Hayt Engineering Circuit Analysis 8th Edition, Chapter 5 - PDF

CHAPTER Handy Circuit 5 Analysis Techniques KEY CONCEPTS Superposition: Determining INTRODUCTION...

CHAPTER Handy Circuit 5 Analysis Techniques KEY CONCEPTS Superposition: Determining INTRODUCTION the Individual Contributions The techniques of nodal and mesh analysis described in Chap. 4 of Different Sources to Any are reliable and extremely powerful methods. However, both Current or Voltage require that we develop a complete set of equations to describe a particular circuit as a general rule, even if only one current, Source Transformation as a voltage, or power quantity is of interest. In this chapter, we Means of Simplifying Circuits investigate several different techniques for isolating specific parts of a circuit in order to simplify the analysis. After examining each Thévenin’s Theorem of these techniques, we focus on how one might go about selecting one method over another. Norton’s Theorem Thévenin and Norton 5.1 LINEARITY AND SUPERPOSITION Equivalent Networks All of the circuits which we plan to analyze can be classified as lin- ear circuits, so this is a good time to be more specific in defining Maximum Power Transfer exactly what we mean by that. Having done this, we can then con- sider the most important consequence of linearity, the principle of superposition. This principle is very basic and will appear repeat-  ↔ Y Transformations for edly in our study of linear circuit analysis. As a matter of fact, the Resistive Networks nonapplicability of superposition to nonlinear circuits is the very reason they are so difficult to analyze! Selecting a Particular The principle of superposition states that the response (a desired Combination of Analysis current or voltage) in a linear circuit having more than one indepen- Techniques dent source can be obtained by adding the responses caused by the separate independent sources acting alone. Performing dc Sweep Simulations Using PSpice Linear Elements and Linear Circuits We define a linear element as a passive element that has a linear voltage-current relationship. By a “linear voltage-current relationship’’ 123 124 CHAPTER 5 HANDY CIRCUIT ANALYSIS TECHNIQUES we simply mean that multiplication of the current through the element by a constant K results in the multiplication of the voltage across the element by the same constant K. At this time, only one passive element has been defined (the resistor), and its voltage-current relationship v(t) = Ri(t) is clearly linear. As a matter of fact, if v(t) is plotted as a function of i(t), the result is a straight line. We define a linear dependent source as a dependent current or voltage The dependent voltage source vs  0.6i1  14v2 is source whose output current or voltage is proportional only to the first linear, but vs  0.6i 12 and vs  0.6i1v2 are not. power of a specified current or voltage variable in the circuit (or to the sum of such quantities). We now define a linear circuit as a circuit composed entirely of inde- pendent sources, linear dependent sources, and linear elements. From this definition, it is possible to show1 that “the response is proportional to the source,’’ or that multiplication of all independent source voltages and cur- rents by a constant K increases all the current and voltage responses by the same factor K (including the dependent source voltage or current outputs). The Superposition Principle The most important consequence of linearity is superposition. 5⍀ Let us explore the superposition principle by considering first the circuit v1 v2 of Fig. 5.1, which contains two independent sources, the current generators that force the currents ia and ib into the circuit. Sources are often called forc- ia 2⍀ 1⍀ ib ing functions for this reason, and the nodal voltages that they produce can be termed response functions, or simply responses. Both the forcing functions Ref. and the responses may be functions of time. The two nodal equations for this FIGURE 5.1 A circuit with two independent current circuit are sources. 0.7v1 − 0.2v2 = i a −0.2v1 + 1.2v2 = i b Now let us perform experiment x. We change the two forcing functions to i ax and i bx ; the two unknown voltages will now be different, so we will call them v1x and v2x. Thus, 0.7v1x − 0.2v2x = i ax −0.2v1x + 1.2v2x = i bx We next perform experiment y by changing the source currents to i ay and i by and measure the responses v1y and v2y : 0.7v1y − 0.2v2y = i ay −0.2v1y + 1.2v2y = i by (1) The proof involves first showing that the use of nodal analysis on the linear circuit can produce only linear equations of the form a1 v1 + a2 v2 + · · · + a N v N = b where the ai are constants (combinations of resistance or conductance values, constants appearing in dependent source expressions, 0, or ±1), the vi are the unknown node voltages (responses), and b is an independent source value or a sum of independent source values. Given a set of such equations, if we multiply all the b’s by K, then it is evident that the solution of this new set of equations will be the node voltages Kv1 , Kv2 ,... , Kv N.  SECTION 5.1 LINEARITY AND SUPERPOSITION 125 These three sets of equations describe the same circuit with three differ- ent sets of source currents. Let us add or “superpose’’ the last two sets of equations. Adding Eqs. and , (0.7v1x + 0.7v1y ) − (0.2v2x + 0.2v2y ) = i ax + i ay 0.7v1 − 0.2v2  ia and adding Eqs. and , −(0.2v1x + 0.2v1y ) + (1.2v2x + 1.2v2y ) = i bx + i by −0.2v1 + 1.2v2  ib where Eq. has been written immediately below Eq. and Eq. below i No voltage drop Eq. for easy comparison. across terminals, + i The linearity of all these equations allows us to compare Eq. with 0V – but current can Eq. and Eq. with Eq. and draw an interesting conclusion. If we flow select i ax and i ay such that their sum is ia and select i bx and i by such that their sum is ib, then the desired responses v1 and v2 may be found by adding v1x (a) to v1y and v2x to v2y , respectively. In other words, we can perform experi- ment x and note the responses, perform experiment y and note the No current responses, and finally add the two sets of responses. This leads to the fun- + flows, but a + voltage can damental concept involved in the superposition principle: to look at each 0A v appear across v independent source (and the response it generates) one at a time with the – the terminals – other independent sources “turned off’’ or “zeroed out.’’ If we reduce a voltage source to zero volts, we have effectively created (b) a short circuit (Fig. 5.2a). If we reduce a current source to zero amps, we FIGURE 5.2 (a) A voltage source set to zero acts have effectively created an open circuit (Fig. 5.2b). Thus, the superposition like a short circuit. (b) A current source set to zero acts theorem can be stated as: like an open circuit. In any linear resistive network, the voltage across or the current through any re- sistor or source may be calculated by adding algebraically all the individual voltages or currents caused by the separate independent sources acting alone, with all other independent voltage sources replaced by short circuits and all other independent current sources replaced by open circuits. Thus, if there are N independent sources, we must perform N experi- ments, each having only one of the independent sources active and the others inactive/turned off/zeroed out. Note that dependent sources are in general active in every experiment. There is also no reason that an independent source must assume only its given value or a zero value in the several experiments; it is necessary only for the sum of the several values to be equal to the original value. An inac- tive source almost always leads to the simplest circuit, however. The circuit we have just used as an example should indicate that a much stronger theorem might be written; a group of independent sources may be made active and inactive collectively, if we wish. For example, suppose there are three independent sources. The theorem states that we may find a given response by considering each of the three sources acting alone and adding the three results. Alternatively, we may find the response due to the first and second sources operating with the third inactive, and then add to this the response caused by the third source acting alone. This amounts to treating several sources collectively as a sort of “supersource.” 126 CHAPTER 5 HANDY CIRCUIT ANALYSIS TECHNIQUES EXAMPLE 5.1 For the circuit of Fig. 5.3a, use superposition to determine the unknown branch current ix. 6⍀ ix + vs = 3 V – 9⍀ is = 2 A (a) 6⍀ 6⍀ ix' ix" + – 3V 9⍀ 9⍀ 2A (b) (c) FIGURE 5.3 (a) An example circuit with two independent sources for which the branch current ix is desired; (b) same circuit with current source open-circuited; (c) original circuit with voltage source short-circuited. First set the current source equal to zero and redraw the circuit as shown in Fig. 5.3b. The portion of ix due to the voltage source has been designated i x to avoid confusion and is easily found to be 0.2 A. Next set the voltage source in Fig. 5.3a to zero and again redraw the circuit, as shown in Fig. 5.3c. Current division lets us determine that i x (the portion of ix due to the 2 A current source) is 0.8 A. Now compute the total current ix by adding the two individual components: i x = i x|3 V + i x|2 A = i x + i x or   3 6 ix = +2 = 0.2 + 0.8 = 1.0 A 6+9 6+9 Another way of looking at Example 5.1 is that the 3 V source and the 2 A source are each performing work on the circuit, resulting in a total cur- rent ix flowing through the 9  resistor. However, the contribution of the 3 V source to ix does not depend on the contribution of the 2 A source, and vice versa. For example, if we double the output of the 2 A source to 4 A, it will now contribute 1.6 A to the total current ix flowing through the 9  resistor. However, the 3 V source will still contribute only 0.2 A to ix, for a new total current of 0.2 + 1.6 = 1.8 A.  SECTION 5.1 LINEARITY AND SUPERPOSITION 127 P R ACTICE 5.1 For the circuit of Fig. 5.4, use superposition to compute the current ix. ix 15 ⍀ 7⍀ 5⍀ + 2A – 3.5 V 3⍀ FIGURE 5.4 Ans: 660 mA. As we will see, superposition does not generally reduce our workload when considering a particular circuit, since it leads to the analysis of several new circuits to obtain the desired response. However, it is particularly use- ful in identifying the significance of various parts of a more complex circuit. It also forms the basis of phasor analysis, which is introduced in Chap. 10. EXAMPLE 5.2 Referring to the circuit of Fig. 5.5a, determine the maximum positive current to which the source Ix can be set before any resistor exceeds its power rating and overheats. i '100 ⍀ 100 ⍀ 1 4 W 100 ⍀ + 64 ⍀ Ix + i '64 ⍀ – 6V 1 – 6V 64 ⍀ 4 W (a) (b) i "100 ⍀ 100 ⍀ i"64 ⍀ 64 ⍀ Ix (c) FIGURE 5.5 (a) A circuit with two resistors each rated at 14 W. (b) Circuit with only the 6 V source active. (c) Circuit with the source Ix active.  Identify the goal of the problem. Each resistor is rated to a maximum of 250 mW. If the circuit allows this value to be exceeded (by forcing too much current through either resistor), excessive heating will occur—possibly leading to (Continued on next page) 128 CHAPTER 5 HANDY CIRCUIT ANALYSIS TECHNIQUES an accident. The 6 V source cannot be changed, so we are looking for an equation involving Ix and the maximum current through each resistor.  Collect the known information. Based on its 250 mW power rating, the maximum current the 100  resistor can tolerate is   Pmax 0.250 = = 50 mA R 100 and, similarly, the current through the 64  resistor must be less than 62.5 mA.  Devise a plan. Either nodal or mesh analysis may be applied to the solution of this problem, but superposition may give us a slight edge, since we are primarily interested in the effect of the current source.  Construct an appropriate set of equations. Using superposition, we redraw the circuit as in Fig. 5.5b and find that the 6 V source contributes a current  6 i 100  = = 36.59 mA 100 + 64  to the 100  resistor and, since the 64  resistor is in series, i 64  = 36.59 mA as well.  Recognizing the current divider in Fig. 5.5c, we note that i 64  will    add to i 64  , but i 100  is opposite in direction to i 100 . Therefore, IX can safely contribute 62.5 − 36.59 = 25.91 mA to the 64  resistor current, and 50 − (−36.59) = 86.59 mA to the 100  resistor current. The 100  resistor therefore places the following constraint on Ix:   100 + 64 Ix < (86.59 × 10−3 ) 64 and the 64  resistor requires that   100 + 64 Ix < (25.91 × 10−3 ) 100  Attempt a solution. Considering the 100  resistor first, we see that Ix is limited to Ix < 221.9 mA. The 64  resistor limits Ix such that Ix < 42.49 mA. In order to satisfy both constraints, Ix must be less than 42.49 mA. If the value is increased, the 64  resistor will overheat long before the 100  resistor does.  Verify the solution. Is it reasonable or expected? One particularly useful way to evaluate our solution is to perform a dc sweep analysis in PSpice as described after the next example. An in- teresting question, however, is whether we would have expected the 64  resistor to overheat first. Originally we found that the 100  resistor has a smaller maximum current, so it might be reasonable to expect it to limit Ix. However, because Ix opposes the current sent by the 6 V source through the 100  resistor but adds to the 6 V source’s contribution to the current through the 64  resistor, it turns out to work the other way—it’s the 64  resistor that sets the limit on Ix.  