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AC Electrical Circuit Analysis AC Electrical Circuit Analysis A Practical Approach A Practical Approach James M. Fiore James M. Fiore
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AC Electrical Circuit Analysis AC Electrical Circuit Analysis A Practical Approach A Practical Approach by James M. Fiore Version 1.1.10, 30 March 2023 3
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This AC Electrical Circuit Analysis, by James M. Fiore is copyrighted under the terms of a Creative Commons license: This work is freely redistributable for non-commercial use, share-alike with attribution Device data sheets and other product information are copyright by their respective owners and have been obtained through publicly accessible manufacturer's web sites. Published by James M. Fiore via dissidents ISBN13: 979-8605022282 For more information or feedback, contact: James Fiore, Professor Electrical Engineering Technology Mohawk Valley Community College 1101 Sherman Drive Utica, NY 13501 jfiore@mvcc.edu oer@jimfiore.org For the latest revisions, related titles, and links to low cost print versions, go to: www.mvcc.edu/jfiore or my mirror sites www.dissidents.com and www.jimfiore.org YouTube Channel: Electronics with Professor Fiore Cover art, Chapman's Contribution Redux, by the author 4
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Introduction Introduction Welcome to the AC Electrical Circuit Analysis, an open educational resource (OER). The goal of this text is to introduce the theory and practical application of analysis of AC electrical circuits. It assumes familiarity with DC circuit analysis. If you have not studied DC circuit analysis, it is strongly recommended that you read the companion OER text, DC Electrical Circuit Analysis before continuing. Both texts are offered free of charge under a Creative Commons non-commercial, share-alike with attribution license. For your convenience, along with the free pdf and odt files, print copies are available at a very modest charge. Check my web sites for links. This text is based on the earlier Workbook for AC Electrical Circuits, which it replaces. The original expository text has been greatly expanded and includes many examples along with computer simulations. For the convenience of those who used the Workbook, many of the problem sets are the same, with some re-ordering depending on the chapter. If you are already familiar with DC Electrical Circuit Analysis, the format of this title is similar. This text picks up where the DC text leaves off; beginning with AC concepts such as sinusoidal waveforms, basic Fourier decomposition of complex waveforms, complex numbers and the like. Also, reactance and impedance are introduced along with phasor diagrams. Chapters on series, parallel and series-parallel RLC circuits commence. Following these, network theorems along with nodal and mesh analysis are discussed for the AC case. The text completes with chapters on AC power, resonance, and introductions to polyphase systems and magnetic circuits. Each chapter begins with a set of learning objectives and concludes with practice exercises that are generally divided into four major types: analysis, design, challenge and simulation. Many SPICE-based circuit simulators are available, both free and commercial, that can be used with this text. The answers to most odd-numbered exercises can be found in the Appendix. A table of standard resistor sizes is also in the Appendix, which is useful for real-world design problems. If you have any questions regarding this workbook, or are interested in contributing to the project, do not hesitate to contact me. This text is part of a series of OER titles in the areas of electricity, electronics, audio and computer programming. It includes three other textbooks covering semiconductor devices, operational amplifiers, and embedded programming using the C language with the Arduino platform. There is a text covering DC electrical circuits similar to this one, and also seven laboratory manuals; one for each of the five texts plus individual titles covering computer programming using the Python language, and the science of sound. The most recent versions of all of my OER texts and manuals may be found at my MVCC web site as well as my mirror site: www.dissidents.com This text was created using several free and open software applications including Open Office, Dia, SciDAVis, and XnView. Special thanks to the following individuals for their efforts in reviewing and proofreading the DC and AC Electrical Circuit Analysis texts: Glenn Ballard, John Markham, João Nuno Carvalho, Mark Steffka and Jim Noon. 5
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For the Have-nots and the Has-beens “Well, the people, I would say. There is no patent. Could you patent the sun?” — Jonas Salk, inventor of the polio vaccine, when asked who owns the patent to it. 6
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Table of Contents Table of Contents Chapter 1: Fundamentals . . . . . . . 10 1.0 Chapter Objectives . . . . . . . . 10 1.1 Introduction . . . . . . . . . 10 1.2 Sinusoidal Waveforms . . . . . . . 11 1.3 Basic Fourier Analysis . . . . . . . . 25 1.4 Complex Numbers . . . . . . . . 29 1.5 Reactance and Impedance . . . . . . . 31 1.6 Phasor Diagrams . . . . . . . . 37 Summary . . . . . . . . . 38 Exercises . . . . . . . . . 39 Chapter 2: Series RLC Circuits . . . . . . 44 2.0 Chapter Objectives . . . . . . . . 44 2.1 Introduction . . . . . . . . . 44 2.2 The Series Connection . . . . . . . 44 2.3 Series Impedance . . . . . . . . 45 2.4 Series Circuit Analysis. . . . . . . . 48 2.5 Concerning Practical Inductors . . . . . . 61 Summary . . . . . . . . . 63 Exercises . . . . . . . . . 64 Chapter 3: Parallel RLC Circuits . . . . . 80 3.0 Chapter Objectives . . . . . . . . 80 3.1 Introduction . . . . . . . . . 80 3.2 The Parallel Connection . . . . . . 81 3.3 Parallel Impedance . . . . . . . . 81 3.4 Parallel Circuit Analysis . . . . . . . 85 Summary . . . . . . . . . 100 Exercises . . . . . . . . . 101 Chapter 4: Series-Parallel RLC Circuits . . . . 110 4.0 Chapter Objectives . . . . . . . . 110 4.1 Introduction . . . . . . . . . 110 4.3 The Series-Parallel Connection . . . . . . 110 4.4 Series-Parallel Impedance . . . . . . . 111 4.5 Series-Parallel Circuit Analysis . . . . . . . 114 Summary . . . . . . . . . 132 Exercises . . . . . . . . . 133 7
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Chapter 5: Analysis Theorems and Techniques . . . . 144 5.0 Chapter Objectives . . . . . . . . 144 5.1 Introduction . . . . . . . . . 144 5.2 Source Conversions . . . . . . . . 145 5.3 Superposition Theorem . . . . . . . 153 5.4 Thévenin's and Norton's Theorems . . . . . . 160 5.5 Maximum Power Transfer Theorem . . . . . . 170 5.6 Delta-Y (Pi-T) Conversions . . . . . . . 179 Summary . . . . . . . . . 185 Exercises . . . . . . . . . 187 Chapter 6: Nodal and Mesh Analysis . . . . . 204 6.0 Chapter Objectives . . . . . . . . 204 6.1 Introduction . . . . . . . . . 204 6.2 Nodal Analysis . . . . . . . . 205 6.3 Mesh Analysis . . . . . . . . 222 6.4 Dependent Sources . . . . . . . . 234 Summary . . . . . . . . . 240 Exercises . . . . . . . . . 242 Chapter 7: AC Power . . . . . . . 260 7.0 Chapter Objectives . . . . . . . . 260 7.1 Introduction . . . . . . . . . 260 7.2 Power Waveforms . . . . . . . . 261 7.3 Power Triangle . . . . . . . . 273 7.4 Power Systems . . . . . . . . 280 Summary . . . . . . . . . 286 Exercises . . . . . . . . . 287 Chapter 8: Resonance . . . . . . . 294 8.0 Chapter Objectives . . . . . . . . 294 8.1 Introduction . . . . . . . . . 294 8.2 Series Resonance . . . . . . . . 295 8.3 Parallel Resonance . . . . . . . . 310 Summary . . . . . . . . . 329 Exercises . . . . . . . . . 331 8
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Chapter 9: Polyphase Power . . . . . . 338 9.0 Chapter Objectives . . . . . . . . 338 9.1 Introduction . . . . . . . . . 338 9.2 Polyphase Definition . . . . . . . . 339 9.3 Three-Phase Connections . . . . . . . 341 9.4 Power Factor Correction . . . . . . . 354 Summary . . . . . . . . . 364 Exercises . . . . . . . . . 365 Chapter 10: Decibels and Bode Plots . . . . . 372 10.0 Chapter Objectives . . . . . . . . 372 10.1 Introduction . . . . . . . . 372 10.2 The Decibel . . . . . . . . . 372 10.3 Bode Plots . . . . . . . . . 383 10.4 Combining the Elements - Multi-Stage Effects . . . . . 393 Summary . . . . . . . . . 395 Exercises . . . . . . . . . 396 Appendices A: Standard Component Sizes . . . . . . . 399 B: Methods of Solution of Linear Simultaneous Equations . . . 400 C: Equation Proofs . . . . . . . . 405 D: Answers to Selected Odd-Numbered Problems . . . . 408 E: Questions for Selected Odd Answers . . . . . 422 9
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1 1 Fundamentals Fundamentals 1.0 Chapter Learning Objectives 1.0 Chapter Learning Objectives After completing this chapter, you should be able to: • Describe sinusoidal waveforms in mathematical terms. • Express complex waveforms using basic Fourier analysis. • Perform mathematical operations using complex numbers. • Draw phasor diagrams for complex numbers. • Compute and plot capacitive and inductive reactance as a function of frequency. • Compute and plot complex impedance as a function of frequency. • Describe the relationships between resistance, reactance, impedance, conductance, susceptance and admittance 1.1 Introduction 1.1 Introduction In this chapter we begin our study of alternating current, or AC, electrical circuit analysis. As such, this chapter is filled with foundational material that will be used in the remainder of this text. It is critical that the concepts presented here be fully understood in order to achieve success in later chapters. We start with the mathematical description of the most simple AC waveform, the sine wave. This includes parameters such as amplitude, frequency, period, phase and DC offset. From there we discover how to describe other waveforms in terms of combinations of sine waves, and also how to determine the effective, or DC equivalent, value of a sine wave. AC analysis is practically impossible to perform without the use of complex numbers, and discussion follows with their description and proper manipulation. Finally, we introduce the concepts of reactance and impedance, and how they relate to resistance. This includes examination using both frequency domain plots and phasor diagrams. Phasor diagrams are vector plots and can be used to show the relationships between various voltages in a circuit, as well as between currents or resistive/reactive values. Many of the topics in this text will echo your studies in DC circuit analysis, such as Ohm's law, Kirchhoff's voltage and current laws, series-parallel analysis, nodal analysis, and the like. Thus many concepts will be familiar. The major practical difference is that all quantities in DC systems are scalars, that is, they have only magnitude. In contrast, AC quantities are vectors, and thus have both magnitude and direction (or more properly, phase). Consequent;y, mathematical operations such as addition or multiplication have to follow vector rules instead of scalar rules. Forgetting to do so is a common error for students new to the subject and one that will bring no end of grief. These rules will be reviewed in the section covering complex numbers. Make sure that you have this material mastered before proceeding. 10
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1.2 Sinusoidal Waveforms 1.2 Sinusoidal Waveforms AC, or alternating current, is so named because the current alternates or flips back and forth between two polarities. In other words, the current (and consequently the voltage) is a function of time. This is fundamentally different from direct current that is fixed in polarity and generally constant over time. A laboratory DC voltage source, for example, ideally maintains a set voltage across its terminals and does not vary over time. In contrast, as an AC waveform swings back and forth through time, its shape can exhibit wide variations ranging from the simple, regular paths of laboratory standards such as sine waves, triangle waves and square waves, to the far more complex and undulating waveforms produced by musical instruments and the human voice. The sine wave is the simplest wave that may be created. It represents the motion of a simple vector rotating at a constant speed, such as the vertical displacement of the second hand of a clock. An example is shown in Figure 1.1. The horizontal axis plots time. It increases as we move from left to right (i.e., if point A is to the right of point B, then A occurs later in time than does B). The vertical axis is represented here in general as a percentage of maximum but would ordinarily be a measurement of voltage, current, sound pressure, or the like. Note the smooth variation that starts at zero, rises to a positive peak one quarter way through, falls back to zero when halfway through, continues to a negative peak three quarters through, and then rises again to where it started. This process then repeats. Each repeat is referred to as a cycle. In Figure 1.1, one complete cycle is shown. 11 Figure 1.1 A sine wave. Sine Wave Percentage of Peak -100 -80 -60 -40 -20 0 20 40 60 80 100 Time (cycles) 0 0.25 0.5 0.75 1
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Sine waves exhibit quarter wave symmetry. That is, each quarter (in time) of the wave is identical to any other if you simply flip it around the horizontal axis and/or rotate it upright about its peak. The time it takes to complete one cycle is called the period and is denoted with the symbol T (for Time). The reciprocal of the period is the frequency, f. f = 1 T (1.1) The frequency indicates how many cycles exist in one second. To honor one of the 19th century researchers in the field, instead of calling the unit “cycles per second”, we use Hertz, named after Heinrich Hertz and abbreviated Hz. In Figure 1.2 three sine waves are shown with different frequencies; the initial wave (green), a wave at twice the frequency (blue), and a third at half the frequency or twice the period (red). The amplitude (vertical) of the wave can be expressed as a peak quantity, which is the change from the center zero line up to the most positive value. Amplitude may also be expressed as peak-to-peak; the distance from the most negative to the most positive. For a sine wave this will always be twice the peak value, although that might not be the case for other waves which may be asymmetrical. A series of three sine waves with differing amplitudes are shown in Figure 1.3. Along side the initial (green) are double amplitude (blue) and half amplitude (red) versions. 12 Variation in Frequency Percentage of Peak -100 -80 -60 -40 -20 0 20 40 60 80 100 Time 0 0.2 0.4 0.6 0.8 1 Double period Initial Double frequency Figure 1.2 Sine wave frequency variation.
