5
9 10
FIG. 9.15 Plotting I versus V for the 6- resistor.
R1
6 k
R3
14 k
35 k
12 k
9 V R2
R4E
6 mAI
I2
FIG. 9.16 Example 9.4.
This results because 2 A 6 A 8 A, but (2 A)2 (6 A)2 (8 A)2
As mentioned previously, the superposition principle is not applicable to power effects since power is proportional to the square of the current or voltage (I2R or V2/R). Figure 9.14 is a plot of the power delivered to the 6- resistor versus current.
EXAMPLE 9.4 Using the principle of superposition, ﬁnd the current I2 through the 12-k resistor of Fig. 9.16.
Solution: Considering the effect of the 6-mA current source (Fig. 9.17):
FIG. 9.14 Plotting the power delivered to the 6- resistor versus current through the resistor.
400
300
200
100
x
012345678 I6Ω (A)
P (W)
y
z
{
Nonlinear curve
SUPERPOSITION THEOREM  327
Current divider rule:
I′2 2 mA
Considering the effect of the 9-V voltage source (Fig. 9.18):
I″2 0.5 mA 9 V 6 k 12 k E R1 R2
(6 k )(6 mA) 6 k 12 k
R1I R1 R2
Th
R1 6 k
R3 14 k
R2 12 k
R4 35 k
6 mAI
I′ 2
6 mA
6 mA
I′ 2
I
R4 35 k R 3 14 k
R2 12 k R 1 6 k
6 mA
FIG. 9.17 The effect of the current source I on the current I2.
R1
6 k
R2
12 k
R3
14 k
R4
35 k
9 V
E
R1 6 k
R3 14 k
R2 12 k
R4 35 k
+ – 9 V
+ – 9 V
9 V
E
I 2
I 2
FIG. 9.18 The effect of the voltage source E on the current I2.
R24
R1 2
I1
I 3 A
6 V12 V +–
E1
+ –
E2
FIG. 9.19 Example 9.5.
Since I′2 and I″2 have the same direction through R2, the desired current is the sum of the two: I2 I′2 I″2 2 mA 0.5 mA 2.5 mA
EXAMPLE 9.5 Find the current through the 2- resistor of the network of Fig. 9.19. The presence of three sources will result in three different networks to be analyzed.
328  NETWORK THEOREMS
Solution: Considering the effect of the 12-V source (Fig. 9.20):
I′1 2 A
Considering the effect of the 6-V source (Fig. 9.21):
I″1 1 A
Considering the effect of the 3-A source (Fig. 9.22): Applying the current divider rule,
I
1 2 A
The total current through the 2- resistor appears in Fig. 9.23, and
I1 I 1 I"1 I"1 ' ' 1A 1A 2A 2A
Same direction as I1 in Fig. 9.19
Opposite direction to I1 in Fig. 9.19
12 A 6
(4 )(3 A) 2 4
R2I R1 R2
6 V 6
6 V 2 4
E2 R1 R2
12 V 6
12 V 2 4
E1 R1 R2
Th
ETh
+
–
a
b
RTh
FIG. 9.24 Thévenin equivalent circuit.
9.3 THE ′VENIN’S THEOREM
Thévenin’s theorem states the following:
Any two-terminal, linear bilateral dc network can be replaced by an equivalent circuit consisting of a voltage source and a series resistor, as shown in Fig. 9.24.
In Fig. 9.25(a), for example, the network within the container has only two terminals available to the outside world, labeled a and b. It is possible using Thévenin’s theorem to replace everything in the container with one source and one resistor, as shown in Fig. 9.25(b), and maintain the same terminal characteristics at terminals a and b. That is, any load connected to terminals a and b will not know whether it is hooked up to the network of Fig. 9.25(a) or Fig. 9.25(b). The load will receive the same current, voltage, and power from either conﬁguration of Fig. 9.25. Throughout the discussion to follow, however, always keep in mind that
the Thévenin equivalent circuit provides an equivalence at the terminals only—the internal construction and characteristics of the original network and the Thévenin equivalent are usually quite different.
R24
R12
E1
12 V +–
I 1
I 1
I 1
FIG. 9.20 The effect of E1 on the current I.
R1 2 R1 2 I 1 = 2 A I 1 = 1 A I
1 = 2 A I1 = 1 A
I1
FIG. 9.23 The resultant current I1.
R24
R1
2
6 V
+
– E2I 1 I 1
I 1
FIG. 9.21 The effect of E2 on the current I1.
R24
R12 3 A I I
1
FIG. 9.22 The effect of I on the current I1.
THÉVENIN’S THEOREM  329Th
12 V
4 V
E2
8 V
a
b
+
–
a
b
E1
6
10
4
(a) (b)
FIG. 9.25 The effect of applying Thévenin’s theorem.
R2
R3R 1
a
b
(a) (b)
+
– RLE
a IL IL
RLE Th
RTh
b
+
–
FIG. 9.26 Substituting the Thévenin equivalent circuit for a complex network.
Although active in the study and design of telegraphic systems (including underground transmission), cylindrical condensers (capacitors), and electromagnetism, he is best known for a theorem ﬁrst presented in the French Journal of Physics—Theory and Applications in 1883. It appeared under the heading of “Sur un nouveau théorème d’électricité dynamique” (“On a new theorem of dynamic electricity”) and was originally referred to as the equivalent generator theorem. There is some evidence that a similar theorem was introduced by Hermann von Helmholtz in 1853. However, Professor Helmholtz applied the theorem to animal physiology and not to communication or generator systems, and therefore he has not received the credit in this ﬁeld that he might deserve. In the early 1920s AT&T did some pioneering work using the equivalent circuit and may have initiated the reference to the theorem as simply Thévenin’s theorem. In fact, Edward L. Norton, an engineer at AT&T at the time, introduced a current source equivalent of the Thévenin equivalent currently referred to as the Norton equivalent circuit. As an aside, Commandant Thévenin was an avid skier and in fact was commissioner of an international ski competition in Chamonix, France, in 1912.
LEON-CHARLES THÉVENIN
French (Meaux, Paris) (1857–1927) Telegraph Engineer, Commandant and Educator École Polytechnique and École Supérieure de Télégraphie
Courtesy of the Bibliothèque École Polytechnique, Paris, France
For the network of Fig. 9.25(a), the Thévenin equivalent circuit can be found quite directly by simply combining the series batteries and resistors. Note the exact similarity of the network of Fig. 9.25(b) to the Thévenin conﬁguration of Fig. 9.24. The method described below will allow us to extend the procedure just applied to more complex conﬁgurations and still end up with the relatively simple network of Fig. 9.24. In most cases, other elements will be connected to the right of terminals a and b in Fig. 9.25. To apply the theorem, however, the network to be reduced to the Thévenin equivalent form must be isolated as shown in Fig. 9.25, and the two “holding” terminals identiﬁed. Once the proper Thévenin equivalent circuit has been determined, the voltage, current, or resistance readings between the two “holding” terminals will be the same whether the original or the Thévenin equivalent circuit is connected to the left of terminals a and b in Fig. 9.25. Any load connected to the right of terminals a and b of Fig. 9.25 will receive the same voltage or current with either network. This theorem achieves two important objectives. First, as was true for all the methods previously described, it allows us to ﬁnd any particular voltage or current in a linear network with one, two, or any other number of sources. Second, we can concentrate on a speciﬁc portion of a network by replacing the remaining network with an equivalent circuit. In Fig. 9.26, for example, by ﬁnding the Thévenin equivalent circuit for the network in the shaded area, we can quickly calculate the change in current through or voltage across the variable resistor RL for the various values that it may assume. This is demonstrated in Example 9.6.
330  NETWORK THEOREMS
Before we examine the steps involved in applying this theorem, it is important that an additional word be included here to ensure that the implications of the Thévenin equivalent circuit are clear. In Fig. 9.26, the entire network, except RL,istobereplaced by a single series resistor and battery as shown in Fig. 9.24. The values of these two elements of the Thévenin equivalent circuit must be chosen to ensure that the resistor RL will react to the network of Fig. 9.26(a) in the same manner as to the network of Fig. 9.26(b). In other words, the current through or voltage across RL must be the same for either network for any value of RL. The following sequence of steps will lead to the proper value of RTh and ETh. Preliminary:
1. Remove that portion of the network across which the Thévenin equivalent circuit is to be found. In Fig. 9.26(a), this requires that the load resistor RL be temporarily removed from the network. 2. Mark the terminals of the remaining two-terminal network. (The importance of this step will become obvious as we progress through some complex networks.)
RTh:
3. Calculate RTh by ﬁrst setting all sources to zero (voltage sources are replaced by short circuits, and current sources by open circuits) and then ﬁnding the resultant resistance between the two marked terminals. (If the internal resistance of the voltage and/or current sources is included in the original network, it must remain when the sources are set to zero.)
ETh:
4. Calculate ETh by ﬁrst returning all sources to their original position and ﬁnding the open-circuit voltage between the marked terminals. (This step is invariably the one that will lead to the most confusion and errors. In all cases, keep in mind that it is the open-circuit potential between the two terminals marked in step 2.)
Conclusion:
5. Draw the Thévenin equivalent circuit with the portion of the circuit previously removed replaced between the terminals of the equivalent circuit. This step is indicated by the placement of the resistor RL between the terminals of the Thévenin equivalent circuit as shown in Fig. 9.26(b).
EXAMPLE 9.6 Find the Thévenin equivalent circuit for the network in the shaded area of the network of Fig. 9.27. Then ﬁnd the current through RL for values of 2 , 10 , and 100 .
Solution:
Steps 1 and 2 produce the network of Fig. 9.28. Note that the load resistor RL has been removed and the two “holding” terminals have been deﬁned as a and b. Step 3: Replacing the voltage source E1 with a short-circuit equivalent yields the network of Fig. 9.29(a), where
RTh R1 R2 2 (3 )(6 ) 3 6
Th
R2 6
R1
3
E1 9 V
a
b
+
–
R1
3
R2 6
b
E1 9 V RL
a
+
–
FIG. 9.27 Example 9.6.
FIG. 9.28 Identifying the terminals of particular importance when applying Thévenin’s theorem.
