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or more commonly, solved for the voltage and expressed as: V = I×R (3.3) This is called Ohm's law, and is named in honor of Georg Ohm, a researcher from the early nineteenth century. It is, along with Kirchhoff's laws that we shall see shortly, one of the most important and useful equations available for the analysis of DC electrical circuits. For the sake of completeness, Ohm's law can also be visualized in terms of resistance: R = V I All three forms of this equation are useful. For example, we can use the first version if we have a known voltage applied across a resistor and we want to determine the resulting current. Similarly, Equation 3.3 can be used to find the voltage drop across a resistor if we know the current through it. Finally, we can use the last version if we need to set the current to a certain value and have to find the resistance required to do so. Example 3.3 A 9 volt battery is used in series to power a 40 Ω resistor as shown in Figure 3.11. Determine the circulating current. Also determine the minimum power rating for the resistor. I = V R I = 9 V 40 Ω I =0.225amps From power law, Equation 2.7, P = I×V P = 0.225A×9 V P =2.025watts Power could also be determined using V2/R or I2R. Try it. You should get the same result. 79 Figure 3.11 Circuit for Example 3.3. | DCElectricalCircuitAnalysis_Page_79_Chunk1001 |
Computer Simulation Computer Simulation To verify the results of the preceding example, the circuit is entered into a simulator, as shown in Figure 3.12. Multisim™ is used here, although any quality circuit simulator will do. Virtual DMMs are used to measure the current and voltage. Remember, current is the rate of charge flow and is measured at a single point; the middle of a connecting wire if you will. Consequently, the ammeter is inserted between the battery and resistor. In contrast, voltage is a potential difference which implies two points for the measurement, and thus, the voltmeter is placed across the resistor. Note the voltage is exactly 9 volts as expected. The current is just shy of the expected 225 milliamps. This is due to the fact that all real world ammeters exhibit some finite internal resistance. This extra resistance, though very much smaller than the resistor, adds to the total resistance and thus slightly decreases the current. The simulator has been programmed to mimic this behavior and thus we see a very slight decrease in the current in the simulation, which is precisely the effect we would see in a proper physical lab. This simulation uses virtual instruments because they echo the layout of a physical lab and offer some familiarity. Unfortunately, they become cumbersome in larger circuits. We shall look at other, more direct, simulation techniques later in this chapter. 80 Figure 3.12 The circuit of Example 3.3 in a simulator. | DCElectricalCircuitAnalysis_Page_80_Chunk1002 |
Example 3.4 Refer to the battery and lamp circuit shown back in Figure 3.3. Determine the resistance of the lamp if the current flowing through it is 300 mA and the battery voltage is 6 volts. Also determine the power dissipated by the lamp. R = V I R = 6 V 0.3A R =20Ω From power law, P = I 2×R P = 0.3A 2×20Ω P =1.8watts Let's continue with a couple of examples utilizing current sources. Example 3.5 Determine the voltage developed across the resistor in the circuit of Figure 3.13. Also determine the power dissipated. V = I×R V = 10 mA×2.2k Ω V =22 volts From power law, P = I 2×R P = (10 mA)2×2.2 kΩ P =0.22 watts Computational hint: don't forget the “milli” part of the current (milli squared is micro). 81 Figure 3.13 Circuit for Example 3.5. | DCElectricalCircuitAnalysis_Page_81_Chunk1003 |
Example 3.6 Determine the resistor value required in the circuit of Figure 3.14 in order for the source to generate 24 watts of power. First, note that power generated must equal power dissipated. Thus, the power generated by the source must equal the power dissipated by the resistor. From power law, P = I 2×R R = P I 2 R = 24 W (2 A)2 R = 6Ω As a crosscheck, note that Ohm's law indicates that a 6 Ω resistor would produce a 12 volt drop given a 2 amp current, and that 12 volts times 2 amps yields the expected 24 watts. It is now time to move onto series circuits using multiple resistors and/or current sources. 3.6 Kirchhoff's Voltage Law (KVL) 3.6 Kirchhoff's Voltage Law (KVL) Along with Ohm's law, the key law governing series circuits is Kirchhoff's voltage law, or KVL. Named after nineteenth century German physicist Gustav Kirchhoff, this law states that the sum of voltage rises and voltage drops around a series loop must equal zero (the rises and drops having opposite polarities). Alternately, it may be reworded as the sum of voltage rises around a series loop must equal the sum of voltage drops. As a pseudo formula: ∑ V↑ = ∑ V↓ (3.4) The Voltage Divider Rule (VDR) The Voltage Divider Rule (VDR) An outgrowth of KVL is the voltage divider rule (VDR). In a series connection, the current is the same through each component. Thus, the voltage drops in a series connection must be directly proportional to the size of the resistances: the larger the resistor, the larger its voltage, and the larger its share of the total voltage applied to the series connection. Thus, the voltage across any resistor must equal the net 82 Figure 3.14 Circuit for Example 3.6. | DCElectricalCircuitAnalysis_Page_82_Chunk1004 |
supplied voltage times the ratio of the resistor of interest to the total resistance: VRx = E ∙ RX /RTOTAL (3.5) In fact, Equation 3.5 is just a combination of two Ohm's law calculations into a single formula. The circulating current is equal to E/RTOTAL. This current is then multiplied by the resistor of interest (RX) to arrive at the voltage across that resistor (VRx). It is worth pointing out that “the resistor of interest” can, in fact, be the sum of multiple resistors in series. While VDR is not required for any particular analysis, it serves two purposes: first, it saves some time because it skips over the computation of current, and second, it reinforces the ideal of a proportional division of voltage in a series connection. For example, if there are two resistors in series and one of them is twice the size of the other, then it must be the case that the larger resistor sees twice the voltage of the smaller resistor. 3.7 Series Analysis 3.7 Series Analysis Ohm's law, Kirchhoff's voltage law, the voltage divider rule and series component combinations are the tools we will use to solve general series circuit problems. There are multiple techniques for analyzing these circuits: • If all voltage source and resistor values are given, the circulating current can be found by dividing the equivalent voltage by the sum of the resistances. Once the current is found, Ohm's law can be used to find the voltage drops across individual resistors. Alternately, the individual voltage drops can be found using the voltage divider rule. At that point, power law can be used to find the power dissipation in each resistor or the power developed by the source(s). • If the circuit uses a current source instead of a voltage source, then the circulating current is known and the voltage drop across any resistor may be determined directly using Ohm's law. • If the problem concerns determining resistance values, the basic idea will be to use these rules in reverse. For example, if a resistor value is needed to set a specific current, the total required resistance can be determined from this current and the given voltage supply. The values of the other series resistors can then be subtracted from the total, yielding the required resistor value. Similarly, if the voltages across two resistors are known, as long as one of the resistance values is known, the other resistance can be determined using either the voltage divider rule or Ohm's law. 83 | DCElectricalCircuitAnalysis_Page_83_Chunk1005 |
Often, there will be more than one way to solve a given problem. We shall refer to these as solution paths. No particular path is “more correct” than any other, although some paths may be shorter or easier in a given case, as will be illustrated next. Example 3.7 Determine the voltage Vb (i.e., the voltage across the 100 Ω resistor) in the circuit of Figure 3.15. Also determine the circulating current. First, note that there is a single voltage source that will create a clockwise current flowing through the two resistors. KVL says that the combined drops across those resistors must equal the rise of the source, or 12 volts. One solution path involves first finding the total resistance. Dividing this into the source voltage will yield the current. Ohm's law can then be used to determine the resistor voltage. RT = R1+R2 RT = 200Ω+100Ω RT = 300Ω I = E RT I = 12 V 300Ω I = 40 mA V b = I×R2 V b = 40 mA×100Ω V b = 4 volts A second solution path involves first finding the resistor voltage via the voltage divider rule. The current can then be found via Ohm's law. V b = E R2 RT V b = 12V 100Ω 300Ω V b = 4 volts I = V b R2 I = 4V 100Ω I = 40 mA 84 Figure 3.15 Circuit for Example 3.7. | DCElectricalCircuitAnalysis_Page_84_Chunk1006 |
Computer Simulation Computer Simulation Let's verify the preceding example using a computer simulation. Instead of using virtual instruments, we shall use a more direct route, namely a listing of node voltages. A node voltage is the voltage at a given point with respect to some other point, normally ground, as in Vb in the prior example. Note that as we treat connecting wires ideally (having no resistance), a node is the entire connection wire, not a specific location on it. Normally, simulators will automatically number each node in a circuit with ground being node 0. This method skips the entire business of inserting virtual meters in the circuit and reduces schematic clutter. Note that ground is node 0, the power supply connection to the first resistor is node 1 and the connection between the two resistors is node 2. Thus, node 2 is Vb. A DC operating point analysis is performed. The results are shown in a separate output window, as depicted in Figure 3.17. The values are as expected and with no loading effects as might be seen with virtual instruments. 85 Figure 3.16 The circuit of Example 3.7 in the simulator. Figure 3.17 Simulation results for the circuit of Figure 3.16. | DCElectricalCircuitAnalysis_Page_85_Chunk1007 |
Example 3.8 Determine the voltage Va (i.e., the voltage across the current source) in the circuit of Figure 3.18. Given the direction of the current source, the current will flow counterclockwise. This produces voltage drops (plus to minus) left to right across the 10 Ω, bottom to top across the 3 Ω, and right to left across the 1 Ω resistor. This means that node a is negative with respect to ground (i.e., the voltage keeps dropping across the resistors as we travel counterclockwise from ground, indicating that the resulting potential keeps getting more and more negative with respect to ground). To find Va, we may simply sum the resistors and then use Ohm's law to determine their combined voltage drop, which of course, must equal the voltage rise from the current source. RT = R1+R2+R3 RT = 1Ω+3Ω+10Ω RT = 14Ω V a =−I×RT V a =−5 A×14 Ω V a =−70 volts One practical point when designing circuits is to appreciate the variations in current and voltage produced by component tolerances. For example, in the preceding problem, we determined that there will be 70 volts across the current source, but this assumes that the resistor values are exactly as stated. In the previous chapter we discovered that all resistor values will vary somewhat, and we grade this variation as a percent tolerance. Thus, if the true resistor values are off a bit, we would expect this to impact the associated voltages. Often, we would like to determine worst case values such as a maximum voltage or a minimum current. Sometimes these conditions occur when all resistors are at their maximum or minimum value. For example, the maximum value of source voltage in the previous problem occurs when all of the resistor values are at their maximum. This is not always the case, though. Sometimes the minima and maxima are achieved with odd combinations, such as one resistor being at maximum and another being at minimum. For instance, in the circuit used for Example 3.7, Vb will be at minimum when R2 is at minimum and when R1 is at maximum. This can be verified easily using VDR. Computer Simulation Computer Simulation Along with worst case, we might also like to know the sort of typical spread we get on a particular parameter using normal component variations. This can be performed 86 Figure 3.18 Circuit for Example 3.8. | DCElectricalCircuitAnalysis_Page_86_Chunk1008 |
in a simulator and it is usually called Monte Carlo Analysis or something similar. Basically, the simulator computes a series of simulation runs, each with component values chosen randomly within the specified tolerances of the chosen components, mimicking the effect of pulling the components out of a parts bins. To illustrate the Monte Carlo technique, the circuit of Example 3.8 is entered in a simulator as shown in Figure 3.19. The resistors are given 10% tolerances. Five runs are specified for node voltage 1 which is equivalent to Va in the original circuit. The results are shown in Figure 3.20. Note that the nominal value run indicates −70 volts as expected. The other runs show variations on this voltage, some above and some below the nominal result. 87 Figure 3.19 The circuit of Example 3.8 in the simulator. Figure 3.20 Simulation results for the circuit of Figure 3.19. | DCElectricalCircuitAnalysis_Page_87_Chunk1009 |
The Importance of Polarity The Importance of Polarity A key item of importance when analyzing series circuits, or indeed any electrical circuit, is noting the polarity of the voltages. This was apparent in Example 3.8. Determining polarity for voltage sources is easy as their polarity is fixed (positive is always at the “long bar” and negative at the “short bar”). The polarity of any resistor will depend on the direction of current flow. To avoid confusion we will use the following standard: • Where conventional current enters a resistor, we label this point with a plus sign, and where the current leaves, a minus sign. • During analysis, moving from plus to minus is a voltage drop. That is, we are moving from a more positive or higher potential to a less positive or lower potential, thus “dropping” voltage. • Similarly, traversing from minus to plus is a voltage rise. Remember, Kirchhoff's voltage law states that the sum of voltage rises around a series loop must equal the sum of voltage drops. Along with determining the voltage across single resistors, we are often interested in determining the voltage between arbitrary points in a circuit. These may span several components. It is imperative that we know the polarities of the individual voltages in order to determine the voltage between any two circuit points. The foregoing is illustrated using the series circuit schematic of Figure 3.21. Here, the two voltage sources, E1 and E2, aid each other because they are both trying to establish a clockwise current. This produces a net voltage of E1+E2. Their polarities are fixed. The current direction will be as drawn because conventional current flows out of the positive terminal of a voltage source. The value of this current will be (E1+E2) / (R1+R2) via Ohm's law. Knowing the direction of current, the polarities of the voltage drops across the two resistors may be found. The point where the current enters the resistor is positive, and negative where it exits. Once these are labeled, either Ohm's law or the voltage divider rule can be used to determine the voltages developed across the two resistors. Note that KVL states that the sum of these two resistor voltages must equal the net supplied voltage, or E1+E2 in this case. To determine the voltage from point a to point c, or Vac, we start at point a and sum the voltage drops and rises until we get to point c. Here, the voltage across R1 shows up as a drop (+ to − and taken as positive) and the voltage across E2 shows up as a rise (− to +, and thus negative). The result is the voltage across R1 minus the voltage across E2. Depending on the specific values, the sum may wind up being either a positive or negative value. If it's positive, this signifies that point a is at a higher potential than is point c. Conversely, if it's negative, this signifies that point a is at a lower potential than is point c. If we reverse the order, starting at point c and moving 88 Figure 3.21 A multi-source, multi-resistor series circuit. | DCElectricalCircuitAnalysis_Page_88_Chunk1010 |
to point a, or Vca, we will wind up with the same voltage magnitude but the sign will flip. In the laboratory, the first point letter (the a in Vac) is where the red lead of the voltmeter is connected and the second letter is where the black lead is connected. As a reminder, if a single connection point is used, as in Va, the second letter is assumed to be ground. Thus, Va is the voltage from point a to ground. To illustrate the importance of voltage source polarity, consider the series circuit shown in Figure 3.22. This circuit is identical to the previous circuit with the exception that the polarity of source E2 has been flipped. This can have a drastic change on the resulting current and voltage drops. For example, if E2 is larger than E1, the net supplied voltage will be E2−E1 and the direction of conventional current will be as drawn. Effectively, E2 will be charging E1. Note that this is opposite to the prior circuit. Because both the direction and the magnitude of the current have changed, the voltage drops and polarities of R1 and R2 will change. Consequently, any point-to-point voltage, such as Vac, will also change. Example 3.9 Determine the voltages Vb and Vca in the circuit of Figure 3.23. The first step here would be to note that the two sources are in opposition. The result is an effective source of 10.5 volts with the same polarity as the 12 volt source, meaning it creates a clockwise current (rather the opposite of Figure 3.22). The total resistance will be the sum of the resistors, or 500 Ω. From this we can compute the current and then the voltage drops across each resistor. The polarity across the 100 Ω will be + to − left to right, and across the 400 Ω it will be + to − right to left. The polarities for both sources are top to bottom + to −. This is shown in Figure 3.24. I = ETotal RTotal I = 10.5 V 500Ω I = 21 mA V 100 = I×R100 V 100 = 21 mA×100Ω V 100 = 2.1 volts V 400 = I×R400 V 400 = 21 mA×400Ω V 400 = 8.4 volts To find Vb we start at node b and work our way to ground, summing the potentials as we go. First we see the 1.5 volt source. Even though this is a 89 Figure 3.22 Series circuit with altered polarity. Figure 3.23 Circuit for Example 3.9. Figure 3.24 Circuit for Example 3.9 with current direction and voltage polarities shown. | DCElectricalCircuitAnalysis_Page_89_Chunk1011 |