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Combining these parameters, consider the voltage waveform shown in Figure 1.4. Here we see two cycles of an AC voltage waveform. 13 Variation in Amplitude Percentage of Peak -200 -150 -100 -50 0 50 100 150 200 Time 0 0.2 0.4 0.6 0.8 1 One half amplitude Initial Double amplitude Figure 1.3 Sine wave amplitude variation. Figure 1.4 Basic sine wave example. Volts -5 -4 -3 -2 -1 0 1 2 3 4 5 Time (seconds) 0 0.1 0.2 0.3 0.4
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The peak value is 4 volts and the peak-to-peak value is 8 volts (typically abbreviated as “8 V pp”). The period of one cycle is 0.2 seconds, or T = 200 milliseconds. Further, the frequency, f = 1/200 milliseconds, or 5 Hz (5 cycles in one second). AC waveforms may also be combined with a DC offset. Adding a positive DC level shifts the wave up vertically, while a negative DC level shifts the wave down vertically. This does not alter the frequency or AC portion of the amplitude (although the absolute peaks would shift by the DC value). Figure 1.5 shows the effect of various DC offsets. Above the initial wave (green) is an otherwise identical wave with a positive DC offset equal to 20% of the original peak value (blue). Below the original is a third wave (red) that exhibits a negative DC offset equal to half of the peak value of the original. Further, it is possible for a sine wave to be shifted in time compared to some other sine wave or reference. While it is possible to indicate this shift as an absolute time, it is more common to do so as a phase shift, that is, the time expressed as a portion of the period in degrees. For example, if one sine is ahead of another by one quarter of the period, it is said to be leading by 90° (i.e., ¼ of 360°). If it is behind by ½ of the period, it is said to be lagging by 180° (i.e., later in time by ½ cycle). Another way of stating this is that leading waveforms start earlier in time and thus are drawn to the left of the reference, while lagging waveforms start later in time and are drawn to the right. 14 Effect of DC Offset Percentage of Peak -150 -100 -50 0 50 100 150 Time 0 0.2 0.4 0.6 0.8 1 Negative DC offset, 50% Initial Positive DC offset, 20% Figure 1.5 Sine wave DC offset variation.
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Figure 1.6 illustrates the effect of phase shift. Note that in this plot, t = 0 has been moved to the center of the horizontal axis. The middle curve is the initial, or reference, wave (green). To the left (red) is a wave leading the initial wave by one- eighth cycle, or 45°. To the right (blue), is a lagging wave of half as much, or −22.5°. Combining the foregoing elements allows us to develop a general format for a sine wave (voltage shown): v(t)=V DC+V P sin(2π f t +θ) (1.2) Where v(t) is the voltage at some time t, VDC is the DC offset, if any, VP is the peak value, f is the frequency, θ is the phase shift (+ if leading and drawn to the left, − if lagging and drawn to the right). For a quick and practical example, the waveform shown in Figure 1.4 has an amplitude of 4 volts peak, a frequency of 5 Hz, and no DC offset or phase shift. Thus, its expression is v(t)=4sin(2 π5t) 15 Effect of Phase Shift Percentage of Peak -100 -80 -60 -40 -20 0 20 40 60 80 100 Time -1 -0.5 0 0.5 1 Leading by 45 degrees Initial Lagging by 22.5 degrees Figure 1.6 Sine wave phase variation.
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To compute a phase shift, first determine the time differential between the waveform and the reference, which we'll call Δt. The reference may be a fixed point in time (e.g., t = 0) or another waveform. Generally, the easy way to do this is to measure the difference at the zero-crossings, assuming there is no DC offset. If there is an offset, make the measurement where the zero crossing has been shifted to (i.e., at the DC offset level). Once the difference is found, divide it by the period to represent the shift as a fraction of a period. As one cycle represents one rotation of the vector, or 360 degrees, simply multiply the fraction by 360 degrees to find the phase shift in degrees. Expressed as a formula: θ = 360° Δt T (1.3) Remember, if the wave is shifted to the left then it is leading and positive, while a shift to the right is lagging or delayed in time, and thus negative. Example 1.1 Write the expression for the waveform shown in Figure 1.7. This waveform superficially may look like the one in Figure 1.4 but don't let this fool you. First of all, the time scale is different. For this waveform, one cycle completes in 10 milliseconds. Therefore, the frequency is 16 Figure 1.7 Waveform for Example 1.1. Amps -3 -2 -1 0 1 2 3 4 5 Time (milliseconds) 0 5 10 15 20
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f = 1 T f = 1 10 ms f = 100Hz The second issue is the DC offset. Note that the positive peak occurs at 4 amps while the negative peak occurs at −2 amps. This indicates a peak-to- peak value of 6 amps. Without an offset, the positive peak would be at 3 amps, therefore there is a +1 amp DC offset. The vertical center of the waveform is shifted up from 0 amps to +1 amp. This point is at t = 0, therefore, there is no phase shift. The resulting expression is: i(t)=1+3sin(2 π100t) Example 1.2 Write the expression for the waveform shown in Figure 1.8. First off, the positive peak is 2 volts and the peak-to-peak value is 4 volts. Therefore there is no DC offset. The vertical center of the wave does not start at t = 0, thus there must be a phase shift. The value at t = 0 is 1.2 volts. The wave hits this same amplitude at t = 2 milliseconds and begins to repeat another cycle. Consequently the period must be 2 milliseconds. The frequency is the reciprocal of this value, and thus f = 500 Hz. 17 Figure 1.8 Waveform for Example 1.2. Volts -3 -2 -1 0 1 2 3 Time (milliseconds) 0 1 2 3 4
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The waveform is shifted to the left which indicates a positive or leading phase shift. If we examine the second cycle, we see that it hits zero volts at 1.8 milliseconds. Therefore the shift is 0.2 milliseconds. Expressed in degrees this is: θ = 360° Δt T θ = 360° 0.2ms 2 ms θ = 36° The final expression is: v(t)=2sin(2π500 t+36 °) Example 1.3 Draw the waveform corresponding to the following expression. v(t)=−3+5sin(2π40000 t−72°) First, note that the −3 volt offset pushes the positive peak down from 5 volts to 2 volts, and the negative peak down from −5 volts to −8 volts. The frequency of 40 kHz dictates a period of: T = 1 f T= 1 40 kHz T= 25μs The phase shift of −72° represents 72/360, or 0.2 cycles. This corresponds to a time delay (shifted right because it's negative) of 0.2 times 25 μs, or 5 μs. Initially, it is often best to construct the plot via a series of discrete steps rather than trying to draw the entire thing in one go. First, draw a sine wave with a 5 volt peak amplitude and a period of 25 μs. Now, push the waveform down 3 volts so that the positive peak is only 2 volts and the negative peak is down at −8 volts. Finally, push the newly shifted waveform to the right by 5 μs. The result is shown in Figure 1.9. 18
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Laboratory Measurements Laboratory Measurements In the laboratory, a function generator is used to generate sines and other waveshapes. These devices will allow precise control over both the amplitude and frequency of the wave along with adding a DC offset, if desired. An example is shown in Figure 1.10. The corresponding measurement tool is the oscilloscope, or just scope, for short. The oscilloscope is perhaps the most useful and versatile measurement device in the laboratory. Typically, they feature either two or four input channels, although more are possible. Each input channel has its own sensitivity adjustment and all channels share a common time reference. The display draws waveforms in the same manner as those seen in Figures 1.1 – 1.9. Also, they can plot one voltage versus another (X – Y mode). Modern oscilloscopes have additional features such as the automatic measurement of frequency, amplitude, phase shift, etc., cursor-based measurements, and the ability to save display images as graphics files. An example of a four channel digital oscilloscope is shown in Figure 1.11. 19 Volts -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 Time (microseconds) 0 5 10 15 20 25 30 35 40 45 50 Figure 1.9 Waveform for Example 1.3. Figure 1.10 Laboratory signal generator.
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Schematic Symbols Schematic Symbols As far as schematics are concerned, the symbols for AC voltage and current sources are shown in Figure 1.12. The polarity and direction markings are not absolute; after all, these are AC sources whose polarity and directions flip back and forth. The markers are instead used to establish a timing reference, especially in circuits employing multiple sources. It is worth remembering that negating a source is the same as flipping its polarity. This was true for DC sources and remains true for AC sources. This is illustrated in Figure 1.13. Sometimes flipping or negating source will make analysis a little more obvious or easier to visualize. 20 Figure 1.11 A digital oscilloscope. Figure 1.12 Schematic symbols for AC voltage source (left) and current source (right). Figure 1.13 Polarity/sign equivalence.
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Example 1.4 Assume an oscilloscope displays two waves as depicted in Figure 1.14. Determine the phase shift of the smaller 20 volt peak (blue) waveform relative to the larger 25 volt peak (red) waveform. First, note that neither wave exhibits a DC offset. If one or both them had an offset, the wave(s) would have to be shifted vertically so that their normal zero-crossing points would be at the same level. Measuring either wave, the period is found to be 1 millisecond. The time shift most easily can be found at any of the zero-crossings (there are four locations to choose from). The delay is one small deviation, or 0.1 milliseconds, with the smaller wave delayed in time, or lagging the larger wave. This indicates a negative phase shift. θ = 360° Δt T θ = 360° −0.1 ms 1ms θ =−36 ° 21 Figure 1.14 Waveforms for Example 1.4. Volts -25 -20 -15 -10 -5 0 5 10 15 20 25 Time (milliseconds) 0 0.5 1 1.5 2
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Sines and Cosines Sines and Cosines There are a handful of specific phase shifts that are worth a closer look. If a sine wave is inverted, that is, flipped upside down, it is indistinguishable from a sine wave that has been shifted either +180 or −180 degrees. In other words, such a wave can be written three different ways: −sin(2πft), sin(2πft − 180°), or sin(2πft + 180°). Further, if a sine wave is shifted by +90 degrees (i.e., leading and to the left), it may also be referred to as a cosine wave. Thus sin(2πft + 90°) = cos(2πft). Finally, if a sine wave is shifted by −90 degrees (i.e., lagging and to the right), it may be referred to as a negative or inverted cosine wave. Thus sin(2πft − 90°) = −cos(2πft). The relationships of these four waves are illustrated in Figure 1.15. It is also worth noting that the cosine wave represents the first derivative, or slope, of the sine wave. As you may recall from other studies, the slope or “steepness” of a line is the ratio of the vertical change to the horizontal change, sometimes called “the rise over the run”. For a voltage, it would be the change in voltage over the change in time, or ΔV/Δt. For a smooth, continuously changing curve like a sine wave, the slope at a given point is defined properly as the first derivative, or dv/dt in this case. To verify that this is true visually, note that the steepest part of the sine wave (green) is where it crosses zero amplitude. As it crosses zero while moving positive (at t = 0 or t = 1 in Figure 1.15), the cosine (blue) is at its positive peak. As the sine cross zero while moving negative (at t = 0.5), the cosine is at its negative peak. Further, the sine wave is flat with a slope of zero at its positive and negative peaks (at at t = 0.25 and t = 0.75, respectively), and at those times the cosine's 22 Sines and Cosines Percentage of Peak -100 -80 -60 -40 -20 0 20 40 60 80 100 Time 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 -Sine Sine Cosine -Cosine Figure 1.15 Time relationships between sines and cosines.