THÉVENIN’S THEOREM  331
The importance of the two marked terminals now begins to surface. They are the two terminals across which the Thévenin resistance is measured. It is no longer the total resistance as seen by the source, as determined in the majority of problems of Chapter 7. If some difﬁculty develops when determining RTh with regard to whether the resistive elements are in series or parallel, consider recalling that the ohmmeter sends out a trickle current into a resistive combination and senses the level of the resulting voltage to establish the measured resistance level. In Fig. 9.29(b), the trickle current of the ohmmeter approaches the network through terminal a, and when it reaches the junction of R1 and R2, it splits as shown. The fact that the trickle current splits and then recombines at the lower node reveals that the resistors are in parallel as far as the ohmmeter reading is concerned. In essence, the path of the sensing current of the ohmmeter has revealed how the resistors are connected to the two terminals of interest and how the Thévenin resistance should be determined. Keep the above in mind as you work through the various examples of this section. Step 4: Replace the voltage source (Fig. 9.30). For this case, the opencircuit voltage ETh is the same as the voltage drop across the 6- resistor. Applying the voltage divider rule,
ETh 6 V
It is particularly important to recognize that ETh is the open-circuit potential between points a and b. Remember that an open circuit can have any voltage across it, but the current must be zero. In fact, the current through any element in series with the open circuit must be zero also. The use of a voltmeter to measure ETh appears in Fig. 9.31. Note that it is placed directly across the resistor R2 since ETh and VR2 are in parallel. Step 5 (Fig. 9.32):
IL
RL 2 : IL 1.5 A
RL 10 : IL 0.5 A
RL 100 : IL 0.059 A 6 V 2 100
6 V 2 10
6 V 2 2
ETh RTh RL
54 V 9
(6 )(9 V) 6 3
R2E1 R2 R1
Th
FIG. 9.29 Determining RTh for the network of Fig. 9.28.
R2 6
(a) (b)
R1
3
RTh R2
bb
a a
I
R1
+ –
R2 6 9 VE 1 ETh + – +
–
a
b
R1
3
FIG. 9.30 Determining ETh for the network of Fig. 9.28.
+ –
R2 6
R1
3
9 VE 1
+
–
ETh
FIG. 9.31 Measuring ETh for the network of Fig. 9.28.
RL
a
RTh = 2
ETh = 6 V
b
IL
+
–
FIG. 9.32 Substituting the Thévenin equivalent circuit for the network external to RL in Fig. 9.27.
332  NETWORK THEOREMS
If Thévenin’s theorem were unavailable, each change in RL would require that the entire network of Fig. 9.27 be reexamined to ﬁnd the new value of RL.
EXAMPLE 9.7 Find the Thévenin equivalent circuit for the network in the shaded area of the network of Fig. 9.33.
Solution:
Steps 1 and 2 are shown in Fig. 9.34.
Step 3 is shown in Fig. 9.35. The current source has been replaced with an open-circuit equivalent, and the resistance determined between terminals a and b. In this case an ohmmeter connected between terminals a and b would send out a sensing current that would ﬂow directly through R1 and R2 (at the same level). The result is that R1 and R2 are in series and the Thévenin resistance is the sum of the two. RTh R1 R2 4 2 6
Step 4 (Fig. 9.36): In this case, since an open circuit exists between the two marked terminals, the current is zero between these terminals and through the 2- resistor. The voltage drop across R2 is, therefore, V2 I2R2 (0)R2 0 V and ETh V1 I1R1 IR1 (12 A)(4 ) 48 V
Th
R3 7
a
b
RTh = 6
ETh = 48 V
+
–
FIG. 9.37 Substituting the Thévenin equivalent circuit in the network external to the resistor R3 of Fig. 9.33.
Step 5 is shown in Fig. 9.37.
EXAMPLE 9.8 Find the Thévenin equivalent circuit for the network in the shaded area of the network of Fig. 9.38. Note in this example that
R2
2
R1 4 I12 A
a
b
FIG. 9.34 Establishing the terminals of particular interest for the network of Fig. 9.33.
R1 4
a
b
RTh
R2
2
FIG. 9.35 Determining RTh for the network of Fig. 9.34. R1 4 R2 = 2 I I = 12 A + –
I = 0 +
–
+ V2 = 0 V – a
b
ETh
FIG. 9.36 Determining ETh for the network of Fig. 9.34.
R3 7
R2
2
R1 4
a
b
12 A I =
FIG. 9.33 Example 9.7.
R2
4
R1 6 R2 4 R1 6
a
b
RTh
“Short circuited”
R3 2
Circuit redrawn:
RTh
a
b
RT = 0 2 = 0
FIG. 9.40 Determining RTh for the network of Fig. 9.39.
THÉVENIN’S THEOREM  333Th
R3 2 8 VE 1 – +
R4 3 R 1 6
R2
4 a
b
FIG. 9.38 Example 9.8.
R1 6
R2
4
R3 2 E 1 8 V
a
b
–
+
FIG. 9.39 Identifying the terminals of particular interest for the network of Fig. 9.38.
there is no need for the section of the network to be preserved to be at the “end” of the conﬁguration.
Solution:
Steps 1 and 2: See Fig. 9.39.
Step 3: See Fig. 9.40. Steps 1 and 2 are relatively easy to apply, but now we must be careful to “hold” onto the terminals a and b as the Thévenin resistance and voltage are determined. In Fig. 9.40, all the remaining elements turn out to be in parallel, and the network can be redrawn as shown.
RTh R1 R2 2.4 24 10 (6 )(4 ) 6 4
334  NETWORK THEOREMS Th
Solution:
Steps 1 and 2 are shown in Fig. 9.45.
Step 3: See Fig. 9.46. In this case, the short-circuit replacement of the voltage source E provides a direct connection between c and c′ of Fig. 9.46(a), permitting a “folding” of the network around the horizontal line of a-b to produce the conﬁguration of Fig. 9.46(b). RTh Ra-b R1 R3 R2 R4 6 3 4 12 2 3 5
R1
6
R2
12
4
R4R 3
3
b a72 VE +
–
FIG. 9.45 Identifying the terminals of particular interest for the network of Fig. 9.44.
Step 4: See Fig. 9.41. In this case, the network can be redrawn as shown in Fig. 9.42, and since the voltage is the same across parallel elements, the voltage across the series resistors R1 and R2 is E1, or 8 V. Applying the voltage divider rule,
ETh 4.8 V
Step 5: See Fig. 9.43.
The importance of marking the terminals should be obvious from Example 9.8. Note that there is no requirement that the Thévenin voltage have the same polarity as the equivalent circuit originally introduced.
EXAMPLE 9.9 Find the Thévenin equivalent circuit for the network in the shaded area of the bridge network of Fig. 9.44.
48 V 10
(6 )(8 V) 6 4
R1E1 R1 R2
R1 6 12
4
R2
RL R3 R4
3
baE 72 V + –
FIG. 9.44 Example 9.9.
ETh R1 6
R2
4
R3 2 E Th E1 8 V – + – + – + a b
FIG. 9.41 Determining ETh for the network of Fig. 9.39.
R4 3
RTh = 2.4
a
b
ETh = 4.8 V – +
FIG. 9.43 Substituting the Thévenin equivalent circuit for the network external to the resistor R4 of Fig. 9.38.
R2 4
ETh R1 6 R3 2 – + – + E1 8 V
FIG. 9.42 Network of Fig. 9.41 redrawn.
THÉVENIN’S THEOREM  335Th
R1
3
R1
6
R2
R3 R4
12
3 4
R2 R4
4
RTh ba
R3
12 6
(b)(a)
ab RTh
c,c′c ′
c
FIG. 9.46 Solving for RTh for the network of Fig. 9.45.
Step 4: The circuit is redrawn in Fig. 9.47. The absence of a direct connection between a and b results in a network with three parallel branches. The voltages V1 and V2 can therefore be determined using the voltage divider rule:
V1 48 V
V2 54 V 864 V 16 (12 )(72 V) 12 4 R2E R2 R4
432 V 9
(6 )(72 V) 6 3
R1E R1 R3
V1 R1 6
R3 3
R2 12
R4 4
KVL
+
–
72 V
+
– +
V2
ba ETh –
+
E E
–
Assuming the polarity shown for ETh and applying Kirchhoff’s voltage law to the top loop in the clockwise direction will result in
V ETh V1 V2 0 and ETh V2 V1 54 V 48 V 6 V
Step 5 is shown in Fig. 9.48.
Thévenin’s theorem is not restricted to a single passive element, as shown in the preceding examples, but can be applied across sources, whole branches, portions of networks, or any circuit conﬁguration, as shown in the following example. It is also possible that one of the methods previously described, such as mesh analysis or superposition, may have to be used to ﬁnd the Thévenin equivalent circuit.
EXAMPLE 9.10 (Two sources) Find the Thévenin circuit for the network within the shaded area of Fig. 9.49.

FIG. 9.47 Determining ETh for the network of Fig. 9.45.
R4
1.4 k
R3 6 k RLR 1 0.8 k
R2 4 k
E2 + 10 V
E1 – 6 V
RL
RTh = 5
ETh = 6 V
–
+
a
b
FIG. 9.48 Substituting the Thévenin equivalent circuit for the network external to the resistor RL of Fig. 9.44.
FIG. 9.49 Example 9.10.
336  NETWORK THEOREMS
Solution: The network is redrawn and steps 1 and 2 are applied as shown in Fig. 9.50.
Th
RTh
2 k
RL3 VE Th
FIG. 9.54 Substituting the Thévenin equivalent circuit for the network external to the resistor RL of Fig. 9.49.
R1 0.8 k
R4
1.4 k R 2 4 k R3 6 k E1 6 V E2 10 V + – – +
a
b
FIG. 9.50 Identifying the terminals of particular interest for the network of Fig. 9.49.
R2 4 k
R3 6 k
E2 10 V
I4 = 0
E ThV 3
R4
1.4 k
V4 +–
+
–
+
–
R1 0.8 k
FIG. 9.53 Determining the contribution to ETh from the source E2 for the network of Fig. 9.50.
R2
4 k
R3 6 k
R1 0.8 k
RTh
2.4 k
a
b
R4
1.4 k
FIG. 9.51 Determining RTh for the network of Fig. 9.50.
R3 6 k
V4
1.4 k
E1
0.8 k R2 4 k
R1
R4
6 V
I4 = 0
–
+
– +
V3 + –
E Th
FIG. 9.52 Determining the contribution to ETh from the source E1 for the network of Fig. 9.50.
