Useful Circuit Analysis Techniques PDF

# Useful Circuit Analysis Techniques ## Chapter 5: Useful Circuit Analysis Techniques ## Introduction The techniques of nodal and mesh analysis described in Chapter 4 are reliable and extremely powerful methods. However, both require that we develop a complete set of equations to describe a parti...

# Useful Circuit Analysis Techniques ## Chapter 5: Useful Circuit Analysis Techniques ## Introduction The techniques of nodal and mesh analysis described in Chapter 4 are reliable and extremely powerful methods. However, both require that we develop a complete set of equations to describe a particular circuit as a general rule, even if only one current, voltage, or power quantity is of interest. In this chapter, we investigate several different techniques for isolating specific parts of a circuit in order to simplify the analysis. After examining the usage of these techniques, we focus on how one might go about selecting one method over another. ## 5.1 Linearity and Superposition **Key Concepts:** - Superposition as a Means of Determining the Individual Contributions of Different Sources to Any Current or Voltage - Source Transformation as a Means of Simplifying Circuits - Thévenin's Theorem - Norton's Theorem - Thévenin and Norton Equivalent Networks - Maximum Power Transfer - ▲→ Y Transformations for Resistive Networks - Selecting a Particular Combination of Analysis Techniques - Performing dc Sweep Simulations Using PSpice All of the circuits which we plan to analyze can be classified as linear circuits, so this is a good time to be more specific in defining exactly what we mean by that. Having done this, we can then consider the most important consequence of linearity, the principle of superposition. This principle is very basic and will appear repeatedly in our study of linear circuit analysis. As a matter of fact, the nonapplicability of superposition to nonlinear circuits is the very reason they are so difficult to analyze! The principle of superposition states that the response *(a desired current or voltage)* in a linear circuit having more than one independent source can be obtained by adding the responses caused by the separate independent sources acting alone. ### Linear Elements and Linear Circuits Let us first define a **linear element** as a passive element that has a linear voltage-current relationship. By a "linear-voltage current relationship," 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 must also define a **linear dependent source** as a dependent current or voltage source whose output current or voltage is proportional only to the first power of a specified current or voltage variable in the circuit *(or to the sum of such quantities)*. For example, the dependent voltage source $v_s = 0.6i_1 - 14i_2^2$ is linear, but $v_s = 0.6i$ and $v_s = 0.6i_1^{102}$ are not. We may now define a **linear circuit** as a circuit composed entirely of independent sources, linear dependent sources, and linear elements. From this definition, it is possible to show that "the response is proportional to the source," or that multiplication of all independent source voltages and currents 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. Let us develop the superposition principle by considering first the circuit of Figure 5.1, which contains two independent sources, the current generators that force the currents $i_a$ and $i_b$ into the circuit. Sources are often called forcing functions for this reason, and the nodal voltages that they produce can be termed response functions, or simply responses. Both the forcing functions and the responses may be functions of time. The two nodal equations for this circuit are $0.7v_1 - 0.2v_2 = i_a$ $-0.2v_1 + 1.2v_2 = i_b$ 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 $v_{1x}$ and $v_{2x}$. Thus, $0.7v_{1x} - 0.2v_{2x} = i_{ax}$ $-0.2v_{1x} + 1.2v_{2x} = i_{bx}$ We next perform experiment $y$ by changing the source currents to $i_{ay}$ and $i_{by}$ and measure the responses $v_{1y}$ and $v_{2y}$: $0.7v_{1y} - 0.2v_{2y} = i_{ay}$ $-0.2v_{1y} + 1.2v_{2y} = i_{by}$ These three sets of equations describe the same circuit with three different sets of source currents. Let us add or “superpose” the last two sets of equations. Adding Equations [3] and [5]: $0.7(v_{1x} + v_{1y}) - 0.2(v_{2x} + v_{2y}) = i_{ax} + i_{ay}$ $0.7v_1 - 0.2v_2 = i_a$. and adding Equations [4] and [6]: $-0.2(v_{1x} + v_{1y}) + 1.2(v_{2x} + v_{2y}) = i_{bx} + i_{by}$ $-0.2v_1 + 1.2v_2 = i_b$. Where Equation [1] has been written immediately below Equation [7] and Equation [2] below Equation [8] for easy comparison. The linearity of all these equations allows us to compare Equation [7] with Equation [1] and Equation [8] with Equation [2] and draw an interesting conclusion. If we select $i_{ax}$ and $i_{ay}$ such that their sum is $i_a$ and select $i_{bx}$ and $i_{by}$ such that their sum is $i_b$, then the desired responses $v_1$ and $v_2$ may be found by adding $v_{1x}$ to $v_{1y}$ and $v_{2x}$ to $v_{2y}$, respectively. In other words, we can perform experiment $x$ and note the responses, perform experiment $y$ and note the responses, and finally add the two sets of responses. This leads to the fundamental concept involved in the superposition principle: to look at each independent source *(and the response it generates)* one at a time with the other independent sources "turned off" or "zeroed out." If we reduce a **voltage source** to zero volts, we have effectively created a **short circuit**. If we reduce a **current source** to zero amps, we have effectively created an **open circuit**. Thus, the superposition theorem can be stated as: In any linear resistive network, the voltage across or the current through any resistor 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 experiments, 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. The circuit we have just used as an example, however, 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.” 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 inactive source almost always leads to the simplest circuit, however. ## Example 5.1 For the circuit of Figure 5.3a, use superposition to write an expression for the unknown branch current $i_x$. We first set the current source equal to zero and redraw the circuit as shown in Figure 5.3b. The portion of $i_x$ due to the voltage source has been designated $i'$ to avoid confusion and is easily found to be 0.2A. We next set the voltage source in Figure 5.3a to zero and again redraw the circuit, as shown in Figure 5.3c. Routine application of current division allows us to determine that $i''$ *(the portion of $i_x$ due to the 2 A current source)* is 0.8A.. We may now compute the complete current $i_x$ as the sum of the two individual components: $i_x = i_x|_{3V} + i_x|_{2_A} = i' + i''$ or $i_x = \frac{3}{6 + 9} + 2\frac{6}{(6+9)} = 0.2 + 0.8 = 1.0A$ 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 current $i_x$ flowing through the 9 resistor. However, the contribution of the 3 V source to $i_x$ 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 $i_x$ flowing through the 9 resistor. However, the 3 V source would still contribute only 0.2 A to $i_x$, for a new total current of 0.2 + 1.6 = 1.8 A. ## Practice 5.1 For the circuit of Figure 5.4, use superposition to compute the current $i_x$. 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 useful in identifying the significance of various parts of a more complex circuit. It also forms the basis of phasor analysis, which is introduced in Chapter 10. ## Example 5.2 Referring to the circuit of Figure 5.5a, determine the maximum positive current to which the source $I_x$ can be set before any resistor exceeds its power rating and overheats. **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 an accident. The 6 V source cannot be changed, so we are looking for an equation involving $I_x$ 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 $\frac{P_{max}}{V_R}=\frac{0.250}{100}=50 mA$ 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 Figure 5.5b and find that the 6 V source contributes a current $\frac{100}{100 + 64}\times 6 = 36.59 mA$ to the 100 resistor and, since the 64 resistor is in series, $i_{64}^6 = 36.59 mA$ as well. Recognizing the current divider in Figure 5.5c, we note that $I_x$ will add to $i_{64}^6$, but $i_{100}^\prime$ is opposite in direction to $i_{100}^6$. $I_x$ can therefore 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 $I_x$: $I_x < (86.59 × 10^{-3}) \times \frac{100 + 64}{64}$. and the 64 resistor requires that $I_x < (25.91 × 10^{-3}) \times \frac{100 + 64}{100}$. **Attempt a solution:** Considering first the 100 resistor, we see that $I_x$ is limited to $I_x < 221.9 mA$. The 64 resistor limits $I_x$ such that $I_x < 42.49 mA$. **Verify the solution. Is it reasonable or expected?