source, it is still considered a voltage drop because the polarity is + to −. The polarity on the 400 Ω resistor is the same, so we add in its 8.4 volts for 9.9 volts total. That is, node b is 9.9 volts above ground. For Vca we proceed similarly. Starting at node c we move through the 1.5 volt source, but this time the polarity is − to +, or a rise of 1.5 volts (i.e., taken as negative). Now at node b, we continue through the 100 Ω resistor, also seeing a − to + polarity, or a rise of 2.1 volts. That adds up to 3.6 volts in magnitude, however, the starting node is lower than the ending node, thus Vca = −3.6 volts. We could also say Vac = 3.6 volts. Back in Chapter 2 the height analogy for voltage was presented. This can be applied nicely in the prior example. You can think of ground as literally ground: the earth under your feet. Voltage rises would be like climbing up a ladder and voltage drops would be like climbing down a ladder. Think of each volt as being analogous to one foot or one meter. Thus, for Vb, you climb up the ladder 8.4 feet for the 400 Ω and another 1.5 feet for the 1.5 volt source, leaving you 9.9 feet (volts) above the ground. Therefore, Vb must be positive with respect to ground by 9.9 volts. This analogy works for voltages below ground as well: just think in terms of digging a hole. For voltages that are not referenced to ground, think of climbing up and down ladders, and then comparing the height of where you stopped to the height of where you started. Precisely labeling every voltage polarity may seem like unnecessary work, but as circuit complexity increases so does the opportunity for polarity errors. This important step should not be omitted. Consider the following example which is only slightly more complex than the one preceding. Example 3.10 Determine the voltage Vbe in the circuit of Figure 3.25. The 24 volt and 6 volt sources are aiding each other while the 4 volt source is in opposition. The result is an effective supplied potential of 26 volts with a current circulating counterclockwise. The current direction and voltage polarities are shown in Figure 3.26. First, find the circulating current and then find the voltage drops across the resistors of interest. The total resistance is the sum of the three, or 13 kΩ. 90 Figure 3.25 Circuit for Example 3.10. | DCElectricalCircuitAnalysis_Page_90_Chunk1012 |
I = ETotal RTotal I = 26 V 13 k Ω I = 2 mA V 10k = I×R1 V 10k = 2 mA×10 k Ω V 10k = 20 volts Similarly, V2k = 4 volts and V1k = 2 volts with the polarities shown. To determine Vbe, we can start at node b and move to node e by either going to the left or right side. We shall do both and compare the results. Going to the left, we see + to − 20 volts across the 10 kΩ, − to + 24 across the left side source, and then + to − 4 volts across the 2 kΩ. This adds up to 0 volts, meaning that nodes b and e are at the same potential! Double checking by going down the right side we see + to − 6 volts for the top source, − to + 4 volts for the right side source, and finally − to + 2 volts for the 1 kΩ. This is also 0 volts, as expected. It is important to understand that saying Vbe = 0 volts does not imply that Vb or Ve are 0 volts. In fact, we can see that Ve is below ground by the voltage dropped across the 2 kΩ, or −4 volts. Similarly, Vb is above node a by 20 volts, and node a itself is 24 volts below ground given the polarity of the left side source. The result, once again, is −4 volts. One more example to better illustrate the voltage divider rule in a larger circuit. Example 3.11 Determine the voltages, Vb, Vd, and Vbd in the circuit of Figure 3.27. While we could find the circulating current, then use Ohm's law to find the individual voltage drops, and finally add these pieces to find the potentials we're looking for, we shall instead focus on using VDR which may be faster. The voltage divider rule states that the voltage across any resistor or group of resistors in a series loop is proportional to their resistance compared to the total resistance in that loop. The total resistance in this circuit is 12 kΩ. To find the voltage across any resistor or group of resistors, we simply make a ratio of the resistance of interest to the total resistance and multiply by the applied source voltage. Numbering the resistors R1 at top to R4 at bottom: 91 Figure 3.26 Circuit for Example 3.10 with current direction and voltage polarities shown. Figure 3.27 Circuit for Example 3.11. | DCElectricalCircuitAnalysis_Page_91_Chunk1013 |
V d = E R4 RT V d = 36 V 4.5 kΩ 12 kΩ V d = 13.5 volts V b = E R2+R3+R4 RT V b = 36V 8k Ω 12 k Ω V b = 24 volts V bd = E R2+R3 RT V bd = 36 V 3.5k Ω 12 kΩ V bd = 10.5 volts There is another way to find Vbd here, and that's to notice that by definition Vbd = Vb − Vd. Therefore, Vbd = 24 volts − 13.5 volts, or 10.5 volts. 3.8 Potentiometers and Rheostats 3.8 Potentiometers and Rheostats Some applications require the use of variable resistances. One example is a basic volume control, but other examples include equipment calibration and user adjustable parameters such as the bass, treble and balance controls on a home audio system. Adjustable resistances fall under two categories, potentiometers and rheostats. Generally, these would be the same physical device, with the application and usage determining its name. A potentiometer is a resistor with three leads. Along with the typical end leads, there is a third lead referred to as the wiper or tap. This third lead is connected to a small metallic arm that makes contact with the resistive material and which can be moved along the length of the material. Potentiometers, or pots for short, often are formed in a circular shape (rotary travel) but they can also be in the form of a straight line (linear travel). Several different types are shown in Figure 3.28. The larger circular units with shafts and the straight unit at the top are designed to be adjusted by the end user (appropriately stylish knobs being attached to said shafts, of course), while the smaller units are intended to be adjusted only by a technician (note the small adjustment screw slots intended to receive fine screwdrivers). The smaller units are designed to be mounted directly on a printed circuit board while the user adjustable units are mounted on a front panel, and have wiring lugs instead of straight leads. 92 | DCElectricalCircuitAnalysis_Page_92_Chunk1014 |
A cutaway view of a rotary panel mount pot with its back casing removed is shown in Figure 3.29. Most circular potentiometers utilize a 270° rotation from end to end, also known as “3/4 turn”. Some precision pots used for calibration may use an internal worm gear to move the wiper and achieve 20 or more turns to shuttle the wiper from one end to the other. This results in the ability to make very fine adjustments. The rectangular units in the right-center of Figure 3.28 are examples of this type. The schematic symbol for a basic potentiometer is shown in Figure 3.30. For convenience, its terminals are labeled A and B for the two ends, and W for the wiper arm. A pot can be visualized as two resistors in sequence with a tap between them. The total resistance never changes. What does change is the ratio of the two pieces. RAB = RAW + RWB (3.6) At the extreme left of travel, RAW will be 0 and RWB will be equal to RAB. At the extreme right, RAW will equal RAB and RWB will be 0. If the wiper is precisely in the middle, then RAW = RWB = RAB/2, and so on11. 11 This 50/50 division is true if the pot has a linear taper, meaning that the resistance change is proportional to the wiper position. While this is typical, it is possible to have a taper that is not linear for special applications, such as audio taper. In these devices the resistance ratio at mid-position might be something like 10/90 instead of 50/50. This would be desired for a volume control because the human ear responds to changes in sound pressure in a logarithmic fashion. Using a linear taper pot would result in a volume control that is too sensitive at one end and yet produces hardly no change at the other end. 93 Figure 3.28 A variety of potentiometers. Figure 3.29 Cutaway view of a rotary potentiometer. | DCElectricalCircuitAnalysis_Page_93_Chunk1015 |
As a result, potentiometers are very useful as controllable voltage dividers (hence the name). Sometimes it is desirable to control two signals simultaneously from a single knob (e.g., one volume knob for a hi-fi stereo controls both the left and right channels). For this application, multiple resistive elements can be stacked and use a common shaft to control all of the wiper arms in tandem. These are known as ganged potentiometers. The tall, round unit in the left-center of Figure 3.28 is a dual gang pot. Note that it has a total of six solder lugs, three for the top element and three for the back element. Finally, unlike fixed resistors, pots are available in limited sizes. Typical values use multiples of 1 or 5 for the total resistance (for example, 1 kΩ, 5 kΩ, 10 kΩ, etc.) although a limited selection of other sizes are also available. If only the wiper and only one of the two end terminals is used, then the device is referred to as a rheostat. Rheostats are often used to limit current. The value of the rheostat ranges from 0 up to the nominal value of the potentiometer. The schematic symbols for a rheostat are shown in Figure 3.31. The ANSI symbol is shown at the left and the IEC version at the right. Time for a couple of examples. Example 3.12 For the circuit depicted in Figure 3.32, determine the maximum and minimum voltages for Vc, as well as its value when the pot is adjusted to midway (assume linear taper). The voltage divider rule states that the voltage across any resistor or group of resistors in a series loop is proportional to their resistance compared to the total resistance in that loop. The total resistance in this circuit is 8 kΩ. The voltage at node c will depend on the setting of the pot. The maximum voltage occurs when the wiper is all the way up, contacting node b. V c(max) = E Ri RT V c(max) = 16 V 6 kΩ 8 kΩ V c(max) = 12 volts For the minimum case, the wiper is set all the way down, contacting node d. 94 Figure 3.30 Schematic symbol for a potentiometer, with labels. Figure 3.31 Schematic symbols for a rheostat: ANSI (right) and IEC (left). Figure 3.32 Circuit for Example 3.12. | DCElectricalCircuitAnalysis_Page_94_Chunk1016 |
V c(min) = E Ri RT V c(min) = 16V 1 kΩ 8 kΩ V c(min) = 2 volts For the midway case, the pot can be treated as two 2.5 kΩ resistors. V c(mid )= E Ri RT V c(mid )= 16 V 3.5k Ω 8 kΩ V c(mid )= 7 volts Note that the mid-position achieves a voltage midway between the extremes. Thus the pot allows a range of voltage adjustment between a minimum of 2 volts and a maximum of 12 volts. The upper and lower fixed resistors serve to set these limits. The larger the values of these resistors compared to the value of the pot, the more narrow the range of adjustment. If these two resistors were excluded, the pot would allow the entire range of adjustment, from ground (0 volts) up to the power supply value of 16 volts. Example 3.13 The potentiometer in the circuit of Figure 3.33 is connected as a rheostat. Determine the maximum and minimum current in the circuit. The rheostat will present a resistance value between 0 Ω and 10 kΩ. The maximum current case occurs when the wiper is set to the extreme left, producing 0 Ω. I (max) = E RT I (max) = 50 V 1 kΩ I (max) = 50 mA For the minimum case, the wiper is set all the way to the right, producing the full 10 kΩ. 95 Figure 3.33 Circuit for Example 3.13. | DCElectricalCircuitAnalysis_Page_95_Chunk1017 |
I (min)= E RT I (min) = 50 V 11 kΩ I (min) = 4.545 mA 3.9 Summary 3.9 Summary Electrons are the charge carrier in metals. Their direction of travel, called electron flow, runs from the negative terminal of a voltage source to the positive terminal. The more commonly used conventional current flow runs from the positive terminal to the negative terminal. As conventional current enters a resistor, we mark this point as positive, and where it exits, negative. Thus, any voltage with a + to − polarity is deemed a voltage drop (which dissipates energy in a resistor). A polarity of − to + is deemed a voltage rise. A series connection is any connection in which the current through one component must be identical to the current flowing through any other component in that connection. In other words, the current exiting any component must be the only current flowing into the next component in line; there can no intervening connections for current flow to or from other parts of the circuit. The equivalent resistance of a string of resistors placed in series is simply the sum of their resistance values. Consequently, the larger resistors dominate the equivalent value. Ohm's law states that the voltage across a resistor must equal the product of its resistance and the current flowing through it. Kirchhoff's voltage law (KVL) states that the sum of voltage rises around any series loop must equal the sum of voltage drops around that loop. Alternately, it may be stated as the sum of voltage rises and drops around a series loop must be zero, as the rises and drops have opposite polarity. The voltage divider rule (VDR) is a convenient computational shortcut based on Ohm's law and KVL. It states that the voltage across any given resistor or group of resistors in a series connection is proportional to their resistance as a percentage of the total resistance. For example, if a resistor makes up one quarter of the total resistance in a series string of resistors, then that resistor will get one quarter of the total voltage applied to that string. A potentiometer, or pot, is a three terminal variable resistance. The resistance of the entire device is fixed, however, a third lead is attached internally to an adjustable wiper, effectively creating two resistors. The sum of the two parts always equals the nominal value of the pot with the values of each piece being determined by the position of the wiper. Pots may be either rotary or linear (straight line) in motion and 96 | DCElectricalCircuitAnalysis_Page_96_Chunk1018 |
are used typically to adjust a voltage level. If the wiper and only one of the end terminals is used, the device is called a rheostat. A common application for a rheostat is the control of current. Finally, computer simulation tools may be used to build virtual circuits and determine various parameters such as voltages and currents. They may use virtual instruments that echo real world instruments or they may use more direct means, offering results in the form of tables. Review Questions Review Questions 1. Describe the differences between electron flow and conventional current flow. 2. Define the term series connection. 3. How is the equivalent resistance for a string of series connected resistors computed? Is the process identical for series connected voltage sources? 4. Why are current sources generally not placed in series? 5. Define Ohm's law. 6. Define Kirchhoff's voltage law. 7. Define the voltage divider rule. 8. What is meant by a Monte Carlo analysis? 9. What is the difference between a potentiometer and a rheostat? 3.10 Exercises 3.10 Exercises Analysis Analysis 1. For the circuit of Figure 3.34, determine the circulating current. 2. For the circuit of Figure 3.35, determine the circulating current. 97 Figure 3.34 Figure 3.35 | DCElectricalCircuitAnalysis_Page_97_Chunk1019 |
3. For the circuit of Figure 3.34, determine the power dissipated in the resistor. 4. For the circuit of Figure 3.35, determine the power dissipated in the resistor. 5. Determine the voltage at the open terminals of Figure 3.36. 6. Determine the voltage at the open terminals of Figure 3.37. 7. Determine the voltage at the open terminals of Figure 3.38. 8. Determine the equivalent resistance of circuit shown in Figure 3.39. 98 Figure 3.36 Figure 3.37 Figure 3.38 Figure 3.39 | DCElectricalCircuitAnalysis_Page_98_Chunk1020 |
9. Determine the equivalent resistance of circuit shown in Figure 3.40. 10. Determine the equivalent resistance of circuit shown in Figure 3.41. 11. For the circuit of Figure 3.42, determine the circulating current. 12. For the circuit of Figure 3.42, determine the voltages across each resistor and find Vab. 13. Given the circuit of Figure 3.42, determine the power dissipated by each resistor and the power delivered by the source. 99 Figure 3.40 Figure 3.41 Figure 3.42 | DCElectricalCircuitAnalysis_Page_99_Chunk1021 |
14. For the circuit of Figure 3.43, determine the circulating current. 15. Given the circuit of Figure 3.43, determine the voltages across each resistor and find Vba. 16. For the circuit of Figure 3.43, determine the power dissipated by each resistor and the power delivered by the source. 17. For the circuit of Figure 3.44, determine the circulating current. 18. For the circuit of Figure 3.44, determine the voltages across each resistor and find Va. 19. For the circuit of Figure 3.45, determine the circulating current and indicate all voltage polarities. 20. Given the circuit of Figure 3.45, determine the voltages across each resistor and find Vb, Vbc, and Vca. 21. For the circuit of Figure 3.45, determine the power delivered by the source. 100 Figure 3.43 Figure 3.44 Figure 3.45 | DCElectricalCircuitAnalysis_Page_100_Chunk1022 |
22. For the circuit of Figure 3.46, determine the circulating current and indicate all voltage polarities. 23. Given the circuit of Figure 3.46, determine the voltages across each resistor and find Vc, Vac, and Va. 24. For the circuit of Figure 3.46, determine the power dissipated by the 10 Ω resistor. 25. For the circuit of Figure 3.47, determine the circulating current and indicate all voltage polarities. 26. For the circuit of Figure 3.47, determine the voltages across each resistor and find Vb, Vc, and Vca. 27. Referring to the circuit of 3.48, determine the voltages across each resistor and find Vb, Vc, and Vac. 28. Referring to the circuit of Figure 3.48, determine the circulating current and indicate all voltage polarities. 101 Figure 3.46 Figure 3.47 Figure 3.48 | DCElectricalCircuitAnalysis_Page_101_Chunk1023 |
29. Given the circuit of 3.49, determine the voltages across each resistor and find Vb, Vc, and Vac. 30. Referring to the circuit of Figure 3.49, determine the circulating current and indicate all voltage polarities. 31. Given the circuit of 3.50, determine the circulating current and indicate all voltage polarities. 32. Referring to the circuit of Figure 3.50, determine the voltages across each resistor and find Vb, Vc, and Vac. 33. Given the circuit of 3.51, determine the circulating current and indicate all voltage polarities. 34. Referring to the circuit of Figure 3.51, determine the voltages across each resistor and find Vb, Vc, and Vac. 102 Figure 3.49 Figure 3.50 Figure 3.51 | DCElectricalCircuitAnalysis_Page_102_Chunk1024 |
35. Using the voltage divider rule, determine the voltages Vb, Vc and Vac for the circuit shown in Figure 3.52. 36. Using the voltage divider rule, determine the voltages Vb, Vc and Vbd for the circuit shown in Figure 3.53. 37. For the circuit of Figure 3.53, determine Vb if the 4 kΩ resistor is accidentally shorted. How does this compare to the original circuit? 38. For the circuit of Figure 3.53, determine Vb if the 4 kΩ resistor is accidentally opened. How does this compare to the original circuit? 103 Figure 3.52 Figure 3.53 | DCElectricalCircuitAnalysis_Page_103_Chunk1025 |
39. Given the circuit shown in Figure 3.54, find the voltage drop across the resistor. 40. Given the circuit shown in Figure 3.55, find the voltage drops across the resistor. 41. Find the voltage drops across the resistors in the circuit of Figure 3.56. 42. Find the voltage drops across the resistors in the circuit of Figure 3.57. 104 Figure 3.54 Figure 3.55 Figure 3.56 Figure 3.57 | DCElectricalCircuitAnalysis_Page_104_Chunk1026 |
43. Find the voltage drops across the resistors in the circuit of Figure 3.58. 44. Find the voltage drops across the resistors in the circuit of Figure 3.59. 45. The circuit of Figure 3.60 uses a linear taper potentiometer. Determine Vb when the wiper arm is at position a, position b, and at the halfway point. 46. What is the maximum current flowing through the potentiometer of Figure 3.60? At what position(s) does this occur? Design Design 47. Redesign the circuit of Figure 3.34 using a new resistor such that the current from the 12 volt battery is 0.1 A. 48. Redesign the circuit of Figure 3.35 using a new resistor such that the current from the 9 volt battery is 2 mA. 49. For the circuit of Figure 3.39, find the value of a series voltage source that would generate 1 mA of current if it was connected across the terminals. 105 Figure 3.58 Figure 3.59 Figure 3.60 | DCElectricalCircuitAnalysis_Page_105_Chunk1027 |