Step 3: See Fig. 9.51. RTh R4 R1 R2 R3 1.4 k 0.8 k 4 k 6 k 1.4 k 0.8 k 2.4 k 1.4 k 0.6 k 2 k
Step 4: Applying superposition, we will consider the effects of the voltage source E1 ﬁrst. Note Fig. 9.52. The open circuit requires that V4 I4R4 (0)R4 0 V,and
E′Th V3 R′T R2 R3 4 k 6 k 2.4 k
Applying the voltage divider rule,
V3 R R ′T ′ E T 1 R1 4.5 V E′Th V3 4.5 V
For the source E2, the network of Fig. 9.53 will result. Again, V4 I4R4 (0)R4 0 V, and
E″Th V3 R′T R1 R3 0.8 k 6 k 0.706 k
and V3 R R ′T ′T E R 2 2 1.5 V E″Th V3 1.5 V
Since E′Th and E″Th have opposite polarities, ETh E′Th E″Th 4.5 V 1.5 V 3 V (polarity of E′Th)
Step 5: See Fig. 9.54.
Experimental Procedures
There are two popular experimental procedures for determining the parameters of a Thévenin equivalent network. The procedure for measuring the Thévenin voltage is the same for each, but the approach for determining the Thévenin resistance is quite different for each.
7.06 V 4.706
(0.706 k )(10 V) 0.706 k 4 k
14.4 V 3.2
(2.4 k )(6 V) 2.4 k 0.8 k
FIG. 9.56 Determining RTh experimentally.
+ –
+ –
RL = RTh
RTh
b
a
ETh
+– ETh 2 ETh 2 +
–
(a) (b)
b
a
+
–
Network
ETh 2
V

RTh
(c)
THÉVENIN’S THEOREM  337
Direct Measurement of ETh and RTh For any physical network, the value of ETh can be determined experimentally by measuring the open-circuit voltage across the load terminals, as shown in Fig. 9.55; ETh Voc Vab. The value of RTh can then be determined by completing the network with a variable RL such as the potentiometer of Fig. 9.56(b). RL can then be varied until the voltage appearing across the load is one-half the open-circuit value, or VL ETh/2. For the series circuit of Fig. 9.56(a), when the load voltage is reduced to one-half the open-circuit level, the voltage across RTh and RL must be the same. If we read the value of RL [as shown in Fig. 9.56(c)] that resulted in the preceding calculations, we will also have the value of RTh, since RL RTh if VL equals the voltage across RTh.
Th
+ –
Network
RTh + V = 0 V –
ETh
I = 0
a
b
(a)
ETh
+
–
Open circuit
a
b
+
–
(b)
ETh
V
FIG. 9.55 Determining ETh experimentally.
Measuring Voc and Isc The Thévenin voltage is again determined by measuring the open-circuit voltage across the terminals of interest;
American (Rockland, Maine; Summit, New Jersey) (1898–1983) Electrical Engineer, Scientist, Inventor Department Head: Bell Laboratories Fellow: Acoustical Society and Institute of Radio Engineers
Although interested primarily in communications circuit theory and the transmission of data at high speeds over telephone lines, Edward L. Norton is best remembered for development of the dual of Thévenin’s equivalent circuit, currently referred to as Norton’s equivalent circuit. In fact, Norton and his associates at AT&T in the early 1920s are recognized as some of the ﬁrst to perform pioneering work applying Thévenin’s equivalent circuit and who referred to this concept simply as Thévenin’s theorem. In 1926 he proposed the equivalent circuit using a current source and parallel resistor to assist in the design of recording instrumentation that was primarily current driven. He began his telephone career in 1922 with the Western Electric Company’s Engineering Department, which later became Bell Laboratories. His areas of active research included network theory, acoustical systems, electromagnetic apparatus, and data transmission. A graduate of MIT and Columbia University, he held nineteen patents on his work.
EDWARD L. NORTON
338  NETWORK THEOREMS
that is, ETh Voc. To determine RTh, a short-circuit condition is established across the terminals of interest, as shown in Fig. 9.57, and the current through the short circuit is measured with an ammeter. Using Ohm’s law, we ﬁnd that the short-circuit current is determined by
Isc
and the Thévenin resistance by
RTh
However, ETh Voc resulting in the following equation for RTh:
(9.2)
9.4 NORTON’S THEOREM
It was demonstrated in Section 8.3 that every voltage source with a series internal resistance has a current source equivalent. The current source equivalent of the Thévenin network (which, you will note, satisﬁes the above conditions), as shown in Fig. 9.58, can be determined by Norton’s theorem. It can also be found through the conversions of Section 8.3. The theorem states the following:
Any two-terminal linear bilateral dc network can be replaced by an equivalent circuit consisting of a current source and a parallel resistor, as shown in Fig. 9.58.
The discussion of Thévenin’s theorem with respect to the equivalent circuit can also be applied to the Norton equivalent circuit. The steps leading to the proper values of IN and RN are now listed.
Preliminary:
1. Remove that portion of the network across which the Norton equivalent circuit is found. 2. Mark the terminals of the remaining two-terminal network.
RN:
3. Calculate RN by ﬁrst setting all sources to zero (voltage sources are replaced with short circuits, and current sources with open circuits) and then ﬁnding the resultant resistance between the two marked terminals. (If the internal resistance of the voltage and/or current sources is included in the original network, it must remain when the sources are set to zero.) Since RN RTh, the procedure and value obtained using the approach described for Thévenin’s theorem will determine the proper value of RN.
IN:
4. Calculate IN by ﬁrst returning all sources to their original position and then ﬁnding the short-circuit current between the marked terminals. It is the same current that would be measured by an ammeter placed between the marked terminals.
RTh V Is o c c
ETh Isc
ETh RTh
Th
+ –
RTh
ETh
a
b
I
Isc
FIG. 9.57 Measuring Isc.
RNI N
a
b
FIG. 9.58 Norton equivalent circuit.
Courtesy of AT&T Archives
NORTON’S THEOREM  339Th
RTh = RN
ETh RTh RN = RThE Th = INRN
+
–
IN
FIG. 9.59 Converting between Thévenin and Norton equivalent circuits.
R2 6
R1
3
RL9 VE + –
a
b
FIG. 9.60 Example 9.11.
R1
3
R2 6 9 VE + –
a
b
FIG. 9.61 Identifying the terminals of particular interest for the network of Fig. 9.60.
R2 6
R1
3
RN
a
b
FIG. 9.62 Determining RN for the network of Fig. 9.61.
Conclusion:
5. Draw the Norton equivalent circuit with the portion of the circuit previously removed replaced between the terminals of the equivalent circuit.
The Norton and Thévenin equivalent circuits can also be found from each other by using the source transformation discussed earlier in this chapter and reproduced in Fig. 9.59.
V2 R2 6
R1
3
Short circuited
E 9 V
Short
+
–
+
–
a
b
I1 IN IN
IN
I2 = 0
FIG. 9.63 Determining IN for the network of Fig. 9.61.
EXAMPLE 9.11 Find the Norton equivalent circuit for the network in the shaded area of Fig. 9.60.
Solution:
Steps 1 and 2 are shown in Fig. 9.61.
Step 3 is shown in Fig. 9.62, and
RN R1 R2 3 6 2
Step 4 is shown in Fig. 9.63, clearly indicating that the short-circuit connection between terminals a and b is in parallel with R2 and eliminates its effect. IN is therefore the same as through R1, and the full battery voltage appears across R1 since V2 I2R2 (0)6 0 V
Therefore,
IN 3 A 9 V 3 E R1
18 9
(3 )(6 ) 3 6
340  NETWORK THEOREMS Th
EXAMPLE 9.12 Find the Norton equivalent circuit for the network external to the 9- resistor in Fig. 9.66.
Solution:
Steps 1 and 2: See Fig. 9.67.
Step 5: See Fig. 9.64. This circuit is the same as the ﬁrst one considered in the development of Thévenin’s theorem. A simple conversion indicates that the Thévenin circuits are, in fact, the same (Fig. 9.65).
FIG. 9.69 Determining IN for the network of Fig. 9.67.
10 A
R2 4
R1
5
a
b
IN
R1
5
ba
IR 2 4
I
IN
10 A
RTh = RN = 2
IN RN = 2
3 A
a
b
ETh = INRN = (3 A)(2 ) = 6 V
a
b
FIG. 9.65 Converting the Norton equivalent circuit of Fig. 9.64 to a Thévenin equivalent circuit.
R1
5
10 A
a
b
RL 9 R 2 4
I
R2 4
R1
5
10 A
a
b
I
FIG. 9.66 Example 9.12.
FIG. 9.67 Identifying the terminals of particular interest for the network of Fig. 9.66.
R2 4
R1
5
a
b
RN
FIG. 9.68 Determining RN for the network of Fig. 9.67.
RLR N = 2 I N = 3 A
a
b
FIG. 9.64 Substituting the Norton equivalent circuit for the network external to the resistor RL of Fig. 9.60.
Step 3: See Fig. 9.68, and
RN R1 R2 5 4 9
Step 4: As shown in Fig. 9.69, the Norton current is the same as the current through the 4- resistor. Applying the current divider rule,
IN 5.556 A 50 A 9 (5 )(10 A) 5 4 R1I R1 R2
Th NORTON’S THEOREM  341
Step 5: See Fig. 9.70.
9 RL 9 I N 5.556 A RN
a
b
EXAMPLE 9.13 (Two sources) Find the Norton equivalent circuit for the portion of the network to the left of a-b in Fig. 9.71.
FIG. 9.70 Substituting the Norton equivalent circuit for the network external to the resistor RL of Fig. 9.66.
R3 9 R4 10 R 2 6
R1 4
E1 7 V
I 8 A
E2 12 V
b
a
FIG. 9.71 Example 9.13.
R1 4
R2 6
E1 7 V
I 8 A
a
b
FIG. 9.72 Identifying the terminals of particular interest for the network of Fig. 9.71.
R1 4
R2 6
a
b
RN
FIG. 9.73 Determining RN for the network of Fig. 9.72.
R2 6
R1 4
Short circuited
E1 7 V
I N
a
b
I N I N
FIG. 9.74 Determining the contribution to IN from the voltage source E1.
Solution:
Steps 1 and 2: See Fig. 9.72.
Step 3 is shown in Fig. 9.73, and
RN R1 R2 4 6 2.4 24 10 (4 )(6 ) 4 6
Step 4: (Using superposition) For the 7-V battery (Fig. 9.74),
342  NETWORK THEOREMS
I′N 1.75 A
For the 8-A source (Fig. 9.75), we ﬁnd that both R1 and R2 have been “short circuited” by the direct connection between a and b, and
I″N I 8 A
The result is
IN I″N I′N 8 A 1.75 A 6.25 A
Step 5: See Fig. 9.76.