SECTION 5.1 LINEARITY AND SUPERPOSITION 129 EXAMPLE 5.3 In the circuit of Fig. 5.6a, use the superposition principle to deter- mine the value of ix. 2⍀ 1⍀ + ix + + 10 V – v 3A 2ix – – (a) 2⍀ 1⍀ 2⍀ 1⍀ + ix' + ix" + 10 V +– 2ix' 3A v" 2ix" – – – (b) (c) FIGURE 5.6 (a) An example circuit with two independent sources and one dependent source for which the branch current ix is desired. (b) Circuit with the 3 A source open-circuited. (c) Original circuit with the 10 V source short-circuited. First open-circuit the 3 A source (Fig. 5.6b). The single mesh equation is −10 + 2i x + i x + 2i x = 0 so that i x = 2 A Next, short-circuit the 10 V source (Fig. 5.6c) and write the single- node equation v  v  − 2i x + =3 2 1 and relate the dependent-source-controlling quantity to v  : v  = 2(−i x ) Solving, we find i x = −0.6 A and, thus, i x = i x + i x = 2 + (−0.6) = 1.4 A Note that in redrawing each subcircuit, we are always careful to use some type of notation to indicate that we are not working with the original variables. This prevents the possibility of rather disastrous errors when we add the individual results. v1 15 ⍀ v2 P R ACTICE i 7⍀ 5.2 For the circuit of Fig. 5.7, use superposition to obtain the voltage 2A 5⍀ 4i across each current source. + – 3V Ans: v1|2A = 9.180 V, v2|2A = −1.148 V, v1|3V = 1.967 V, v2|3V = −0.246 V; v1 = 11.147 V, v2 = −1.394 V. FIGURE 5.7 130 CHAPTER 5 HANDY CIRCUIT ANALYSIS TECHNIQUES Summary of Basic Superposition Procedure 1. Select one of the independent sources. Set all other indepen- dent sources to zero. This means voltage sources are replaced with short circuits and current sources are replaced with open circuits. Leave dependent sources in the circuit. 2. Relabel voltages and currents using suitable notation (e.g., v  , i 2 ). Be sure to relabel controlling variables of dependent sources to avoid confusion. 3. Analyze the simplified circuit to find the desired currents and/or voltages. 4. Repeat steps 1 through 3 until each independent source has been considered. 5. Add the partial currents and/or voltages obtained from the separate analyses. Pay careful attention to voltage signs and current directions when summing. 6. Do not add power quantities. If power quantities are required, calculate only after partial voltages and/or currents have been summed. Note that step 1 may be altered in several ways. First, independent sources can be considered in groups as opposed to individually if it simpli- fies the analysis, as long as no independent source is included in more than one subcircuit. Second, it is technically not necessary to set sources to zero, although this is almost always the best route. For example, a 3 V source may appear in two subcircuits as a 1.5 V source, since 1.5 + 1.5 = 3 V just as 0 + 3 = 3 V. Because it is unlikely to simplify our analysis, however, there is little point to such an exercise. COMPUTER-AIDED ANALYSIS Although PSpice is extremely useful in verifying that we have analyzed a complete circuit correctly, it can also assist us in determining the contribution of each source to a particular response. To do this, we employ what is known as a dc parameter sweep. Consider the circuit presented in Example 5.2, when we were asked to determine the maximum positive current that could be obtained from the current source without exceeding the power rating of either resistor in the circuit. The circuit is shown redrawn using the Orcad Capture CIS schematic tool in Fig. 5.8. Note that no value has been assigned to the current source. After the schematic has been entered and saved, the next step is to specify the dc sweep parameters. This option allows us to specify a range of values for a voltage or current source (in the present case, the current source Ix ), rather than a specific value. Selecting New Simulation Profile under PSpice, we provide a name for our profile and are then provided with the dialog box shown in Fig. 5.9.  SECTION 5.1 LINEARITY AND SUPERPOSITION 131 FIGURE 5.8 The circuit from Example 5.2. FIGURE 5.9 DC Sweep dialog box shown with Ix selected as the sweep variable. Under Analysis Type, we pull down the DC Sweep option, specify the “sweep variable’’ as Current Source, and then type in Ix in the Name box. There are several options under Sweep Type: Linear, Logarithmic, and Value List. The last option allows us to specify each value to assign to Ix. In order to generate a smooth plot, however, we choose to perform a Linear sweep, with a Start Value of 0 mA, an End Value of 50 mA, and a value of 0.01 mA for the Increment. After we perform the simulation, the graphical output package Probe is automatically launched. When the window appears, the horizontal axis (corresponding to our variable, Ix ) is displayed, but the vertical axis variable must be chosen. Selecting Add Trace from the Trace menu, we click on I(R1), then type an asterisk in the Trace Expression box, click on I(R1) once again, insert yet another asterisk, and finally type in 100. This asks Probe to plot the power absorbed by the 100  resistor. In a similar fashion, we repeat the process to add the power (Continued on next page) 132 CHAPTER 5 HANDY CIRCUIT ANALYSIS TECHNIQUES (a) (b) FIGURE 5.10 (a) Probe output with text labels identifying the power absorbed by the two resistors individually. A horizontal line indicating 250 mW has also been included, as well as text labels to improve clarity. (b) Cursor dialog box. absorbed by the 64  resistor, resulting in a plot similar to that shown in Fig. 5.10a. A horizontal reference line at 250 mW was also added to the plot by typing 0.250 in the Trace Expression box after selecting Add Trace from the Trace menu a third time. We see from the plot that the 64  resistor does exceed its 250 mW power rating in the vicinity of Ix = 43 mA. In contrast, however, we also see that regardless of the value of the current source Ix (provided that it is between 0 and 50 mA), the 100  resistor will never dissipate 250 mW; in fact, the absorbed power decreases with increasing current from the current source. If we desire a more precise answer, we can make use of the cursor tool, which is invoked by selecting Trace, Cursor, Display from the menu bar. Figure 5.10b shows the result of dragging cursor 1 to 42.52 A, where the 64  resistor is dissipating just over its maximum rated power of 250 mW. Increased precision can be obtained by decreasing the increment value used in the dc sweep. This technique is very useful in analyzing electronic circuits, where we might need, for example, to determine what input voltage is required  SECTION 5.2 SOURCE TRANSFORMATIONS 133 to a complicated amplifier circuit in order to obtain a zero output voltage. We also notice that there are several other types of parameter sweeps that we can perform, including a dc voltage sweep. The ability to vary temperature is useful only when dealing with component models that have a temperature parameter built in, such as diodes and transistors. Unfortunately, it usually turns out that little if any time is saved in ana- lyzing a circuit containing one or more dependent sources by use of the superposition principle, for there must always be at least two sources in operation: one independent source and all the dependent sources. We must constantly be aware of the limitations of superposition. It is applicable only to linear responses, and thus the most common nonlinear response—power—is not subject to superposition. For example, consider two 1 V batteries in series with a 1  resistor. The power delivered to the re- sistor is 4 W, but if we mistakenly try to apply superposition, we might say that each battery alone furnished 1 W and thus the calculated power is only 2 W. This is incorrect, but a surprisingly easy mistake to make. 5.2 SOURCE TRANSFORMATIONS Practical Voltage Sources So far, we’ve only worked with ideal sources—elements whose terminal voltage is independent of the current flowing through them. To see the relevance of this fact, consider a simple independent (“ideal”) 9 V source connected to a 1  resistor. The 