50. For the circuit of Figure 3.41, find the value of a series voltage source that would generate 1 mA of current if it was connected across the terminals. 51. Determine values for the resistors in Figure 3.61 such that R1 is four times the size of R2 and R2 is three times the size of R3, with the total resistance equaling 8 kΩ. 52. For the circuit shown in Figure 3.62, determine values for R1 and R2 such that Vab is 6 volts if E is a 9 volt battery and the total current draw is 20 mA. 53. Consider the circuit shown in Figure 3.63. If all resistors have the same value, determine that value if E, a 24 volt source, generates a total power of 10 watts. 106 Figure 3.61 Figure 3.62 Figure 3.63 | DCElectricalCircuitAnalysis_Page_106_Chunk1028 |
54. For the circuit shown in Figure 3.64, determine values for R1 and R2 such that Vab is 6 volts if I is a 2 mA source and the total voltage drop is 24 volts. 55. Consider the circuit of Figure 3.53. Is it possible to add a fifth resistor such that the circulating current is 0.1 mA? If so, what is that resistor value? 56. Consider the circuit of Figure 3.53. Is it possible to add a fifth resistor such that the circulating current is 2 mA? If so, what is that resistor value? Challenge Challenge 57. Assume that two AA cells, E1 and E2, rated at 900 mAh each are used to drive a 2 watt lamp as shown in Figure 3.65. Determine the expected life of the batteries. 58. Given the circuit of Figure 3.66, determine the required values of E, R1, R2 and R3 if there is one volt across R3, the total current draw is 10 mA, the voltage across R1 is twice the size of voltage across R2 and the power dissipation in R2 is 100 mW. 107 Figure 3.64 Figure 3.65 Figure 3.66 | DCElectricalCircuitAnalysis_Page_107_Chunk1029 |
59. Given the circuit of Figure 3.66, determine the required source voltage if R1 is 1 kΩ, R2 is 1 kΩ, the power dissipation in R1 is 4 mW and the power dissipation in R3 is 2 mW. 60. Given the circuit of Figure 3.67, determine Vc, Vdb and Vce. 61. Given the circuit of Figure 3.67, determine Vac, Veb and Vd. 62. Refer to the circuit of Figure 3.53. Assuming each resistor has a 10% tolerance, determine the maximum and minimum values for Vc. Simulation Simulation 63. Simulate the solution of design problem 43 and determine if the values produce the required results. 64. Perform a DC simulation on the circuit of Figure 3.46 and find all of the node voltages along with the circulating current. 65. Perform a DC simulation on the circuit of Figure 3.47 and find all of the node voltages along with the circulating current. 66. Perform a DC simulation on the circuit of Figure 3.59 and find all of the node voltages. 67. Perform a DC simulation on the circuit of Problem 34 and find all of the node voltages. 68. Perform a DC simulation on the circuit of Problem 36 and find all of the node voltages. 69. Simulate the solution of Challenge problem 58 and determine if the values produce the required results. 70. Simulate the circuit of Figure 3.67 (Challenge problems 60 and 61) and determine if the node voltages produced match the expected results. 71. Perform a Monte Carlo simulation on the circuit of Figure 3.52. Set each resistor to 5% tolerance and run at least ten variations for Vb to determine a typical spread of values. 108 Figure 3.67 | DCElectricalCircuitAnalysis_Page_108_Chunk1030 |
Source: www.xkcd.com This page intentionally left not blank. 109 | DCElectricalCircuitAnalysis_Page_109_Chunk1031 |
4 4 Parallel Resistive Circuits Parallel Resistive Circuits 4.0 Chapter Learning Objectives 4.0 Chapter Learning Objectives After completing this chapter, you should be able to: • Identify parallel resistive circuits that include a single voltage source or one or more current sources. • Compute equivalent resistance of parallel resistive networks. • Determine the equivalent of multiple parallel current sources. • Compute component and total current for parallel resistive networks. • Compute system voltage and component powers for parallel resistive networks. • Utilize Ohm's law, Kirchhoff's current law (KCL) and the current divider rule (CDR) to aid in the analysis of parallel resistive circuits. • Describe the function of fuses and circuit breakers. 4.1 Introduction 4.1 Introduction In the preceding chapter we examined series circuits, starting with series configurations of resistors and voltage sources. In this chapter we shall examine the “mirror twin” of series circuits, namely parallel circuits. Again, we shall begin by defining the parallel configuration, and explore how to combine sources and resistors in parallel. From there we shall introduce new laws and rules unique to parallel configurations. For everything said about series circuits, there are corresponding statements regarding parallel circuits. Just as there were specific laws and techniques useful in the series case, such as Kirchhoff's voltage law and the voltage divider rule, there are corresponding laws and techniques useful in the series case (e.g., Kirchhoff's current law and the current divider rule). A parallel circuit may contain any number of resistors and current sources, or in place of the current sources, a single voltage source. We shall examine how to determine the current flow through each component, the voltage across each component and the power either dissipated or generated by each component. As usual, other practical issues will also be examined along with appropriate computer simulations. Parallel circuits are in many ways the complement of series circuits. The hallmark of a series connection is that all components in that connection see the same current. Similarly, the most notable characteristic of a parallel connection is that all components see the same voltage. This implies that parallel connections have only two nodes, or two points of connection upon which everything is connected. Just as a series connection can be envisioned as a sort of chain with every element a link in that chain, a parallel connection can be envisioned as a sort of ladder with every element a rung on that ladder. 110 | DCElectricalCircuitAnalysis_Page_110_Chunk1032 |
4.2 The Parallel Connection 4.2 The Parallel Connection Consider the generic layout of Figure 4.1, each component represented by a box. Although it may be drawn oddly, there are only two common connection nodes in this configuration: one along the top and the other along the bottom (if you're having difficulty seeing the bottom node, simply slide box E to the right and then rotate it counterclockwise 180 degrees so that its bottom winds up on top). Each of these blocks is like a rung on a ladder. If we were to take a voltmeter and place it across any of these blocks, the meter would read the same voltage across each because the probes are always contacting the same two nodes (remember, ideally the connecting wires have negligible resistance). Thus, we come to the first observation of parallel circuits: In a parallel connection the voltage is the same across each component. (4.1) A more complex configuration is shown in Figure 4.2. In this layout some elements are in series and some are in parallel. Only elements D and E are strictly in parallel here as they are the only two that connect to the same two nodes, and thus must see the same voltage. Items C and F might at first appear to be in parallel with them but upon closer inspection it should be noted that C and F are in series with each other. That is, C and F will split the voltage that D or E sees, in accordance with the voltage divider rule. In other words, it is not true that the voltage across C or across F must be the same as the voltage across D or E. The same can be said for items A and B.12 4.3 Combining Parallel Components 4.3 Combining Parallel Components Our first step is to determine how to combine parallel components in order to create a single equivalent component. Unlike series connections, this can be a little more time consuming. Sources in Parallel Sources in Parallel First, voltage sources are not placed in parallel as a general rule, see Figure 4.3. The reason is because a parallel connection requires the same voltage across each component. This would be impossible to achieve with each source trying to maintain a different voltage across the same two nodes. This may result in damage to the sources (e.g., exploding batteries). The main exception to this rule is if the sources have the same potential and the goal is to extend operational life (e.g., multiple D cell batteries in parallel). 12 What can be said is that the series combination of C and F together is in parallel with D which is, in turn, in parallel with E. This is the basic idea behind series-parallel networks, the theme of the next chapter. 111 Figure 4.1 A parallel configuration. A B C D E Figure 4.2 A more complex configuration. A B C D F E Figure 4.3 Do not place voltage sources in parallel. | DCElectricalCircuitAnalysis_Page_111_Chunk1033 |
When it comes to current sources, they may simply be added together, however, like series voltage sources, polarity is important. Referring to the three parallel current sources in Figure 4.4, the middle and right sources are pumping current into the top node while the left source is feeding the bottom node. You can also think of the left source as draining the top node. Thus, these three sources together behave like a single source of five amps feeding the top node (eight amps in, three amps out). Resistors in Parallel Resistors in Parallel When placed in parallel, resistor values do not add they way they do in series connections. The reason for this is obvious if we look at the basic resistance equation from Chapter 2, Equation 2.11: R =ρl A If we consider two identical resistors placed in parallel, side by side as in Figure 4.5, the effective area would double while keeping the resistivity and length unchanged. The combined result would be a halving of the resistance of just one of them. Thus we see that placing resistors in parallel results in a decrease in net resistance, the opposite of the series case. This situation is simplified if we consider conductance instead of resistance. Recalling that conductance, G, is the reciprocal of resistance, we can rewrite the resistance equation: G = A ρl Now we can see that the increase in area creates a proportional increase in conductivity. If we generalize this for N resistors we find that the equivalent conductance of a group of parallel resistors is their sum: GTotal = G 1+G 2+G3+...+G N (4.2) As we deal normally with component resistance values instead of conductance values, Equation 4.2 may be rewritten in terms of equivalent resistance: 1 RParallel = 1 R1 + 1 R2 + 1 R3 +...+ 1 RN RParallel = 1 1 R1 + 1 R2 + 1 R3 +...+ 1 RN (4.3) 112 Figure 4.4 Current sources in parallel. Figure 4.5 Resistors in parallel. A B | DCElectricalCircuitAnalysis_Page_112_Chunk1034 |
Among other things, Equation 4.3 tells us that the equivalent resistance of a group of parallel resistances will always be smaller than the smallest resistance in that group. It also indicates that if we have N identical resistors of value R, the equivalent will be R/N (e.g., 3 parallel 3.6 kΩ resistors are equivalent to a single 1.2 kΩ resistor). As a convenience, it is common to use two parallel lines, ||, as a shortcut way of saying “in parallel with”, as in R1 || R2. Also, || has higher operational precedence than +. Product-Sum Rule Product-Sum Rule A handy shortcut for two parallel resistors may be created from Equation 4.3. We start by writing the version for just two parallel resistors, RParallel = 1 1 R1 + 1 R2 Multiply the fraction through by R1 R2 and simplify: RParallel = R1 R2 R1+R2 (4.4) For obvious reasons, Equation 4.4 is referred to as the product-sum rule. It is only applicable to two resistors, however, it can be used repeatedly on pairs of resistors within larger groups instead of using Equation 4.3. For example, a group of four resistors can be reduced to two pairs, and then those two equivalents can be reduced as a third pair. This can be a handy technique for obtaining quick estimates. We can take the product-sum rule a step further by writing R2 as a multiple of R1: R2 = N R1 Substituting this into Equation 4.4 yields: RParallel = R1 N R1 R1+N R1 RParallel = N R1 2 (N +1)R1 RParallel = N N +1 R1 (4.5) Equation 4.5 is useful for quick estimates of parallel resistor pairs. For example, if we have a 24 kΩ resistor in parallel with an 8 kΩ resistor, that's a ratio (N) of 3:1. 113 | DCElectricalCircuitAnalysis_Page_113_Chunk1035 |
Thus the equivalent will be N/(N+1), or 3/4ths, of the smaller resistor, yielding 6 kΩ. Similarly, a 180 Ω in parallel with a 90 Ω would yield 2/3rds of 90 Ω, or 60 Ω. Example 4.1 A group of resistors is placed in parallel as shown in Figure 4.6. Determine the equivalent parallel value. There are several solution paths. We'll try them all and compare. First, let's try the conductance form: RParallel = 1 1 R1 + 1 R2 + 1 R3 +...+ 1 RN RParallel = 1 1 2k Ω+ 1 6k Ω+ 1 12 kΩ RParallel ≈1.333kΩ We could also use the product-sum rule twice. We'll use the middle and right-most resistor first (although any pairing would do, but this pairing is convenient): R23 = R2 R3 R2+R3 R23 = 6k Ω12 k Ω 6 kΩ+12k Ω R23 = 4 k Ω RParallel = R1 R23 R1+R23 RParallel = 2 kΩ4 k Ω 2 kΩ+4k Ω RParallel ≈1.333kΩ Finally, we could use the ratio technique. First, consider the 6 kΩ and 12 kΩ resistor pair. That's a 2:1 ratio, so the result will be 2/3rds of the smaller resistor, and 2/3rds of 6 kΩ is 4 kΩ. The ratio between this and the 2 kΩ resistor is also 2:1, and 2/3rds of 2 kΩ is approximately 1.333 kΩ. The ratio technique happened to work out well here because of the convenient resistor values with perfect integer ratios. That will not always be the case. 114 Figure 4.6 Resistor arrangement for Example 4.1. | DCElectricalCircuitAnalysis_Page_114_Chunk1036 |
4.4 Kirchhoff's Current Law (KCL) 4.4 Kirchhoff's Current Law (KCL) Just as Kirchhoff's voltage law is a key element in understanding series circuits, Kirchhoff's current law (KCL) is the operative rule for parallel circuits. It states that the sum of all currents entering and exiting a node must sum to zero. Alternately, it can be stated as the sum of currents entering a node must equal the sum of currents exiting that node. As a pseudo formula: ∑ I→ = ∑ I← (4.6) Recalling that a node is a connection area wherein the voltage is the same (ignoring the resistance of connecting wires), we see that there are two nodes in a parallel configuration. For each of these nodes, whatever currents flow in must be balanced by the same total current flowing out. For example, referring to the circuit shown in Figure 4.7, there is a current source feeding the top node. At the three-way connection with R1 and R2, KCL along with Ohm's law dictate that the incoming current from the source must split with one portion exiting by flowing down through R1 and the remainder exiting by flowing down through R2. At the bottom node (ground) these two resistor currents must combine and equal the current that is being drawn back into the current source, which must be the same as the current originally produced by the source. This is perfectly sensible because it would be impossible for the current source to establish two different currents; one at its input and the other at its output. To assert otherwise would be a little like saying you're two different heights depending on whether you measure from your toes up to your head versus from your head down to your toes. Thus, a simplistic way of stating KCL is “What goes in must come out”.13 The Current Divider Rule (CDR) The Current Divider Rule (CDR) Just as series circuits follow the voltage divider rule (voltage dividing in proportion to resistance), parallel circuits follow the current divider rule which states that current divides in reverse proportion to resistance (i.e., in direct proportion to conductance). This can be reduced to a simple formula when only two resistors are involved. Consider two parallel resistors, R1 and R2, fed by a current, ITotal, as in Figure 4.7. First, we note that the voltage across the pair must equal the entering current times the effective resistance of the pair. V = I Total R1R2 R1+R2 13 For details on this corollary to Fudd's First Law of Opposition, see Firesign Theatre, I Think We're All Bozos On This Bus. Remember, the future can't wait. 115 Figure 4.7 A simple parallel circuit. | DCElectricalCircuitAnalysis_Page_115_Chunk1037 |
The current through each of the resistors must be this voltage divided by the corresponding resistance. For the current through R1: I R1 = V R1 I R1 = I Total R1 R2 R1(R1+R2) I R1 = I Total R2 R1+R2 Similarly, for the current through R2: I R2 = V R2 I R2 = I Total R1 R2 R2(R1+R2) I R2 = I Total R1 R1+R2 Thus, the current through one of the resistors will equal the total current times the ratio of the opposite resistor over the sum of the two resistors. In general: I X = ITotal RY R X+RY (4.7) This rule is convenient in that you don't have to compute the parallel equivalent resistance, but remember, it is valid only when there are just two resistors involved. A more general version that can be used for any number of resistors is Ii = ITotal ∙ RP/Ri, where RP is the total equivalent parallel resistance and Ri is the resistor of interest. This is, in essence, merely a rewriting of the fact that all components must see the same voltage: Ii ∙ Ri = ITotal ∙ RP. This is not as much of a shortcut as the two resistor version because the parallel resistance still must be calculated. Example 4.2 A simple parallel network is shown in Figure 4.8. Determine the current through each resistor. The current divider rule can be used to find the currents through the two resistors. At the top node the total entering current is 10 amps. 116 Figure 4.8 Circuit for Example 4.2. | DCElectricalCircuitAnalysis_Page_116_Chunk1038 |
I R1 = I Total R2 R1+R2 I 15 = 10 A 5Ω 15Ω+5Ω I 15 = 2.5 A I R2 = ITotal R1 R1+R2 I 5 = 10 A 15Ω 15Ω+5Ω I 5 = 7.5 A At this point it is worth taking a moment to verify the voltage polarities and current directions. Given the direction of the source, the currents through the two resistors must be flowing from top to bottom. Consequently, the voltage polarities must be + to − from top to bottom. This is illustrated in Figure 4.9. At the top node, entering current is in red with exiting current in blue. Note that KCL is satisfied as 10 amps enters while a total of 10 amps (2.5 amps plus 7.5 amps) exits. 4.5 Parallel Analysis 4.5 Parallel Analysis Keeping in mind that the voltage across each element in a parallel configuration is constant, Ohm's law dictates that currents divide among parallel resistors in proportion to their conductance (i.e., in inverse proportion to their resistance). As a consequence, Ohm's law, Kirchhoff's current law, the current divider rule and parallel component combinations are the tools we will use to solve general parallel circuit problems. There are multiple techniques for analyzing these circuits: • If the circuit uses a voltage source and its value along with the resistor values are given, the resistor currents can be found by dividing the source voltage by each resistance. Once these currents are found, KCL can be used to determine the source current. At that point, power law can be used to find the power dissipation in each resistor or the power developed by the source, as needed. • If the circuit uses current sources, then the total circulating current can be found by combining their values (taking into account the current directions, of course). The individual resistor branch currents then can be found using the current divider rule (repeatedly, if needed). Alternately, the effective parallel resistance may be found first. Multiplying this by the total source current via Ohm's law will determine the system voltage, and from there the individual resistor branch currents may be found using Ohm's law. 117 Figure 4.9 Polarities and directions for Example 4.2. | DCElectricalCircuitAnalysis_Page_117_Chunk1039 |