7 V 4
E1 R1
Th
Experimental Procedure
The Norton current is measured in the same way as described for the short-circuit current for the Thévenin network. Since the Norton and Thévenin resistances are the same, the same procedures can be employed as described for the Thévenin network.
9.5 MAXIMUM POWER TRANSFER THEOREM
The maximum power transfer theorem states the following:
A load will receive maximum power from a linear bilateral dc network when its total resistive value is exactly equal to the Thévenin resistance of the network as “seen” by the load.
For the network of Fig. 9.77, maximum power will be delivered to the load when
(9.3)
From past discussions, we realize that a Thévenin equivalent circuit can be found across any element or group of elements in a linear bilateral dc network. Therefore, if we consider the case of the Thévenin equivalent circuit with respect to the maximum power transfer theorem, we are, in essence, considering the total effects of any network across a resistor RL, such as in Fig. 9.77. For the Norton equivalent circuit of Fig. 9.78, maximum power will be delivered to the load when
RL RTh
R1 4
R2 6 I 8 A I N
a
b
I N I N
I N
Short circuited
FIG. 9.75 Determining the contribution to IN from the current source I.
IN 6.25 A
R3 9 RN = 2.4 E2 12 V
R4 10
a
b
FIG. 9.76 Substituting the Norton equivalent circuit for the network to the left of terminals a-b in Fig. 9.71.
RL
IR Th
ETh + –
FIG. 9.77 Deﬁning the conditions for maximum power to a load using the Thévenin equivalent circuit.
MAXIMUM POWER TRANSFER THEOREM  343Th
RL
IL
RTh
ETh
+
–
9
60 V VL
PL
FIG. 9.79 Thévenin equivalent network to be used to validate the maximum power transfer theorem.
RL
I
RNI N
FIG. 9.78 Deﬁning the conditions for maximum power to a load using the Norton equivalent circuit.
(9.4)
This result [Eq. (9.4)] will be used to its fullest advantage in the analysis of transistor networks, where the most frequently applied transistor circuit model employs a current source rather than a voltage source. For the network of Fig. 9.77,
I and PL I2RL 2 RL
so that PL
Let us now consider an example where ETh 60 V and RTh 9 , as shown in Fig. 9.79.
E2ThRL (RTh RL)2
ETh RTh RL
ETh RTh RL
RL RN
The power to the load is determined by
PL
with IL
and VL
A tabulation of PL for a range of values of RL yields Table 9.1. A plot of PL versus RL using the data of Table 9.1 will result in the plot of Fig. 9.80 for the range RL 0.1 to 30 .
RL(60 V) 9 RL
RL(60 V) RTh RL
60 V 9 RL
ETh RTh RL
3600RL (9 RL)2
E2 ThRL (RTh RL)2
344  NETWORK THEOREMS Th
TABLE 9.1
RL ( ) PL (W) IL (A) VL (V)
0.1 4.35 6.59 0.66 0.2 8.51 6.52 1.30 0.5 19.94 6.32 3.16 1 36.00 6.00 6.00 2 59.50 5.46 10.91 3 75.00 5.00 15.00 4 85.21 4.62 18.46 5 91.84 Increase 4.29 Decrease 21.43 Increase 6 96.00 4.00 24.00 7 98.44 3.75 26.25 8 99.65 3.53 28.23 9 (RTh) 100.00 (Maximum) 3.33 (Imax/2) 30.00 (ETh/2) 10 99.72 3.16 31.58 11 99.00 3.00 33.00 12 97.96 2.86 34.29 13 96.69 2.73 35.46 14 95.27 2.61 36.52 15 93.75 2.50 37.50 16 92.16 2.40 38.40 17 90.53 Decrease 2.31 Decrease 39.23 Increase 18 88.89 2.22 40.00 19 87.24 2.14 40.71 20 85.61 2.07 41.38 25 77.86 1.77 44.12 30 71.00 1.54 46.15 40 59.98 1.22 48.98 100 30.30 0.55 55.05 500 6.95 0.12 58.94 1000 3.54 0.06 59.47 555 5 5 5
PL
PL (W)
0 59 10 15 20 25 30 RL ( )
10
20
30
40
50
60
70
80
90
100
RL = RTh = 9
FIG. 9.80 PL versus RL for the network of Fig. 9.79.
MAXIMUM POWER TRANSFER THEOREM  345
Note, in particular, that PL is, in fact, a maximum when RL RTh 9 . The power curve increases more rapidly toward its maximum value than it decreases after the maximum point, clearly revealing that a small change in load resistance for levels of RL below RTh will have a more dramatic effect on the power delivered than similar changes in RL above the RTh level. If we plot VL and IL versus the same resistance scale (Fig. 9.81), we ﬁnd that both change nonlinearly, with the terminal voltage increasing with an increase in load resistance as the current decreases. Note again that the most dramatic changes in VL and IL occur for levels of RL less than RTh. As pointed out on the plot, when RL RTh, VL ETh/2 and IL Imax/2, with Imax ETh/RTh.
Th
FIG. 9.81 VL and IL versus RL for the network of Fig. 9.79.
50
IL
40
30
20
10
0 0 5 10 15 20 25 30 9 RL ( )
1
2
3
4
5
6
7
8
VL (V) IL (A)
Imax = ETh /RL = 6.67 A
ETh /2
Imax /2
VL
RL = RTh = 9
The dc operating efﬁciency of a system is deﬁned by the ratio of the power delivered to the load to the power supplied by the source; that is,
(9.5)
For the situation deﬁned by Fig. 9.77,
h% 100% 100% I2 LRL I2 LRT PL Ps
h% P P L s 100%
346  NETWORK THEOREMS
and h% 100%
For RL that is small compared to RTh,RTh k RL and RTh RL RTh, with
RL RTh RL
Th
FIG. 9.82 Efﬁciency of operation versus increasing values of RL.
100
Approaches 100%
75
50
25
0 20 40 60 80 100 RL ( )
RL = RTh
h% ≅ kRL 100%
h%
10
h% 100% RL 100% kRL 100% RL RTh 1 RTh
Constant
The resulting percentage efﬁciency, therefore, will be relatively low (since k is small) and will increase almost linearly as RL increases. For situations where the load resistance RL is much larger than RTh, RL k RTh and RTh RL RL.
h% 100% 100%
The efﬁciency therefore increases linearly and dramatically for small levels of RL and then begins to level off as it approaches the 100% level for very large values of RL, as shown in Fig. 9.82. Keep in mind, however, that the efﬁciency criterion is sensitive only to the ratio of PL to Ps and not to their actual levels. At efﬁciency levels approaching 100%, the power delivered to the load may be so small as to have little practical value. Note the low level of power to the load in Table 9.1 when RL 1000 , even though the efﬁciency level will be
h% 100% 100% 99.11% 1000 1009 RL RTh RL
RL RL
MAXIMUM POWER TRANSFER THEOREM  347Th
When RL RTh,
h% 100% 100% 50%
Under maximum power transfer conditions, therefore, PL is a maximum, but the dc efﬁciency is only 50%; that is, only half the power delivered by the source is getting to the load. A relatively low efﬁciency of 50% can be tolerated in situations where power levels are relatively low, such as in a wide variety of electronic systems. However, when large power levels are involved, such as at generating stations, efﬁciencies of 50% would not be acceptable. In fact, a great deal of expense and research is dedicated to raising powergenerating and transmission efﬁciencies a few percentage points. Raising an efﬁciency level of a 10-mega-kW power plant from 94% to 95% (a 1% increase) can save 0.1 mega-kW, or 100 million watts, of power—an enormous saving! Consider a change in load levels from 9 to 20 . In Fig. 9.80, the power level has dropped from 100 W to 85.61 W (a 14.4% drop), but the efﬁciency has increased substantially to 69% (a 38% increase), as shown in Fig. 9.82. For each application, therefore, a balance point must be identiﬁed where the efﬁciency is sufﬁciently high without reducing the power to the load to insigniﬁcant levels. Figure 9.83 is a semilog plot of PL and the power delivered by the source Ps EThIL versus RL for ETh 60 V and RTh 9 . A semilog graph employs one log scale and one linear scale, as implied by the preﬁx semi, meaning half. Log scales are discussed in detail in Chapter 23. For the moment, note the wide range of RL permitted using the log scale compared to Figs. 9.80 through 9.82. It is now quite clear that the PL curve has only one maximum (at RL RTh), whereas Ps decreases for every increase in RL. In particular, note that for low levels of RL, only a small portion of the power delivered by the source makes it to the load. In fact, even when RL RTh, the source is generating twice the power absorbed by the load. For values of RL greater than RTh, the two curves approach each other until eventually they are essentially the same at high levels of RL. For the range RL RTh 9 to RL 100 , PL and Ps are relatively close in magnitude, suggesting that this would be an appropriate range of operation, since a majority of the power delivered by the source is getting to the load and the power levels are still signiﬁcant. The power delivered to RL under maximum power conditions (RL RTh) is
I ETh 2RTh ETh RTh RL
RL 2RL
RL RTh RL
PL I2RL 2 RTh
and (watts, W) (9.6)
For the Norton circuit of Fig. 9.78,
(W) (9.7)P Lmax I2 N 4 RN
PLmax 4 E R 2 T T h h
E2ThRTh 4R2Th
ETh 2RTh
348  NETWORK THEOREMS Th
400
350
300
250
200
150
100
90
80
70
60
50
40
30
20
10
0.1 0.5 1 23 4567810 20 3040 100 1000 RL( ) RL = RTh = 9
RL = RTh = 9
0.2
PL
PL max
Ps = 2PL max (at RL = RTh)
Ps = ETh IL
P (W)
FIG. 9.83 Ps and PL versus RL for the network of Fig. 9.79.
MAXIMUM POWER TRANSFER THEOREM  349Th
EXAMPLE 9.14 A dc generator, battery, and laboratory supply are connected to a resistive load RL in Fig. 9.84(a), (b), and (c), respectively.
RL
2.5 R int
–
+ E
RL
0.5 R int
–
+
E
RL
40 R int
–
+
E
(a) dc generator (b) Battery (c) Laboratory supply
FIG. 9.84 Example 9.14.
a. For each, determine the value of RL for maximum power transfer to RL. b. Determine RL for 75% efﬁciency.