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amplitude is also zero. It is also true that the sine wave is the slope of the negative cosine wave, the negative cosine is the slope of the negative sine, and the negative sine is the slope of the cosine. Moving in the reverse direction, we can say that the anti-derivative (indefinite integral) of a cosine wave is a sine wave, the integral of a sine wave is a negative cosine wave, and so forth. These relationships will prove most useful when we turn our attention to the response of capacitors and inductors in AC circuits. RMS – Root Mean Square Measurement RMS – Root Mean Square Measurement Along with peak and peak-to-peak, amplitude may be given as an RMS (Root Mean Square) value. In fact, if peak or peak-to-peak is not specified, the measurement is assumed to be RMS. RMS is a special calculation used for finding equivalent DC power (very common, for example, with audio power amplifiers). In other words, if we are interested in finding the power in a resistor, the calculation must be performed using RMS values for voltage or current, not peak or peak-to-peak values. Failure to do so will result in erroneous powers. This is true regardless of the waveshape; be it a sine wave, triangle wave or the complex waves of music signals. If a voltage is specified as RMS, it can be treated for power calculations just like an equivalently sized DC voltage. For example, a 1 volt RMS sine will produce the same power dissipation and heating in a given resistor as will 1 volt DC. For this reason, RMS is sometimes referred to as the effective value (i.e., effective DC value). The name root-mean-square describes the process of determining the effective value. First, recall that power is proportional to the square of the voltage or current. Thus, our first step will be to square the input waveform. Of course, the waveform is a function of time and its square will yield some new shape. At this point, we need to find the average value of this new shape. The reason for this is simple, yet not necessarily obvious. Electrical and electronic components have mass, and thus do not heat or cool instantly. They exhibit a thermal time constant. Therefore, they respond to the average input over time. While we could compute some manner of “instantaneous peak power” at some specific instant in time, it does not represent the equivalent DC power. Once we have obtained the mean value of this squared waveform, the corresponding DC value is just the square root of the mean. The result is a fractional value between zero and one that is used as a scaling factor to turn a peak value into an RMS value. The value will be unique to the specific waveshape. That is, all sines (regardless of phase) have the same factor, all regular triangle waves have the same factor, and so on. As we mostly concern ourselves with sines, let's take a closer look at determining the RMS factor for them. We begin with the basic expression for a sine wave without DC offset or phase shift, and with an amplitude of one: v(t)=sin(2π f t) 23
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The first step is to square this waveform. A useful trigonometric identity is (sin x) 2=1 2 −1 2 cos 2 x Applying this to our wave yields: v(t) 2= 1 2 −1 2 cos(2π2 f t) This expression describes an inverted cosine wave at twice the original frequency and half of the original amplitude, riding on a DC offset equal to its peak value. In other words, the negative peak of the cosine is at zero and the positive peak is at 1. The next step is to find the average or mean value of this intermediate result. The mean is equal to the offset of 0.5. This can be visualized as the area above the offset perfectly filling the “dip” below the offset. The final step is to take the square root of the mean. The square root of 0.5 is equal to one over the square root of two, or approximately 0.707. Therefore the RMS value is 0.707 times the peak. Alternately, you could divide the peak by square root of two, or approximately 1.414. This process is shown graphically in Figure 1.16. In summation, for sine waves, RMS is always the peak value times 0.707. We could also say the RMS value of any sine wave is its peak divided by approximately 1.414. Again, these ratios would not necessarily be true of non-sine waves. Details regarding other common shapes can be found in Appendix C. Finally, the ratio of the peak value to the RMS value is called the crest ratio. This is a fixed value for sine waves (again, about 1.414), but can be over 10:1 for some kinds of audio signals. 24 Figure 1.16 Process to find RMS factor for sines. Fraction of Peak -1.00 -0.75 -0.50 -0.25 0.00 0.25 0.50 0.75 1.00 Time 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 Original sine Sine squared Mean
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Wavelength Wavelength Another item of interest is the speed of propagation of the wave. This varies widely. In the case of light in a vacuum (or to a close approximation, an electrical current in a wire), the velocity is about 3E8 meters per second (i.e., 300,000 km/s) or about 186,000 miles per second. Given a velocity and a period, we can imagine how far apart the peaks of the wave are. This distance is called the wavelength and is denoted by the Greek letter lambda λ. Wavelength is equal to the velocity divided by the frequency, λ = v/f. Thus, for a loudspeaker producing a 100 Hz sine, as the velocity of sound in air is 344 m/s, then λ = 344 m/s / 100 Hz, or 3.44 meters (a little over 11 feet). Notice that the higher the frequency, the shorter the wavelength. Also, note that the faster the velocity, the longer the wavelength. Wavelength calculations are of particular importance in the fields of telecommunications and acoustics. 1.3 Basic Fourier Analysis 1.3 Basic Fourier Analysis The Fourier theorem, named after the French mathematician Jean-Baptiste Joseph Fourier, states that any repetitive waveform can be represented as a collection of sine and cosine waves of the proper amplitude and frequency. Alternately, it may be represented as a series of sine waves each with the proper amplitude, frequency and phase. This includes complex signals such as the human voice and musical instruments. Consequently, if a system is linear, the response of a system to a complex wave may be understood in terms of its response to individual sine waves, via superposition. In this collection of waves, each component is known as a partial with the lowest frequency component known as the fundamental. All other partials are grouped together and referred to as overtones. “Regular” waveforms such as square waves and triangle waves feature a harmonic overtone sequence meaning that these overtones are integer multiples of the fundamental. As a shortcut, they are often referred to as just harmonics. It might be hard to visualize initially, but like all waves, waves in the shape of a square or triangle are made up of a series of sines. The equation for a square wave is: ) 2 )1 2 sin(( 1 2 1 ) ( 1 ft n n t v n (1.4) This says that a square wave of frequency f is made up of an infinite series of sines at odd integer multiples of f, with an inverse amplitude characteristic. For example, a 100 Hz square consists of a 100 Hz sine, plus a 300 Hz sine at 1/3 amplitude, plus a 500 Hz sine at 1/5 amplitude, plus a 700 Hz sine at 1/7 amplitude, and so on. 25
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A triangle wave is similar: ) 2 )1 2 cos(( )1 2 ( 1 ) ( 1 2 ft n n t v n (1.5) Thus a triangle wave of frequency f is made up of an infinite series of cosines (sines with a 90 degree or one quarter cycle phase shift) at odd integer multiples of f, with an inverse square amplitude characteristic. For example, a 100 Hz triangle consists of a 100 Hz cosine, plus a 300 Hz cosine at 1/9 amplitude, plus a 500 Hz cosine at 1/25 amplitude, plus a 700 Hz cosine at 1/49 amplitude, and so on. A series of graphs showing the construction of a square wave and a triangle wave follow. The square wave sequence begins with the fundamental and the first harmonic in Figure 1.17. The result is an oddly bumpy wave. The second graph of Figure 1.18 adds the next two harmonics. As more harmonics are added, the sides get steeper and the top/bottom start to flatten. They flatten because each additional harmonic partially cancels some of the peaks and valleys from the previous summation. This gives rise to a greater number of undulations with each undulation being smaller in amplitude. The sequence finishes with Figure 1.19 showing seven harmonics being added with the result approaching a reasonable square wave. As ever more harmonics are added, the wave would approach a flat top and bottom with vertical sides, the idealized square wave. 26 Building a Square Wave Voltage -1.0 -0.8 -0.6 -0.4 -0.2 0.0 0.2 0.4 0.6 0.8 1.0 Time 0 0.2 0.4 0.6 0.8 1 Fundamental plus third harmonic Fundamental Third harmonic Figure 1.17 Building a square wave, part 1.
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27 Building a Square Wave Voltage -1.0 -0.8 -0.6 -0.4 -0.2 0.0 0.2 0.4 0.6 0.8 1.0 Time 0 0.2 0.4 0.6 0.8 1 Fundamental plus third, fifth and seventh harmonics Fundamental Third harmonic Fifth harmonic Seventh harmonic Figure 1.18 Building a square wave, part 2. Figure 1.19 Building a square wave, part 3. Building a Square Wave Voltage -1.0 -0.5 0.0 0.5 1.0 Time 0 0.2 0.4 0.6 0.8 1 Fundamental plus seven harmonics: 3, 5, 7, 9, 11, 13, 15 Fundamental
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The triangle sequence begins with a fundamental and the first harmonic as shown in Figure 1.20. The resulting combination is already trending away from a simple sine shape. The second and final graph, Figure 1.21, shows a total of seven harmonics. The result is very close to a triangle, the only obvious deviation is the slight rounding at the very peaks. The addition of more harmonics would cause these to sharpen further. 28 Building a Triangle Wave One Harmonic Voltage -1.5 -1.0 -0.5 0.0 0.5 1.0 1.5 Time 0 0.5 1 1.5 2 Fundamental plus one harmonic Fundamental Harmonic at three times fundamental Figure 1.20 Building a triangle wave, part 1. Building a Triangle Wave Seven Harmonics Voltage -1.5 -1.0 -0.5 0.0 0.5 1.0 1.5 Time 0 0.5 1 1.5 2 Fundamental plus seven harmonics Fundamental Figure 1.21 Building a triangle wave, part 2.
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1.4 Complex Numbers 1.4 Complex Numbers In AC circuits, parameters such as voltage and current are vectors, that is, they have both a magnitude and a phase shift or angle. For example, a voltage might be “12 volts at an angle of 30 degrees” (or more compactly, 1230º). This is known as polar form or magnitude-angle form. Alternately, a vector can be broken into rectangular form, that is, its right angle components. This can be visualized as a right triangle where the magnitude is the hypotenuse, the angle is the rotation above or below the horizontal, the horizontal component is the side adjacent to the angle and the vertical component is the side opposite of the angle. This is shown in Figure 1.22. Properly, voltage and current vectors are complex numbers that lay on a complex plane consisting of a real part and an imaginary part. The horizontal axis is the real number axis and the vertical axis is the imaginary number axis. The imaginary axis denotes values times the imaginary operator j (and often referred to as i outside of electrical analysis). The j operator is the square root of −1. An example of such a value is 3 + j4, in other words, 3 units along the horizontal real axis and 4 units up the vertical imaginary axis. This is depicted in Figure 1.23. 29 Horizontal component Vertical component Magnitude θ Figure 1.22 Polar and rectangular forms. Figure 1.23 Polar and rectangular forms on the complex plane. +Real +Imaginary (+j) Real component Imaginary component Magnitude θ -Real -Imaginary (-j)
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Converting from one form to another relies on basic trigonometric relations. For convenience, the relationships between the magnitude, angle, real component and imaginary component are reproduced below: Magnitude = √Real 2+Imaginary 2 θ = tan−1 Imaginary Real Real = Magnitude cosθ jImaginary = Magnitude sinθ (1.6a – d) To add or subtract complex quantities, first put them into rectangular form and then combine the reals with the reals and the j terms with the j terms as in (3 + j5) + (13 − j1) = 16 + j4. These real and imaginary terms must be kept separate. 3 + j5 does not equal 8 (or even j8). That would be like saying that moving 3 feet to your right and 5 feet forward puts you in the same location as moving 8 feet to your right (or 8 feet forward). The direct way to multiply or divide complex values is to first put them in polar form and then multiply or divide the magnitudes. The angles are added together for multiplication and subtracted for division. For example, 1230º times 245º is 2475º while dividing them yields 6−15º. The need for complex numbers will become more obvious as we move through the upcoming material. It is imperative that you have mastered the manipulation of complex values before moving on to subsequent chapters. Example 1.5 Convert 15 + j20 and 1 k − j2 k into polar form, and 1045º and 0.2−30º into rectangular form. For the first two conversions, use Equations 1.6a and 1.6b. Magnitude = √15 2+20 2 = 25 θ = tan −1( 20 15) = 53.1º Magnitude = √1 k2+(−2 k)2 = 2.236k θ = tan−1( −2k 1 k )= −63.4º The answers for the first part are 2553.1º and 2.236 k−63.4º. For the second pair of conversions use Equations 1.6c and 1.6d. 30