9 volt source will force a current of 9 amperes through the 1  resistor (perhaps this seems reasonable enough), but the same source would apparently force 9,000,000 amperes through a 1 m resistor 12 V + – (which hopefully does not seem reasonable). On paper, there's nothing to stop us from reducing the resistor value all the way to 0  … but that would lead to a contradiction, as the source would be “trying” to maintain 9 V across a (a) dead short, which Ohm’s law tells us can’t happen (V = 9 = R I = 0?). What happens in the real world when we do this type of experiment? For 0.01 ⍀ example, if we try to start a car with the headlights already on, we most likely notice the headlights dim as the battery is asked to supply a large (∼100 A or + 12 V more) starter current in parallel with the current running to the headlights. If – we model the 12 V battery with an ideal 12 V source as in Fig. 5.11a, our observation cannot be explained. Another way of saying this is that our model (b) breaks down when the load draws a large current from the source. FIGURE 5.11 (a) An ideal 12 V dc voltage source To better approximate the behavior of a real device, the ideal voltage used to model a car battery. (b) A more accurate source must be modified to account for the lowering of its terminal voltage model that accounts for the observed reduction in when large currents are drawn from it. Let us suppose that we observe ex- terminal voltage at large currents. perimentally that our car battery has a terminal voltage of 12 V when no current is flowing through it, and a reduced voltage of 11 V when 100 A is flowing. How could we model this behavior? Well, a more accurate model might be an ideal voltage source of 12 V in series with a resistor across which 1 V appears when 100 A flows through it. A quick calculation shows that the resistor must be 1 V/100 A  0.01 , and the ideal voltage source and this series resistor constitute a practical voltage source (Fig. 5.11b). 134 CHAPTER 5 HANDY CIRCUIT ANALYSIS TECHNIQUES 0.01 ⍀ IL Thus, we are using the series combination of two ideal circuit elements, an independent voltage source and a resistor, to model a real device. + We do not expect to find such an arrangement of ideal elements inside 12 V + – VL RL our car battery, of course. Any real device is characterized by a certain current-voltage relationship at its terminals, and our problem is to develop – some combination of ideal elements that can furnish a similar current-voltage (a) characteristic, at least over some useful range of current, voltage, or power. In Fig. 5.12a, we show our two-piece practical model of the car battery now connected to some load resistor RL. The terminal voltage of the practical Ideal source 12 source is the same as the voltage across RL and is marked2 VL. Figure 5.12b shows a plot of load voltage VL as a function of the load current IL for this Source voltage VL (V) 10 practical source. The KVL equation for the circuit of Fig. 5.12a may be 8 written in terms of IL and VL: 6 Practical source 12 = 0.01I L + VL 4 2 and thus 0 0 200 400 600 800 1000 1200 VL = −0.01I L + 12 Load current IL (A) This is a linear equation in IL and VL, and the plot in Fig. 5.12b is a straight (b) line. Each point on the line corresponds to a different value of RL. For exam- FIGURE 5.12 (a) A practical source, which approximates the behavior of a certain 12 V ple, the midpoint of the straight line is obtained when the load resistance is automobile battery, is shown connected to a load equal to the internal resistance of the practical source, or R L = 0.01 . Here, resistor RL. (b) The relationship between IL and VL the load voltage is exactly one-half the ideal source voltage. is linear. When R L = ∞ and no current whatsoever is being drawn by the load, the practical source is open-circuited and the terminal voltage, or open-circuit voltage, is VLoc = 12 V. If, on the other hand, R L = 0, thereby short-circuiting the load terminals, then a load current or short-circuit cur- Rs iL rent, I Lsc = 1200 A, would flow. (In practice, such an experiment would probably result in the destruction of the short circuit, the battery, and any + measuring instruments incorporated in the circuit!) vs + vL Since the plot of VL versus IL is a straight line for this practical voltage – RL source, we should note that the values of VLoc and I Lsc uniquely determine – the entire VL –I L curve. (a) The horizontal broken line of Fig. 5.12b represents the VL –I L plot for an ideal voltage source; the terminal voltage remains constant for any value of iL load current. For the practical voltage source, the terminal voltage has a value near that of the ideal source only when the load current is relatively small. iLsc = Ideal Let us now consider a general practical voltage source, as shown in vs /Rs source Fig. 5.13a. The voltage of the ideal source is vs, and a resistance Rs, called an internal resistance or output resistance, is placed in series with it. Again, we must note that the resistor is not really present as a separate component Practical but merely serves to account for a terminal voltage that decreases as the source load current increases. Its presence enables us to model the behavior of a 0 vL 0 vLoc = vs physical voltage source more closely. (b) The linear relationship between vL and iL is FIGURE 5.13 (a) A general practical voltage v L = vs − Rs i L source connected to a load resistor RL. (b) The terminal voltage of a practical voltage source decreases as iL (2) From this point on we will endeavor to adhere to the standard convention of referring to strictly dc increases and RL  vL /iL decreases. The terminal quantities using capital letters, whereas lowercase letters denote a quantity that we know to possess some voltage of an ideal voltage source (also plotted) time-varying component. However, in describing general theorems which apply to either dc or ac, we will remains the same for any current delivered to a load. continue to use lowercase to emphasize the general nature of the concept.  SECTION 5.2 SOURCE TRANSFORMATIONS 135 and this is plotted in Fig. 5.13b. The open-circuit voltage (R L = ∞, so i L = 0) is v Loc = vs and the short-circuit current (R L = 0, so v L = 0) is vs i Lsc = Rs Once again, these values are the intercepts for the straight line in Fig. 5.13b, and they serve to define it completely. Practical Current Sources An ideal current source is also nonexistent in the real world; there is no physical device that will deliver a constant current regardless of the load re- sistance to which it is connected or the voltage across its terminals. Certain iL transistor circuits will deliver a constant current to a wide range of load re- + sistances, but the load resistance can always be made sufficiently large that the current through it becomes very small. Infinite power is simply never is Rp vL RL available (unfortunately). – A practical current source is defined as an ideal current source in paral- (a) lel with an internal resistance Rp. Such a source is shown in Fig. 5.14a, and the current iL and voltage vL associated with a load resistance RL are indi- iL cated. Application of KCL yields vL iLsc = is Ideal source i L = is − Rp which is again a linear relationship. The open-circuit voltage and the short- Practical source circuit current are v Loc = R p i s vL and vLoc = Rpis i Lsc = i s (b) FIGURE 5.14 (a) A general practical current The variation of load current with changing load voltage may be inves- source connected to a load resistor RL. (b) The load tigated by changing the value of RL as shown in Fig. 5.14b. The straight line current provided by the practical current source is is traversed from the short-circuit, or “northwest,’’ end to the open-circuit shown as a function of the load voltage. termination at the “southeast’’ end by increasing RL from zero to infinite ohms. The midpoint occurs for R L = R p. The load current iL and the ideal source current are approximately equal only for small values of load volt- age, which are obtained with values of RL that are small compared to Rp. Equivalent Practical Sources It may be no surprise that we can improve upon models to increase their accuracy; at this point we now have a practical voltage source model and also a practical current source model. Before we proceed, however, let’s take a moment to compare Fig. 5.13b and Fig. 5.14b. One is for a circuit with a voltage source and the other, with a current source, but the graphs are indistinguishable! It turns out that this is no coincidence. In fact, we are about to show that a practical voltage source can be electrically equivalent to a practical cur- rent source—meaning that a load resistor RL connected to either will have 136 CHAPTER 5 HANDY CIRCUIT ANALYSIS TECHNIQUES Rs iL the same vL and iL. This means we can replace one practical source with the other and the rest of the circuit will not know the difference. + Consider the practical voltage source and resistor RL shown in Fig. 5.15a, vs + vL RL and the circuit composed of a practical current source and resistor RL shown – in Fig. 5.15b. A simple calculation shows that the voltage across the load RL – of Fig. 5.15a is (a) RL v L = vs Rs + R L iL A similar calculation shows that the voltage across the load RL in + Fig. 5.15b is   is Rp vL RL Rp vL = is · RL Rp + RL – The two practical sources are electrically equivalent, then, if (b) FIGURE 5.15 (a) A given practical Rs = R p voltage source connected to a load RL. and (b) The equivalent practical current source connected to the same load. vs = R p i s = Rs i s where we now let Rs represent the internal resistance of either practical source, which is the conventional notation. Let’s try this with the practical current source shown in Fig. 5.16a. Since 3A 2⍀ its internal resistance is 2 , the internal resistance of the equivalent practi- cal voltage source is also 2 ; the voltage of the ideal voltage source con- tained within the practical voltage source is (2)(3) = 6 V. The equivalent (a) practical voltage source is shown in Fig. 5.16b. 2⍀ To check the equivalence, let us visualize a 4  resistor connected to each source. In both cases a current of 1 A, a voltage of 4 V, and a power of 4 W are associated with the 4  load. However, we should note very care- + fully that the ideal current source is delivering a total power of 12 W, while 6V – the ideal voltage source is delivering only 6 W. Furthermore, the internal resistance of the practical current source is absorbing 8 W, whereas the in- (b) ternal resistance of the practical voltage source is absorbing only 2 W. Thus FIGURE 5.16 (a) A given practical we see that the two practical sources are equivalent only with respect to current source. (b) The equivalent practical what transpires at the load terminals; they are not equivalent internally! voltage source. EXAMPLE 5.4 Compute the current through the 4.7 k resistor in Fig. 5.17a after transforming the 9 mA source into an equivalent voltage source. It’s not just the 9 mA source at issue, but also the resistance in parallel with it (5 k). We remove these components, leaving two terminals “dan- gling.” We then replace them with a voltage source in series with a 5 k resistor. The value of the voltage source must be (0.09)(5000)  45 V. Redrawing the circuit as in Fig. 5.17b, we can write a simple KVL equation 45  5000I  4700I  3000I  3  0 which is easily solved to yield I  3.307 mA. We can check our answer of course by analyzing the circuit of Fig. 5.17a using either nodal or mesh techniques.  SECTION 5.2 SOURCE TRANSFORMATIONS 137 4.7 k⍀ 3 k⍀ 5 k⍀ 4.7 k⍀ 3 k⍀ I + + + 9 mA 5 k⍀ – 3V – 45 V 3V – I (a) (b) FIGURE 5.17 (a) A circuit with both a voltage source and a current source. (b) The circuit after the 9 mA source is transformed into an equivalent voltage source. P R ACTICE 5.3 For the circuit of Fig. 5.18, compute the current IX through the 47 k resistor after performing a source transformation on the voltage source. 5 k⍀ + IX 5V – 47 k⍀ 1 mA FIGURE 5.18 Ans: 192 μA. EXAMPLE 5.5 Calculate the current through the 2  resistor in Fig. 5.19a by making use of source transformations to first simplify the circuit. We begin by transforming each current source into a voltage source (Fig. 5.19b), the strategy being to convert the circuit into a simple loop. We must be careful to retain the 2  resistor for two reasons: first, the dependent source controlling variable appears across it, and second, we desire the current flowing through it. However, we can combine the 17  and 9  resistors, since they appear in series. We also see that the 3  and 4  resistors may be combined into a single 7  resistor, which can then be used to transform the 15 V source into a 15/7 A source as in Fig. 5.19c. Finally, we note that the two 7  resistors can be combined into a single 3.5  resistor, which may be used to transform the 15/7 A current source into a 7.5 V voltage source. The result is a simple loop circuit, shown in Fig. 5.19d. The current I can now be found using KVL: −7.5 + 3.5I − 51Vx + 28I + 9 = 0 where Vx = 2I Thus, I = 21.28 mA (Continued on next page) 138 CHAPTER 5 HANDY CIRCUIT ANALYSIS TECHNIQUES 17 ⍀ 4⍀ + Vx – I 2⍀ 3Vx 5A 3⍀ 7⍀ 9⍀ 1A (a) 3⍀ 4⍀ 17 ⍀ + Vx – 9⍀ – + 2⍀ + 51Vx I + 15 V – 7⍀ – 9V (b) 26 ⍀ + Vx – – + 2⍀ 15 51Vx I + 7 A 7⍀ 7⍀ – 9V (c) 3.5 ⍀ 26 ⍀ + Vx – – + 2⍀ + 51Vx I + 7.5 V – – 9V (d ) FIGURE 5.19 (a) A circuit with two independent current sources and one dependent source. (b) The circuit after each source is transformed into a voltage source. (c) The circuit after further combinations. (d) The final circuit. P R ACTICE 5.4 For the circuit of Fig. 5.20, compute the voltage V across the 1 M resistor using repeated source transformations. 40 ␮A 6 M⍀ + V – 1 M⍀ 75 ␮A + 3V 4 M⍀ – 200 k⍀ FIGURE 5.20 Ans: 27.2 V.  