Solutions:
a. For the dc generator,
RL RTh Rint 2.5
For the battery,
RL RTh Rint 0.5
For the laboratory supply,
RL RTh Rint 40
b. For the dc generator,
h (h in decimal form)
h
h(RTh RL) RL hRTh hRL RL RL(1 h) hRTh
and (9.8)
RL 7.5
For the battery,
RL 1.5 0.75(0.5 ) 1 0.75
0.75(2.5 ) 1 0.75
RL 1 hR Th h
RL RTh RL
Po Ps
350  NETWORK THEOREMS
For the laboratory supply,
RL 120
The results of Example 9.14 reveal that the following modiﬁed form of the maximum power transfer theorem is valid:
For loads connected directly to a dc voltage supply, maximum power will be delivered to the load when the load resistance is equal to the internal resistance of the source; that is, when
(9.9)
EXAMPLE 9.15 Analysis of a transistor network resulted in the reduced conﬁguration of Fig. 9.85. Determine the RL necessary to transfer maximum power to RL, and calculate the power of RL under these conditions.
Solution: Eq. (9.4):
RL Rs 40 k
Eq. (9.7):
PLmax 1 W
EXAMPLE 9.16 For the network of Fig. 9.86, determine the value of R for maximum power to R, and calculate the power delivered under these conditions.
Solution: See Fig. 9.87.
RTh R3 R1 R2 8 8 2
and R RTh 10
See Fig. 9.88.
ETh 4 V
and, by Eq. (9.6),
PLmax 0.4 W
EXAMPLE 9.17 Find the value of RL in Fig. 9.89 for maximum power to RL, and determine the maximum power.
Solution: See Fig. 9.90.
RTh R1 R2 R3 3 10 2 15 and RL RTh 15
(4 V)2 4(10 )
E2Th 4RTh
36 V 9
(3 )(12 V) 3 6
R2E R2 R1
(6 )(3 ) 6 3
(10 mA)2(40 k ) 4
I2 NRN 4
RL Rint
0.75(40 ) 1 0.75
Th
I 10 mA Rs 40 k RL
FIG. 9.85 Example 9.15.
R1
6
R2 3
R3
8
RE 12 V + –
FIG. 9.86 Example 9.16.
R2 3
R3
8
R1
6
RTh
FIG. 9.87 Determining RTh for the network external to the resistor R of Fig. 9.86.
R2 3
R3
8
R1
6
ETh EThE 12 V + – + – + –
+ V3 = 0 V –
FIG. 9.88 Determining ETh for the network external to the resistor R of Fig. 9.86.
R2 10
R1
3
6 AI RL
E1
68 V –+
R3
2
FIG. 9.89 Example 9.17.
MILLMAN’S THEOREM  351Th
R1
E1
R2
E2
R3
E3
RL
Req
Eeq
RL
FIG. 9.92 Demonstrating the effect of applying Millman’s theorem.
Note Fig. 9.91, where
V1 V3 0 V and V2 I2R2 IR2 (6 A)(10 ) 60 V
Applying Kirchhoff’s voltage law,
V V2 E1 ETh 0 and ETh V2 E1 60 V 68 V 128 V
Thus, PLmax 273.07 W
9.6 MILLMAN’S THEOREM
Through the application of Millman’s theorem, any number of parallel voltage sources can be reduced to one. In Fig. 9.92, for example, the three voltage sources can be reduced to one. This would permit ﬁnding the current through or voltage across RL without having to apply a method such as mesh analysis, nodal analysis, superposition, and so on. The theorem can best be described by applying it to the network of Fig. 9.92. Basically, three steps are included in its application.
(128 V)2 4(15 )
E2Th 4RTh

I1 E1G1 G1 I2 E2G2 I3 G2 E3G3 G3 RL ( ) E3 R3( )E 2 R2( )E 1 R1
FIG. 9.93 Converting all the sources of Fig. 9.92 to current sources.
FIG. 9.90 Determining RTh for the network external to the resistor RL of Fig. 9.89.
R2 10
R1
3
R3
2
RTh
V2 R2 = 10
E1
68 V –+R 1 = 3
I = 0
I = 6 A
I = 0
R3 = 2 + V3 = 0 V –
ETh
–
+
– V1 = 0 V +
–
+
6 A
I =
6 A
FIG. 9.91 Determining ETh for the network external to the resistor RL of Fig. 9.89.
Step 1: Convert all voltage sources to current sources as outlined in Section 8.3. This is performed in Fig. 9.93 for the network of Fig. 9.92.
352  NETWORK THEOREMS
Step 2: Combine parallel current sources as described in Section 8.4. The resulting network is shown in Fig. 9.94, where
IT I1 I2 I3 and GT G1 G2 G3
Step 3: Convert the resulting current source to a voltage source, and the desired single-source network is obtained, as shown in Fig. 9.95.
In general, Millman’s theorem states that for any number of parallel voltage sources,
Eeq
or Eeq (9.10)
The plus-and-minus signs appear in Eq. (9.10) to include those cases where the sources may not be supplying energy in the same direction. (Note Example 9.18.) The equivalent resistance is
Req (9.11)
In terms of the resistance values,
Eeq (9.12)
and Req (9.13)
The relatively few direct steps required may result in the student’s applying each step rather than memorizing and employing Eqs. (9.10) through (9.13).
EXAMPLE 9.18 Using Millman’s theorem, ﬁnd the current through and voltage across the resistor RL of Fig. 9.96.
Solution: By Eq. (9.12),
Eeq
The minus sign is used for E2/R2 because that supply has the opposite polarity of the other two. The chosen reference direction is therefore
E R 1 1 E R 2 2 E R 3 3 —— R 1 1 R 1 2 R 1 3
1 ——— R 1 1 R 1 2 R 1 3 ••• R 1 N

E R 1 1
E R 2 2
E R 3 3
•••
E R N N ———— R 1 1 R 1 2 R 1 3 ••• R 1 N
1 G1 G2 G3 ••• GN
1 GT

E1G1
E2G2
E3G3
•••
ENGN G1 G2 G3 ••• GN

I1
I2
I3
•••
IN G1 G2 G3 ••• GN
IT GT
Th
GTI T RL
FIG. 9.94 Reducing all the current sources of Fig. 9.93 to a single current source.
–
+
Req
1 GT
Eeq
IT GT
RL
FIG. 9.95 Converting the current source of Fig. 9.94 to a voltage source.
R1 5 R2 4 R3 2
E1
10 V
E2
16 V
E3
8 V
RL 3
IL
VL + –
FIG. 9.96 Example 9.18.
MILLMAN’S THEOREM  353Th
that of E1 and E3. The total conductance is unaffected by the direction, and
Eeq
2.105 V
with Req 1.053
The resultant source is shown in Fig. 9.97, and
IL 0.519 A
with VL ILRL (0.519 A)(3 ) 1.557 V
EXAMPLE 9.19 Let us now consider the type of problem encountered in the introduction to mesh and nodal analysis in Chapter 8. Mesh analysis was applied to the network of Fig. 9.98 (Example 8.12). Let us now use Millman’s theorem to ﬁnd the current through the 2- resistor and compare the results.
Solutions:
a. Let us ﬁrst apply each step and, in the (b) solution, Eq. (9.12). Converting sources yields Fig. 9.99. Combining sources and parallel conductance branches (Fig. 9.100) yields
IT I1 I2 5 A A A A A
GT G1 G2 1 S S S S S 7 6 1 6 6 6 1 6
20 3
5 3
15 3
5 3
2.105 V 4.053
2.105 V 1.053 3
1 0.95 S
1 ——— 5 1 4 1 2 1
2 A 0.95 S
2 A 4 A 4 A ——— 0.2 S 0.25 S 0.5 S
1 5 0 V 1 4 6 V 2 8 V ——— 5 1 4 1 2 1
Req 1.053
Eeq
RL 3 VL – +
2.105 V
IL
FIG. 9.97 The result of applying Millman’s theorem to the network of Fig. 9.96.
R1 1 R2 6
E1 5 V E2 10 V
R3 2
FIG. 9.98 Example 9.19.
I1
R1
5 A
1 R2 6 I2 5 3
R3 2
A
FIG. 9.99 Converting the sources of Fig. 9.98 to current sources.
R3 2
6 7 Req
Eeq 40 7 V
FIG. 9.101 Converting the current source of Fig. 9.100 to a voltage source.
IT 20 3
7 6 R3 2 SA GT
FIG. 9.100 Reducing the current sources of Fig. 9.99 to a single source.
Converting the current source to a voltage source (Fig. 9.101), we obtain
Eeq V V 40 — 7 (6)(20) (3)(7) 2 3 0 A — 7 6 S IT GT
354  NETWORK THEOREMS Th
and Req
so that
I2 2 A
which agrees with the result obtained in Example 8.18. b. Let us now simply apply the proper equation, Eq. (9.12):
Eeq V
and
Req
which are the same values obtained above.
The dual of Millman’s theorem (Fig. 9.92) appears in Fig. 9.102. It can be shown that Ieq and Req, as in Fig. 9.102, are given by
6 — 7
1 — 7 6 S
1 —— 6 6 6 1
1 —— 1 1 6 1
40 — 7
3 6 0 V 1 6 0 V —— 6 6 6 1
1 5 V 1 6 0 V —— 1 1 6 1
40 V 20
4 7 0 V —— 6 7 1 7 4
4 7 0 V —— 6 7 2
Eeq Req R3
6 — 7
1 — 7 6 S
1 GT
Ieq (9.14)
and (9.15)
The derivation will appear as a problem at the end of the chapter.
9.7 SUBSTITUTION THEOREM
The substitution theorem states the following:
If the voltage across and the current through any branch of a dc bilateral network are known, this branch can be replaced by any
Req R1 R2 R3
I1R1 I2R2 I3R3 R1 R2 R3
FIG. 9.102 The dual effect of Millman’s theorem.
R1
I2
R2
I3
R3
I1
RL Req
Ieq
RL
FIG. 9.103 Demonstrating the effect of the substitution theorem.
R2 4
R1
6
a
b
12 V – +
3 A
E 30 V
SUBSTITUTION THEOREM  355Th
combination of elements that will maintain the same voltage across and current through the chosen branch.
More simply, the theorem states that for branch equivalence, the terminal voltage and current must be the same. Consider the circuit of Fig. 9.103, in which the voltage across and current through the branch a-b are determined. Through the use of the substitution theorem, a number of equivalent a-a′ branches are shown in Fig. 9.104.