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Real = 10 cos 45º = 7.07 jImaginary = 10sin 45º = j 7.07 Real = 0.2cos−30º = 0.173 jImaginary = 0.2sin−30º =−j0.1 The answers for the second part are 7.07 + j7.07 and 0.173 − j0.1. Example 1.6 Perform the following: a) Add 7 + j8 to + j6, b) Subtract 553.1º from 10−45º, c) Divide 2090º by 4−50º, d) Multiply 90 + j75 by 6 + j10. a) Add real to real and imaginary to imaginary: 7 + j14. b) First convert the values into rectangular: 3 + j4 and 7.07 − j7.07. Now subtract the first pair from the second pair: 4.07 − j11.07 (or 11.8−69.8º). c) Divide the magnitudes and subtract the angles: 5140º. d) First convert the values into polar: 117.1539.8º and 11.6659º. Now multiply the magnitudes and add the angles: 1.366 k98.8º. 1.5 Reactance and Impedance 1.5 Reactance and Impedance Unlike a resistor, the voltage and current will not be in phase for an ideal capacitor or for an ideal inductor. For the capacitor, the current leads the voltage across the capacitor by 90 degrees. Recall that the voltage across a capacitor cannot change instantaneously, i = C dv/dt. For an inductor, the voltage leads the current by 90 degrees. Similarly, the current through an inductor cannot change instantaneously due to v = L di/dt. While ideal capacitors and inductors do not exhibit resistance, the voltage does react to the current. Unsurprisingly, we call this characteristic reactance and denote it with the letter X. Reactance, like resistance, is a ratio of voltage to current. We define capacitive reactance as: X C = v c ic (1.7) Using a simple sine wave with a peak value of unity, vc = sin 2π f t, ic can be found from i = C dv/dt. Recall from our prior work that the derivative or slope of a sine wave is a cosine wave, which is in turn equivalent to a sine wave shifted by +90º. 31
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ic = C dvc dt ic = C d sin(2 π f t) dt ic = C sin(2π f t+90º) Substituting this result into Equation 1.7 yields: X C = vc ic X C = sin(2π f t) 2 π f C sin(2π f t+90º) X C = 1 2π f C −90º Or more directly: X C =−j 1 2π f C (1.8) In summation, if we divide the capacitor's sinusoidal voltage by its current (Ohm's law), we obtain a value with a phase angle of −90°. While the resultant is an ohmic value, it cannot be classified as a resistance. Instead, it is referred to as a reactance and denoted with the letter X. Thus we can refer to a capacitor's reactance, XC, as some number at an angle of −90º, or more conveniently, we simply prepend −j, as in XC = −j75 Ω. The case for the inductor is similar and left as an exercise. The inductive reactance, XL, can be found using: X L=+j 2π f L (1.9) An example would be XL = j68 Ω. As stated, while a resistance cannot be added directly to a reactance, reactances can be added together so long as we heed the signs. For example, if we have a capacitive reactance of −j60 Ω in series with an inductive reactance of j100 Ω, the result is j40 Ω. This is due to the fact that these two items are 180 degrees out of phase with each other, and so they partially cancel. Remember, always add or subtract like items: real (resistance) to real, and imaginary (reactance) to imaginary. Equations 1.8 and 1.9 are notable because the reactance is not just a function of the capacitance or inductance, but also a function of frequency. The reactance of an inductor is directly proportional to frequency while the reactance of a capacitor is inversely proportional to frequency. The ohmic variations of a 20 Ω resistor, a 500 μF capacitor and a 500 μH inductor across frequency are shown in Figure 1.24. 32
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We can see that the value of resistance does not change with frequency while the inductive reactance increases with frequency and the capacitive reactance decreases. The linear frequency scale makes the capacitor change difficult to see. If this is plotted again but using a logarithmic frequency scale as in Figure 1.25, the symmetry becomes apparent. 33 Resistance & Reactance vs. Frequency Resistance or Reactance (ohms) 0 5 10 15 20 25 30 35 Frequency (Hz) 2,000 4,000 6,000 8,000 1e+04 20 ohm resistor 500 μH Inductor 500 μF capacitor Figure 1.24 Resistance and reactance versus frequency (linear axis). Figure 1.25 Resistance and reactance versus frequency (log axis). Resistance & Reactance vs. Frequency Resistance or Reactance (ohms) 0 5 10 15 20 25 30 35 Frequency (Hz) 10 100 1,000 1e+04 20 ohm resistor 500 μH Inductor 500 μF capacitor
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The effect of both capacitor size and frequency is shown in Figure 1.26 using a log frequency axis: the smaller the capacitor, the larger the capacitive reactance at any particular frequency. Similarly, the effect of both inductor size and frequency is shown in Figure 1.27 using a linear frequency axis: the larger the inductor, the larger the inductive reactance at any given frequency. 34 Figure 1.26 Variation of capacitive reactance with capacitance and frequency. Capacitive Reactance vs. Frequency Reactance (ohms) 0 5 10 15 20 25 30 35 40 Frequency (Hz) 100 1,000 1e+04 1e+05 10 μF 200 μF 50 μF 1 μF 100 nF Figure 1.27 Variation of inductive reactance with inductance and frequency. Inductive Reactance vs. Frequency Reactance (ohms) 0 5 10 15 20 25 30 35 40 45 50 Frequency (Hz) 0 2,000 4,000 6,000 8,000 10,000 50 μH 400 μH 3 mH 1 mH
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It is worth noting that the plots of Figures 1.26 and 1.27 are for ideal components. In reality, all components exhibit some resistive, capacitive and inductive effects due to their construction. For example, eventually, inductive and resistive effects will cause the capacitive reactance curves of Figure 1.26 to begin to rise at high frequencies. Similarly, resistive and capacitive effects will cause the curves Figure 1.27 to flatten at very low and very high frequencies. An interesting observation to remember is that capacitors and inductors are a bit like a Chimera. That is, they look like different things to different sources, all mashed together at once. It would be improper to think of, say, a particular inductor as being “so many ohms”. If the source signal is comprised of multiple sine waves, such as a square wave or a music waveform, the inductor “looks like” a different ohmic value to each of the different frequency components, simultaneously. This is an important concept and one that we can take great advantage of, for example, in the design of filter circuits. Impedance Impedance We now arrive at impedance. Impedance is a mixture of resistance and reactance, and is denoted by Z. This can be visualized as a series combination of a resistor and either a capacitor or an inductor. Examples include Z = 100 − j50 Ω, i.e., 100 ohms of resistance in series with 50 ohms of capacitive reactance; and Z = 60045º Ω, i.e., a magnitude of 600 ohms that includes resistance and inductive reactance (it must be inductive reactance and not capacitive reactance because the sign of the angle is positive). To complete this system, we have susceptance and admittance. Susceptance, B, is the reciprocal of reactance. Admittance, Y, is the reciprocal of impedance. These are similar to the relation between conductance and resistance, and are convenient for parallel circuit combinations. Example 1.7 Determine the reactances of a 1 mH inductor and a 2 μF capacitor to a sine wave of 2 kHz. Repeat for a frequency of 50 kHz. Use Equations 1.8 and 1.9. For the capacitor at 2 kHz we have: X C =−j 1 2π f C X C =−j 1 2π 2kHz 2μ F X C =−j39.8Ω 35
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For the second source, 50 kHz is 25 times larger than the original and capacitive reactance is inversely proportional to frequency. Therefore XC is 25 times smaller, or −j1.59 Ω. For the inductor at 2 kHz, X L = +j 2π f L X L = +j 2π 2kHz1 mH X L = +j 12.57Ω Inductive reactance is directly proportional to frequency. Thus, increasing f by a factor of 25 increases XL by the same factor, resulting in j314.2 Ω. Example 1.8 Determine the susceptance of an inductor whose reactance is j400 Ω. Further, if this inductor is placed in series with a 1000 Ω resistor, determine the resulting impedance in polar form, as well as the admittance. Susceptance is the reciprocal of reactance. BL = 1 X L BL = 1 j 400Ω BL =−j 2.5 millisiemens The impedance, Z, is the vector sum, or 1000 + j400 Ω. Converting this to polar form: Magnitude = √Real 2+Imaginary 2 Magnitude = √10002+4002 Magnitude = 1077 θ = tan−1( Imaginary Real ) θ = tan −1( 400 1000) θ = 21.8º The result is Z = 107721.8º Ω. The admittance is the reciprocal, yielding Y = 928E-6−21.8º μS. 36
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1.6 Phasor Diagrams 1.6 Phasor Diagrams A time domain representation of sine waves as drawn earlier tells us everything we need to know about the waves, however it is not the most compact method of displaying them, nor are they easy to sketch by hand. Consider the the plot of Figure 1.28. Here, two sine waves (green and red) combine to create a third sine wave (blue). While the general idea of the two waves adding is apparent in the graph, it takes a moment of inspection to determine the wave magnitudes and the precise phase relationships between them. In contrast, phasor diagrams can be used to show the relationships of multiple sines waves in a simple and easy to read format. They can also be used to show how various voltages or currents combine. The phasor diagram is based on the complex plane discussed previously where the horizontal is the real axis and the vertical is the imaginary (j) axis. The magnitude and phase of each wave can then be drawn as a vector, and the relationships between the waves is shown directly. For manual plotting, it is convenient to convert from polar form to rectangular form. In the example phasor diagram of Figure 1.29, two vectors are shown: 8 + j6 and 5 − j3 (equivalent to 1036.9º and 5.83−31º). Phasor diagrams can be used to plot voltages, currents and impedances. We shall make considerable use of them in upcoming chapters. 37 Adding Sines Percentage of Peak -150 -100 -50 0 50 100 150 Time -1 -0.5 0 0.5 1 Additional wave Initial Resultant wave Figure 1.28 A time domain plot.
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1.7 Summary 1.7 Summary The most simple of AC waveforms is the sine wave. It can be thought of as the vertical displacement of a vector rotating at a constant speed, like the second hand of a clock. The length of the second hand represents the height or amplitude of the sine wave and the speed of rotation represents its period. As the speed tends to be very fast, it is more convenient for us to use frequency, which is just the reciprocal of the period. Sines can be displaced vertically, which is also called a DC offset, as well as having a horizontal or time shift. When expressed relative to a single cycle, this change is referred to as the phase shift. Finally, waveforms such as square waves, triangle waves and even more complex waveforms such as voice or music can be described in terms of combinations of sounds. Indeed, Fourier analysis tells us that any repeating waveform can be described as a series of sines each with the appropriate frequency, amplitude and phase shift. In order to determine the “effective DC value” of a sine wave, that is, the value that produces the same power dissipation, RMS values are used. The RMS value of a sine wave is its peak value divided by the square root of two (approximately equal to 0.707 of peak). Complex numbers are used to describe AC voltages and currents, among other things. They consist of two parts: a real part and an imaginary part that is plotted perpendicular to the real part. All mathematical operations on complex numbers must follow vector rules. This includes basic trigonometric operations. Reactance can be thought of as the imaginary axis version of resistance. That is, it restricts current flow. The difference is that there is a 90 degree phase shift between the current and voltage through a reactive element while the two are in phase for resistive elements. Capacitive reactance is inversely proportional to frequency while inductive reactance is directly proportional to frequency. The combination of resistance and reactance is known as impedance. Phasor diagrams may be used to plot the components of a complex impedance as well as show the relations between voltages or currents in a circuit. 38 Figure 1.29 A phasor diagram.