SECTION 5.2 SOURCE TRANSFORMATIONS 139 Several Key Points We conclude our discussion of practical sources and source transformations with a few observations. First, when we transform a voltage source, we must be sure that the source is in fact in series with the resistor under con- sideration. For example, in the circuit of Fig. 5.21, it is perfectly valid to perform a source transformation on the voltage source using the 10  resis- tor, as they are in series. However, it would be incorrect to attempt a source transformation using the 60 V source and the 30  resistor—a very common type of error. In a similar fashion, when we transform a current source and resistor combination, we must be sure that they are in fact in parallel. Consider the current source shown in Fig. 5.22a. We may perform a source transforma- tion including the 3  resistor, as they are in parallel, but after the transfor- mation there may be some ambiguity as to where to place the resistor. In such circumstances, it is helpful to first redraw the components to be trans- formed as in Fig. 5.22b. Then the transformation to a voltage source in series with a resistor may be drawn correctly as shown in Fig. 5.22c; the resistor may in fact be drawn above or below the voltage source. It is also worthwhile to consider the unusual case of a current source in series with a resistor, and its dual, the case of a voltage source in parallel 60 V 10 ⍀ – + 4A 20 ⍀ 30 ⍀ 0.4i1 i1 FIGURE 5.21 An example circuit to illustrate how to determine if a source transformation can be performed. 7⍀ 7⍀ + + + + 5V – 2⍀ 1A 3⍀ – 3V 5V – 2⍀ 1A 3⍀ – 3V (a) (b) 7⍀ 3⍀ + + 5V – 2⍀ – 3V – + 3V (c) FIGURE 5.22 (a) A circuit with a current source to be transformed to a voltage source. (b) Circuit redrawn so as to avoid errors. (c) Transformed source/resistor combination. 140 CHAPTER 5 HANDY CIRCUIT ANALYSIS TECHNIQUES with a resistor. Let’s start with the simple circuit of Fig. 5.23a, where we are interested only in the voltage across the resistor marked R2. We note that re- gardless of the value of resistor R1 , VR2 = Ix R2. Although we might be tempted to perform an inappropriate source transformation on such a cir- cuit, in fact we may simply omit resistor R1 (provided that it is of no interest to us itself). A similar situation arises with a voltage source in parallel with a resistor, as depicted in Fig. 5.23b. Again, if we are only interested in some quantity regarding resistor R2, we may find ourselves tempted to perform some strange (and incorrect) source transformation on the voltage source and resistor R1. In reality, we may omit resistor R1 from our circuit as far as resistor R2 is concerned—its presence does not alter the voltage across, the current through, or the power dissipated by resistor R2. R1 + Ix R2 Vx + R1 R2 VR2 – – (a) (b) FIGURE 5.23 (a) Circuit with a resistor R1 in series with a current source. (b) A voltage source in parallel with two resistors. Summary of Source Transformation 1. A common goal in source transformation is to end up with either all current sources or all voltage sources in the circuit. This is especially true if it makes nodal or mesh analysis easier. 2. Repeated source transformations can be used to simplify a circuit by allowing resistors and sources to eventually be combined. 3. The resistor value does not change during a source transfor- mation, but it is not the same resistor. This means that currents or voltages associated with the original resistor are irretrievably lost when we perform a source transformation. 4. If the voltage or current associated with a particular resistor is used as a controlling variable for a dependent source, it should not be included in any source transformation. The original resistor must be retained in the final circuit, untouched. 5. If the voltage or current associated with a particular element is of interest, that element should not be included in any source transformation. The original element must be retained in the final circuit, untouched. 6. In a source transformation, the head of the current source arrow corresponds to the “+” terminal of the voltage source. 7. A source transformation on a current source and resistor requires that the two elements be in parallel. 8. A source transformation on a voltage source and resistor requires that the two elements be in series.  SECTION 5.3 THÉVENIN AND NORTON EQUIVALENT CIRCUITS 141 5.3 THÉVENIN AND NORTON EQUIVALENT CIRCUITS Now that we have been introduced to source transformations and the super- position principle, it is possible to develop two more techniques that will greatly simplify the analysis of many linear circuits. The first of these theo- rems is named after L. C. Thévenin, a French engineer working in telegra- phy who published the theorem in 1883; the second may be considered a corollary of the first and is credited to E. L. Norton, a scientist with the Bell Telephone Laboratories. Let us suppose that we need to make only a partial analysis of a circuit. For example, perhaps we need to determine the current, voltage, and power delivered to a single “load” resistor by the remainder of the circuit, which may consist of a sizable number of sources and resistors (Fig. 5.24a). Or, perhaps we wish to find the response for different values of the load resis- tance. Thévenin’s theorem tells us that it is possible to replace everything except the load resistor with an independent voltage source in series with a resistor (Fig. 5.24b); the response measured at the load resistor will be un- changed. Using Norton’s theorem, we obtain an equivalent composed of an independent current source in parallel with a resistor (Fig. 5.24c). RTH Complex + RL VTH RL IN RN RL network – (a) (b) (c) FIGURE 5.24 (a) A complex network including a load resistor RL. (b) A Thévenin equivalent network connected to the load resistor RL. (c) A Norton equivalent network connected to the load resistor RL. Thus, one of the main uses of Thévenin’s and Norton’s theorems is the replacement of a large part of a circuit, often a complicated and uninter- esting part, with a very simple equivalent. The new, simpler circuit enables us to make rapid calculations of the voltage, current, and power which the original circuit is able to deliver to a load. It also helps us to choose the best value of this load resistance. In a transistor power amplifier, for example, the Thévenin or Norton equivalent enables us to determine the maximum power that can be taken from the amplifier and delivered to the speakers. EXAMPLE 5.6 Consider the circuit shown in Fig. 5.25a. Determine the Thévenin equivalent of network A, and compute the power delivered to the load resistor RL. The dashed regions separate the circuit into networks A and B; our main interest is in network B, which consists only of the load resistor R L. Net- work A may be simplified by making repeated source transformations. (Continued on next page) 142 CHAPTER 5 HANDY CIRCUIT ANALYSIS TECHNIQUES 3⍀ 7⍀ 7⍀ + RL – 12 V 6⍀ 4A 3⍀ 6⍀ RL Network A Network B Network A (a) (b) 7⍀ 2⍀ 7⍀ RL + RL 4A 2⍀ – 8V Network A Network A (c) (d ) 9⍀ + RL – 8V Network A (e) FIGURE 5.25 (a) A circuit separated into two networks. (b)–(d) Intermediate steps to simplifying network A. (e) The Thévenin equivalent circuit. We first treat the 12 V source and the 3  resistor as a practical volt- age source and replace it with a practical current source consisting of a 4 A source in parallel with 3  (Fig. 5.25b). The parallel resistances are then combined into 2  (Fig. 5.25c), and the practical current source that results is transformed back into a practical voltage source (Fig. 5.25d). The final result is shown in Fig. 5.25e. From the viewpoint of the load resistor R L , this network A (the Thévenin equivalent) is equivalent to the original network A; from our viewpoint, the circuit is much simpler, and we can now easily compute the power delivered to the load:  2 8 PL = RL 9 + RL Furthermore, we can see from the equivalent circuit that the maxi- mum voltage that can be obtained across R L is 8 V and corresponds to R L = ∞. A quick transformation of network A to a practical current source (the Norton equivalent) indicates that the maximum current that may be delivered to the load is 8/9 A, which occurs when R L = 0. Neither of these facts is readily apparent from the original circuit. 