** In order to satisfy both constraints, $I_x$ must be less than 42.49 mA. If the value is increased, the 64 resistor will overheat long before the 100 resistor does. One particularly useful way to evaluate our solution is to perform a dc sweep analysis in PSpice as described after the next example. An interesting 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 $I_x$. However, because $I_x$ 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 $I_x$. ## Example 5.3 In the circuit of Figure 5.6a, use the superposition principle to determine the value of $i_x$. We first open-circuit the 3A source (Figure 5.6b). The single mesh equation is: $-10 + 2i' + li' + 2i' = 0$ so that: $i' = 2A$ Next, we short-circuit the 10 V source (Figure 5.6c) and write the single-node equation: $\frac{v''-v''}{2}+\frac{v''-v''}{1}=3$ and relate the dependent-source-controlling quantity to $v''$: $v'' = 2(-i'')$ We find: $i'' = - 0.6A$ and, thus, $i_x = i' + i'' = 2+(-0.6) = 1.4A$ 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. ## Practice 5.2 For the circuit of Figure 5.7, use superposition to obtain the voltage across each current source. Ans: $V_{2A} = 9.180 V$, $V_{2/2A} = -1.148 V$, $V_{3V} = 1.967 V$, $V_{2/3V} = -0.246 V$; $V_1 = 11.147 V$, $V_2 = -1. 394 V$. ## Summary of Basic Superposition Procedure 1. Select one of the independent sources. Set all other independent sources to zero. This means voltage sources are replaced with short circuits and current sources are replaced with open circuits. Leave dependent sources alone. 2. Relabel voltages and currents using suitable notation *(e.g. v’, i”)*. 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 simplifies 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 Figure 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 I₁)*, 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 Figure 5.9. Under Analysis Type, we pull down the DC Sweep option, specify the "sweep variable" as Current Source, and then type in I₁ 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 I₁. 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, I₁) 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 absorbed by the 64 resistor, resulting in a plot similar to that shown in Figure 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 $I_x = 43 mA$. In contrast, however, we see that regardless of the value of the current source $I_x$ *(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 both cursors to 42.53 mA; the 64 resistor has just barely exceeded its rating at this current level. 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 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. ## 5.2 Source Transformations Unfortunately, it usually turns out that little if any time is saved in analyzing 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 1V batteries in series with a 1 Ω resistor. The power delivered to the resistor is obviously 4 W, but if we mistakenly try to apply superposition we might say that each battery alone furnished 1 W and thus the total power is 2 W. This is incorrect, but a surprisingly easy mistake to make. ## Practical Voltage Sources Up to now we have been working exclusively with ideal voltage and current sources; it is now time to take a step closer to reality by considering practical sources. These sources will enable us to make more realistic representations of physical devices. Once we have defined practical sources, we will see that practical current and voltage sources may be interchanged without affecting the remainder of the circuit. Such sources will be called equivalent sources. Our methods will be applicable to both independent and dependent sources, although we will find that they are not as useful with dependent sources. The ideal voltage source was defined as a device whose terminal voltage is independent of the current through it. A 1 V dc source produces a current of 1 A through a 1 Ω resistor, and a current of 1,000,000 A through a 1 μΩ resistor; it can provide an unlimited amount of power. No such device exists practically, of course, and we agreed previously that a real physical voltage source could be represented by an ideal voltage source only as long as relatively small currents, or powers, were drawn from it. For example, a car battery may be approximated by an ideal 12 V dc voltage source if its current is limited to a few amperes. However, anyone who has ever tried to start an automobile with the headlights on must have observed that the lights dimmed perceptibly when the battery was asked to deliver the heavy starter current, 100 A or more, in addition to the headlight current. Under these conditions, an ideal voltage source is not really an adequate representation of the battery. To better approximate the behavior of a real device, the ideal voltage source must be modified to account for the lowering of its terminal voltage when large currents are drawn from it. Let us suppose that we observe experimentally 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 Ω, and the ideal voltage source and this series resistor constitute a practical voltage source. 