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Review Questions Review Questions 1. What is a sine wave? Describe its constituent parameters (amplitude, frequency, offset, etc.). 2. What is the relationship between the frequency, period and phase of a sine wave? 3. What is the mathematical relationship between a sine wave and a cosine wave? 4. What is meant by the term RMS (root-mean-square) and what does it have to do with AC versus DC voltages, currents and powers? 5. Describe the difference between a scalar and a vector. 6. What is a complex number? 7. Detail the relationships between resistance, reactance, impedance, conductance, susceptance and admittance. 8. Describe how capacitive reactance varies with frequency. 9. Describe how inductive reactance varies with frequency. 10. Describe a phasor diagram. 1.8 Exercises 1.8 Exercises Analysis Analysis 1. Determine the AC peak and RMS voltages, DC offset, frequency, period and phase shift for the following expression: v(t) = 10 sin 2π 1000 t 2. Determine the AC peak and RMS voltages, DC offset, frequency, period and phase shift for the following expression: v(t) = 0.4 sin 2π 5000 t 3. Determine the peak AC portion voltage, DC offset, frequency, period and phase shift for the following expression: v(t) = −3 + 20 sin 2π 50 t 4. Determine the peak AC portion voltage, DC offset, frequency, period and phase shift for the following expression: v(t) = 12 + 2 sin 2π 20000 t 5. Determine the AC peak and RMS voltages, DC offset, frequency, period and phase shift for the following expression: v(t) = 10 sin (2π 100 t + 45°) 6. Determine the AC peak and RMS voltages, DC offset, frequency, period and phase shift for the following expression: v(t) = 5 sin (2π 1000 t − 90°) 7. Determine the peak AC portion voltage, DC offset, frequency, period and phase shift for the following expression: v(t) = 10 + 1 sin (2π 400 t − 45°) 8. Determine the peak AC portion voltage, DC offset, frequency, period and phase shift for the following expression: v(t) = 10 + 10 sin (2π 5000 t + 30°) 39
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9. A 1 kHz sine wave has a phase of 72°. Determine the time delay. Repeat for a 20 kHz sine wave. 10. A 2 kHz sine wave has a phase of 18°. Determine the time delay. Repeat for a 100 kHz sine wave. 11. An oscilloscope measures a time delay of 0.2 milliseconds between a pair of 500 Hz sine waves. Determine the phase shift. 12. An oscilloscope measures a time delay of −10 microseconds between a pair of 20 kHz sine waves. Determine the phase shift. 13. Convert the following from rectangular to polar form: a) 10 + j10 b) 5 − j10 c) −100 + j20 d) 3k + j4k 14. Convert the following from rectangular to polar form: a) 2k + j1.5k b) 8 − j8 c) −300 + j300 d) −1k − j1k 15. Convert these from polar to rectangular form: a) 1045° b) 0.490° c) −960° d) 100−45° 16. Convert these from polar to rectangular form: a) −460° b) −0.930° c) 5120° d) 6−135° 17. Perform the following computations: a) (10 + j10) + (5 + j20) b) (5 + j2) + (−5 + j2) c) (80 − j2) − (100 + j2) d) (−65 + j50) − (5 − j200) 18. Perform the following computations: a) (100 + j200) + (75 + j210) b) (−35 + j25) + (15 + j8) c) (500 − j70) − (200 + j30) d) (−105 + j540) − (5− j200) 19. Perform the following computations: a) (100 + j200) ∙ (75 + j210) b) (−35 + j25) ∙ (15 + j8) c) (500 − j70) / (200 + j30) d) (−105 + j540) / (5− j200) 20. Perform the following computations: a) (10 + j10) ∙ (5 + j20) b) (5 + j2) ∙ (−5 + j2) c) (80 − j2) / (100 + j2) d) (−65 + j50) / (5− j200) 21. Perform the following computations: a) (100°) ∙ (100°) b) (545°) ∙ (−220°) c) (20135°) / (40−10°) d) (80°) / (3245°) 22. Perform the following computations: a) (0.30°) ∙ (3180°) b) (5−45°) ∙ (−420°) c) (0.0595°) / (0.04−20°) d) (5000°) / (60225°) 23. Perform the following computations: a) (0.30°) + (3180°) b) (5−45°) + (−420°) c) (0.0595°) − (0.04−20°) d) (5000°) − (60225°) 24. Perform the following computations: a) (100°) + (100°) b) (545°) + (−220°) c) (20135°) − (40−10°) d) (80°) − (3245°) 40
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25. Determine the capacitive reactance of a 1 μF capacitor at the following frequencies: a) 10 Hz b) 500 Hz c) 10 kHz d) 400 kHz e) 10 MHz 26. Determine the capacitive reactance of a 220 pF capacitor at the following frequencies: a) 10 Hz b) 500 Hz c) 10 kHz d) 400 kHz e) 10 MHz 27. Determine the capacitive reactance at 50 Hz for the following capacitors: a) 10 pF b) 470 pF c) 22 nF d) 33 μF 28. Determine the capacitive reactance at 1 MHz for the following capacitors: a) 22 pF b) 560 pF c) 33 nF d) 4.7 μF 29. Determine the inductive reactance of a 100 mH inductor at the following frequencies: a) 10 Hz b) 500 Hz c) 10 kHz d) 400 kHz e) 10 MHz 30. Determine the inductive reactance of a 100 mH inductor at the following frequencies: a) 10 Hz b) 500 Hz c) 10 kHz d) 400 kHz e) 10 MHz 31. Determine the inductive reactance at 1 kHz for the following inductors: a) 10 mH b) 500 mH c) 10 μH d) 400 μH 32. Determine the inductive reactance at 500 kHz for the following inductors: a) 1 mH b) 40 mH c) 2 μH d) 50 μH 33. Draw phasor diagrams for the following: a) 5 + j2 b) −10 −j20 c) 845° d) 2−35° 34. Draw phasor diagrams for the following: a) 60j−20 b) −40 + j500 c) 0.05−45° d) −1560° 35. The fundamental of a certain square wave is a 5 volt peak, 1 kHz sine. Determine the amplitude and frequency of each of the next five harmonics. 36. The fundamental of a certain triangle wave is a 10 volt peak, 100 Hz sine. Determine the amplitude and frequency of each of the next five harmonics. Design Design 37. Determine the capacitance required for the following reactance values at 1 kHz: a) 560 Ω b) 330 kΩ c) 470 kΩ d) 1.2 kΩ e) 750 Ω 38. Determine the capacitance required for the following reactance values at 20 Hz: a) 56 kΩ b) 330 kΩ c) 470 kΩ d) 1.2 kΩ e) 750 Ω 39. Determine the inductance required for the following reactance values at 100 MHz: a) 560 Ω b) 330 kΩ c) 470 kΩ d) 1.2 kΩ e) 750 Ω 40. Determine the inductance required for the following reactance values at 25 kHz: a) 56 Ω b) 33 kΩ c) 470 kΩ d) 1.2 kΩ e) 750 Ω 41
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41. Which of the following have a reactance of less than 100 Ω for all frequencies below 1 kHz? a) 2 mH b) 99 mH c) 470 pF d) 10000 μF 42. Which of the following have a reactance of less than 8 Ω for all frequencies above 10 kHz? a) 10 nH b) 5 mH c) 56 pF d) 470 μF 43. Which of the following have a reactance of at least 1k Ω for all frequencies above 20 kHz? a) 2 mH b) 200 mH c) 680 pF d) 33 μF 44. Which of the following have a reactance of at least 75 Ω for all frequencies below 5 kHz? a) 680 μH b) 10 mH c) 82 pF d) 33 nF Challenge Challenge 45. Determine the negative and positive peak voltages, RMS voltage, DC offset, frequency, period and phase shift for the following expression: v(t) = −10 sin (2π 250 t + 180°) 46. Determine the negative and positive peak voltages, DC offset, frequency, period and phase shift for the following expression: v(t) = 1 − 100 sin 2π 50000 t 47. Assume you have a DC coupled oscilloscope set as follows: time base = 100 microseconds/division, vertical sensitivity = 1 volt/division. Sketch the display of this waveform: v(t) = 2 + 3 sin 2π 2000 t 48. Assume you have a DC coupled oscilloscope set to the following: time base = 20 microseconds/division, vertical sensitivity = 200 millivolts/division. Sketch the display of this waveform: v(t) = −0.2 + 0.4 sin 2π 10000 t 49. A 200 Ω resistor is in series with a 1 mH inductor. Determine the impedance of this combination at 200 Hz and at 20 kHz. 50. A 1 kΩ resistor is in series with an inductor. If the combined impedance at 10 kHz is 1.41 k45°, determine the inductance in mH. 42
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Notes Notes ♫♫ ♫♫ 43
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2 2 Series RLC Circuits Series RLC Circuits 2.0 Chapter Learning Objectives 2.0 Chapter Learning Objectives After completing this chapter, you should be able to: • Utilize Kirchhoff's voltage law, the voltage divider rule and Ohm's law to find node and component voltages in series RLC networks that utilize voltage sources or a single current source. • Compute complex impedance and system current in series RLC circuits. • Draw phasor diagrams for impedance and component voltages in series RLC circuits. 2.1 Introduction 2.1 Introduction In this chapter we introduce series RLC circuits for the AC case. There is much here that will be familiar from your prior studies with DC series circuits, however, there will be a few notable changes and perhaps a surprise or two lurking. The key to most of this is to remember that all computations involve vector quantities. In fact, DC can be thought of as a special case of AC; namely, the frequency drops to zero hertz causing XC to approach an open and XL to drop to zero. This leaves us with just resistors and scalar quantities because the phase angles in the remainder of the circuit collapse to zero for DC steady state. Many of the solution techniques from DC analysis will be applicable here. This includes the use of Ohm's law and Kirchhoff's voltage law, along with the voltage divider rule. Generally, reactance values will need to be computed from capacitor and inductor values before the main analysis may begin. In this chapter, as in the most of the remaining chapters, we shall be concerned with determining the circuit response based on a source with a single frequency of excitation, in other words, a simple sine wave. To clarify our analyses, we shall make considerable use of both time domain voltage plots as well as phasor diagrams. 2.2 The Series Connection 2.2 The Series Connection A series connection is always characterized by a single loop or path for current flow. There are no junctions from which current can flow out of or into. Consequently, The current is the same everywhere in a series connection. (2.1) Each component in such a loop will see the same current, regardless of whether it is a resistor, capacitor or inductor. Before component voltages can be determined, the capacitive and inductive reactance values must be 44
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computed from the capacitor and inductor values, based on the frequency of the driving source. Consequently, if this frequency were to change, the reactances would change, and this would in turn cause changes in the circulating current and component voltages. 2.3 Series Impedance 2.3 Series Impedance Perhaps the first practical issue we face is determining the effective impedance of an RLC series loop. For starters, resistors in series simply add. Reactances also add but we must be careful of the sign. Inductive reactance and capacitive reactance will partially cancel each other. Thus, the impedance in rectangular form is the sum of the resistive components for the real portion, plus the sum of the reactances for the imaginary (j) portion. We will often find it convenient to express this value in polar form. Example 2.1 What is the impedance of the network shown in Figure 2.1 at a frequency of 15 kHz? First we need to find the capacitive reactance value. X C =−j 1 2π f C X C =−j 1 2π15kHz 910pF X C =−j11.66k Ω As there is only one resistor and one capacitor, the result in rectangular form is 47 k −j11.66 kΩ. In polar form this is: Magnitude = √Real 2+Imaginary 2 Magnitude = √47k 2+(−11.66k) 2 Magnitude = 48.42k θ = tan−1( Imaginary Real ) θ = tan −1( −11.66k 47 k ) θ =−13.9º That is, in polar form Z = 48.42E3−13.9° Ω. 45 Figure 2.1 Circuit for Example 2.1.
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Example 2.2 Determine the effective impedance of the circuit shown in Figure 2.2 if the source frequency is 2 kHz. Repeat this for source frequencies of 200 Hz and 20 kHz. Finally, express the results in both rectangular and polar form. The first step is to find the reactance values at 2 kHz. X L = j 2π f L X L = j 2π2000Hz 15mH X L = j188.5Ω X C =−j 1 2π f C X C =−j 1 2π2000 Hz 270 nF X C =−j294.7 Ω Combine reals with reals and j terms with j terms. Z = R+ j X L −j X C Z = 500+ j188.5 −j 294.7Ω Z ≈500−j 106.2Ω = 511.2−12°Ω At 200 Hz XC will ten times larger and XL will be ten times smaller. XC = −j2947 Ω XL = j18.85 Ω Z = 500 + j18.85 −j2947 Ω Z ≈ 500 −j2928 Ω = 2970−80.3° Ω At 20 kHz XC will ten times smaller and XL will be ten times larger. XC = −j29.47 Ω XL = j1885 Ω Z = 500 + j1885 −j29.47 Ω Z ≈ 500 + j1856 Ω = 192274.9° Ω To help visualize this complex impedance, it is useful to construct a phasor plot as shown in Figure 2.3. We will do this for the initial case of 2 kHz. The resistive component is the horizontal vector of length 500 (yellow). XL is straight up (blue) at 188.590°, and XC is straight down (red) at 294.7−90°. By copying the XL vector and then shifting it down and next to XC, the difference between the two reactive components can be seen (purple component directly above the XL copy). This 46 Figure 2.2 Circuit for Example 2.2.
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reactive sum is then copied and shifted right to join the resistive component to show the final result. This is 511.2−12° Ω (green), as expected. Example 2.3 Determine the impedance of the network shown in Figure 2.4. If the input frequency is 1 kHz, determine the capacitor and inductor values. The reactance values are already given, so we simply add them to determine the impedance in rectangular form. Combine reals with reals and j terms with j terms, and then convert to polar form. Z = R+j X L −j X C Z = 750+j600 −j200 Ω Z = 750+j400 Ω= 85028.1°Ω To find the capacitance and inductance, we simply rearrange the reactance formulas and solve. First, the inductor: X L = j 2π f L L = | X L| 2π f L = 600Ω 2π1kHz L ≈95.5 mH 47 Figure 2.3 Construction of the impedance plot for the network of Figure 2.2. R jX -jX -400 300 600 jX -jX R Z jX + jX L C L C L -jX Figure 2.4 Circuit for Example 2.3.
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And now for the capacitor: X C =−j 1 2 π f C C = 1 2π f | X C| C = 1 2π1 kHz 200Ω C = 796nF A plot of the impedance vector summation is shown in Figure 2.5. Note how the three components combine graphically to arrive at Z. 2.4 Series Circuit Analysis 2.4 Series Circuit Analysis The techniques employed for series AC circuit analysis are the same as those used for DC. The key item to remember for series circuits, whether AC or DC, as that the current will be the same everywhere in a series connection. The major analysis tools are Ohm's law, Kirchhoff's voltage law (KVL), and optionally, the voltage divider rule. 48 Figure 2.5 Impedance plot for the network of Example 2.3.