• If the problem concerns determining resistance values, the basic idea will be to use these rules in reverse. For example, if a resistor value is needed to set a specific voltage, the equivalent parallel resistance can be determined from this voltage and the given current source. The conductance values of the other parallel resistors can then be subtracted from the total conductance (i.e., the reciprocal of the equivalent parallel resistance), yielding the remaining required conductance value, the reciprocal being the required resistance. Similarly, if the currents through two parallel resistors are known, as long as one of the resistance values is known, the other resistance can be determined using either the current divider rule in reverse (i.e., the currents divide in inverse proportion to the resistances) or via Ohm's law. Example 4.3 A simple parallel network is shown in Figure 4.10. Determine the current through each resistor as well as the total current exiting the voltage source. The most direct solution is to use Ohm's law to determine each of the resistor branch currents. KCL can then be used to determine the current flowing from the source. Recalling the voltage is the same across parallel branches: I 600 = E R1 I 600 = 12 V 600Ω I 600 = 20 mA I 400 = E R2 I 400 = 12 V 400Ω I 400 = 30 mA KCL dictates that the entering current must equal the sum of the exiting currents, or 50 mA. An alternate technique would be to determine the parallel resistance and divide this into the source voltage to determine the exiting source current. RParallel = R1 R2 R1+R2 RParallel = 400Ω600Ω 400Ω+600Ω RParallel = 240Ω 118 Figure 4.10 Circuit for Example 4.3. | DCElectricalCircuitAnalysis_Page_118_Chunk1040 |
I Total = E RParallel I Total = 12 V 240Ω I Total = 50 mA Now CDR can be used to find the currents through the two resistors. I R1 = I Total R2 R1+R2 I 400 = 50 mA 600Ω 400Ω+600Ω I 400 = 30 mA I R2 = ITotal R1 R1+R2 I 600 = 50 mA 400Ω 400Ω+600Ω I 600 = 20 mA The voltage polarities and current directions are illustrated in Figure 4.11. The voltage polarity is + to − from top to bottom, as set by the voltage source. With these polarities, the currents through the two resistors must be flowing from top to bottom and the current from the source is flowing right, from the positive terminal. At the top node, entering current is in red with exiting current in blue. Computer Simulation Computer Simulation The circuit of Example 4.3 is entered into a simulator as shown in Figure 4.12. In this example, virtual instruments are used. Three ammeters are inserted in line to measure the current exiting the voltage source as well as the currents flowing through the two resistors. The polarities of the ammeters are set to match those illustrated in Figure 4.11. As a consequence, we expect to see all positive currents. The result is pretty much as expected. The current through the 600 Ω resistor is precisely 20 mA and the source current is precisely the sum of the two resistor currents. The only wrinkle is that the current through the 400 Ω resistor is very slightly less than expected, 29.999 mA versus 30 mA calculated. This is due to meter loading effects. In Chapter 3 we saw that ammeters present a very low internal resistance, but this cannot always be ignored, especially when measuring in series with resistors having very small values. This slight increase in resistance forces a slight decrease in current due to Ohm's law. 119 Figure 4.11 Polarities and directions for Example 4.3. | DCElectricalCircuitAnalysis_Page_119_Chunk1041 |
It turns out that a similar situation exists for voltmeters. Ideally, voltmeters present a very high internal resistance which, when placed across a resistor, has minimal impact. This effect cannot always be ignored, especially when measuring across resistors having large values, as the current divider rule will come into play. Fortunately, the internal resistance of many virtual instruments is adjustable and can be set to extreme values to minimize any impact on measurements. In the world of physical instruments this is not possible so the internal resistance of any real world meter is something that must always be kept in mind. Example 4.4 A parallel network is shown in Figure 4.13. Determine the current through each resistor. The current divider rule can be used to find the currents through the two resistors. At the bottom node (ground) the total entering current is 2 mA. I R1 = I Total R2 R1+R2 I 12k = 2mA 100kΩ 12 kΩ+100kΩ I 12k ≈1.7857 mA The current through the 100 kΩ can be found via KCL as follows: 120 Figure 4.12 The circuit of Example 4.3 in a simulator. Figure 4.13 Circuit for Example 4.4. | DCElectricalCircuitAnalysis_Page_120_Chunk1042 |
I R2 = I Total −I R1 I 100k = 2 mA −1.7857 mA I 100k ≈0.2143 mA The polarities and directions are shown in Figure 4.14. Note that the currents are flowing up through the resistors producing voltage drops of + to − from ground up to the top. This means that the voltage of the top node is negative with respect to ground. Computer Simulation Computer Simulation In order to get a better handle on the idea of meter loading, the circuit shown in Example 4.4 is entered into a simulator. This is shown in Figure 4.15 with the simulation using virtual instruments. Using Ohm's law, the voltage drop across the system should be 100 kΩ times its current, or approximately 21.4286 volts if we carry the digits out a little further. The virtual voltmeter has been set for an internal resistance of 1 GΩ. The results are nearly the same as those calculated previously. To test the impact of the voltmeter's internal resistance the simulation is run a second time with the internal resistance set to 1 MΩ, a value seen commonly in general purpose digital multimeters. The result is shown in Figure 4.16. The voltage magnitude has dropped more than 200 mV, entirely due to meter loading. 121 Figure 4.14 Polarities and directions for Example 4.4. Figure 4.15 The circuit of Example 4.4 in a simulator. | DCElectricalCircuitAnalysis_Page_121_Chunk1043 |
Depending on component values, the results could be much worse or hardly noticeable at all. As a general rule, make sure the voltmeter's internal resistance is at least 10 times (and preferably, 100 times) larger than any resistance it is placed across in order to avoid loading errors caused by unwanted current division. In the case of an ammeter, its internal resistance should be at least 10 times (and preferably, 100 times) smaller than any resistance with which it is placed in series. Example 4.5 A parallel network is shown in Figure 4.17. Determine the current through the 200 Ω resistor. The first step is to simplify the circuit. The two current sources aid each other as they both feed current into the top node. Thus they are equivalent to a single 1.4 amp current source with the same direction. Further, the 60 Ω and 120 Ω resistors are in parallel and can be treated as a single unit. Note this is a 2:1 ratio, and thus the result will be 2/3rds of 60 Ω, or 40 Ω. The basic product-sum rule or conductance formula could also be used. We have simplified the circuit down to a single 1.4 amp source feeding a 200 Ω resistor in parallel with a 40 Ω resistor. The current divider rule can be used to find the current through the 200 Ω resistor as follows: 122 Figure 4.16 Simulation repeated using a voltmeter internal resistance of 1 MΩ. Figure 4.17 Circuit for Example 4.5. | DCElectricalCircuitAnalysis_Page_122_Chunk1044 |
I R1 = I Total R2 R1+R2 I 200 = 1.4 A 40Ω 200Ω+40Ω I 200 ≈233.33 mA As a crosscheck, Ohm's law indicates the system voltage must be 200 Ω times 233.33 mA, or approximately 46.667 volts. Via Ohm's law, the current through the 60 Ω must be 46.667 volts divided by 60 Ω, or approximately 777.78 mA. Similarly, the current through the 120 Ω can be determined to be approximately 388.89 mA. Via KCL, these three currents should add up to the total supplied current of 1.4 amps, which they do. An alternate solution path would be to find the equivalent of the three parallel resistors, or 200 || 60 || 120, which is 33.333 Ω. This is fed by the combined sources yielding 1.4 amps. Ohm's law is then used to find the system voltage of approximately 46.667 volts (1.4 amps times 33.333 Ω). From there Ohm's law is used again to find the current flowing through the 200 Ω (46.667 volts divided by 200 Ω, or approximately 233.33 mA). Let's continue with the prior example, this time with a design twist. Example 4.6 Using the circuit of Figure 4.17, determine the value of an additional fourth parallel resistor such that the system voltage will drop to 42 volts. The total source current in this circuit is fixed at 1.4 amps. If the resistance it feeds is reduced, then by Ohm's law, the system voltage must also be reduced. One way to solve this problem is to first determine the required parallel equivalent resistance. Then the conductances of the three known resistors can be subtracted out, leaving the conductance of the new resistor, from which we determine its resistance. RParallel = V I RParallel = 42 V 1.4A RParallel = 30 Ω G4 = GParallel −G1 −G2 −G3 G4 = 1 30 Ω − 1 200Ω −1 60 Ω − 1 120Ω G4 ≈3.333mS 123 | DCElectricalCircuitAnalysis_Page_123_Chunk1045 |
And finally, R4 = 1/G4, thus R4 = 300 Ω. The smaller the value of this fourth resistor, the more current it will siphon off from the other three resistors, thus reducing the system voltage further. Time for an example involving power. Example 4.7 A parallel network is shown in Figure 4.18. Determine the total power dissipated by the four resistors. While the powers for each resistor can be calculated individually and then added, it is worthwhile to note that the total power dissipated must be equal to the power generated. The power generated by the battery is dictated by power law, in other words, the battery voltage times its output current. To find this current, we can determine the effective resistance and then apply Ohm's law. The total resistance can be found directly by using the conductance formula, however, the values can be divided into convenient pairs first. The pair of 200 Ω resistors is equivalent to a single 100 Ω. Similarly, the two 600 Ω resistors are equivalent to a single 300 Ω. Product-sum rule may be used next or 100 || 300 can be determined using the ratio rule. This is a 3:1 ratio, so the result will 3/4ths of the smaller resistor, or 75 Ω. I Total = E RTotal I Total = 9 V 75Ω I Total = 120 mA P = E×I P = 9V×120 mA P = 1.08 W Slightly quicker, we could also use the following approach: P = E 2 R P = (9 V) 2 75Ω P = 1.08 W 124 Figure 4.18 Circuit for Example 4.7. | DCElectricalCircuitAnalysis_Page_124_Chunk1046 |
4.6 Current Limiting: Fuses and Circuit Breakers 4.6 Current Limiting: Fuses and Circuit Breakers Most electrical and electronic devices are designed to run off of a constant voltage source. Examples include battery powered devices and the numerous items that connect to standard residential and commercial electrical systems (e.g., the 120 VAC system in North America). As such, parallel connections are ideal when a variable number of devices are to be powered because, unlike a series connection, each device will see the same voltage rather than having that voltage split among them. There is another advantage to a parallel connection. Consider the two lighting schemes presented in Figure 4.19; one utilizing a series configuration and the other a parallel configuration. Typically, when a lamp fails, it fails creating an open. If any of the lamps in the series connection fails, current in the loop is interrupted and all of the lamps go out. In contrast, if one of the lamps in the parallel connection fails, the remaining lights will stay on. Further, in the series connection, there is no immediate way to determine which lamp has failed as none of them will be lit. Each lamp has to be tested in turn to determine the faulty unit. In the parallel system, it is obvious which lamp has failed because that's the one that's out. In a home, school or commercial setting, any number of devices may be plugged into wall sockets with the expectation that each unit will receive the same voltage and operate correctly due to the parallel nature of the system. As convenient as this is, there is a practical problem. Every device that is added to the system presents another path for current flow. Due to KCL, this means that the total current drawn from the source must increase. This creates a potential problem because if too much current is drawn, there can be an appreciable voltage drop across the wires feeding the outlets as they can no longer be assumed to be zero ohms. This lowered voltage may present a problem for the attached devices, but more importantly, the large current results in unexpected heating of the wires because of the power dissipation in them (I2R) and creates a possible fire hazard. Consequently, we must include some method to limit the current to a safe maximum value. 125 Figure 4.19 Lamp connections: Series (left) and parallel (right). Figure 4.20 Various styles of fuses. | DCElectricalCircuitAnalysis_Page_125_Chunk1047 |
While some electronic systems use active current limiting schemes14, the more common approach is a fuse or circuit breaker. These devices are placed between the voltage source and the various loads, and see their combined current. If this current becomes too great, the device interrupts the current flow by opening the circuit. Nothing will operate but nothing catches fire, either. Fuses are used typically in electronic systems, motors, automotive subsystems and the like. An array of various types of fuses is shown in Figure 4.20. Whether they're the cylindrical type found in electronic instrumentation or the blade type common in automobiles, fuses are one-shot devices. They consist of a fine wire or metal link that will heat up and melt if it experiences too high of a current. Upon melting, circuit continuity is lost and current flow stops. This is called a “blown” fuse. The fuse will then need to be replaced once the problem has been fixed (i.e., whatever it was that caused excessive current to flow in the first place). Fuses have a nominal current rating, for example, two amps. This does not mean that as soon as two amps is reached, the fuse blows. It will take some time for the fuse to heat up. The higher the current is above the nominal rating, the faster the fuse will blow. For example, it might take this nominal two amp fuse several seconds before it blows if the current is, say, 2.5 amps, but less than a second if the current is five amps. Fuses are also available in fast-blow and slow-blow forms. Fast-blow fuses will activate many times faster than standard fuses while slow-blow types will require much long to activate (this prevents the fuse from activating accidentally with certain kinds of loads that exhibit a high initial startup current but that falls back to a more modest level once running smoothly, such as a motor). The schematic symbols for the fuse are shown in Figure 4.21. The top two are IEC standard and the bottom version is the ANSI (North American) standard. Fuses are generally inexpensive and effective. The obvious downside to a fuse is that they need to be replaced once they do their job. A circuit breaker solves this problem. A circuit breaker, or just breaker for short, can be thought of as a resettable fuse. Breakers do not have a melting link that interrupts current flow. Instead, a breaker operates like an intelligent switch: if the current is too large, the switch is thrown open. The operator can then fix the problem and reset the breaker (i.e., return the switch to the operating position). A selection of breakers is shown in Figure 4.22. Circuit breakers are available in a variety of sizes. For residential wiring, 15 and 20 amp sizes are common for general use although higher sizes such as 30 and 50 amps are available for circuits feeding high power appliances such as electric stoves, water heaters and clothes dryers. 14 For details, see Fiore, J., Semiconductor Devices as well as Operational Amplifiers and Linear Integrated Circuits, both OER titles. 126 Figure 4.21 Schematic symbols for fuses: IEC (top and middle), ANSI (bottom). Figure 4.22 Residential circuit breakers including cutaway view. | DCElectricalCircuitAnalysis_Page_126_Chunk1048 |
4.7 Summary 4.7 Summary In this chapter we have introduced and examined parallel circuits using either one or more current sources or a single voltage source, along with two or more resistors. The hallmark of a parallel configuration is that all components are connected to just two nodes. This means that all elements in a parallel configuration see the same voltage. If multiple current sources are present, they may be combined into a single equivalent current source by adding their values, taking care to watch for the directions of source currents; like directions add while opposing directions subtract, in much the same manner as series-connected voltage sources. Voltage sources generally are not placed in parallel as it would be a practical impossibility to establish different voltages across the same two nodes. The major exception to this rule is if the sources use the same voltage and the goal is to extend battery life. Dissimilar parallel voltage sources will most likely cause excessive heat and an unstable voltage. The equivalent resistance of a group of parallel resistors will always be smaller than the smallest resistor in the group because each resistor adds another path for current flow, thus enhancing conductivity and reducing resistance. In general, the effective resistance is found by summing the individual conductances to find the total conductance of the group and then taking the reciprocal of this value. Two shortcut techniques when just two resistors are involved are the product-sum rule and the ratio rule (which is particularly handy when one resistor is the integer multiple of the other). Kirchhoff's current law (KCL) states that the sum of currents entering a node must equal the sum of currents leaving that node. Another way of stating this is that the sum of all currents entering and exiting a node must be zero (obeying polarities). The current divider rule (CDR) is useful for determining the division of current between two parallel resistors. It states that the current through one resistor must equal the current entering the pair times the ratio of the other resistance value to the summed resistance of the pair. To determine individual branch currents in a circuit driven by a voltage source, Ohm's law may be used by dividing the source voltage by the individual resistor values. These branch currents must sum to the total current delivered by the voltage source due to KCL. If the parallel network is driven by current sources, the individual branch currents can be found by determining the effective parallel resistance and then using Ohm's law to find the system voltage. Once the voltage is known, Ohm's law is used again on each resistor to find the associated branch current. Alternately, the resistors can be simplified into successive pairings and then CDR can be applied repeatedly. Real world ammeters and voltmeters exhibit internal resistances that can load down circuit elements and cause errors in measurement. It is best if voltmeters have an 127 | DCElectricalCircuitAnalysis_Page_127_Chunk1049 |
internal resistance that is lat east an order of magnitude (preferably two) larger than the resistors they are placed across in order to limit undesired current divider effects. Similarly, ammeters should have an internal resistance at least an order of magnitude (preferably two) smaller than the resistors with which they are placed in-line to limit undesired Ohm's law effects. Fuses and circuit breakers are used to limit potentially damaging current in a system. Both devices will activate by opening the circuit if the current passing through them exceeds their rated value for a specified period of time. This results in zero current flow. Fuses are one-shot devices that must be replaced after being activated. In contrast, circuit breakers can be activated numerous times and are simply reset via a switch or button after being tripped. Review Questions Review Questions 1. Define the term parallel connection. 2. How is the equivalent resistance for a group of parallel connected resistors computed? 3. How is the equivalent value for parallel connected current sources computed? 4. Why are voltage sources generally not placed in parallel? 5. Define the product-sum rule and its use. 6. Define Kirchhoff's current law. 7. Define the current divider rule. 8. Explain the basic operation and usage of fuses and circuit breakers. 4.8 Exercises 4.8 Exercises Analysis Analysis 1. Determine the effective resistance of the network shown in Figure 4.23. 128 Figure 4.23 | DCElectricalCircuitAnalysis_Page_128_Chunk1050 |