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 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 characteristic, at least over some useful range of current, voltage, or power. In Figure 5.12a, we show our two-piece practical model of the car battery now connected to some load resistor R₁. The terminal voltage of the practical source is the same as the voltage across R₁ and is marked V₁. Figure 5.12b shows a plot of load voltage V₁ as a function of the load current I for this practical source. The KVL equation for the circuit of Figure 5.12a may be written in terms of I₁ and V₁: $12 = 0.01 I_L + V_L$ and thus $V_L = -0.01 I_L + 12$ This is a linear equation in I₁ and V₁, and the plot in Figure 5.12b is a straight line. Each point on the line corresponds to a different value of R₁. For example, the midpoint of the straight line is obtained when the load resistance is equal to the internal resistance of the practical source, or R₁ = 0.01 Ω. Here, the load voltage is exactly one-half the ideal source voltage. When R₁ = ∞ 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 V_ {Loc} = 12 V. If, on the other hand, R₁ = 0, thereby short-circuiting the load terminals, then a load current or short-circuit current, I_ {Lsc} = 1,200 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!)* Since the plot of V₁ versus I₁ is a straight line for this practical voltage source, we should note that the values of V_ {Loc} and I_ {Lsc} uniquely determine the entire V₁-I₁ curve. The horizontal broken line of Figure 5.12b represents the V_L-I_L plot for an ideal voltage source; the terminal voltage remains constant for any value of 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. Let us now consider a general practical voltage source, as shown in Figure 5.13a. The voltage of the ideal source is u_s, and a resistance R_s, 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 but merely serves to account for a terminal voltage that decreases as the load current increases. Its presence enables us to model the behavior of a physical voltage source more closely. The linear relationship between u_L and i_L is $u_L = u_s - R_s i_L$ and this is plotted in Figure 5.13b. The open-circuit voltage (R₁ = ∞, so i_L = 0) is $V_{Loc} = u_s$ and the short-circuit current (R₁ = 0, so u_L = 0) is $I_{Lsc} = \frac{u_s}{R_s}$. Once again, these values are the intercepts for the straight line in Figure 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 resistance to which it is connected or the voltage across its terminals . Certain transistor circuits will deliver a constant current to a wide range of load resistances, but the load resistance can always be made sufficiently large that the current through it becomes very small . Infinite power is simply never available (unfortunately). A practical current source is defined as an ideal current source in parallel with an internal resistance R_p. Such a source is shown in Figure 5.14a, and the current i_L and voltage u_L associated with a load resistance R₁ are indicated. Application of KCL yields $i_L = i_s - \frac{u_L}{R_p}$ which is again a linear relationship. The open-circuit voltage and the short-circuit current are $U_{Loc} = R_p i_s$ and $I_{Lsc} = i_s$ The variation of load current with changing load voltage may be investigated by changing the value of R₁ as shown in Figure 5.14b. The straight line is traversed from the short-circuit, or “northwest”, end to the open-circuit termination at the “southeast” end by increasing R₁ from zero to infinite ohms. The midpoint occurs for R₁ = R_p. The load current i_L and the ideal source current i_s are approximately equal only for small values of load voltage, which are obtained with values of R₁ that are small compared to R_p. ## Equivalent Practical Sources Having defined both practical sources, we are now ready to discuss their equivalence. We will define two sources as being equivalent if they produce identical values of u_L and i_L when they are connected to identical values of R₁, no matter what the value of R₁ may be. Since R₁ = ∞ and R₁ = 0 are two such values, equivalent sources provide the same open-circuit voltage and short-circuit current. In other words, if we are given two equivalent sources, one a practical voltage source and the other a practical current source, each enclosed in a black box with only a single pair of terminals, then there is no way in which we can tell which source is in which box by measuring current or voltage in a resistive load. Consider the practical voltage source and resistor R₁ shown in Figure 5.15a, and the circuit composed of a practical current source and resistor R₁ shown in Figure 5.15b. A simple calculation shows that the voltage across the load R_L of Figure 5.15a is $u_L = \frac{u_s}{R_s + R_L}R_L$ A similarly simple calculation shows that the voltage across the load R_L in Figure 5.15b is $u_L = \frac{R_p}{R_p + R_L}i_sR_L$. The two practical sources are electrically equivalent, then, if: $R_s = R_p$ and $u_s = R_p i_s = R_s i_s$ where we now let R_s represent the internal resistance of either practical source, which is the conventional notation. As an illustration of the use of these ideas, consider the practical current source shown in Figure 5.16a. Since its internal resistance is 2 Ω, the internal resistance of the equivalent practical voltage source is also 2 Ω; the voltage of the ideal voltage source contained within the practical voltage source is (2)(3) = 6 V. The equivalent practical voltage source is shown in Figure 5.16b. To check the equivalence, let us visualize a 4 Ω 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 Ω load. However, we should note very carefully that the ideal current source is delivering a total power of 12 W, while the ideal voltage source is delivering only 6 W. Furthermore, the internal resistance of the practical current source is absorbing 8 W, whereas the internal resistance of the practical voltage source is absorbing only 2 W. Thus we see that the two practical sources are equivalent only with respect to what transpires at the load terminals; they are not equivalent internally! ## Example 5.4 Compute the current through the 4.7 ΚΩ resistor in Figure 5.17a after transforming the 9 mA source into an equivalent voltage source. The equivalent source consists of an independent voltage source of (9 mA) × (5 ΚΩ) = 45 V in series with a 5 kΩ resistor, as depicted in Figure 5.17b. A simple KVL equation around the loop yields: -45 + 5000I + 4700I + 3000I + 3 = 0 which can be easily solved to find that the current I = 3. 307 mA. ## Practice 5.3 For the circuit of Figure 5.18, compute the current I_x through the 47 ΚΩ resistor after performing a source transformation on the voltage source. Ans: 192 μA. ## Example 5.5 Calculate the current through the 2 Ω resistor in Figure 5.19a on the next page by making use of source transformations to first simplify the circuit. We begin by transforming each current source into a voltage source (Figure 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 Ω resistors may be combined into a single 7 Ω resistor, which can then be used to transform the 15 V source into a 15/7 A source as in Figure 5.19c. As a final simplification, we note that the two 7 Ω resistors can be combined into a single 3.5 Ω 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 Figure 5.19d. The current I can now be found using KVL: -7.5 + 3.5I - 5I_V_x + 28I + 9 = 0 $V_x = 2I$ where: Thus, $I = 21.28 mA$ ## Practice 5.4 For the circuit of Figure 5.20, compute the voltage V across the 1 ΜΩ resistor using repeated source transformations. Ans: 27.23 V. ## Several Key Points We conclude our discussion of practical sources and source transformations with a few specialized observations. First, when we transform a voltage source, we must be sure that the source is in fact in series with the resistor under consideration. For example, in the circuit shown in Figure 5.21, it is perfectly valid to perform a source transformation on the voltage source using the 10 Ω resistor, as they are in series. However, it would be incorrect to attempt a source transformation using the 60 V source and the 30 Ω 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 Figure 5.22a. We may perform a source transformation including the 3 Ω resistor, as they are in parallel, but after the transformation 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 transformed as in Figure 5.22b. Then, the transformation to a voltage source in series with a resistor may be drawn correctly as shown in Figure 5.22c; the resistor