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KVL states that the sum of voltages around a loop must be zero, or that the sum of voltage rises around a loop must equal the sum of voltage drops: ∑ v↑ = ∑ v↓ (2.2) The voltage divider rule for AC states that the voltage across any component or group of components is proportional to the ratio of the impedance of said component and the total series impedance. For some component A driven by source e, v A = e Z A Z Total (2.3) If current is the desired quantity, the equivalent impedance of the network is found and then divided into the equivalent source voltage. If the voltages across specific components are desired, they can be found using the circulating current and Ohm's law. Alternately, they can be computed using the voltage divider rule. If a current source is driving the series circuit then the circulating current is the source current. Individual component voltages may be found using the source current and Ohm's law. The source voltage may be found by summing the component voltages via KVL, or by determining the equivalent series impedance and multiplying it by the source current. Example 2.4 For the circuit of Figure 2.6, find the circulating current and the voltages across the capacitor and resistor. The first step is to determine XC and then find Ztotal. Then use Ohm's law to find the circulating current. X C =−j 1 2π f C X C =−j 1 2π1000Hz 33nF X C =−j4823Ω ZTotal = R+(−j X C) ZTotal = 2200Ω −j 4823Ω ZTotal = 5301−65.5°Ω i = v Z i = 1V 5301−65.5°Ω i = 188.665.5°μ A 49 Figure 2.6 Circuit for Example 2.4.
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Now use Ohm's law to find the component voltages. v R = i×R v R = 188.665.5°μ A×22000°Ω v R = 0.414965.5° V vC = i× X C vC = 188.665.5° μA×4823−90°Ω vC = 0.9098−24.5° V To complete this Example, we shall generate a series of plots: First off, an impedance plot showing the vector combination of R and XC leading to ZTotal. This is shown in Figure 2.7. Next, Figure 2.8 illustrates the phasor diagram of the three voltages. It may not be immediately apparent but an examination of this plot shows that it is just a counterclockwise rotation of the impedance plot. In fact, it is rotated by 65.5 degrees, the phase angle of the current. This should come as no surprise as the current is multiplied by each resistance, reactance or impedance to develop the voltage, and a vector multiplication simply adds the angles. Also, note that if we simply summed the magnitudes of the component voltages, the result would be considerably larger than the input voltage, seeming to violate KVL. In contrast, once a proper vector summation is performed, all is right in paradise. 50 Figure 2.7 Impedance plot for Example 2.4.
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For further clarity, Figure 2.9 plots the voltages in the time domain. Note how the component voltages add graphically resulting in the input voltage. 51 Figure 2.8 Voltage phasor diagram for Example 2.4. Time Domain Graph Voltage (volts) -1.00 -0.75 -0.50 -0.25 0.00 0.25 0.50 0.75 1.00 Time (milliseconds) 0 0.25 0.5 0.75 1 1.25 1.5 vc vin vR Figure 2.9 Time domain response for Example 2.4.
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The time shifts may be computed for verification as follows: θ = 360° Δt T ΔtR = T θR 360° Δt R = 1 ms 65.5° 360° Δt R = 181.9μ s Δt C = T θC 360° ΔtC = 1ms−24.5° 360° ΔtC =−68μ s Thus we verify the resistor voltage leading (to the left) by 181.9 μs and the capacitor voltage lagging (to the right) by 68 μs, as seen in Figure 2.9. Computer Simulation Computer Simulation The circuit of Example 2.4 is captured into a simulator as shown in Figure 2.10. Node voltage 1 corresponds to the input voltage and node 2 corresponds to the capacitor voltage. The resistor voltage is merely the difference between these two, or node 1 minus node 2. A transient or time domain simulation is performed. The results of the simulation are shown in Figure 2.11. Note the tight agreement with the computed results as plotted in Figure 2.9. 52 Figure 2.10 The circuit of Example 2.4 in a simulator.
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We will now examine a variety of different series circuits. To ease the computations, we will simply state values for the component reactances rather than specify a frequency along with inductor and/or capacitor values. Example 2.5 Determine the component voltages for the circuit shown in Figure 2.12. Given the reference direction of the source (which produces a counterclockwise reference current), the voltage across the resistor will be defined as vb − va. The first step is to find the equivalent series impedance. By inspection, this is 180 + j360 Ω, which is equivalent to 402.563.4° Ω. The voltages can be found via the voltage divider rule. v L = e X L Z v L = 60° V 36090°Ω 402.563.4°Ω v L = 5.36626.6° Vp 53 Figure 2.11 Time domain response of Example 2.4 from a simulator. Figure 2.12 Circuit for Example 2.5.
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v R = e R Z v R = 60° V 1800° Ω 402.563.4 °Ω v R = 2.683−63.4° Vp The rectangular versions of vL and vR are 4.8 + j2.4 and 1.2 − j2.4 respectively, summing up to the source of 60°. This is illustrated in the phasor voltage plot of Figure 2.13. Compare the plot of Figure 2.13 to that of the RC circuit in Figure 2.8. In this circuit, the impedance is inductive. This causes the current to lag the voltage, the opposite of the case for the RC circuit. Thus, the resistor voltage winds up below the real axis instead of above it (remember, the current and voltage for a resistor are in phase, and therefore the voltage across the resistor will have the same phase angle as the current through it). Once again, simply summing the magnitudes of the voltages yields a greater value than the source, however, a proper vector addition shows that KVL is satisfied. A time domain plot of the source and component voltages is shown in Figure 2.14. Note the similarity with the plot of Figure 2.9. Further, note how the time positions of the component voltages have swapped. 54 Figure 2.13 Voltage phasor diagram for Example 2.5.
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Finally, as a source frequency was not specified for this circuit, the time scale of Figure 2.14 is calibrated in terms of cycles rather than seconds. Example 2.6 Find the voltages va and vb in the circuit of Figure 2.15. Assume the current source is the reference of 0°. Ohm's law may be used directly as the current is set by the source. Given the reference direction of the 20 mA source, the reference polarity for the capacitor voltage will be + to − from bottom to top. Thus, va is negative. v a = vC = −i×X C v a =−20E-30°A×1000−90°Ω v a =−20−90° V Recalling that a negative magnitude is the same as a 180° phase shift, we may remove the magnitude's negative sign by adding 180° to the phase angle. Therefore, va may also be written as 2090°. The node b voltage is found by multiplying the current by the impedance seen from node b to ground. 55 Figure 2.14 Time domain response for Example 2.5. Time Domain Graph Voltage (volts) -6.0 -4.0 -2.0 0.0 2.0 4.0 6.0 Time (cycles) 0 0.5 1 1.5 vL vin vR Figure 2.15 Circuit for Example 2.6.
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vb = i×(R+jX L) vb = 20 0°mA×(4300+j 150Ω) vb = 86.052° V Notice the small size of the phase angle. This is expected given that the inductive reactance is so much smaller than the resistive component. Finally, these voltage are RMS as the current is assumed to be RMS (not having been labeled as peak or peak-to-peak). Example 2.7 Find the voltage vbd in the circuit of Figure 2.16. E1 = 20° volts peak and E2 = 560° volts peak. The first step is to find the equivalent source voltage. Voltage sources in series add, however, note the reference polarity for E2. If we take E1 as the system reference, then the combined source is E1 − E2. Using E1 as the reference source creates a clockwise reference current and thus a positive reference polarity for vbd. If E2 was taken as the reference then vbd would be assumed to be negative due to the assumed counterclockwise reference direction of the resulting current. Either way will work. ETotal = E1 −E2 ETotal = 20° Vp −560 ° Vp ETotal = 4.359−96.6° Vp We can now use a voltage divider between the impedance seen from node b to node d versus the total series impedance. The impedance of the circuit is 2 k + j7.5 k − j800 Ω, or 2 k + j6.7 k Ω. This is equivalent to 699273.4° Ω. The impedance between nodes b and d is j7.5 k − j800 Ω, or j6.7 k Ω. In polar form this is 670090° Ω. 56 Figure 2.16 Circuit for Example 2.7.
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The voltage divider yields: vbd = e Z bd Z Total vbd = 4.359−96.6°V 670090°Ω 699273.4 °Ω vbd = 4.177−80° Vp Although we have answered the question, at this point attempting to visualize the waveforms in your head to verify the results may be a little challenging. No problem! If we also find the voltage across the resistor, we can check to see if KVL is verified using a phasor diagram. First, the resistor voltage may be found using the voltage divider rule. v R = e R Z Total v R = 4.359−96.6° V 20000°Ω 699273.4°Ω v R = 1.247−170° Vp We now plot all four voltages as seen in Figure 2.17 (note the unequal horizontal versus vertical scaling for ease of visual analysis). 57 Figure 2.17 Voltage phasor diagram for Example 2.7.
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KVL tells us that the sum of the voltage rises must equal the sum of the voltage drops. We had used E1 as the reference and treated it as a rise. This made the reference direction clockwise for the circulating current. As a consequence, vR and vbd are voltage drops. E2 is also seen as a drop given its reference polarity markings. In other words, E1 should equal the sum of vR, vbd and E2. Let's see if this is borne out graphically in the phasor diagram. Looking at imaginary parts first, E1 is strictly real so the other three vectors should sum to zero for their imaginary parts. The imaginary part of E2 is about +4.3 while the other two come in around −4.1 and −0.2, so they do balance. For the real parts, E2 is +2.5, vbd is about +0.7 and vR is about −1.2. These add up to +2, the real component of E1. Everything balances. For more exacting results, you could convert each of the polar form results into rectangular form and add the reals to the reals, and the imaginaries to the imaginaries, and get the same result. In fact, using the prior recorded values, the three drops sum to 1.9970.0013° volts versus precisely 20° volts. This small deviation is due to the accumulated rounding and truncation errors of the intermediate results. Sometimes a problem concerns determining resistance or reactance values. The solution paths will require use of the analysis rules in reverse. For example, if a resistor value is needed to set a specific current, the total required impedance can be determined from this current and the given voltage supply. The values of the other series components can then be subtracted from the total (using rectangular form), yielding the required resistor or reactance value. Another possibility is determining a specific capacitance or inductance to meet certain requirements. An example of this would be a simple crossover network for a loudspeaker system. Generally, a single transducer is incapable of reproducing music signals of all frequencies at sufficiently loud levels while maintaining low distortion. Consequently, the frequency range is split into two or more bands with each being covered by a transducer optimized to reproduce those frequencies. These transducers, along with a few electrical components, are then packaged into an appropriate enclosure commonly referred to as a loudspeaker1. In a basic system, the audible spectrum is split into two parts, as seen in Figure 2.18. The low frequencies are handled by a large transducer called a woofer, while the high frequencies are handled by a small, light transducer called a tweeter that can follow the rapidly changing high frequency content with accuracy. Woofers are not capable of reproducing high frequencies, and excessive low frequency content can physically damage a tweeter. Thus, some means of “steering” the high frequency content to the tweeter and the low frequency content to the woofer must be employed. One way to do this is to place an inductor or capacitor in series with the transducer. 1 To add some confusion, the term “loudspeaker” can refer to either the individual transducers (engineer-speak) or the entire finished system (audiophile-speak). 58 Figure 2.18 A two-way bookshelf loudspeaker system with tweeter (top) and woofer (bottom).
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Figure 2.19 shows a simple way to block low frequencies from a tweeter. In this diagram, the loudspeaker is modeled as a resistor. Although this is not perfectly accurate, it is sufficient to illustrate the idea. To understand this concept, simply think of the circuit as a voltage divider between the loudspeaker and the reactance of the capacitor. Capacitive reactance is inversely proportional to frequency. The circuit is designed such that XC is much larger than than the impedance of the loudspeaker at very low frequencies. This means that very little of a low frequency signal will reach the loudspeaker as most of that voltage will drop across the capacitor. In contrast, at high frequencies XC is much smaller than the tweeter's impedance and thus virtually all of the source signal reaches the loudspeaker. The transition point between these two regions is called the crossover frequency and is usually defined as the point where XC equals the magnitude of the loudspeaker impedance. For a woofer, the capacitor would be replaced by an inductor. At low frequencies the inductive reactance would be minimal, thereby allowing all of the low frequency content to reach the woofer. The opposite happens at high frequencies. The inductive reactance would rise with frequency and eventually prevent or “choke off” the signal to the woofer (hence, the slang term “choke” for an inductor). Example 2.8 Referring to the circuit of Figure 2.19, assume the loudspeaker impedance is 80° Ω. Determine a capacitor value for a crossover frequency of 2.5 kHz. At this frequency the magnitude of XC should be the same as that of the loudspeaker. For a 10 volt RMS input signal, determine the loudspeaker voltage at frequencies of 100 Hz, 2.5 kHz and 15 kHz. The first step is to determine the value of the capacitor such that at 2.5 kHz XC equals 8 Ω. To do this, simply rearrange the capacitive reactance formula: 59 Figure 2.19 A basic loudspeaker crossover network.