8⍀ 5A RL P R ACTICE 2⍀ 10 ⍀ 5.5 Using repeated source transformations, determine the Norton equivalent of the highlighted network in the circuit of Fig. 5.26. FIGURE 5.26 Ans: 1 A, 5 .  SECTION 5.3 THÉVENIN AND NORTON EQUIVALENT CIRCUITS 143 Thévenin’s Theorem Using the technique of source transformation to find a Thévenin or Norton equivalent network worked well enough in Example 5.6, but it can rapidly become impractical in situations where dependent sources are present or the circuit is composed of a large number of elements. An alternative is to employ Thévenin’s theorem (or Norton’s theorem) instead. We will state the theorem3 as a somewhat formal procedure and then consider various ways to make the approach more practical depending on the situation we face. A Statement of Thévenin’s Theorem 1. Given any linear circuit, rearrange it in the form of two networks, A and B, connected by two wires. Network A is the network to be simplified; B will be left untouched. 2. Disconnect network B. Define a voltage voc as the voltage now appearing across the terminals of network A. 3. Turn off or “zero out” every independent source in network A to form an inactive network. Leave dependent sources unchanged. 4. Connect an independent voltage source with value voc in series with the inactive network. Do not complete the circuit; leave the two terminals disconnected. 5. Connect network B to the terminals of the new network A. All currents and voltages in B will remain unchanged. Note that if either network contains a dependent source, its control variable must be in the same network. Let us see if we can apply Thévenin’s theorem successfully to the circuit we considered in Fig. 5.25. We have already found the Thévenin equivalent of the circuit to the left of R L in Example 5.6, but we want to see if there is an easier way to obtain the same result. EXAMPLE 5.7 Use Thévenin’s theorem to determine the Thévenin equivalent for that part of the circuit in Fig. 5.25a to the left of RL. We begin by disconnecting R L , and note that no current flows through the 7  resistor in the resulting partial circuit shown in Fig. 5.27a. Thus, Voc appears across the 6  resistor (with no current through the 7  resistor there is no voltage drop across it), and voltage division enables us to determine that   6 Voc = 12 =8V 3+6 (3) A proof of Thévenin’s theorem in the form in which we have stated it is rather lengthy, and therefore it has been placed in Appendix 3, where the curious may peruse it. 144 CHAPTER 5 HANDY CIRCUIT ANALYSIS TECHNIQUES 3⍀ 7⍀ 7⍀ + + 12 V – 6⍀ Voc 3⍀ 6⍀ RTH – (a) (b) FIGURE 5.27 (a) The circuit of Fig. 5.25a with network B (the resistor RL) disconnected and the voltage across the connecting terminals labeled as Voc. (b) The independent source in Fig. 5.25a has been killed, and we look into the terminals where network B was connected to determine the effective resistance of network A. Turning off network A (i.e., replacing the 12 V source with a short circuit) and looking back into the dead network, we see a 7  resistor connected in series with the parallel combination of 6  and 3  (Fig. 5.27b). Thus, the inactive network can be represented here by a 9  resistor, referred to as the Thévenin equivalent resistance of network A. The Thévenin equivalent then is Voc in series with a 9  resistor, which agrees with our previous result. 4⍀ 5⍀ I2⍀ P R ACTICE + 9V – 4⍀ 2⍀ 5.6 Use Thévenin’s theorem to find the current through the 2  resistor 6⍀ in the circuit of Fig. 5.28. (Hint: Designate the 2  resistor as network B.) FIGURE 5.28 Ans: VTH = 2.571 V, RTH = 7.857 , I2  = 260.8 mA. A Few Key Points The equivalent circuit we have learned how to obtain is completely inde- pendent of network B: we have been instructed to first remove network B and then measure the open-circuit voltage produced by network A, an operation that certainly does not depend on network B in any way. The B network is mentioned only to indicate that an equivalent for A may be obtained no mat- ter what arrangement of elements is connected to the A network; the B net- work represents this general network. There are several points about the theorem which deserve emphasis. The only restriction that we must impose on A or B is that all dependent sources in A have their control variables in A, and similarly for B. No restrictions are imposed on the complexity of A or B; either one may contain any combination of independent voltage or current sources, linear dependent voltage or current sources, resistors, or any other circuit elements which are linear. The dead network A can be represented by a single equivalent resis- tance RTH , which we will call the Thévenin equivalent resistance.  SECTION 5.3 THÉVENIN AND NORTON EQUIVALENT CIRCUITS 145 This holds true whether or not dependent sources exist in the inactive A network, an idea we will explore shortly. A Thévenin equivalent consists of two components: a voltage source in series with a resistance. Either may be zero, although this is not usually the case. Norton’s Theorem Norton’s theorem bears a close resemblance to Thévenin’s theorem and may be stated as follows: A Statement of Norton’s Theorem 1. Given any linear circuit, rearrange it in the form of two networks, A and B, connected by two wires. Network A is the network to be simplified; B will be left untouched. As before, if either network contains a dependent source, its controlling variable must be in the same network. 2. Disconnect network B, and short the terminals of A. Define a current isc as the current now flowing through the shorted terminals of network A. 3. Turn off or “zero out” every independent source in network A to form an inactive network. Leave dependent sources unchanged. 4. Connect an independent current source with value isc in parallel with the inactive network. Do not complete the circuit; leave the two terminals disconnected. 5. Connect network B to the terminals of the new network A. All currents and voltages in B will remain unchanged. The Norton equivalent of a linear network is the Norton current source i sc in parallel with the Thévenin resistance RTH. Thus, we see that in fact it is possible to obtain the Norton equivalent of a network by performing a source transformation on the Thévenin equivalent. This results in a direct relationship between voc , i sc , and RTH : voc = RTH i sc In circuits containing dependent sources, we will often find it more con- venient to determine either the Thévenin or Norton equivalent by finding both the open-circuit voltage and the short-circuit current and then deter- mining the value of RTH as their quotient. It is therefore advisable to be- come adept at finding both open-circuit voltages and short-circuit currents, even in the simple problems that follow. If the Thévenin and Norton equiv- alents are determined independently, Eq. can serve as a useful check. Let’s consider three different examples of the determination of a Thévenin or Norton equivalent circuit. 