Note that for each equivalent, the terminal voltage and current are the same. Also consider that the response of the remainder of the circuit of Fig. 9.103 is unchanged by substituting any one of the equivalent branches. As demonstrated by the single-source equivalents of Fig. 9.104, a known potential difference and current in a network can be replaced by an ideal voltage source and current source, respectively. Understand that this theorem cannot be used to solve networks with two or more sources that are not in series or parallel. For it to be applied, a potential difference or current value must be known or found using one of the techniques discussed earlier. One application of the theorem is shown in Fig. 9.105. Note that in the ﬁgure the known potential difference V was replaced by a voltage source, permitting the isolation of the portion of the network including R3, R4, and R5. Recall that this was basically the approach employed in the analysis of the ladder network as we worked our way back toward the terminal resistance R5.
R2
R3a
b
E V
–
+
R1
R4 R5
R3
b
E′ = V
+
R5
–
R4
a
FIG. 9.105 Demonstrating the effect of knowing a voltage at some point in a complex network.
The current source equivalence of the above is shown in Fig. 9.106, where a known current is replaced by an ideal current source, permitting the isolation of R4 and R5.
FIG. 9.104 Equivalent branches for the branch a-b of Fig. 9.103.
2 A 12 12 V
b
3 A
a
–
+
a
b
–
+
2
6 V
3 A
12 V
a
b
–
+
3 A 12 V
a
b
12 V
–
+
3 A
356  NETWORK THEOREMS Th
You will also recall from the discussion of bridge networks that V 0 and I 0 were replaced by a short circuit and an open circuit, respectively. This substitution is a very speciﬁc application of the substitution theorem.
9.8 RECIPROCITY THEOREM
The reciprocity theorem is applicable only to single-source networks. It is, therefore, not a theorem employed in the analysis of multisource networks described thus far. The theorem states the following:
The current I in any branch of a network, due to a single voltage source E anywhere else in the network, will equal the current through the branch in which the source was originally located if the source is placed in the branch in which the current I was originally measured.
In other words, the location of the voltage source and the resulting current may be interchanged without a change in current. The theorem requires that the polarity of the voltage source have the same correspondence with the direction of the branch current in each position. In the representative network of Fig. 9.107(a), the current I due to the voltage source E was determined. If the position of each is interchanged as shown in Fig. 9.107(b), the current I will be the same value as indicated. To demonstrate the validity of this statement and the theorem, consider the network of Fig. 9.108, in which values for the elements of Fig. 9.107(a) have been assigned. The total resistance is
FIG. 9.107 Demonstrating the impact of the reciprocity theorem.
I
E
a
b
c
d
I
E
a
b
c
d
(a) (b)
R1
E
R2 R4
R3
R5
ba
I
R4 R5
ba
I
FIG. 9.106 Demonstrating the effect of knowing a current at some point in a complex network.
APPLICATION  357Th
RT R1 R2 (R3 R4) 12 6 (2 4 ) 12 6 6 12 3 15
and Is 3 A
with I 1.5 A
For the network of Fig. 9.109, which corresponds to that of Fig. 9.107(b), we ﬁnd
RT R4 R3 R1 R2 4 2 12 6 10
and Is 4.5 A
so that I 1.5 A
which agrees with the above. The uniqueness and power of such a theorem can best be demonstrated by considering a complex, single-source network such as the one shown in Fig. 9.110.
4.5 A 3
(6 )(4.5 A) 12 6
45 V 10
E RT
3 A 2
45 V 15
E RT
c
d
a
b
I
c
d
a
b
I
E
E
FIG. 9.110 Demonstrating the power and uniqueness of the reciprocity theorem.
FIG. 9.108 Finding the current I due to a source E.
R1
12
R3
2
R2 6 R4 4 E 45 V
Is
I
R1
12
R3
2
R2 6
R4 4
E 45 V I RT
Is
FIG. 9.109 Interchanging the location of E and I of Fig. 9.108 to demonstrate the validity of the reciprocity theorem.
9.9 APPLICATION
Speaker System
One of the most common applications of the maximum power transfer theorem introduced in this chapter is to speaker systems. An audio ampliﬁer (ampliﬁer with a frequency range matching the typical range
4" Woofer 8 Ω 5 W(rms) 10 W (max)(c)
Audio amplifier
Ri
Speaker
a
b
a
b
8 Ω
a
b
8 Ω
Vs + –
(a)
a
b
(b)
Ri
Ro
Ro
FIG. 9.111 Components of a speaker system: (a) ampliﬁer; (b) speaker; (c) commercially available unit.
358  NETWORK THEOREMS Th
of the human ear) with an output impedance of 8 is shown in Fig. 9.111(a). Impedance is a term applied to opposition in ac networks— for the moment think of it as a resistance level. We can also think of impedance as the internal resistance of the source which is normally shown in series with the source voltage as shown in the same ﬁgure. Every speaker has an internal resistance that can be represented as shown in Fig. 9.111(b) for a standard 8- speaker. Figure 9.111(c) is a photograph of a commercially available 8- woofer (for very low frequencies). The primary purpose of the following discussion is to shed some light on how the audio power can be distributed and which approach would be the most effective. Since the maximum power theorem states that the load impedance should match the source impedance for maximum power transfer, let us ﬁrst consider the case of a single 8- speaker as shown in Fig. 9.112(a) with an applied ampliﬁer voltage of 12 V. Since the applied voltage will split equally, the speaker voltage is 6 V, and the power to the speaker is a maximum value of P V2/R (6 V)2/8 4.5 W. If we have two 8- speakers that we would like to hook up, we have the choice of hooking them up in series or parallel. For the series conﬁguration of Fig. 9.112(b), the resulting current would be I E/R 12 V/24 500 mA, and the power to each speaker would be P I2R (500 mA)2(8 ) 2W ,which is a drop of over 50% from the maximum output level of 4.5 W. If the speakers are hooked up in parallel as shown in Fig. 9.112(c), the total resistance of the parallel combination is 4 , and the voltage across each speaker as determined by the voltage divider rule will be 4 V. The power to each speaker is P V2/R (4 V)2/8 2 W which, interestingly enough, is the same power delivered to each speaker whether in series or parallel. However, the parallel arrangement is normally chosen for a variety of reasons. First, when the speakers are connected in parallel, if a wire should become disconnected from one of the speakers due simply to the vibration caused by the emitted sound, the other speakers will still be operating—perhaps not at maximum efﬁciency, but they will still be operating. If in series they would all fail to operate. A second reason relates to the general wiring procedure. When all of the speakers are in parallel, from various parts of a room all the red wires can be connected together and all the black wires together. If the speakers are in series, and if you
(a)
8 Ω
Ro
Amplifier
+
– 12 V
+
– 6 V Ri 8-Ω speaker
8 Ω 8 Ω
Ro Rspeaker 1
I = 500 mA
Amplifier Series speakers
+
– 12 V Rspeaker 2 8 Ω
(b)
Ro Rspeaker 1
8 Ω
Rspeaker 2
+
– 12 V
+
– 4 V 8 Ω 8 Ω
Amplifier Parallel speakers (c)
FIG. 9.112 Speaker connections: (a) single unit; (b) in series; (c) in parallel.
COMPUTER ANALYSIS  359Th
In any case, always try to match the total resistance of the speaker load to the output resistance of the supply. Yes, a 4- speaker can be placed in series with a parallel combination of 8- speakers for maximum power transfer from the supply since the total resistance will be 8 . However, the power distribution will not be equal, with the 4- speaker receiving 2.25 W and the 8- speakers each 1.125 W for a total of 4.5 W. The 4- speaker is therefore receiving twice the audio power of the 8- speakers, and this difference may cause distortion or imbalance in the listening area. All speakers have maximum and minimum levels. A 50-W speaker is rated for a maximum output power of 50 W and will provide that level on demand. However, in order to function properly, it will probably need to be operating at least at the 1- to 5-W level. A 100-W speaker typically needs between 5 W and 10 W of power to operate properly. It is also important to realize that power levels less than the rated value (such as 40 W for the 50-W speaker) will not result in an increase in distortion, but simply in a loss of volume. However, distortion will result if you exceed the rated power level. For example, if we apply 2.5 W to a 2-W speaker, we will deﬁnitely have distortion. However, applying 1.5 W will simply result in less volume. A rule of thumb regarding audio levels states that the human ear can sense changes in audio level only if you double the applied power [a 3-dB increase; decibels (dB) will be introduced in Chapter 23]. The doubling effect is always with respect to the initial level. For instance, if the original level were 2 W, you would have to go to 4 W to notice the change. If starting at 10 W, you would have to go to 20 W to appreciate the increase in volume. An exception to the above is at very low power levels or very high power levels. For instance, a change from 1 W to 1.5 W may be discernible, just as a change from 50 W to 80 W may be noticeable.
9.10 COMPUTER ANALYSIS
Once the mechanics of applying a software package or language are understood, the opportunity to be creative and innovative presents itself. Through years of exposure and trial-and-error experiences, professional
8 Ω 4 Ω
Series 4-Ω speakers
+
– Vs Rspeaker 2 4 Ω Ri = 8 Ω
Ro Rspeaker 1 Ro Rspeaker 1
8 Ω
Rspeaker 2
+
– Vs 16 Ω 16 Ω
Parallel 16-Ω speakers
Ri = 8 Ω
FIG. 9.113 Applying 4- and 16- speakers to an ampliﬁer with an output impedance of 8 .
are presented with a bundle of red and black wires in the basement, you would ﬁrst have to determine which wires go with which speakers. Speakers are also available with input impedances of 4 and 16 . If you know that the output impedance is 8 , purchasing either two 4- speakers or two 16- speakers would result in maximum power to the speakers as shown in Fig. 9.113. The 16- speakers would be connected in parallel and the 4- speakers in series to establish a total load impedance of 8 .
360  NETWORK THEOREMS Th
programmers develop a catalog of innovative techniques that are not only functional but very interesting and truly artistic in nature. Now that some ofthebasicoperationsassociatedwithPSpicehavebeenintroduced,afew innovative maneuvers will be made in the examples to follow.
PSpice
Superposition Let us now apply superposition to the network of Fig 9.114, which appeared earlier as Fig. 9.10 in Example 9.3, to permit a comparison of resulting solutions. The current through R2 is to be determined. Using methods described in earlier chapters for the application of PSpice, the network of Fig. 9.115 will result to determine the effect of the 36-V voltage source. Note that both VDC and IDC (ﬂipped vertically) appear in the network. The current source, however, was set to zero simply by selecting the source and changing its value to 0 A in the Display Properties dialog box.