2. Determine the effective resistance of the network shown in Figure 4.24. 3. Determine the effective resistance of the network shown in Figure 4.25. 4. Find the effective source current of the network shown in Figure 4.26. 5. Determine the effective source current of the network shown in Figure 4.27. 129 Figure 4.24 Figure 4.25 Figure 4.26 Figure 4.27 | DCElectricalCircuitAnalysis_Page_129_Chunk1051 |
6. Find the source and resistor currents for the circuit of Figure 4.28. 7. Determine the source and resistor currents for the circuit of Figure 4.29. 8. Find the source and resistor currents for the circuit of Figure 4.30. 9. For the circuit of Figure 4.31, determine the source and resistor currents. 130 Figure 4.28 Figure 4.29 Figure 4.30 Figure 4.31 | DCElectricalCircuitAnalysis_Page_130_Chunk1052 |
10. Determine the current through each resistor in the circuit of Figure 4.32. Also determine the total power generated by the source. 11. Consider the circuit shown in Figure 4.32. Assume that the 100 kΩ is replaced with another resistor ten times as large. Will this have a major impact on the current exiting source? Why/why not? 12. Consider the circuit shown in Figure 4.32. Assume that the 1 kΩ is replaced with another resistor ten times smaller. Will this have a major impact on the current exiting source? Why/why not? 13. Find the current through each resistor in the circuit of Figure 4.33. 14. Determine the current through each resistor in the circuit of Figure 4.34. Also determine the total current exiting by the source. 131 Figure 4.32 Figure 4.33 Figure 4.34 | DCElectricalCircuitAnalysis_Page_131_Chunk1053 |
15. Determine the current through each resistor in the circuit of Figure 4.35. 16. For the circuit shown in Figure 4.36, determine the current through each resistor and the source voltage. 17. For the circuit shown in Figure 4.37, determine the current through each resistor and the source voltage. 18. For the circuit shown in Figure 4.38, determine the current through each resistor and the source voltage. 132 Figure 4.35 Figure 4.36 Figure 4.37 Figure 4.38 | DCElectricalCircuitAnalysis_Page_132_Chunk1054 |
19. For the circuit shown in Figure 4.39, determine the current through each resistor and the source voltage. 20. For the circuit shown in Figure 4.40, determine the current through each resistor and the source voltage. 21. For the circuit shown in Figure 4.41, determine the current through each resistor and the source voltage. 22. For the circuit shown in Figure 4.42, determine the current through each resistor and the source voltage. 133 Figure 4.39 Figure 4.40 Figure 4.41 Figure 4.42 | DCElectricalCircuitAnalysis_Page_133_Chunk1055 |
23. Referring to the circuit of Figure 4.42, determine the resistor currents if the right-most 75 kΩ resistor is accidentally opened (i.e., unconnected). How do these results compare to those of problem 22? 24. Referring to the circuit of Figure 4.42, determine the resistor currents if the right-most 75 kΩ resistor is accidentally shorted. How do these results compare to those of problems 22 and 23? 25. For the circuit shown in Figure 4.43, determine the current through each resistor and the source voltage. 26. Given the circuit of Figure 4.44, find the currents through the two resistors. 27. If the 5 mA current source shown in Figure 4.44 is accidentally wired in upside down, does the voltage across the 12 k Ω resistor become more positive or more negative with respect to ground? 28. Given the circuit of Figure 4.45, find the voltages across the three resistors. 29. Find the currents through the three resistors in Figure 4.45. 134 Figure 4.43 Figure 4.44 Figure 4.45 | DCElectricalCircuitAnalysis_Page_134_Chunk1056 |
Design Design 30. For the network shown in Figure 4.46, determine a for values for R1 given that R2 is 12 kΩ and the equivalent combination is 8 kΩ. 31. Add a third parallel resistor to the circuit of Figure 4.30 such that the source current is 10 mA. 32. Add a fourth parallel resistor to the circuit of Figure 4.32 such that the source current is 20 mA. 33. Consider the circuit shown in Figure 4.36. Determine a new value for the current source such that the source voltage equals 10 volts. 34. Consider the circuit shown in Figure 4.38. Determine a new value for the current source such that the source voltage equals 20 volts. 35. Given the circuit of Figure 4.47, if the source is 6 volts and R1 is 2 kΩ, what must be the value of R2 if the total current exiting the source is 10 mA? 36. For the circuit shown in Figure 4.48, determine values for resistors R2 and R3 such that the current through R2 is twice the current through R1 and the current through R3 is half the current through R1. The source is 6 volts and R1 is 2 kΩ. 135 Figure 4.46 Figure 4.47 Figure 4.48 | DCElectricalCircuitAnalysis_Page_135_Chunk1057 |
37. For the circuit shown in Figure 4.49, determine values for resistors R1 and R2 such that the current through R2 is twice the current through R1. The source is 10 mA and R1 is 6 kΩ. Challenge Challenge 38. For the circuit shown in Figure 4.34, determine a new value for the 11 kΩ resistor such that the supply current is 50 mA. 39. For the circuit shown in Figure 4.36, determine a new value for the 2 kΩ resistor such that the voltage drop across the 6 kΩ is 15 volts. 40. Consider the circuit shown in Figure 4.43. If the current source was replaced with a voltage source, what value is needed so that the same currents flow through the resistors as in the original circuit? 41. For the network shown in Figure 4.46, determine values for R1 and R2 such that R2 is twice the size of R1 and the equivalent combination is 6 kΩ. 42. Given the network of Figure 4.25, is it possible to replace the 60 Ω resistor with another value such that the equivalent combination of the three resistors is 25 Ω? If so, what is that value? 43. Given the network of Figure 4.33, is it possible to add a fifth parallel resistor such that the source current is 1 mA? If so, what is that value? 44. For the circuit shown in Figure 4.48, determine values for the three resistors such that the current through R1 is twice the current through R2 and four times the current through R3. The source is 12 volts and should produce a total of 9 mA of current. 45. For the circuit shown in Figure 4.49 determine values for the two resistors such that the current through R1 is half the current through R2. The source is 24 mA and should produce a drop of 16 volts across R1. 136 Figure 4.49 | DCElectricalCircuitAnalysis_Page_136_Chunk1058 |
46. Given three current sources with values of 1 mA, 2 mA and 7 mA; how would they need to be connected in order to deliver 4 volts across a 1 kΩ load resistor? 47. Consider the circuit of Figure 4.50. Assume I is a 4 mA source. Using only 5% standard resistor values (see Appendix A), pick values for the three resistors such that the voltage across R1 is within 10% of 10 volts as long as the resistors are within tolerance. Simulation Simulation 48. Verify the currents found in problem 11 via a DC simulation. 49. Verify the currents found in problem 15 via a DC simulation. 50. Verify the currents and voltages found in problem 17 via a DC simulation. 51. Verify the results found in problem 25 via a DC simulation. 52. Verify the results found in problem 27 via a DC simulation. 53. Perform a DC simulation on the design of problem 44 to verify its performance. 54. Perform a DC simulation on the design of problem 45 to verify its performance. 55. Perform a DC simulation on the design of problem 46 to verify its performance. 56. Perform a Monte Carlo or worst-case simulation on the design of problem 47 to verify its performance. 137 Figure 4.50 | DCElectricalCircuitAnalysis_Page_137_Chunk1059 |
5 5 Series-Parallel Resistive Circuits Series-Parallel Resistive Circuits 5.0 Chapter Learning Objectives 5.0 Chapter Learning Objectives After completing this chapter, you should be able to: • Identify series-parallel resistive circuits that include a single effective voltage or current source. • Compute component and node voltages for single-source series-parallel resistive networks. • Compute branch currents for single-source series-parallel resistive networks. • Break down series-parallel networks into smaller series-only and parallel-only sub-units, and utilize KVL, KCL and other previously established techniques to aid in the analysis of these circuits. 5.1 Introduction 5.1 Introduction In the preceding two chapters, series circuits and parallel circuits were examined. Each offered unique laws and rules for their configuration, such as Kirchhoff's voltage law and the current divider rule. Distinct techniques of analysis were offered to determine system current and voltage, and the currents and voltages associated with individual components. This almost mechanical approach was successful mainly because there are so few variations on the theme. In the case of a series circuit, more resistors might be added in the “daisy chain”, and likewise multiple voltage sources, but once you have experience with a few of these types of circuits, there is not much concern when simply adding more of the same because a pattern becomes apparent. The same holds true with parallel circuits: more resistors might be added as further “rungs on the ladder”, as well as multiple current sources similarly arranged, but again, once familiarity has been gained, obvious patterns emerge. This is not the case with series-parallel circuits, and thus, there are no simple recipes for our solution paths. Fortunately, laws and rules such as Ohm's law, KCL and KVL still hold true for these circuits, thus the focus shifts to their appropriate application. To begin with, there are infinite variations of series-parallel circuits. This chapter deals with a subset, namely those that are driven by a single current or voltage source (after reducing trivial combinations), and which may be simplified using series and parallel resistor combinations. More complex configurations using multiple sources await in subsequent chapters. The key to analyzing basic series-parallel circuits is in recognizing portions of the circuit, or sub-circuits, that exhibit a series or parallel configuration by themselves, and then applying the series and parallel analysis rules to those sections. Ohm's law, KVL and KCL may be used in turn to “chip away” at the problem until all currents and voltages are found. As individual voltages and currents are determined, this makes it easier to apply these rules to determine other values. Given this observation, the number of potential solution paths tends to grow exponentially as the number of components increases. As a consequence, when faced with the same circuit, six people may solve it six different ways, no particular way being more or less correct than any other. The only thing we can say is that some solution paths might be more computationally efficient than others, meaning they take less work. Do not let this bother you. The fact that these 138 | DCElectricalCircuitAnalysis_Page_138_Chunk1060 |
circuits can be solved in a diverse number of ways is a strength, not a weakness. After all, you only need to recognize one of those ways, not all of them, in order to be successful. In this chapter's examples, various competing methods will be explored, but not every solution path will be spelled out for each one. Flexibility of thought and view will prove to be an asset. 5.2 Series-Parallel Connections 5.2 Series-Parallel Connections A basic series-parallel system is shown in Figure 5.1. It should be obvious that this system is neither a simple series connection nor a simple parallel connection, but rather some combination of the two. In order for it to be series, the current through each component would have to be the same. This does not have to be true here. Consider the connection node directly above block C. This also connects to items B, D and E. All we really know about that node is that there are four paths leading to it and that the four associated currents must sum to zero, thanks to KCL. There is nothing that requires the current through, say, block D to be the same as that through block E. Similarly, there is nothing that stipulates that the voltage across block C must be the same as that across block F. After all, that is the hallmark of a parallel configuration, and that would require those two blocks to be connected to the same two nodes, which they aren't. What we can say about the configuration of Figure 5.1 is that some of the components are in series among themselves, and some are in parallel among themselves. We might recognize these groupings as sub-circuits. For example, it is true that blocks A and B are in series with each other because the current through one of them must be the same as the current through the other. The same is true of blocks C and F; they are in series with each other. Similarly, we can say that blocks D and E are in parallel with each other because they must see the same voltage as they are each connected to the same two nodes. 5.3 Simplifying Series-Parallel Components 5.3 Simplifying Series-Parallel Components We can go beyond the simple observations presented above. Precisely how we approach this will depend on the components represented by the blocks (most often resistors, but possibly a voltage source or a current source) and how they are wired together. For starters, if all of the items in a sub-circuit are of the same type, they can just be combined using the techniques discussed in the previous chapters. Resistors in series are added, as are voltage sources in series-being wary of polarity. Resistors in parallel are combined using conductance or the product-sum rule while parallel current sources are added, again being wary of polarity. Things get a little trickier when there is a mix, such as two resistors in series with a voltage source, or a resistor in parallel with a couple of current sources. Precisely how this sort of 139 A B C D F E Figure 5.1 A series-parallel configuration. | DCElectricalCircuitAnalysis_Page_139_Chunk1061 |
situation is handled will become apparent shortly. To illustrate, let us consider the resistor configuration presented in Figure 5.2. Imagine that an ohmmeter is connected to the two open terminals. What would it read? The ohmmeter would apply a small current to the circuit to determine the effective resistance of the group. The three resistors pictured here are not a simple series configuration because we can identify a current split at the 40 Ω and 80 Ω resistors. Similarly, it is not a parallel network because the 40 Ω and 160 Ω resistors are not both tied to the same two nodes. What we can say is that the 40 Ω and 160 Ω resistors are in series with each other as they must each see the same current. Thus, we can combine the two into a single equivalent resistance of 200 Ω. This equivalent is in parallel with the 80 Ω resistor and 80 || 200 is approximately 57.14 Ω. That's what the ohmmeter should read. Indeed, wherever we see this configuration with these precise values, we can replace it with a single resistor of 57.14 Ω. The simplification technique can be outlined as follows: • Identify subgroups of resistors that exhibit series or parallel configurations within themselves. • Replace the subgroups with a single equivalent resistance. • Repeat the above steps until the circuit is reduced to a single resistance, or alternately, a simple series-only or parallel-only configuration. The looping nature of the technique outlined above may need to be invoked many times, depending on the complexity of the network. In simpler networks, a single pass may be sufficient. Initially, it may be wise to redraw the circuit for each iteration of the loop. Eventually, with practice, this will not be needed except with the most complex networks. Let's look at a somewhat more involved example. Example 5.1 A series-parallel resistor network is shown in Figure 5.3. Determine its equivalent resistance. The first step is to recognize those subgroups that are in series or in parallel with themselves. One obvious candidate is that the two 100 Ω resistors are in parallel with each other. Two resistors of equal value in parallel are equivalent to half the resistance, or 50 Ω in this case. The other candidate is the 40 Ω, 200 Ω pair. These are in series. The equivalent resistance of the pair is the series combination, or 240 Ω. There are no other subsets of resistors that can be reduced (yet). The network is redrawn in Figure 5.4 with the equivalent resistances. The process is now repeated. 140 Figure 5.2 A simple series-parallel resistor network. Figure 5.3 Network for Example 5.1. | DCElectricalCircuitAnalysis_Page_140_Chunk1062 |
In the newly reduced network, the 50 Ω and 30 Ω resistances are in series, yielding 80 Ω. Without redrawing, the process may be repeated again, this time the 240 Ω being in parallel with the new 80 Ω equivalent. 240 || 80 is equal to 60 Ω. This sequence has reduced the three right-most resistances down to one: the new 60 Ω equivalent. Finally, the 60 Ω equivalent is in series with the 10 Ω resistor, yielding a final equivalent resistance of 70 Ω. Ladders Ladders A ladder is a unique series-parallel configuration. It is arranged in a cascade of series and parallel connections. Ladders are used in a variety of applications, one of which will be examined in the next section. The name is obvious based on the appearance of the network. An example resistive ladder network is shown in Figure 5.5. The main thing to note is that the ladder network consists of “sections loading sections” repeatedly. By loading, we mean that this element (the load) draws off current and power from the prior section. As a result, simple two-resistor voltage dividers cannot be used. For example, it is a common error to assume that R1 and R2 create a voltage divider which can be used to determine Vbf if Vaf is known. Similarly, the uninitiated may believe incorrectly that R3 and R4, or R5 and R6 create voltage dividers. In fact, the only proper voltage divider in this configuration is between R7 and R8. The reason for this becomes apparent if we imagine determining the equivalent resistance of the network by placing an ohmmeter at terminals a and f. The ohmmeter will apply a current to node a which passes through R1. At node b this current splits, part going down R2 and part continuing through R3. Clearly then, R1 and R2 are not in series as they do not see the same current. Further, the current that flows through R3 enters node c where it splits again, part down R4 with the remainder flowing through R5. Thus, R3 and R4 are not in series, either. The same sort of thing happens again at node d. This would continue for as many rungs as are on the ladder, the sole exception being the final two resistors which are in series (R7 and R8 here). 141 Figure 5.4 Partially simplified network. Figure 5.5 A resistive ladder network. | DCElectricalCircuitAnalysis_Page_141_Chunk1063 |