146 CHAPTER 5 HANDY CIRCUIT ANALYSIS TECHNIQUES EXAMPLE 5.8 Find the Thévenin and Norton equivalent circuits for the network faced by the 1 k resistor in Fig. 5.29a. 2 k⍀ 3 k⍀ 3 k⍀ + RTH 4V – 2 mA 1 k⍀ 2 k⍀ (a) (b) 5 k⍀ + 8V – 1 k⍀ 1.6 mA 5 k⍀ 1 k⍀ (c) (d) 2 k⍀ 3 k⍀ + Isc 4V – 2 mA (e) FIGURE 5.29 (a) A given circuit in which the 1 k resistor is identified as network B. (b) Network A with all independent sources killed. (c) The Thévenin equivalent is shown for network A. (d) The Norton equivalent is shown for network A. (e) Circuit for determining Isc. From the wording of the problem statement, network B is the 1 k resistor, so network A is everything else. Choosing to find the Thévenin equivalent of network A first, we apply superposition, noting that no current flows through the 3 k resistor once network B is disconnected. With the current source set to zero, Voc|4 V  4 V. With the voltage source set to zero, Voc|2 mA  (0.002)(2000)  4 V. Thus, Voc  4  4  8 V. To find RTH, set both sources to zero as in Fig. 5.29b. By inspection, RTH  2 k  3 k  5 k. The complete Thévenin equivalent, with network B reconnected, is shown in Fig. 5.29c. The Norton equivalent is found by a simple source transformation of the Thévenin equivalent, resulting in a current source of 8/5000  1.6 mA in parallel with a 5 k resistor (Fig. 5.29d). Check: Find the Norton equivalent directly from Fig. 5.29a. Re- moving the 1 k resistor and shorting the terminals of network A, we  SECTION 5.3 THÉVENIN AND NORTON EQUIVALENT CIRCUITS 147 find Isc as shown in Fig. 5.29e by superposition and current division: 4 2 Isc = Isc|4 V + Isc|2 mA = + (2) 2+3 2+3 = 0.8 + 0.8 = 1.6 mA which completes the check. P R ACTICE 5.7 Determine the Thévenin and Norton equivalents of the circuit of Fig. 5.30. 2 k⍀ 1 k⍀ + 3V – 7 mA 5 k⍀ FIGURE 5.30 Ans: −7.857 V, −3.235 mA, 2.429 k. When Dependent Sources Are Present Technically speaking, there does not always have to be a “network B” for us to invoke either Thévenin’s theorem or Norton’s theorem; we could instead be asked to find the equivalent of a network with two terminals not yet con- nected to another network. If there is a network B that we do not want to in- volve in the simplification procedure, however, we must use a little caution if it contains dependent sources. In such situations, the controlling variable and the associated element(s) must be included in network B and excluded from network A. Otherwise, there will be no way to analyze the final circuit because the controlling quantity will be lost. If network A contains a dependent source, then again we must ensure that the controlling variable and its associated element(s) cannot be in net- work B. Up to now, we have only considered circuits with resistors and in- dependent sources. Although technically speaking it is correct to leave a dependent source in the “inactive” network when creating a Thévenin or Norton equivalent, in practice this does not result in any kind of simplifica- tion. What we really want is an independent voltage source in series with a single resistor, or an independent current source in parallel with a single resistor—in other words, a two-component equivalent. In the following examples, we consider various means of reducing networks with dependent sources and resistors into a single resistance. 148 CHAPTER 5 HANDY CIRCUIT ANALYSIS TECHNIQUES EXAMPLE 5.9 Determine the Thévenin equivalent of the circuit in Fig. 5.31a. 2 k⍀ 3 k⍀ 2 k⍀ 3 k⍀ – + + + vx 8V + vx 4V vx vx – 4000 4000 – – (a) (b) 10 k⍀ + 8V – (c) FIGURE 5.31 (a) A given network whose Thévenin equivalent is desired. (b) A possible, but rather useless, form of the Thévenin equivalent. (c) The best form of the Thévenin equivalent for this linear resistive network. To find Voc we note that vx = Voc and that the dependent source current must pass through the 2 k resistor, since no current can flow through the 3 k resistor. Using KVL around the outer loop:  v  x −4 + 2 × 103 − + 3 × 103 (0) + vx = 0 4000 and vx = 8 V = Voc By Thévenin’s theorem, then, the equivalent circuit could be formed with the inactive A network in series with an 8 V source, as shown in Fig. 5.31b. This is correct, but not very simple and not very helpful; in the case of linear resistive networks, we really want a simpler equivalent for the inactive A network, namely, RTH. The dependent source prevents us from determining RTH directly for the inactive network through resistance combination; we therefore seek Isc. Upon short-circuiting the output terminals in Fig. 5.31a, it is apparent that Vx = 0 and the dependent current source is not active. Hence, Isc = 4/(5 × 103 ) = 0.8 mA. Thus, Voc 8 RTH = = = 10 k Isc 0.8 × 10−3 and the acceptable Thévenin equivalent of Fig. 5.31c is obtained. 100 V P R ACTICE – + 5.8 Find the Thévenin equivalent for the network of Fig. 5.32. (Hint: + a quick source transformation on the dependent source might help.) 0.01V1 20 k⍀ V1 Ans: −502.5 mV, −100.5 . – Note: a negative resistance might seem strange—and it is! Such a thing is physically FIGURE 5.32 possible only if, for example, we do a bit of clever electronic circuit design to create something that behaves like the dependent current source we represented in Fig. 5.32.  SECTION 5.3 THÉVENIN AND NORTON EQUIVALENT CIRCUITS 149 As another example, let us consider a network having a dependent source but no independent source. EXAMPLE 5.10 Find the Thévenin equivalent of the circuit shown in Fig. 5.33a. 3⍀ i 3⍀ i + + + 1.5i 2⍀ 1.5i 2 ⍀ vtest 1A – – – (a) (b) 0.6 ⍀ (c) FIGURE 5.33 (a) A network with no independent sources. (b) A hypothetical measurement to obtain RT H. (c) The Thévenin equivalent to the original circuit. The rightmost terminals are already open-circuited, hence i = 0. Consequently, the dependent source is inactive, so voc = 0. We next seek the value of RTH represented by this two-terminal network. However, we cannot find voc and i sc and take their quotient, for there is no independent source in the network and both voc and i sc are zero. Let us, therefore, be a little tricky. We apply a 1 A source externally, measure the voltage vtest that results, and then set RTH = vtest /1. Referring to Fig. 5.33b, we see that i = −1 A. Applying nodal analysis, vtest − 1.5(−1) vtest + =1 3 2 so that vtest = 0.6 V and thus RTH = 0.6  The Thévenin equivalent is shown in Fig. 5.33c. A Quick Recap of Procedures We have now looked at three examples in which we determined a Thévenin or Norton equivalent circuit. The first example (Fig. 5.29) contained only independent sources and resistors, and several different methods could have been applied to it. One would involve calculating RTH for the inactive network and then Voc for the live network. We could also have found RTH and Isc , or Voc and Isc.  PRACTICAL APPLICATION The Digital Multimeter One of the most common pieces of electrical test equip- 1 k⍀ ment is the DMM, or digital multimeter (Fig. 5.34), 4.500 VDC