R2 6
R1
12 I2
I 9 A
E 36 V
FIG. 9.114 Applying PSpice to determine the current I2 using superposition.
FIG. 9.115 Using PSpice to determine the contribution of the 36-V voltage source to the current through R2.
Following simulation, the results appearing in Fig. 9.115 will result. The current through the 6- resistor is 2 A due solely to the 36-V voltage source. Although direction is not indicated, it is fairly obvious in this case. For those cases where it is not obvious, the voltage levels can be displayed, and the direction would be from the point of high potential to the point of lower potential. For the effects of the current source, the voltage source is set to 0 V as shown in Fig. 9.116. The resulting current is then 6 A through R2, with the same direction as the contribution due to the voltage source.
COMPUTER ANALYSIS  361Th
The resulting current for the resistor R2 is the sum of the two currents: IT 2 A 6 A 8 A, as determined in Example 9.3.
Thévenin’s Theorem The application of Thévenin’s theorem requires an interesting maneuver to determine the Thévenin resistance. It is a maneuver, however, that has application beyond Thévenin’s theorem whenever a resistance level is required. The network to be analyzed appears in Fig. 9.117 and is the same one analyzed in Example 9.10 (Fig. 9.49).
FIG. 9.116 Using PSpice to determine the contribution of the 9-A current source to the current through R2.
R1 0.8 k
R4
1.4 k
R2 4 k
R3 6 k
+
–
E2 10 V + –
E1 6 V – +
RTh ETh
FIG. 9.117 Network to which PSpice is to be applied to determine ETh and RTh.
Since PSpice is not set up to measure resistance levels directly, a 1-A current source can be applied as shown in Fig. 9.118, and Ohm’s law
362  NETWORK THEOREMS Th
can be used to determine the magnitude of the Thévenin resistance in the following manner: RTh V Is s 1 V A s Vs In Eq. (9.16), since Is 1 A, the magnitude of RTh in ohms is the same as the magnitude of the voltage Vs (in volts) across the current source. The result is that when the voltage across the current source is displayed, it can be read as ohms rather than volts. When PSpice is applied, the network will appear as shown in Fig. 9.118. The voltage source E1 and the current source are ﬂipped using a right click on the source and using the Mirror Vertically option. Both voltage sources are set to zero through the Display Properties dialog box obtained by double-clicking on the source symbol. The result of the Bias Point simulation is 2 kV across the current source. The Thévenin resistance is therefore 2 k between the two terminals of the network to the left of the current source (to match the results of Example 9.10). In total, by setting the voltage sources to 0 V, we have dictated that the voltage is the same at both ends of the voltage source, replicating the effect of a short-circuit connection between the two points. Forthe open-circuit Thévenin voltage between the terminals of interest, the network must be constructed as shown in Fig. 9.119. The resistance of 1 T ( 1million M )isconsidered large enough to represent an open circuit to permit an analysis of the network using PSpice. PSpice does not recognize ﬂoating nodes and would generate an error signal if a connection were not made from the top right node to ground. Both voltage sources are now set on their prescribed values, and a simulation will
FIG. 9.118 Using PSpice to determine the Thévenin resistance of a network through the application of a 1-A current source.
COMPUTER ANALYSIS  363Th
result in 3V across the 1-T resistor. The open-circuit Thévenin voltage is therefore 3 V which agrees with the solution of Example 9.10.
Maximum Power Transfer The procedure for plotting a quantity versus a parameter of the network will now be introduced. In this case it will be the output power versus values of load resistance to verify the fact that maximum power will be delivered to the load when its value equals the series Thévenin resistance. A number of new steps will be introduced, but keep in mind that the method has broad application beyond Thévenin’s theorem and is therefore well worth the learning process. The circuit to be analyzed appears in Fig. 9.120. The circuit is constructed in exactly the same manner as described earlier except for the value of the load resistance. Begin the process by starting a New Project called MaxPower, and build the circuit of Fig. 9.120. For the moment hold off on setting the value of the load resistance. The ﬁrst step will be to establish the value of the load resistance as a variable since it will not be assigned a ﬁxed value. Double-click on the value of RL to obtain the Display Properties dialog box. For Value, type in {Rval} and click in place. The brackets (not parentheses) are required, but the variable does not have to be called Rval—it is the choice of the user. Next select the Place part keytoobtain the Place Part dialog box. If you are not already in the Libraries list, choose Add Library and add SPECIAL to the list. Select the SPECIAL library and scroll the Part List until PARAM appears. Select it; then click OK to obtain a rectangular box next to the cursor on the
FIG. 9.119 Using PSpice to determine the Thévenin voltage for a network using a very large resistance value to represent the open-circuit condition between the terminals of interest.
364  NETWORK THEOREMS Th
screen. Select a spot near Rval, and deposit the rectangle. The result is PARAMETERS: as shown in Fig. 9.120. Next double-click on PARAMETERS: to obtain a Property Editor dialog box which should have SCHEMATIC1:PAGE1 in the second column from the left. Now select the New Column option from the top list of choices to obtain the Add New Column dialog box. Enter the Name:Rval and Value:1 followed by an OK to leave the dialog box. The result is a return to the Property Editor dialog box but with Rval and its value (below Rval) added to the horizontal list. Now select Rval/1 by clicking on Rval to surround Rval by a dashed line and add a black background around the 1. Choose Display to produce the Display Properties dialog box, and select Name and Value followed by OK. Then exit the Property Editor dialog box (X) to obtain the screen of Fig. 9.120. Note that now the ﬁrst value (1 ) of Rval is displayed. We are now ready to set up the simulation process. Select the New Simulation Proﬁle key to obtain the New Simulation dialog box. Enter DC Sweep under Name followed by Create. The Simulation Settings-DC Sweep dialog box will appear. After selecting Analysis, select DC Sweep under the Analysis type heading. Then leave the Primary Sweep under the Options heading, and select Global parameter under the Sweep variable. The Parameter name should then be entered as Rval. For the Sweep type, the Start value should be 1 ; but if we use 1 , the curve to be generated will start at 1 , leaving a blank from 0 to 1 . The curve will look incomplete. To solve this problem, we will select 0.001 as the Start value (very close to 0 ) and the End value 30.001 with an Increment of 1 . The values of RL will therefore be 0.001 , 1.001 , 2.001 , etc., although the plot
FIG. 9.120 Using PSpice to plot the power to RL for a range of values for RL.
COMPUTER ANALYSIS  365
will look as if the values were 0 , 1 , 2 , etc. Click OK, and select the Run PSpice key to obtain the display of Fig. 9.121. First note that there are no plots on the graph and that the graph extends to 35 rather than 30 as desired. It did not respond with a plot of power versus RL because we have not deﬁned the plot of interest for the computer. This is done by selecting the AddTrace key (the key in the middle of the lower toolbar that looks like a red sawtooth waveform) or Trace-Add Trace from the top menu bar. Either choice will result in the Add Traces dialog box. The most important region of this dialog box is the Trace Expression listing at the bottom. The desired trace can be typed in directly, or the quantities of interest can be chosen from the list of Simulation OutputVariables and deposited in the Trace Expression listing. Since we are interested in the power to RL for the chosen range of values for RL, W(RL) is selected in the listing; it will then appear as the Trace Expression. Click OK, and the plot of Fig. 9.122 will appear. Originally, the plot extended from 0 to 35 .Wereduced the range to 0 to 30 by selecting Plot-Axis Settings-X Axis-User Deﬁned 0 to 30-OK. Select the Toggle cursor key (which looks like a red curve passing through the origin of a graph), and then left-click the mouse. A vertical line and a horizonal line will appear, with the vertical line controlled by the position of the cursor. Moving the cursor to the peak value will result in A1 9.0010 as the x value and 100.000 W as the y value as shown in the Probe Cursor box at the right of the screen. A second cursor can be generated by a right click of the mouse, which was set at RL 30.001 to result in a power of 71.005 W. Notice also that the
Th
FIG. 9.121 Plot resulting from the dc sweep of RL for the network of Fig. 9.120 before deﬁning the parameters to be displayed.
366  NETWORK THEOREMS Th
plot generated appears as a listing at the bottom left of the screen as W(RL). Before leaving the subject, we should mention that the power to RL can be determined in more ways than one from the Add Traces dialog box. For example, ﬁrst enter a minus sign because of the resulting current direction through the resistor, and then select V2(RL) followed by the multiplication of I(RL). The following expression will appear in the Trace Expression box: V2(RL)*I(RL), which is an expression having the basic power format of P V * I. Click OK, and the same power curve of Fig. 9.122 will appear. Other quantities, such as the voltage across the load and the current through the load, can be plotted against RL by simply following the sequence Trace-Delete All Traces-TraceAdd Trace-V1(RL) or I(RL).
FIG. 9.122 A plot of the power delivered to RL in Fig. 9.120 for a range of values for RL extending from 0 to 30 .
PROBLEMS
SECTION 9.2 Superposition Theorem
1. a. Using superposition, ﬁnd the current through each resistor of the network of Fig. 9.123. b. Find the power delivered to R1 for each source. c. Find the power delivered to R1 using the total current through R1. d. Does superposition apply to power effects? Explain.
R1
12
R3
R2 6
6
E1 10 V
5 V E2
FIG. 9.123 Problem 1.
PROBLEMS  367Th
2. Using superposition, ﬁnd the current I through the 10- resistor for each of the networks of Fig. 9.124.
R2 8
R1 10 I = 9 A
E 18 V
I
(a)
R1 18
R2 9 R3 15 R4 10
24 V
I
E2
E1 = + 42 V
(b)
FIG. 9.124 Problems 2 and 41.
R2
3.3 k
R1 2.2 k
5 mA
I R3 4.7 k 8 V
(a)
R1 6
6 A
I
R4 12
8 V
(b)
E1 12 V
R3 30
E2
R2 4
R5
4
IR1
IR1
FIG. 9.125 Problem 3.
* 3.Using superposition, ﬁnd the current through R1 for each network of Fig. 9.125.
4. Using superposition, ﬁnd the voltage V2 for the network of Fig. 9.126.
R1 12 k
I 9 mA
36 VE
V2 + –
R2 6.8 k
FIG. 9.126 Problems 4 and 37.
368  NETWORK THEOREMS Th
7. Find the Thévenin equivalent circuit for the network external to the resistor R in each of the networks of Fig. 9.129.
SECTION 9.3 The ′venin’s Theorem
5. a. Find the Thévenin equivalent circuit for the network external to the resistor R of Fig. 9.127. b. Find the current through R when R is 2 , 30 , and 100 .