How then do we find the equivalent resistance of this network? We begin at the end farthest from the open terminals. We have already noted that R7 and R8 are in series. This pair, if treated as a single resistance, is in parallel with R6, or R6||(R7+R8). This group of three is in series with R5, yielding R5+R6||(R7+R8)15. This new group of four is in parallel with R4, yielding R4||(R5+R6||(R7+R8)). This group is in series with R3, and that new group is in turn in parallel with R2. Finally, that penultimate grouping is in series with R1. Therefore, the equivalent resistance of the network is R1+R2||(R3+ R4||(R5+R6||(R7+R8))). Surely, it is a tad tedious to compute the value, but particularly difficult. A word to the wise: if the goal is determine the voltages at the various nodes or the myriad branch currents, it will be helpful to keep track of the equivalent resistances of the various groupings. For example, if Vcf is known and the goal is to find Vdf, the voltage divider would be R6||(R7+R8) / (R5+R6||(R7+R8)). Example 5.2 A resistive ladder network is shown in Figure 5.6. Determine its equivalent resistance. Starting at the far end from the open terminals, we note that the 7 kΩ and 5 kΩ resistors are in series, yielding 12 kΩ. This is in parallel with the 8 kΩ. 12 k || 8 k is 4.8 kΩ. This is in series with the 1.2 kΩ yielding 6 kΩ. The 6 kΩ equivalent is in parallel with the 1.5 kΩ which yields 1.2 kΩ. Finally, the 1.2 kΩ equivalent is in series with the 3 kΩ, yielding a final equivalent resistance of 4.2 kΩ. Remember, always start at the far end and work toward the open terminals, i.e., the terminals into which you are trying to find the equivalent resistance. Example 5.3 A resistive ladder network is shown in Figure 5.7. Determine its equivalent resistance. The far end from the open terminals is the extreme left. First, the 6 kΩ and 4 kΩ resistors are in series, yielding 10 kΩ. This is in parallel with the 8 kΩ, achieving approximately 4.444 kΩ. In turn, this is in series with the 2 kΩ yielding 6.444 kΩ. That result is in parallel with the 12 kΩ which yields 4.193 kΩ. Placing this in series 15 Recall that || has higher precedence than + or –, therefore || is performed first, rather like multiplication when mixed with addition and subtraction. 142 Figure 5.6 Network for Example 5.2. Figure 5.7 Network for Example 5.3. | DCElectricalCircuitAnalysis_Page_142_Chunk1064 |
with the 10 kΩ resistor yields 14.193 kΩ. This is in parallel with the 20 kΩ yielding 8.302 kΩ. Finally, the 8.302 kΩ equivalent is in series with the 5 kΩ, yielding a final equivalent resistance of approximately 13.3 kΩ. 5.4 Series-Parallel Analysis 5.4 Series-Parallel Analysis Analysis of series-parallel networks involves recognizing those sub-circuits that are in series or that are in parallel among themselves, performing simplifications as needed, and winding up with a simple series-only or parallel-only equivalent. Then the various laws such as Ohm's law, KVL, KCL, VDR and CDR are applied to the various simplified networks to determine the parameters of interest. There is no single solution technique, each circuit being unique. Let's begin by considering the circuit of Figure 5.8. To review, this is neither just a series circuit nor just a parallel circuit. If it was a series circuit then the current through all components would have to be same, that is, there would no nodes where the current could divide. This is clearly not the case as the current flowing through R1 can divide at node b, with one portion flowing down through R2 and the remainder through R3. On the other hand, if it was strictly parallel, then all of the components would have to exhibit the same voltage and therefore there would be only two connection points in the circuit. This is also not the case as there are three such points: a, b and ground. What is true is that resistors R2 and R3 are in parallel. We know this because both components are attached to the same two nodes, b and ground, and must exhibit the same voltage, Vb. As such, we can find the equivalent resistance of this pair and treat the result as a single resistance, let's call it Rx. In this newly simplified circuit, Rx is in series with R1 and the source, E. We have reduced the original circuit down to a simple series circuit and thus the series analysis rules may be applied. There are many solution paths at this point. For example, we could find the total resistance, Rt, by adding R1 to Rx. Dividing this by E yields the total current flowing out of the source, Itotal. This current must flow through R1 so Ohm's law can be used to find the voltage drop across R1. This same current must be flowing through Rx, so Ohm's law can be used to find the associated voltage (Vb). The currents through R2 and R3 may then be found using Ohm's law for each resistor (e.g., the current through R2 must be Vb/R2). Alternately, these currents may be found by using the current divider rule between R2 and R3 (e.g., the current through R2 must be Itotal ∙ R1/(R1 + R2); remember, the current divider rule uses the ratio of the opposite resistor over the sum). 143 Figure 5.8 A simple series-parallel circuit. | DCElectricalCircuitAnalysis_Page_143_Chunk1065 |
Another solution path would be to apply the voltage divider rule to R1 and Rx in order to derive the two voltage drops (or the rule can be applied to find just one of the drops and the other voltage may be found by subtracting that from the source, an application of KVL). Once the voltages are determined, Ohm's law can be used to find the currents. If powers are needed, they are easily determined once the voltages and currents are found. Example 5.4 A series-parallel circuit is shown in Figure 5.9. Determine Vb. As we are only interested in finding one voltage, the voltage divider rule is a good candidate. The 1 kΩ is in parallel with the 3 kΩ, yielding an equivalent resistance of 750 Ω. From here we apply VDR. V b = E Req R1+Req V b = 6V 750Ω 250Ω+750Ω V b = 4.5V To clarify the current directions and voltage polarities, the circuit is redrawn and appropriately labeled in Figure 5.10. We can crosscheck the answer in a variety of ways. For example, Figure 5.10 indicates, via KVL, that the voltage drop across the 250 Ω resistor plus Vb must equal the source voltage of 6 volts. Consequently, V250 must be 6 volts minus 4.5 volts, or 1.5 volts. Ohm's law then indicates that the current through the 250 Ω must be 1.5 V/250 Ω, or 6 milliamps. Based on the earlier resistive equivalence, the total resistance as seen by the source must be 250 Ω in series with 750 Ω, or 1 kΩ. This indicates the source current must be 6 V/1 kΩ, or 6 milliamps. As the source and the 250 Ω are in series, they must see the same current. This crosscheck shows that they do: happy happy joy joy. We can go further and verify the currents via KCL. For the first parallel resistor we find: I 1k = V b 1 kΩ I 1k = 4.5V 1 kΩ I 1k = 4.5 mA 144 Figure 5.9 Circuit for Example 5.4. Figure 5.10 Circuit for Example 5.4 with polarities and directions included. | DCElectricalCircuitAnalysis_Page_144_Chunk1066 |
And for the second resistor we see: I 3k = V b 1 kΩ I 3k = 4.5V 3 kΩ I 3k = 1.5mA Obviously, these sum to the entering current of 6 mA. Example 5.5 Determine Vb and the source current in the circuit of Figure 5.11. In this circuit the 3 kΩ resistor is in parallel with the series combination of the 2 kΩ and 1 kΩ. This leads to an equivalent resistance of 3 kΩ in parallel with 3 kΩ, or 1.5 kΩ. From here we can find the source current. I s = E RTotal I s = 30 V 1.5k Ω I s = 20 mA This current should split evenly down the two vertical paths as they each present 3 kΩ of resistance (10 mA each achieves 30 volts for Va, which is the source). The voltage divider rule is a good choice for Vb as we know the applied voltage. V b = E Req R1+Req V b = 30 V 1k Ω 1kΩ+2k Ω V b = 10 V For clarity, the circuit is redrawn and relabeled with current directions and voltage polarities in Figure 5.12. As a further exercise, try to verify the answers above using alternate means, as illustrated in Exercise 5.4. 145 Figure 5.11 Circuit for Example 5.5. Figure 5.12 Circuit for Example 5.5 with polarities and directions included. | DCElectricalCircuitAnalysis_Page_145_Chunk1067 |
As the circuit grows, more and more solution paths exist. Consider the circuit of Figure 5.13. In this case, R3 and R4 are in parallel. This parallel combination is in series with R2. Finally, this set of three resistors is in parallel with R1 and E, reducing to a parallel circuit. Consequently, we know that the voltage across R1 must be equal to E. Also, the currents through R1 and R2 must add up to the current exiting the source (KCL). Further, the currents through R3 and R4 must add up to the current flowing through R2 (KCL). Also, the voltages across R3 and R4 must be the same (they are in parallel) and that this voltage plus the voltage across R2 must equal E (KVL). One solution path would be to find the total resistance as seen by the source, R1 || (R2 + (R3 || R4)), and use this to find the total current flowing out of the source. The current divider rule can then be used between R1 and the equivalent resistance (R2 + (R3 || R4)). Ohm's law can be used subsequently to find various voltage drops. Alternately, the voltage divider rule can be used between R2 and the parallel equivalent (R3 || R4) as this combination is driven by E. Knowing the voltages, the currents may be determined. With so many possible solution paths for large circuits, it is often worthwhile to take a moment to map out a strategy rather than just “diving in” and hoping it will all work out. Example 5.6 The circuit of Figure 5.14 is the same as the resistive network presented in Example 5.1 but with the addition of a voltage source. Determine Vb, Vc, Vd, the source current Is, and the current flowing through the 40 Ω resistor, I40. Thanks to Example 5.1, the equivalent resistances of various portions have already been determined, saving some time. To reiterate, the pair of 100 Ω resistors are in parallel for 50 Ω. That's in series with the 30 Ω for 80 Ω. In the middle path, the 200 Ω and 40 Ω are in series, yielding 240 Ω. 240 || 80 = 60 Ω. This is in series with the 10 Ω, yielding a total resistance loading the source of 70 Ω. I s = E RTotal I s = 16V 70Ω I s ≈228.6 mA The three voltages can be quickly found via VDR as we know all of the associated sub-circuit resistances. 146 Figure 5.13 A slightly more complex series-parallel circuit. Figure 5.14 Circuit for Example 5.6. | DCElectricalCircuitAnalysis_Page_146_Chunk1068 |
V b = E Rx Rx+Ry V b = 16V 60 Ω 60Ω+10Ω V b ≈13.71V V c = V b Rx Rx+Ry V c = 13.71V 50Ω 50 Ω+30Ω V c ≈8.57 V V d = V b Rx Rx+Ry V d = 13.71V 200Ω 200Ω+40 Ω V d ≈11.43V Lastly, the current through the 40 Ω resistor can be found by dividing its voltage by its resistance. Its voltage is Vb − Vd. A quicker method is to note that because it is in series with the 240 Ω, their currents are the same. I 40 = V b RSeries I 40 = 13.71V 240Ω I 40 ≈57.1mA Once again the circuit is redrawn in Figure 5.15 to illustrate the current directions and voltage polarities. The usefulness of this will become even more apparent in the next example. Example 5.6 Determine Vac and the voltages across each resistor in the circuit of Figure 5.16. This circuit is perhaps the most complex offered so far. It may not be immediately apparent what the voltage polarities are for the various resistors or the proper current directions. Without these data, it will be impossible to determine Vac. To complicate matters, the voltage of interest is not ground referenced, but instead refers to two nodes located in arbitrary points of the circuit. It is worth recognizing that, by definition, Vac = Va − Vc, 147 Figure 5.15 Circuit for Example 5.6 with directions and polarities. Figure 5.16 Circuit for Example 5.6. | DCElectricalCircuitAnalysis_Page_147_Chunk1069 |
but how are those voltages determined? To assist with this minor quandary, the circuit has been redrawn in Figure 5.17. Note that the current source and associated 1 kΩ resistor have been flipped to the side, essentially trading places with the 2 kΩ and 7 kΩ series combination. This is entirely acceptable as both subgroups share node b and ground, meaning they are in parallel. Of course, the order of parallel elements in a situation like this does not matter. This orientation, having the source at one end of the circuit, tends to make visualization of current flows a little easier. In any case, the current exits the source and then splits at node b, some heading through the 10 kΩ with the rest flowing down the 7 kΩ with 2 kΩ series combo. At node c the current that entered the 10 kΩ splits again, flowing down into the parallel subgroup of the 12 kΩ and 24 kΩ. Knowing the current directions, the voltage polarities are determined directly. Now that the circuit has been redrawn, several solution paths might come to mind. For starters, the voltage drop across the 1 kΩ is trivial as both its current and resistance are known. From there, a current divider can be used to find the current flowing down into the series 7 kΩ plus 2 kΩ combo. Ohm's law could then be used to find their voltages. Following that, KCL can be used to find the current through the 10 kΩ (and thus its voltage), and then Ohm's law can be applied to find Vc by treating the two right-most parallel resistors as a single unit. Another solution path would be to determine the equivalent resistance that loads the current source. This allows the source voltage to be determined, after which a voltage divider may be applied between the 1 kΩ and the remaining five resistors to find Vb. From there, two more voltage dividers can be used to find Va and Vc which will allow determination of the remaining resistor voltages via KVL. There are other possible paths as well but for lack of a better reason, we shall chase the first path that was outlined. Before getting too deep, it would be wise to determine the equivalent resistances of a few subgroups. First, the 12 kΩ || 24 kΩ combo reduces down to 8 kΩ. This is in series with the 10 kΩ, leaving 18 kΩ for the three right-most resistors. In the middle branch, the 2 kΩ + 7 kΩ series combo is equivalent to 9 kΩ. The parallel combination of 9 kΩ || 18 kΩ yields 6 kΩ. Finally, this last bit is in series with the 1 kΩ producing 7 kΩ for the equivalent resistance of all resistors. 148 Figure 5.17 Circuit for Example 5.6 redrawn and with polarities and current directions added. | DCElectricalCircuitAnalysis_Page_148_Chunk1070 |
The source voltage and the voltage across the 1 kΩ resistor are determined via Ohm's law: V source = I source×Requivalent V source = 10 mA×7k Ω V source = 70 V V 1k = I source×R V 1k = 10 mA×1k Ω V 1k = 10 V As the source produces 70 volts and the drop on the 1 kΩ is 10 volts, then by KVL, Vb must be 70 volts − 10 volts, or 60 volts (which will be useful for a crosscheck in a moment). Using a current divider, the 10 mA source splits between the middle and right branches. The equivalent resistances of those branches are 9 kΩ and 18 kΩ, as found earlier. I middle = I source Rright Rright+Rmiddle I middle = 10 mA 18k Ω 18 k Ω+9k Ω I middle ≈6.667mA Via KCL, the right branch current must be 10 mA − 6.667 mA, or approximately 3.333 mA. As a crosscheck, if 6.667 mA passes through a total of 9 kΩ, that produces a voltage drop of 60 volts, exactly as expected from the earlier calculation for node b. At this point, the current through each resistor is known (treating the 12 kΩ || 24 kΩ combo as a unit) and therefore Ohm's law may be used to determine their voltage drops. They are summarized below. V7k ≈ 46.67 V V2k ≈ 13.33 V V10k ≈ 33.33 V V12k = V24k ≈ 26.67 V Finally, to determine Vac we know that Vac = Va − Vc, and by observation Va is the drop across the 2 kΩ while Vc is the drop across the 12 kΩ || 24 kΩ combo. V ac = V a−V c V ac ≈13.33V−26.67V V ac ≈−13.33 V 149 | DCElectricalCircuitAnalysis_Page_149_Chunk1071 |
Do not let the negative sign be bothersome. All it indicates is that node a is at a lower potential than node c, i.e., that node a is 13.33 volts below node c. It is also purely coincidental that Vac and Va have the same magnitude. In the lab, if a voltmeter is connected such that the red (positive) lead is attached to node a and the black (negative or reference) lead is attached to node c, the meter will show a negative value. If the leads are connected in reverse, thus measuring Vca, the voltage will show up as positive (meaning that node c is higher in potential than node a, the reverse way of looking at the situation)16. Any Route Works Any Route Works In the example just completed it was noted that, by definition, Vac = Va − Vc, and this dictated the method employed to find said voltage. It turns out that this is just one of many ways of determining a voltage from one point to another. In general: To find some voltage VXY, start at point X and proceed along any convenient route to Y, subtracting voltage rises (polarities of − to +) and adding voltage drops (polarities of + to −) along the way. In the prior example, an obvious route for Vac would be to go down the 2 kΩ and then up the 12 kΩ. That gives a drop of approximately13.33 volts (positive). Going up the 12 kΩ produces a rise of approximately 26.67 volts (negative). The result is approximately −13.33 volts, as previously calculated. The technique states that any route may be used, no matter how inefficient or wacky it might be. To verify this, one odd route might be to go up the 7 kΩ, down through the 1 kΩ and source, and then up through the 24 kΩ. The summation would be − 46.67 − 10 + 70 − 26.67, or approximately −13.33 volts once again. An easy way to remember whether the potential is added or subtracted is to just look at the polarity sign where the component is entered along your chosen route. R-2R Ladder R-2R Ladder As promised earlier in the chapter, it's time to look at a ladder network. While ladders can use any resistance values desired, certain ratios turn out to be eminently practical. Of particular note is the R-2R ladder, an example of which is shown in Figure 5.18, and which is being driven by a current source. In this configuration, only two different resistor values are used; some initial value and a second value of precisely twice that size. As drawn in Figure 5.18, all horizontal resistors use value R (6 kΩ here) while all of the vertical resistors use 2R (12 kΩ here) with the exception of the very last, or terminating, resistor which uses R. 16 Or, if you prefer this situation posed as a question: When standing normally, is your head above your feet or are your feet below your head? Answer: Yes! 150 | DCElectricalCircuitAnalysis_Page_150_Chunk1072 |
It turns out that this arrangement offers something very unique: a halving of successive node voltages and branch currents. This is extremely useful in digital to analog conversion circuits; the sort of hardware that would turn the digital bits of an MP3 or WAV file into listenable analog signals to feed loudspeakers or headphones. Here is how it works. Starting at the far right end, it should be obvious that the last two 6 kΩ resistors comprise a 50% voltage divider, and thus Ve will half of Vd. The combination of the final three resistors works out to 12 kΩ || (6 kΩ + 6 kΩ), or 6 kΩ again. Thus, the divider from c to d is also 50%. This process repeats as we move to the left, each equivalent working out to R and resulting in a 50% voltage divider. It will not matter what value is chosen for R. As long as 2R is used for the other set of resistors, the result will always be successive 50% voltage dividers. Example 5.7 Determine all lettered node voltages in the circuit of Figure 5.18. Continuing the prior analysis, the entire resistor network presents an equivalent resistance of 12 kΩ. The node a voltage is determined by Ohm's law: V a = I×R V a = 24 mA×12 kΩ V a = 288V The prior analysis indicated that the subsequent node voltages should be, from left to right, 144 volts, 72 volts, 36 volts and 18 volts. A quick check follows. The source of 24 mA flows into the left-most 6 kΩ, producing a drop of 144 volts by Ohm's law. This can be subtracted from Va (288 V) to arrive at Vb, or 144 volts, or half of the prior node voltage, as expected. The current through the left-most vertical resistor must be Vb/12 kΩ, or 12 mA. That's half of the entering current, as expected. KCL indicates that the remaining current, 24 mA − 12 mA, or 12 mA, must be flowing into the second 6 kΩ between nodes b and c. This will produce a voltage drop of 72 volts across that resistor. In other words, Vc must be 72 volts less than Vb (144 volts), or 72 volts. This is half of Vb, as expected. Likewise, the current 151 Figure 5.18 An R-2R ladder network. | DCElectricalCircuitAnalysis_Page_151_Chunk1073 |