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X C =−j 1 2π f C C = 1 2π f | X C| C = 1 2π 2500Hz 8Ω C = 7.96μ F To find the loudspeaker voltage at the three specified frequencies, first find the reactance at that frequency and then use the voltage divider rule. We already know that at 2.5 kHz the reactance is −j8 Ω. vloudspeaker = e Z Load ZTotal vloudspeaker = 10 0°Vrms 8Ω 8 −j8Ω vloudspeaker = 7.0745 °Vrms At 100 Hz, the frequency has decreased by a factor of 25 so the reactance goes up by the same factor yielding −j200 Ω. vloudspeaker = e Z Load ZTotal vloudspeaker = 10 0°Vrms 8Ω 8 −j200Ω vloudspeaker = 0.487.7° Vrms At 15 kHz, the frequency has increased by a factor of 6 so the reactance goes down by the same factor yielding −j1.333 Ω. vloudspeaker = e Z Load ZTotal vloudspeaker = 10 0°Vrms 8Ω 8 −j1.333Ω vloudspeaker = 9.869.5° Vrms From this tally, it should be obvious that the lowest bass frequencies will experience attenuation while the highest frequencies will see little reduction in amplitude. This is just what we want. The concept of a variation in output voltage with respect to frequency, or frequency response, will be examined in much greater detail in upcoming chapters. 60
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2.5 Concerning Practical Inductors 2.5 Concerning Practical Inductors Up to this point, inductors have been treated as ideal components, that is, pure inductance. In reality, all inductors have some resistance associated with them due to the resistance of the wire used to make the coil. This is called ESR, or Equivalent Series Resistance. It is also denoted as Rcoil. Ideally, this resistance will be small enough to ignore, but ultimately it will place a limit on the performance of any circuit that utilizes an inductor. While it is possible to measure the resistance of an inductor using a DMM, this will not yield an accurate value at all frequencies. In fact, as frequency increases, ESR will also increase. This is due to skin effect. At higher frequencies, current is not distributed equally throughout the cross-section of a conductor. In fact, it tends to “hug” the outer surface or “skin” of the conductor. This reduces the effective cross- sectional area and thereby increases the resistance2. In general, as frequency rises, so does Rcoil. Unfortunately, the situation is further complicated by distributed capacitance that will become an issue at still higher frequencies. As a consequence, manufacturers will give a “Q plot”, such as the one shown in Figure 2.20. 2 Recall that R = ρ l / A. A decrease in area A results in increased resistance, R. 61 Figure 2.20 Inductor Q graph. Courtesy of TDK
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The Q, or quality factor, of an inductor can be defined in terms of the peak energy stored in the device versus the energy dissipated per cycle. Keeping time constant, we can relate this to the power via the relation i2R. The current necessarily is the same for both the reactive and resistive components, therefore the Q of the coil is equal to the ratio of its reactance to its resistance at the frequency of interest. Qcoil = X L Rcoil (2.4) Where Qcoil is the quality factor of the inductor, XL is the magnitude of the inductive reactance at the frequency of interest, Rcoil is the resistance of the inductor at the frequency of interest. In general, the higher the Qcoil, the better. As can be seen in Figure 2.20, Qcoil is not a constant. Indeed, it increases with frequency until a peak is reached, at which point it begins to fall. When dealing with practical inductors, the effective Qcoil can be determined from a graph if the operating frequency is known. Once the Qcoil is found, the effective value of Rcoil can be found from Equation 2.4. This value can then be placed in series with the ideal inductor to create a more accurate result. For analysis, this pair is sometimes drawn with a box around it to denote that Rcoil is not a separate physical resistor, but is the effective resistance of the inductor. Example 2.9 Find the voltage across the inductor in the circuit of Figure 2.21. Assume the source voltage is 200° peak-to-peak at a frequency of 20 kHz. L is equal to 10 mH, Qcoil is 50, and R1 is 600 Ω. Remember, the inductor consists of both elements within the dashed box. First, find the magnitude of the inductive reactance. | X L| = 2π f L | X L| = 2π20 kHz 10 mH | X L| = 1257Ω We can now find Rcoil via Equation 2.4: Qcoil = X L Rcoil Therefore, 62 Figure 2.21 Circuit for Example 2.9.
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Rcoil = X L Qcoil Rcoil = 1257 50 Rcoil = 25.1Ω The impedance of the inductor is 25.1 + j1257 Ω , or 125788.9° Ω. The total impedance is 600 + 25.1 + j1257 Ω , or 140463.6° Ω. We can use a voltage divider to find the inductor's voltage. vind = e Z ind Z Total vind = 20 0°Vpp 125788.9°Ω 140463.6°Ω vind = 17.925.3° Vpp It is worth noting that, due to the coil resistance, the inductor's current and voltage are not precisely 90° out of phase, but rather 88.9°. That's a small but measurable deviation. 2.6 Summary 2.6 Summary 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. This remains true for AC circuits employing resistors, capacitors and inductors. The equivalent impedance of a set of resistors, inductors and capacitors placed in series is the vector sum of their resistance and reactance values. Inductive and capacitive reactance have opposing signs and will partially cancel each other. While resistance remains constant across frequency, reactance does not. Capacitive reactance decreases with frequency while inductive reactance increases. Therefore, the precise impedance value will vary with frequency. If the impedance angle is positive, the circuit is said to be inductive. If the impedance angle is negative, the circuit is said to be capacitive. Once inductance and capacitance values have been turned into their corresponding reactances at the frequency of interest, Ohm's law can be used to find the voltage across any component. The voltage will equal the product of the resistance or reactance and the current flowing through it. This is a vector computation. The current through a resistor will be in phase with the voltage across it. In contrast, the inductor voltage will lead the current and the resistor voltage by 90 degrees. Also, the capacitor voltage will lag the current and the resistor voltage by 90 degrees. From these observations it is apparent that the inductor and capacitor voltages in a series connection must be 180 degrees out of phase. 63
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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. This remains true for AC analysis, however, it is imperative that a vector summation is performed and not just a simple summation of the magnitudes. If the phase angles are ignored, it is possible for the voltage magnitudes to sum to considerably more than the source voltage. The voltage divider rule (VDR) remains true for AC analysis, but again, it is a vector computation. Thus, a rendering using just magnitudes will not produce a correct result. For example, a divider between a resistor and inductive reactance of equal value will not yield the source voltage splitting in half across each component. For highest accuracy, the coil resistance, or Rcoil, of inductors should not be ignored. It is treated as a small resistance in series with and integral to the inductor. Coil resistance is computed from the inductor's Q, or quality factor. Q is defined as the ratio of XL to Rcoil. Review Questions Review Questions 1. How is the equivalent impedance for a string of series connected resistors, capacitors and inductors computed? 2. Will the impedance of an RLC circuit be the same at all frequencies? Why/why not? 3. How are series connected AC voltage sources combined? 4. Does Kirchhoff's voltage law change for AC circuit analysis? 5. Is it possible for the voltage magnitudes of an AC RLC circuit to sum to more than the source voltage? Why/why not? 6. How is a phasor impedance plot for a series network related to its phasor voltage plot? 2.7 Exercises 2.7 Exercises Analysis Analysis 1. Determine the impedance of the circuit of Figure 2.22 for a 1 kHz sine. 2. Determine the impedance of the circuit of Figure 2.22 for a 5 kHz sine. 64 Figure 2.22
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3. Determine the impedance of the circuit of Figure 2.23 for a 10 kHz sine. 4. Determine the impedance of the circuit of Figure 2.23 for a 50 kHz sine. 5. Determine the impedance of the circuit of Figure 2.24 for a 1 kHz sine. 6. Determine the impedance of the circuit of Figure 2.24 for a 500 Hz sine. 7. Determine the impedance of the circuit of Figure 2.25. 8. In the circuit of Figure 2.25, if the input frequency is 100 Hz, what is the value of the inductor, in mH? 9. In the circuit of Figure 2.25, if the input frequency is 200 Hz, what is the value of the capacitor, in μF? 10. Draw the voltage and current waveforms for the circuit of Figure 2.26. 65 Figure 2.23 Figure 2.24 Figure 2.25 Figure 2.26
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11. Draw the voltage and current waveforms for the circuit of Figure 2.27. 12. Draw the voltage and current waveforms for the circuit of Figure 2.28 if E is a one volt peak sine at a frequency of 10 kHz and C = 3.3 nF. 13. Draw the voltage and current waveforms for the circuit of Figure 2.29 if E is a two volt peak-peak sine at a frequency of 40 Hz and L = 33 mH. 14. Draw the voltage and current waveforms for the circuit of Figure 2.30 if I is a 10 μA peak sine at a frequency of 2 kHz and C = 6.8 nF. 66 Figure 2.27 Figure 2.28 Figure 2.29 Figure 2.30
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15. Draw the voltage and current waveforms for the circuit of Figure 2.31 if I is a two amp peak-peak sine at a frequency of 40 Hz and L = 33 mH. 16. Determine the impedance of the circuit of Figure 2.32. 17. Determine the impedance of the circuit of Figure 2.32 using a frequency of 10 kHz. 18. For the circuit of Figure 2.32, determine the circulating current and the voltages across each component. Draw a phasor diagram of the three component voltages. Also find the time delay between the voltages of the components. 19. For the circuit of Figure 2.32 using a frequency of 10 kHz, determine the circulating current and the voltages across each component. Draw a phasor diagram of the three component voltages and determine the time delay between the capacitor and resistor voltages. 20. Determine the impedance of the circuit of Figure 2.33. 21. Determine the impedance of the circuit of Figure 2.33 using a frequency of 10 kHz. 67 Figure 2.31 Figure 2.32 Figure 2.33
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22. For the circuit of Figure 2.33, determine the circulating current and the voltages across each component. Also find the time delay between the voltages of the components. 23. For the circuit of Figure 2.33 with a frequency of 3 kHz, determine the circulating current and the voltages across each component. Also find the time delay between the voltages of the components. 24. For the circuit of Figure 2.34, determine the circulating current. 25. Determine the impedance of the circuit of Figure 2.34 using a frequency of 1.5 kHz. 26. For the circuit of Figure 2.34, determine the circulating current and the voltages across each component. Also find the time delay between the voltages of the components. 27. For the circuit of Figure 2.34 with a frequency of 1.5 kHz, determine the circulating current and the voltages across each component. Also find the time delay between the voltages of the components. 28. For the circuit of Figure 2.35, determine the circulating current and the voltages across each component. Draw a phasor diagram of the three component voltages and determine the time delay between the inductor and resistor voltages.. 68 Figure 2.34 Figure 2.35
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29. For the circuit of Figure 2.36, determine the circulating current and the voltages across each component. 30. For the circuit of Figure 2.37, determine the circulating current and the voltages across each component. 31. For the circuit of Figure 2.38, determine the applied voltage and the voltages across each component. 32. For the circuit of Figure 2.39, determine the applied voltage and the voltages across each component. 69 Figure 2.36 Figure 2.37 Figure 2.38 Figure 2.39
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33. For the circuit of Figure 2.40, determine the circulating current and the voltages across each component. 34. Repeat the previous problem using an input frequency of 10 kHz. 35. For the circuit of Figure 2.41, determine the circulating current and the voltages across each component. The source is a 10 volt peak sine at 20 kHz, R = 200 Ω, C = 100 nF and L = 1 mH. 36. For the circuit of Figure 2.41, find vb and vac. 37. For the circuit of Figure 2.42, find vb and vac. The source is a 50 volt peak- peak sine at 10 kHz, R = 100 Ω, C = 200 nF and L = 1 mH. 38. For the circuit of preceding problem, determine the circulating current and the voltages across each component. 70 Figure 2.40 Figure 2.41 Figure 2.42
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39. For the circuit of Figure 2.43, determine the circulating current and the voltages across each component. E is a 1 volt peak 2 kHz sine. Also, draw a phasor diagram of the four component voltages. 40. For the circuit of Figure 2.43, find vb and vca. E is a 1 volt peak 2 kHz sine. 41. For the circuit of Figure 2.44, determine vb, vc and vac. E is a 10 volt peak 15 kHz sine. 42. For the circuit of Figure 2.45, determine the circulating current and the voltages across each component. E is a 100 millivolt peak 250 Hz sine. Further, draw a phasor diagram of the four component voltages. 71 Figure 2.43 Figure 2.44 Figure 2.45