6. a. Find the Thévenin equivalent circuit for the network external to the resistor R in each of the networks of Fig. 9.128. b. Find the power delivered to R when R is 2 and 100 .
R1
6
R2 3 E R 18 V
R3
4
FIG. 9.127 Problem 5.
(II)
5
E 20 V
R2 R
5 R 1
5
R3
(I)
R1
12
2
3 AR 2 IR
FIG. 9.128 Problems 6, 13, and 19.
R
E1 E2 18 V 72 V
3
6 5.6 k R
2.2 k
16 V8 mA
(b)(a)
FIG. 9.129 Problems 7, 14, and 20.
PROBLEMS  369Th
* 9.Find the Thévenin equivalent circuit for the portions of the networks of Fig. 9.131 external to points a and b.
* 8.Find the Thévenin equivalent circuit for the network external to the resistor R in each of the networks of Fig. 9.130.
*10. Determine the Thévenin equivalent circuit for the network external to the resistor R in both networks of Fig. 9.132.
FIG. 9.130 Problems 8, 15, 21, 38, 39, and 42.
R
20 V
10
3 A 25 6 72 V
6
3
2
R
4
(b)(a)
15 V
25
30
(I)
10 V
60
a
b
4.7 k
2.7 k
(II)
R1
180 V
a
b
E
47 k R
I 18 mA
R2
3.9 k R 3
10 V
FIG. 9.131 Problems 9 and 16.
20 V 5
(a)
R
20 1.1 k
2.2 k
(b)
R2
– 4 V
R2E
R1
16 R 4
12
R3
2
R5 R1
4.7 k R
E2
E1 = +12 V
3.3 k
R3
FIG. 9.132 Problems 10 and 17.
370  NETWORK THEOREMS Th
*11. Forthe network of Fig. 9.133, ﬁnd theThévenin equivalent circuit for the network external to the load resistor RL. *12. For the transistor network of Fig. 9.134: a. Find the Thévenin equivalent circuit for that portion of the network to the left of the base (B) terminal. b. Using the fact that IC IE and VCE 8 V, determine the magnitude of IE. c. Using the results of parts (a) and (b), calculate the base current IB if VBE 0.7 V. d. What is the voltage VC?
SECTION 9.4 Norton’s Theorem
13. Find the Norton equivalent circuit for the network external to the resistor R in each network of Fig. 9.128.
14. a. Find the Norton equivalent circuit for the network external to the resistor R for each network of Fig. 9.129. b. Convert to the Thévenin equivalent circuit, and compare your solution for ETh and RTh to that appearing in the solutions for Problem 7.
15. Find the Norton equivalent circuit for the network external to the resistor R for each network of Fig. 9.130.
16. a. Find the Norton equivalent circuit for the network external to the resistor R for each network of Fig. 9.131. b. Convert to the Thévenin equivalent circuit, and compare your solution for ETh and RTh to that appearing in the solutions for Problem 9.
17. Find the Norton equivalent circuit for the network external to the resistor R for each network of Fig. 9.132.
18. Find the Norton equivalent circuit for the portions of the networks of Fig. 9.135 external to branch a-b.
SECTION 9.5 Maximum Power Transfer Theorem
19. a. For each network of Fig. 9.128, ﬁnd the value of R for maximum power to R. b. Determine the maximum power to R for each network.
20. a. For each network of Fig. 9.129, ﬁnd the value of R for maximum power to R. b. Determine the maximum power to R for each network.
3.3 k
6.8 k
+ 6 V
RL
+ 22 V
5.6 k
– 12 V
2.2 k
1.2 k
FIG. 9.133 Problem 11.
R1 51 k
R2 10 k
RC 2.2 k
RE 0.5 k
IE
IC
20 V20 V
B
C
E
VCE = 8 V + – VC
IB
FIG. 9.134 Problem 12.
a
12 V
6
12
(b)(a)
2 A
72 V
b
100
12
6 V
2
4
4
a
2 V
300
b
4
FIG. 9.135 Problems 18 and 40.
PROBLEMS  371Th
*23. Find the resistance R1 of Fig. 9.137 such that the resistor R4 will receive maximum power. Think! *24. a. For the network of Fig. 9.138, determine the value of R2 for maximum power to R4. b. Is there a general statement that can be made about situations such as those presented here and in Problem 23?
*25. For the network of Fig. 9.139, determine the level of R that will ensure maximum power to the 100- resistor.
21. For each network of Fig. 9.130, ﬁnd the value of R for maximum power to R, and determine the maximum power to R for each network.
22. a. For the network of Fig. 9.136, determine the value of R for maximum power to R. b. Determine the maximum power to R. c. Plot a curve of power to R versus R for R equal to , , , 1, 1 , 1 , 1 , and 2 times the value obtained in part (a). 3 4 1 2 1 4 3 4 1 2 1 4
SECTION 9.6 Millman’s Theorem
26. Using Millman’s theorem, ﬁnd the current through and voltage across the resistor RL of Fig. 9.140. 27. Repeat Problem 26 for the network of Fig. 9.141.
24 V
4 5 A R R2
4 R 1
E
I
FIG. 9.136 Problems 22 and 43.
100 V 50 R4
R1
50 R 2
50
R3
FIG. 9.137 Problem 23.
100 V R4 R2 25
R3
25
R1
E
FIG. 9.138 Problem 24.
12 V
RL
500 Pot.
R
100
FIG. 9.139 Problem 25.
40 V
6 R 2
E1
RL 3
42 VE 2 + –
+
–
10 R 1 5 V 8.2 k R2E 1
RL 5.6 k
20 VE 2 + –
+
–
2.2 k R 1
FIG. 9.140 Problem 26.
FIG. 9.141 Problem 27.
372  NETWORK THEOREMS Th
28. Repeat Problem 26 for the network of Fig. 9.142.
29. Using the dual of Millman’s theorem, ﬁnd the current through and voltage across the resistor RL of Fig. 9.143.
*30. Repeat Problem 29 for the network of Fig. 9.144.
SECTION 9.7 Substitution Theorem
31. Using the substitution theorem, draw three equivalent branches for the branch a-b of the network of Fig. 9.145.
32. Repeat Problem 31 for the network of Fig. 9.146.
*33. Repeat Problem 31 for the network of Fig. 9.147. Be careful!
FIG. 9.142 Problem 28.
FIG. 9.143 Problem 29.
400 V 100 R2E 1
20 VE 2 + –
+
–
200 R 1 RL 200 10 V E3 + – R3
10 k
R1
4.7 RL 2.7
I1 = 4 A
R2
3.3
I2 = 1.6 A
FIG. 9.144 Problem 30.
R2
4.7
8 mA
I2
I1
10 mA R1 2 k I3
4 mA
R3 8.2 k
6.8 k
RL
FIG. 9.145 Problem 31.
FIG. 9.146 Problem 32.
2 k 1.5 k
R2
0.51 k
4 mA
I
10 V
E ba
R1
7 k 15 k 60 VE
8 k 2.5 k a
b
FIG. 9.147 Problem 33.
R2 12 20 VE 1
8 4 a
b
40 VE 2
R3R 1
PROBLEMS  373Th
SECTION 9.8 Reciprocity Theorem
34. a. For the network of Fig. 9.148(a), determine the current I. b. Repeat part (a) for the network of Fig. 9.148(b). c. Is the reciprocity theorem satisﬁed?
35. Repeat Problem 34 for the networks of Fig. 9.149.
36. a. Determine the voltage V for the network of Fig. 9.150(a). b. Repeat part (a) for the network of Fig. 9.150(b). c. Is the dual of the reciprocity theorem satisﬁed?
SECTION 9.10 Computer Analysis
PSpice or Electronics Workbench 37. Using schematics, determine the voltage V2 and its components for the network of Fig. 9.126.
FIG. 9.148 Problem 34.
FIG. 9.149 Problem 35.
FIG. 9.150 Problem 36.
24 VE
4 k 8 k
24 k
20 k
24 k
I
(a)
24 V
E
4 k 8 k
24 k 20 k
24 k
I
(b)
4 k 4 k
4 k 8 k
E
10 V
I
(a)
4 k 4 k
4 k 8 k
E 10 V
I
(b)
R1 3
R2
2 R3 4 I
6 A
+ V –
(a)
R1 3
R2
2
R3 4
I = 6 A
V
(b)
+
–
374  NETWORK THEOREMS Th
38. Using schematics, determine theThévenin equivalent circuit for the network of Fig. 9.130(b).
*39. a. Using schematics, plot the power delivered to the resistor R of Fig. 9.130(a) for R having values from 1 to 50 . b. From the plot, determine the value of R resulting in maximum power to R and the maximum power to R. c. Compare the results of part (a) to the numerical solution. d. Plot VR and IR versus R, and ﬁnd the value of each under maximum power conditions. *40. Change the 300- resistor of Fig. 9.135(b) to a variable resistor, and plot the power delivered to the resistor versus values of the resistor. Determine the range of resis
GLOSSARY
Maximum power transfer theorem A theorem used to determine the load resistance necessary to ensure maximum power transfer to the load. Millman’s theorem A method employing source conversions that will permit the determination of unknown variables in a multiloop network. Norton’s theorem A theorem that permits the reduction of any two-terminal linear dc network to one having a single current source and parallel resistor. Reciprocity theorem A theorem that states that for singlesource networks, the current in any branch of a network, due to a single voltage source in the network, will equal the current through the branch in which the source was originally located if the source is placed in the branch in which the current was originally measured.
tance by trial and error rather than ﬁrst performing a longhand calculation. Determine the Norton equivalent circuit from the results. The Norton current can be determined from the maximum power level.
Programming Language (C ,QBASIC, Pascal, etc.)
41. Write a program to determine the current through the 10- resistor of Fig. 9.124(a) (for any component values) using superposition.
42. Write a program to perform the analysis required for Problem 8, Fig. 9.130(b), for any component values.
*43. Write a program to perform the analysis of Problem 22, and tabulate the power to R for the values listed in part (c).

Substitution theorem A theorem that states that if the voltage across and current through any branch of a dc bilateral network are known, the branch can be replaced by any combination of elements that will maintain the same voltage across and current through the chosen branch. Superposition theorem A network theorem that permits considering the effects of each source independently. The resulting current and/or voltage is the algebraic sum of the currents and/or voltages developed by each source independently. Thévenin’s theorem A theorem that permits the reduction of any two-terminal, linear dc network to one having a single voltage source and series resistor.

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