flowing down through the second 12 kΩ must be Vc/12 kΩ, or 6 mA, once again half of the preceding current, as expected. The process continues as we move to the right, each subsequent node seeing half the voltage of the prior node and each current getting cut in half as well. Computer Simulation Computer Simulation In order to verify the results of Example 5.7, the R-2R ladder circuit is entered into a simulator, as shown in Figure 5.19. The numbered nodes 1 through 5 in the simulation schematic correspond to the lettered nodes of the original circuit, a through e, respectively. This is one example where the use of virtual instruments would result in a very cluttered display with five multimeters. Consequently, the simulation uses a DC operating point analysis with a single output window, as shown in Figure 5.20. Although the node numbers are not in ascending order, the voltages are a perfect match to the results computed previously. Each successive node sees half the voltage of the prior node. Although not explicitly listed here, it should be obvious that the currents will also be cut in half when progressing from left to right, via a quick application of Ohm's law. 152 Figure 5.19 The circuit of Example 5.6 in a simulator. | DCElectricalCircuitAnalysis_Page_152_Chunk1074 |
Bridges Bridges A bridge is a particular series-parallel configuration consisting of two pairs of series connected elements placed in parallel. An example of a resistive bridge being driven by a voltage source is shown in Figure 5.21. In this circuit, R1 and R2 create one series connection while R2 and R3 create the other. The two pairs are then placed in parallel. While this circuit is shown with a voltage source, it could also be driven by a current source. Other elements might also be added. A typical use for a bridge is the measurement of some environmental quantity. As stated in Chapter 2, there are certain resistive devices that are sensitive to environmental change. Examples include photoresistors, whose resistance is a function of light level; and thermistors, whose resistance is a function of temperature. To understand the operation, the voltages Vb and Vc are established by the voltage dividers comprised of R1 and R2; and R3 and R4, respectively. With ordinary resistors, these voltages would be unchanging and therefore Vbc would also be a fixed value. We shall set it up such that Vb and Vc are the same voltage, and thus Vbc will be zero. Now assume that we replace R1 with a photoresistor. As the light level increases, its resistance decreases. This will cause Vb to rise which will cause Vbc to go positive. If the light level decreases, the resistance of R1 will increase, forcing Vb to drop which will cause Vbc to go negative. The greater the change in light, the larger Vbc will be, and its sign indicates whether the light level has increased or decreased compared to the original set point. By simply placing some manner of voltmeter between points b and c, we can create a display that indicates light levels. 153 Figure 5.20 Simulation results of the circuit of Example 5.7. Figure 5.21 A resistive bridge network. | DCElectricalCircuitAnalysis_Page_153_Chunk1075 |
If we want to flip the sign, we would place the photoresistor in the position of R2 instead of R1. Further, it is possible to increase the sensitivity by using sensors at opposite corners (e.g., R1 and R4), and comparative or differential measurements are possible by using both sides (e.g., R1 with R3). Example 5.8 Determine Vbc in the circuit of Figure 5.22. Perhaps the most straightforward way to do this is to treat each half as a voltage divider and then subtract Vc from Vb. V b = E R2 R2+R1 V b = 12V 2k Ω 2k Ω+1k Ω V b = 8V V c = E R4 R4+R3 V c = 12 V 4k Ω 4k Ω+3k Ω V c ≈6.857V Thus, Vbc is 8 volts minus 6.857 volts, or 1.143 volts. Computer Simulation Computer Simulation In order to verify the results of Example 5.8, the bridge network of Figure 5.22 is entered into a simulator, as depicted in Figure 5.23. 154 Figure 5.22 Circuit for Example 5.8. Figure 5.23 The circuit of Example 5.8 in a simulator. | DCElectricalCircuitAnalysis_Page_154_Chunk1076 |
A DC operating point analysis is run and the results are shown in Figure 5.24. The voltages are precisely as expected. Final Comments Final Comments Series-parallel simplification techniques will not work for all circuits. Complex multi-source circuits and some resistive networks such as delta configurations or five-element bridge configurations require other techniques that will be addressed in later chapters. 5.5 Summary 5.5 Summary In this chapter we have determined how to identify basic series-parallel networks driven by a single effective voltage or current source. A key element here is to identify sub-circuits or subgroups of resistors that are comprised of either series-only or parallel-only configurations within themselves. These groupings can then be reduced to equivalent resistances using series and parallel combination techniques examined in earlier chapters. This process may be repeated until the entire circuit is simplified down to either a single series loop or parallel arrangement of resistors driven by a voltage or current source. Once a circuit has been simplified, series and parallel analysis techniques, and laws such as Ohm's law, KVL, KCL, VDR and CDR, may be employed to determine various voltages and currents in the simplified equivalent. Given these results, the circuit may be expanded back into its original form in stages, reapplying these rules and techniques to determine voltages and currents within the sub-circuits. The process may be iterated until every current and voltage in the original circuit is 155 Figure 5.24 Simulation results for the bridge circuit of Example 5.8. | DCElectricalCircuitAnalysis_Page_155_Chunk1077 |
discovered, if desired. Once these values are determined, power calculations are trivial. There is an infinite variety of series-parallel configurations and consequently no single solution technique will work for all of them. Indeed, the more complex the circuit, the more solution paths there are for that circuit. It is therefore wise to plan some strategy for a solution instead of randomly “diving in” in order to ease the ultimate effort. Two series-parallel configurations of note are the ladder network and the bridge network. A resistive ladder may be used to derive several different currents or voltages from a single source; the R-2R variation being particularly useful in that it divides the source down by powers of two. A basic resistive bridge is comprised of four elements arranged as a pair of series resistors in parallel with another pair of series resistors. Bridges can be used as part of a measurement scheme. For example, one or more of the resistors could be environmentally sensitive, such as a photoresistor, force sensing resistor or thermistor. As the associated variable changes, the resistance of the sensor changes which unbalances the bridge, and the resulting voltage indicates both the magnitude and sign of said environmental change. Review Questions Review Questions 1. In general, describe the process of reducing a series-parallel resistive network down to a single equivalent resistance. 2. Do Ohm's law, KVL and KCL still apply in series-parallel networks? Why? 3. Is there a finite number of variations of series-parallel networks? Why/why not? 4. Describe the primary functional characteristic of an R-2R ladder network. 5. What is meant by the terms “load” and “loading”, as in “R1 loads R2”? 6. How might a bridge network be used to sense changes in temperature? 7. Describe a general procedure to find the voltage between two arbitrary points in a series-parallel circuit. 156 | DCElectricalCircuitAnalysis_Page_156_Chunk1078 |
5.6 Exercises 5.6 Exercises Analysis Analysis 1. In the circuit of Figure 5.1, which individual resistors are strictly in series and which are in parallel? 2. In the circuit of Figure 5.2, which individual resistors are strictly in series and which are in parallel? 3. In the circuit of Figure 5.3, which individual resistors are strictly in series and which are in parallel? 157 Figure 5.1 Figure 5.2 Figure 5.3 | DCElectricalCircuitAnalysis_Page_157_Chunk1079 |
4. In the circuit of Figure 5.4, which individual resistors are strictly in series and which are in parallel? 5. In the circuit of Figure 5.5, which individual resistors are strictly in series and which are in parallel? 6. In the circuit of Figure 5.6, which individual resistors are strictly in series and which are in parallel? 7. In the circuit of Figure 5.7, which individual resistors are strictly in series and which are in parallel? 158 Figure 5.4 Figure 5.5 Figure 5.6 Figure 5.7 | DCElectricalCircuitAnalysis_Page_158_Chunk1080 |
8. In the circuit of Figure 5.8, which individual resistors are strictly in series and which are in parallel? 9. Determine the equivalent resistance of the network shown in Figure 5.1 (i.e., as if an ohmmeter is connected to the two terminals). 10. Determine the equivalent resistance of the network shown in Figure 5.2. 11. Determine the equivalent resistance of the network shown in Figure 5.3. 12. Determine the equivalent resistance of the network shown in Figure 5.4. 13. Determine the equivalent resistance of the network shown in Figure 5.5. 14. Determine the equivalent resistance of the network shown in Figure 5.6. 15. Determine the equivalent resistance of the network shown in Figure 5.7. 16. Determine the equivalent resistance of the network shown in Figure 5.8. 17. For the circuit of Figure 5.9, find voltages Va, Vb and Vab. 18. For the circuit of Figure 5.9, find the current through each resistor. 19. For the circuit of Figure 5.10, find the current through each resistor. 20. For the circuit of Figure 5.10, find voltages Va, Vb and Vab. 159 Figure 5.8 Figure 5.9 Figure 5.10 | DCElectricalCircuitAnalysis_Page_159_Chunk1081 |
21. For the circuit of Figure 5.11, find voltages Va, Vb and Vab. 22. For the circuit of Figure 5.11, find the current through each resistor. 23. For the circuit of Figure 5.12, find the current through each resistor. 24. For the circuit of Figure 5.12, find voltages Va, Vb and Vab. 25. In the circuit of Figure 5.13, find voltages Va, Vb and Vab. 26. In the circuit of Figure 5.13, find the current through each resistor. 27. In the circuit of Figure 5.14, find the current through each resistor. 160 Figure 5.11 Figure 5.12 Figure 5.13 | DCElectricalCircuitAnalysis_Page_160_Chunk1082 |
28. In the circuit of Figure 5.14, find voltages Va, Vb and Vab. 29. For the circuit of Figure 5.15, find voltages Vb, Vc and Vcb. 30. For the circuit of Figure 5.15, find the current through the 470 Ω and 3.6 kΩ resistors along with the source current. 31. For the circuit of Figure 5.16, find the current through the 2 kΩ and 5.1 kΩ resistors along with the source current. 32. For the circuit of Figure 5.16, find voltages Vc, Vb, Vbc and Vab. 161 Figure 5.14 Figure 5.15 Figure 5.16 | DCElectricalCircuitAnalysis_Page_161_Chunk1083 |
33. For the circuit of Figure 5.17, find voltages Va and Vb. 34. For the circuit of Figure 5.17, find the current through the 10 kΩ and 4 kΩ resistors along with the source current. 35. For the circuit of Figure 5.18, find the current through the 30 kΩ and the left-most 36 kΩ resistors along with the source current. 36. For the circuit of Figure 5.18, find voltages Va, Vb and Vab. 37. In the circuit of Figure 5.18, must the currents through the two 36 kΩ resistors be the same? Why/why not? 38. In the circuit of Figure 5.18, must the currents through the two 48 kΩ resistors be the same? Further, must these currents be the same as the currents through the two 36 kΩ resistors Why/why not? 39. For the circuit of Figure 5.19, find voltages Va, Vb and Vab. 40. For the circuit of Figure 5.19, find the current through the 3 kΩ, 11 kΩ, and both 9 kΩ resistors. 162 Figure 5.17 Figure 5.18 Figure 5.19 | DCElectricalCircuitAnalysis_Page_162_Chunk1084 |
41. Given the circuit of Figure 5.20, find the current through the 47 kΩ, 5.1 kΩ, and 3.9 kΩ resistors. 42. Given the circuit of Figure 5.20, find voltages Va, Vb and Vab. 43. For the circuit of Figure 5.21, find voltages Vb, Vc and Vd. 44. For the circuit of Figure 5.21, find the current through the 6.8 kΩ, 8.5 kΩ, and 20 kΩ resistors. 45. The circuit of Figure 5.22 is called an R-2R ladder. Find the current through each of the 60 kΩ resistors. 46. Given the circuit of Figure 5.22, find voltages Vb, Vc, Vd and Ve. What is unique about this configuration? 163 Figure 5.20 Figure 5.21 Figure 5.22 | DCElectricalCircuitAnalysis_Page_163_Chunk1085 |
47. For the circuit of Figure 5.23, find voltages Va, Vb and Vab. 48. For the circuit of Figure 5.23, find the current through each of the resistors. 49. For the circuit of Figure 5.24, find the current through each of the resistors. 50. For the circuit of Figure 5.24, find voltages Va, Vb and Vab. 51. Given the circuit of Figure 5.25, find voltages Va, Vb and Vab. 52. Given the circuit of Figure 5.25, find the current through each of the resistors. 53. For the circuit of Figure 5.26, find the current through the 200 Ω, 400 Ω, and 100 Ω resistors. 164 Figure 5.23 Figure 5.24 Figure 5.25 Figure 5.26 | DCElectricalCircuitAnalysis_Page_164_Chunk1086 |
54. For the circuit of Figure 5.26, find voltages Va, Vb and Vab. 55. Given the circuit of Figure 5.27, find voltages Va, Vb and Vab. 56. Given the circuit of Figure 5.27, find the current through the 5 kΩ, and 6 kΩ resistors. 57. For the circuit of Figure 5.28, find the current through the 9 kΩ and right- most 82 kΩ resistors. 58. For the circuit of Figure 5.28, find voltages Va, Vb and Vab. 59. Must the current through the 82 kΩ resistors be identical in the circuit of 4.28? Why/why not? 60. Must the current through the 25 kΩ resistors be identical in the circuit of 4.28? Must they be the same as the currents through the 82 kΩ resistors? Why/why not? 61. For the circuit of Figure 5.29, find voltages Va, Vb and Vab. 62. For the circuit of Figure 5.29, find the current through the 30 kΩ, 5 kΩ, and 1 kΩ resistors. 165 Figure 5.27 Figure 5.28 Figure 5.29 | DCElectricalCircuitAnalysis_Page_165_Chunk1087 |
63. For the circuit of Figure 5.30, find the current through the 1 kΩ, 2.2 kΩ, and 18 kΩ resistors. 64. For the circuit of Figure 5.30, find voltages Va, Vb and Vc. 65. Given the circuit of Figure 5.31, find voltages Va, Vb and Vab. 66. Given the circuit of Figure 5.31, find the current through the 20 kΩ, 15 kΩ, and 3.3 kΩ resistors. 67. The circuit of Figure 5.32 is referred to as an R-2R ladder network. Find the current through each of the 4 kΩ resistors. 68. For the circuit of Figure 5.32, find voltages Va, Vb and Vab. What is the unique characteristic of this configuration? 166 Figure 5.30 Figure 5.31 Figure 5.32 | DCElectricalCircuitAnalysis_Page_166_Chunk1088 |
Design Design 69. Determine a new value for the 33 Ω resistor in Figure 5.9 such that the source current is 10 mA. 70. Determine a new value for the 20 kΩ resistor in Figure 5.12 such that Vb is 4 volts. 71. Determine a new value for the 470 Ω resistor in Figure 5.15 such that Vb is 6 volts. 72. Determine a new value for the 1 kΩ resistor in Figure 5.14 such that the source current is 20 mA. 73. Determine a new value for the 12 kΩ resistor in Figure 5.16 such that Vbc is 0 volts. 74. Given the circuit of Figure 5.24, determine a new value for the 100 k Ω resistor such that Va is 75 volts. 75. Given the circuit of Figure 5.23, determine a new value for the current source such that Vb is 8 volts. 76. Determine a new value for the 2.2 kΩ resistor in Figure 5.30 such that Vbc is 0 volts. 77. Given the circuit of Figure 5.32, determine a new value for the current source such that Ve is 1 volt. Challenge Challenge 78. Utilizing only 1 kΩ resistors, create a series-parallel combination that achieves 1.25 kΩ of total resistance. 79. Utilizing only 12 kΩ resistors, create a series-parallel combination that achieves 9 kΩ of total resistance. 80. Consider the circuit of Figure 5.22. Alter the values of the two right-most 30 kΩ resistors such that Ve is 1.2 volts. The remaining node voltages should be unchanged from the original circuit. 81. Alter the circuit shown in Figure 5.22 such that another “wrung” is added to the ladder creating a new right-most node f. If the former terminating 30 kΩ resistor (i.e., the vertical one) is reset to 60 kΩ, determine the values of the resistors on the new wrung such that Vf is 0.5 volts. 167 | DCElectricalCircuitAnalysis_Page_167_Chunk1089 |
82. Given the circuit of Figure 5.25, determine a new value for the 2 k Ω resistor such that Vb is 12 volts. 83. Given the circuit of Figure 5.32, determine new values for the resistors such that all of the node voltages are twice the value of the original circuit's node voltages. Simulation Simulation 84. Perform a DC simulation on the result of problem 24 to verify the node voltages. 85. Perform a DC simulation on the result of problem 25 to verify the node voltages. 86. Perform a DC simulation on the result of problem 43 to verify the node voltages. 87. Perform a DC simulation on the result of problem 46 to verify the node voltages. 88. Perform a DC simulation on the result of problem 51 to verify the node voltages. 89. Perform a DC simulation on the result of problem 68 to verify the node voltages. 90. Perform a DC simulation on the proposed solution to problem 69 to verify the new design. 91. Perform a DC simulation of the alteration presented in Challenge problem 80. Does the new design meet all of the requirements? 92. Perform a DC simulation of the alteration requested in Challenge problem 83. Does the new design meet all of the requirements? 168 | DCElectricalCircuitAnalysis_Page_168_Chunk1090 |
Notes Notes ♫♫ ♫♫ 169 | DCElectricalCircuitAnalysis_Page_169_Chunk1091 |
6 6 Analysis Theorems and Techniques Analysis Theorems and Techniques 6.0 Chapter Learning Objectives 6.0 Chapter Learning Objectives After completing this chapter, you should be able to: • Define whether or not a circuit is a linear bilateral network. • Find the voltage source equivalent of a current source and vice versa. • Compute voltages and currents in multi-source networks using superposition. • Simplify networks using Thévenin's and Norton's theorems. • Determine conditions for maximum power transfer and compute the maximum power. • Utilize delta-Y and Y-delta conversions for circuit simplification and analysis. 6.1 Introduction 6.1 Introduction As presented in the previous chapter, series-parallel analysis broadened our ability to analyze circuits beyond simple series and parallel networks, and leveraged the laws and techniques learned in earlier chapters. In this chapter we shall examine a number of theorems and techniques to help us analyze yet more complex circuits and address specialized applications. Specifically, we will address a method of analyzing circuits that contain multiple current and/or voltage sources that are connected in a non-trivial fashion (i.e., not just series voltage sources or parallel current sources). It is called the superposition theorem and can be applied to any circuit or parameter that meets certain requirements, including circuits that have both current sources and voltage sources together. We shall also examine several ways of simplifying circuits or making functional equivalents for them. These are more tools to aid in the process of circuit simplification and analysis. The first item in this category is to make more accurate models of our idealized constant voltage and current sources. Once this is completed, it will be possible to convert from one type of source to another, such as creating a current source that is the functional equivalent of a voltage source. By this we mean that if we swap one for the other in any circuit, the remainder of the circuit will behave identically, producing the same component voltage drops and branch currents. This technique is useful in a number of ways, not the least of which is that it can help reduce more complex circuits to ease analysis. The concept of equivalence can be extended beyond just a single source to an entire network. For this we will explore Thévenin's and Norton's theorems. Using these theorems, entire circuits utilizing dozens of components can be modeled as a single source with an associated resistance. When coupled with the maximum power transfer theorem, these theorems will allow us to determine component values that produce the maximum amount of load power. 170 | DCElectricalCircuitAnalysis_Page_170_Chunk1092 |
Finally, we will examine how to find equivalent circuits for certain resistor arrangements that use three connecting points, in other words, resistor arrangements shaped like the letter Y or like a triangle. These are known as delta and Y configurations. These configurations are difficult to address with basic series- parallel simplification techniques but the conversion equivalences will help solve that issue. 6.2 Source Conversions 6.2 Source Conversions We begin by considering a more realistic model for constant voltage and constant current sources. The ideal voltage source produces a stated potential forever, and without regard to what it is connected to. The ideal current source behaves similarly; it will always produce the same current regardless of its load. These expectations are not realistic. For example, if we were to place a solid bar of copper across the terminals of a voltage source, the bar may exhibit a resistance of mere milliohms, implying an output current of thousands of amps. Similarly, if we were to disconnect a current source from any load, its effective load would be the resistance of the air between its terminals and Ohm's law would dictate an output voltage of perhaps thousands or even millions of volts. Real world sources do not behave like this. Realistic Models for Sources Realistic Models for Sources A highly precise model for any voltage source or current source could be fairly complex but for general purpose work, we can greatly improve our ideal sources by simply adding a resistor to them. This resistance is referred to as the internal resistance of the source. It is important to understand that this is not a resistor, as in an internal component that can be altered, but rather a mathematical addition to the source that better predicts how it will behave. Further, the value of this effective resistance can be deduced in a laboratory via proper measurements. The model for a voltage source adds a resistance in series, as shown in Figure 6.1. This resistance sets an upper limit on the source's current output. Even if the output terminals are shorted, the maximum current will be dictated via Ohm's law to be the source voltage divided by the internal resistance, or E/R. Obviously, this internal resistance will create some voltage divider effect with the attached load. To minimize this effect, the internal resistance should be as small as practicably possible. Thus, The ideal internal resistance of a voltage source is zero ohms (a short). Given a zero ohm internal resistance, this improved model reverts back to the original ideal source. In the case of a lab power supply, the internal resistance typically will be a small fraction of an ohm. 171 Figure 6.1 Practical voltage source model. | DCElectricalCircuitAnalysis_Page_171_Chunk1093 |
For a current source, the improved model adds a resistance in parallel, as shown in Figure 6.2. This resistance sets an upper limit on the source's voltage output. If the output terminals are opened, the maximum voltage will no longer produce a huge voltage. Instead, it is dictated by Ohm's law to be the source current times the internal resistance, or I∙R. This internal resistance will create some current divider effect with the attached load. To minimize this effect, the internal resistance should be as large as practicably possible. Thus, The ideal internal resistance of a current source is infinite ohms. Given an infinite ohm internal resistance (i.e., an open), this improved model reverts back to the original ideal source. From here on, whenever we deal with practical voltage and current sources, we understand that these sources have some associated internal resistance, even if they're not shown explicitly in a schematic diagram. Further, whenever we talk about ideal sources, we simply use a short for the internal resistance of a voltage source and an open for the internal resistance of a current source. Source Equivalences Source Equivalences For any simple voltage source consisting of an ideal voltage source with a series internal resistance, an equivalent current source may be created. Similarly, for any simple current source consisting of an ideal current source with a parallel internal resistance, an equivalent voltage source may be created. By “equivalent”, we mean the load currents must be identical for both circuits given any value of load resistance. (Note that if the currents are the same then the voltages must also be the same due to Ohm’s Law.) Consider the simple voltage source of Figure 6.1. Its equivalent current source would be that shown in Figure 6.2. The opposite is also true. For reasons that will become apparent under the section on Thévenin's theorem following, the internal resistances of these two circuits must be identical if they are to behave identically. Knowing that, it is a straightforward process to find the required value of the other source. As the current/voltage characteristic is linear for these circuits, a plot line can be defined by just two points. The two obvious points to use are the opened and shorted load cases. In other words, if it's equivalent for these two situations, it must work for any load. The shorted load case produces the maximum load current with zero load voltage, while in contrast, the opened load case produces the maximum load voltage with zero load current. Given a voltage source, the maximum current is developed when the load is shorted producing a current of E/R. Under that same load condition, all of the current from the current source version must be flowing through the load. Therefore, the value of the equivalent current source must be the maximum current of E/R. It would be nonsensical to use a current source that was larger or smaller than this value. 172 Figure 6.2 Practical current source model. | DCElectricalCircuitAnalysis_Page_172_Chunk1094 |
To continue, if we look at the open load case, for the voltage source the load current would be zero and the load voltage would be the entire source voltage of E. For the current source, the load would also see no current and its voltage would be the voltage appearing across its internal resistance which is R times the current E/R, or just E. Thus, the two behave identically at the load limits. Similarly, if we start with a current source, an open load produces the maximum load voltage of I∙R. Therefore, the equivalent voltage source must have a value of I∙R. For the current source, a shorted load would produce a load current equal to the source value, or I. The voltage source version would produce a current of E/R, where the value of E was just found to be equal to I∙R, and thus the load current would be I∙R/R, or just I. Once again, the two versions behave identically at the load limits. To summarize the process of source conversion: • The internal resistance will be the same for both versions. • If converting from a voltage source to a current source, the value of the current source will be the maximum current available from the voltage source (shorted load case), and is equal to E/R. • If converting from a current source to a voltage source, the value of the voltage source will be the maximum voltage available from the current source (opened load case), and is equal to I∙R. If a multi-source is being converted (i.e., voltage sources in series or current sources in parallel), first combine the sources to arrive at the simplest source and then do the conversion. Do not convert the sources first and then combine them as you will wind up with series-parallel configurations rather than simple sources. Judicious use of source conversions can sometimes simplify multi-source circuits by allowing converted sources to be combined, resulting in a single source. Example 6.1 Determine the current source equivalent for the source of Figure 6.3. First off, the resistance value is simply copied over, therefore the internal resistance of the current source is 50 Ω. The value of the current source is computed using Ohm's law based on the maximum current produced under the shorted load case. Thus, all 15 volts drops across the 50 Ω resistance. I s = E Rs I s = 15V 50Ω I s = 0.3A 173 Figure 6.3 Source for Example 6.1. | DCElectricalCircuitAnalysis_Page_173_Chunk1095 |
The equivalent current source is shown in Figure 6.4. We know that this will work for the shorted an opened cases, but if any doubt is left as to its universal nature, simply substitute any other resistance value and compare the outcomes of the two circuits. For no particular reason, let's try using a load of 200 Ω and see if the load currents are identical. For the original voltage source we can use Ohm's law: I L = E Rs+RL I L = 15 V 50Ω+200Ω I L = 60 mA For the equivalent current source we can use CDR: I L = I s Rs Rs+RL I L = 0.3 A 50 Ω 50Ω+200Ω I L = 60 mA Try this with any other load resistor. The results should always be identical. Now let's try one going the other way. Example 6.2 Determine the voltage source equivalent for the source of Figure 6.5. Once again, the resistance value is simply copied over, therefore the internal resistance of the voltage source is 22 kΩ. The value of the voltage source is based on the maximum voltage produced under the opened load case and is computed using Ohm's law. For the open case, all 15 milliamps flows through the 22 kΩ resistance. Es = I×Rs Es = 15 mA×22k Ω Es = 330V The equivalent voltage source is shown in Figure 6.6. Again, let's try using another value of load resistor to see if the load current and voltage results are identical between the two sources. This time, we'll match the load to the internal resistance of 22 kΩ. 174 Figure 6.4 Equivalent current source for the source of Figure 6.3. Figure 6.5 Source for Example 6.2. | DCElectricalCircuitAnalysis_Page_174_Chunk1096 |
For a voltage source with matched resistances we wind up with a simple 50% voltage divider, thus the load voltage will be half the source voltage, or 165 volts. The original current source sees the current split in half due to the current divider rule. Thus, the load current should be 7.5 mA. Given this current, the load voltage will be, V L = I×Rs V L = 7.5mA×22 k Ω V L = 165V Once again the results turn out to be identical. Now that we have it is possible to substitute one kind of source for another, it is time to investigate how we might put this to use beyond just giving us a different way of driving a circuit. If applied intelligently, source conversions can be used to simplify and reduce complex circuits, and thus ease computational difficulty. For example, consider the circuit of Figure 6.7. This circuit is unlike any circuit we have seen so far. Although we have analyzed circuits using multiple voltage sources, they have always been in a simple series loop. As such, their voltages may be added together to find a single equivalent voltage source. Such is not the case here. In this circuit the voltage sources are part of a series-parallel network and thus their potentials cannot be merely added. In fact, there are no further simplifications to be performed on this circuit using basic series-parallel techniques. We appear to be stuck. But we're not. This circuit can be simplified into a straight all-parallel network through the use of source conversions. The E1, R1 combo can be converted into one current source while the E2, R2 combo can be converted into a second source. Once these are converted, the resulting circuit will consist of two current sources and three resistors, all in parallel. This is the sort of circuit we solved back in Chapter 4. Example 6.3 Determine Vb for the circuit of Figure 6.8. The first step will be to convert the voltage sources into current sources. We will treat the resistors attached to their positive terminals as their internal resistances. In other words, we have a 15 volt source with 1 kΩ resistance and a 6 volt source with a 4 kΩ resistance. 175 Figure 6.6 Equivalent current source for the source of Figure 6.5. Figure 6.7 A dual source circuit. Figure 6.8 Circuit for Example 6.3. | DCElectricalCircuitAnalysis_Page_175_Chunk1097 |
For the first source, the current will be: I s = E Rs I s = 15 V 1k Ω I s = 15mA And for the second source: I s = E Rs I s = 6V 4 kΩ I s = 1.5mA The equivalent converted circuit is shown in Figure 6.9. Before continuing, it is worth noting that connection nodes a and c no longer exist in this circuit. More on this in a moment. This new circuit consists of a pair of current sources that sum to 16.5 mA and which drive three parallel resistors, 1 kΩ || 4 kΩ || 5 kΩ, or approximately 689.7 Ω. Ohm's law tells us Vb is: V b = I Total×REquivalent V b ≈16.5mA×689.7Ω V b = 11.38V We can now take this voltage and apply it back into the original (unconverted) circuit. With this potential known, it is relatively easy to determine the other currents and voltages using KVL and Ohm's law. Looking at the 1 kΩ resistor, the voltage across it must be 15 V − 11.38 V, or 3.62 V. Therefore, the current through it must be 3.62 mA. Similarly, the voltage across the 4 kΩ resistor must be 11.38 V − 6 V, or 5.38 V, which yields a current of 1.345 mA. Both of these currents are flowing left to right. Then third current, flowing down through the 5 kΩ, is 11.38 V/5 kΩ, or 2.276 mA. KCL states that the current entering node b must equal the currents leaving. The entering current is 3.62 mA. The exiting currents are 1.345 mA and 2.276 mA, or 3.62 mA when rounded to three places (like the entering current). As mentioned, nodes a and c disappeared in the converted circuit in Example 6.3. This brings up an important point. Equivalent circuits are equivalent in terms that the items connected to the equivalent behave the same way as they do with the original circuit. That does not mean that the items within the equivalent see the same current or voltage. We do not expect the voltage across the 1 kΩ or 4 kΩ resistors in the converted version to be the same as those in the original version. 176 Figure 6.9 Equivalent current source version of the circuit depicted in Figure 6.8. | DCElectricalCircuitAnalysis_Page_176_Chunk1098 |
6.3 Superposition Theorem 6.3 Superposition Theorem As useful as the source conversion technique proved to be in Example 6.3, it will not work for all circuits. Thus, more general approaches are needed. One of these methods is superposition. Superposition allows the analysis of multi-source series-parallel circuits. Superposition can only be applied to networks that are linear and bilateral. Further, it cannot be used to find values for non-linear functions, such as power, directly. Fortunately, if the circuit contains nothing but resistors, and ordinary voltage sources and current sources, the circuit will be a linear bilateral network. Also, although power is a square law function (i.e., it is proportional to the square of voltage or current), it can be computed from the resulting voltage or current values so this presents no limits to analysis. The basic idea is to determine the contribution of each source by itself, and then adding the results to get the final answer(s). Let's start with an almost trivial example to illustrate the idea. Consider the circuit of Figure 6.10. Here we have two voltage sources driving a pair of resistors. It is a basic series circuit. The way this was approached in Chapter 3 was to combine the sources and the resistors into a single source and resistor. The circulating current could then be found using Ohm's law. As the two sources are in opposition, the net voltage is 10.5 volts while the total resistance is 500 Ω. This yields a circulating current of 21 mA. Instead of this approach, let's consider the contribution of each source by itself. To do that, we will replace all other sources in the circuit with their ideal internal resistance. From earlier work we have discovered that the ideal internal resistance of a voltage source is a short, or zero ohms. Thus, we wind up with two circuits: one with a 12 volt source that drives 100 Ω, 400 Ω and a short; and a second circuit with a 1.5 volt source that drives pretty much the same thing, but with an opposing current direction. The first circuit produces a clockwise current of 12 V/ 500 Ω, or 24 milliamps. Meanwhile, the second circuit produces a counterclockwise current of 1.5 V / 500 Ω, or 3 mA. As these two currents oppose each other, the resulting current is 24 mA − 3 mA, or 21 mA, the same value originally computed. Granted, this second method is less efficient than the original method but it illustrates the process. More importantly, this process can be applied to a variety of multi-source series-parallel circuits. A key element is to remember the directions of the currents and polarities of the voltages created in the sub-circuits. Without these data, it will be impossible to determine whether the various pieces of the puzzle are added to or subtracted from the total. 177 Figure 6.10 A dual source circuit. | DCElectricalCircuitAnalysis_Page_177_Chunk1099 |
To summarize the superposition technique: • For every voltage or current source in the original circuit, create a new sub- circuit. The sub-circuits will be identical to the original except that all sources other than the one under consideration will be replaced by their ideal internal resistance. This means that all remaining voltage sources will be shorted and all remaining current sources will be opened. • Label the current directions and voltage polarities on each of the new sub- circuits, as generated by the source under consideration. • Solve each of the sub-circuits for the desired voltages and/or currents using standard series-parallel analysis techniques. Make sure to note the voltage polarities and current directions for these items. • Add all of the contributions from each of the sub-circuits to arrive at the final values, being sure to account for current directions and voltage polarities in the process. In order to get a better handle on the superposition technique, let's reexamine the the dual source circuit shown in Figure 6.8 (repeated in Figure 6.11 for ease of reference). We will solve this using superposition. As the circuit has two sources, it will require two sub-circuits. Example 6.4 Determine Vb for the circuit of Figure 6.11 using superposition. As this circuit has two voltage sources, two sub-circuits will be needed. The first sub-circuit will use the 15 volt source. Consequently, the 6 volt source will be replaced with its ideal internal resistance, a short. This new circuit is shown in Figure 6.12. The current directions are as follows: current exits the source and travels through the 1 kΩ producing a voltage drop + to − from left to right. At node b the current splits. The 4 kΩ and 5 kΩ are in parallel so they both see the same voltage. Some of the current flows down through the 5 kΩ resistor producing a voltage drop + to − from top to bottom. The remainder of the current flows to the right, through the 4 kΩ, producing a voltage drop + to − from left to right. Vb can be determined via voltage divider between the 1 kΩ and the parallel combo of the 5 kΩ and 4 kΩ. 5 kΩ in parallel with 4 kΩ is approximately 2.222 kΩ. 178 Figure 6.11 Circuit for Example 6.4. Figure 6.12 First sub-circuit for the circuit of Figure 6.11. | DCElectricalCircuitAnalysis_Page_178_Chunk1100 |
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