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43. For the circuit of Figure 2.46, determine the circulating current and the voltages across each component. E is a 2 volt RMS 1 kHz sine. Also, draw a phasor diagram of the four component voltages. 44. For the circuit of Figure 2.47, determine vb, vc and vac. E is a 1 volt peak 25 kHz sine. 45. For the circuit of Figure 2.48, determine the voltages across each component. The source is a 50 mA peak sine at 15 kHz, R = 200 Ω, C = 100 nF and L = 1.5 mH. 72 Figure 2.46 Figure 2.47 Figure 2.48
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46. For the circuit of Figure 2.49, determine vac, vb and vc. The source is a 10 mA peak-peak sine at 50 kHz, R = 2 kΩ, C = 10 nF and L = 800 μH. 47. For the circuit of Figure 2.50, determine the voltages across each component. The source is a 2 mA RMS sine at 1 kHz, R = 1.2 kΩ, C = 750 nF and L = 6.8 mH. 48. For the circuit of Figure 2.51, determine vac, vb and va. The source is a 2 mA peak-peak sine at 300 kHz, R = 560 Ω, C = 6.8 nF and L = 400 μH. 73 Figure 2.49 Figure 2.50 Figure 2.51
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49. For the circuit of Figure 2.52, determine the voltages across each component. 50. For the circuit of Figure 2.53, determine the voltages across each component. Further, draw a phasor diagram of the four component voltages. 51. For the circuit of Figure 2.54, find vac, vb and vc. The source is 5 mA peak at 8 kHz. 74 Figure 2.52 Figure 2.53 Figure 2.54
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52. For the circuit of Figure 2.55, determine the voltages across each component. The source is 20 mA peak at 100 kHz. 53. For the circuit of Figure 2.56, determine the voltages across each component. 54. For the circuit of Figure 2.57, determine the voltages vb and vdb. E1 = 20º and E2 = 590º. 55. Determine the inductance and capacitance values for the circuit of problem 52. 56. For the circuit of Figure 2.57, determine the inductor and capacitor values if the source frequency is 12 kHz. 75 Figure 2.55 Figure 2.56 Figure 2.57
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57. For the circuit of Figure 2.58, determine the voltages across each component. E1 = 10º and E2 = 860º. Design Design 58. Redesign the circuit of Figure 2.32 using a new capacitor such that the current magnitude from the source is 100 μA. 59. Redesign the circuit of Figure 2.33 using a new frequency such that the current magnitude from the source is 200 μA. 60. For the circuit of Figure 2.32, determine a new capacitor such that |XC| = R. 61. For the circuit of Figure 2.33, determine a new frequency such that |XL| = R. Challenge Challenge 62. For the circuit of Figure 2.40, determine a new frequency such that |XC| = |XL|. 63. Determine the output voltage across the capacitor of Figure 2.32 at frequencies of 100 Hz, 5 kHz and 20 kHz. In light of this, if the input signal was a 1 kHz square wave instead of a sine wave as pictured, how would this circuit affect the shape of the output waveform (hint: consider superposition)? 64. Assume that you are troubleshooting a circuit like the one shown in Figure 2.41. E is a 2 volt peak sine at 2 kHz, R = 390 Ω, C = 100 nF and L = 25 mH. The circulating current measures approximately 4 mA with a lagging phase angle of just under −40 degrees. What is the likely problem? 65. Given the circuit shown in Figure 2.41, find the values for C and L if the source is a 6 volt sine wave at 1 kHz, R = 2 kΩ, vR = 4 V and vL = 5 V. 76 Figure 2.58
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66. The circuit of Figure 2.59 can be used as part of a loudspeaker crossover network. The goal of this circuit is to steer the low frequency tones to the low frequency transducer (labeled here as “Loudspeaker” and often referred to as a woofer). A similar network substitutes a capacitor for the inductor to steer the high frequency tones to the high frequency transducer (AKA tweeter). These networks can be pictured as frequency sensitive voltage dividers. At very low frequencies, XC is very large and blocks low frequency tones from reaching the tweeter. A mirror situation occurs with the inductor/woofer variant. The crossover frequency is the frequency at which the reactance magnitude equals the resistance. Assuming simple 8 Ω resistances for the woofer and tweeter, determine capacitor and inductor values that would yield a 1.5 kHz crossover frequency. How might this concept be extended to a mid-range loudspeaker that only produces tones in the middle of the musical frequency spectrum? (Note, this concept will be revisited in the final simulation problem, below, and also in the Simulation portion of Chapter 4 which covers series-parallel circuits.) Simulation Simulation 67. Simulate the solution of design problem 58 and determine if the values produce the required results. 68. Simulate the solution of design problem 59 and determine if the values produce the required results. 69. Simulate the solution of design problem 60 and determine if the values produce the required results. Hint: if the reactance/resistance magnitudes are the same, then the voltage magnitudes will be identical. 70. Simulate the solution of design problem 61 and determine if the values produce the required results. Hint: if the reactance/resistance magnitudes are the same, then the voltage magnitudes will be identical. 77 Figure 2.59
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71. Simulate the solution of challenge problem 62 and determine if the new frequency produces the required results. Hint: if the reactance magnitudes are the same, then the voltage magnitudes will be identical. Further, their phases will cause these voltages to cancel, leaving the resistor voltage equal to the source voltage. 72. Using a transient analysis, crosscheck the crossover design of the final challenge problem, above. Plot the resistor (loudspeaker) voltage across the range of 100 Hz to 20 kHz for both sections. 78
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Notes Notes ♫♫ ♫♫ 79
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3 3 Parallel RLC Circuits Parallel RLC Circuits 3.0 Chapter Learning Objectives 3.0 Chapter Learning Objectives After completing this chapter, you should be able to: • Utilize Kirchhoff's current law, the current divider rule and Ohm's law to find branch currents in parallel RLC networks that utilize current sources or a single voltage source. • Compute complex impedance and system voltage in parallel RLC circuits. • Determine the equivalent value of multiple AC current sources in parallel. • Draw phasor diagrams for impedance/susceptance and component currents in parallel RLC circuits. 3.1 Introduction 3.1 Introduction In this chapter we introduce the analysis of AC RLC circuits configured in parallel. AC parallel circuits echo the characteristic of their DC counterpart and many of the solution techniques from DC analysis will be applicable here. This includes the use of Ohm's law and Kirchhoff's current law, along with the current divider rule. Generally, as with the series circuits presented in the previous chapter, reactance values will need to be computed from capacitor and inductor values before the main analysis may begin. Here, as in the most of the remaining chapters, we shall be concerned with determining the circuit response based on a source with a single frequency of excitation, in other words, a simple sine wave. Parallel circuits are in many ways the complement of series circuits. The most notable characteristic of a parallel circuit is that it has only two nodes and each component is connected from one node to the other. There are no other connections with which to create a voltage divider. Consequently, all components see the same voltage. Currents divide among the components in proportion to their conductance/susceptance (i.e., in inverse proportion to their resistance/reactance). The key to this is to remember that all computations involve vector quantities. This can lead to some surprising results for the uninitiated. For example, due to the 180 degree phase differential between inductors and capacitors, it is possible for an individual branch current to be greater than the source current. This does not violate Kirchhoff's current law, as we shall see. Indeed, it is reminiscent of a similar situation in AC series circuits where an individual component voltage can be greater than the source voltage without violating Kirchhoff's voltage law. To clarify our analyses, we shall make considerable use of both time domain plots of currents as well as phasor diagrams. 80
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3.2 The Parallel Connection 3.2 The Parallel Connection A parallel connection is always characterized by having all components share just two nodes. There are no other junctions from which current can flow out of or into, or for voltage to split. Consequently, The voltage is the same everywhere in a parallel connection. (3.1) Each component in such a connection will see the same voltage, regardless of whether it is a resistor, capacitor or inductor. Before the branch currents can be determined, the capacitive and inductive reactance values must be computed. As seen from prior work, these are obtained from the capacitor and inductor values, and based on the frequency of the driving source. Consequently, if this frequency were to change, the reactances would change, and this would in turn cause changes in the various component currents as well as the total current delivered by the source. 3.3 Parallel Impedance 3.3 Parallel Impedance Perhaps the first order of business is to determine equivalent impedance values for some collection of parallel components. Recall that the reciprocal of reactance is susceptance, B = 1 X (3.2) and that the reciprocal of impedance is admittance, Y = 1 Z (3.3) The units are siemens for each. It is also worth noting that, due to the division, the signs reverse. For example, a capacitive susceptance has an angle of +90 degrees and if a complex admittance has a negative angle, then the associated impedance is inductive. The “conductance rule” for parallel combinations studied in the DC case remains valid for the AC case, although we generalize it for impedances: Ztotal = 1 1 Z1 + 1 Z 2 + 1 Z3 +...+ 1 Z N (3.4) 81
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Each of the individual impedances presented in Equation 3.4 (i.e., Z1, Z2, etc) can represent a simple resistance, a pure reactance or a complex impedance. Further, the product-sum rule shortcut for two components also remains valid for AC components: Ztotal = Z 1×Z 2 Z 1+Z 2 (3.5) There is one special case where Equation 3.5 can be “troublesome”, and that's when the two impedances consist of a pure capacitive reactance and a pure inductive reactance, both of the same magnitude. The two items will effectively cancel each other leaving a denominator of zero and an undefined result. While the theoretical combination “blows up” and approaches infinity, in reality it is limited by associated resistances such as Rcoil, and arrives at some finite value This situation is studied in great depth in Chapter 8, which covers the concept of resonance. Example 3.1 Determine the impedance of the network shown in Figure 3.1. Equation 3.4 would be best here. Ztotal = 1 1 Z1 + 1 Z 2 + 1 Z 3 Z total = 1 1 j12 kΩ + 1 20 k Ω + 1 −j 48 kΩ Ztotal = 12.49E351.3° Ω This result might be a little surprising to the sharp-eyed. Notice that the magnitude of the total is larger than the magnitude of the smallest component (the inductor at j12 kΩ). This would never be the case if these three components were all resistors: the result would have to be smaller than the smallest element in the group. The reason for this is that the capacitive reactance partially cancels the inductive reactance. If the product-sum rule (Equation 3.5) is used with these two components, the result is 16E390° or j16 kΩ. Placing that in parallel with the 20 kΩ resistor (again using Equation 3.5) leads to the result computed above. An admittance diagram is illustrated in Figure 3.2. The vector summation of the component conductance and susceptances is verified nicely. 82 Figure 3.1 Circuit for Example 3.1.
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The individual component values are: BL = 1 j 12 kΩ ≈−j83.33E-6S BC = 1 −j 48k Ω ≈j 20.83E-6S G = 1 20 kΩ = 50E-6S Y total = 1 12.49E351.3°Ω ≈80.1E-6−51.3°S In rectangular form Ytotal = 50E−6 − j62.5E−6 S. 83 Figure 3.2 Admittance diagram for the network of Figure 3.1.
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Example 3.2 Determine the impedance of the network shown in Figure 3.3 at a frequency of 10 kHz. Repeat this for a frequency of 1 kHz. First, find the reactances at 10 kHz. For the inductor we find: X L = j 2π f L X L = j 2π10 kHz680μ H X L≈j 42.73Ω And for the capacitor: X C =−j 1 2 π f C X C =−j 1 2 π10kHz 470 nF X C ≈−j 33.86Ω Now use Equation 3.4 to combine the elements. Z total = 1 1 Z 1 + 1 Z2 + 1 Z3 Ztotal = 1 1 j 42.73Ω + 1 1.8k Ω + 1 −j 33.86Ω Ztotal = 162.6−84.8°Ω In rectangular form this is 14.68 − j161.9 Ω. As the capacitor's reactance is the smallest of the three components, it dominates the equivalent impedance at this frequency. By working the capacitive reactance formula in reverse, it can be shown that the reactive portion of − j161.9 Ω can achieved at this frequency by using a capacitance of 98.3 nF. That means that at 10 kHz, this parallel network has the same impedance as a 14.68 Ω resistor in series with a 98.3 nF capacitor. At any other frequency this will no longer be true, as will be illustrated momentarily. At 1 kHz, the frequency is reduced by a factor of ten. Therefore, XL will be ten times lower, or approximately j4.273 Ω. Further, XC will be ten times higher, or about − j338.6 Ω. The inductive reactance will now dominate. The new impedance is: 84 Figure 3.3 Circuit for Example 3.2.
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Ztotal = 1 1 Z1 + 1 Z 2 + 1 Z 3 Ztotal = 1 1 j 4.273Ω + 1 1.8k Ω + 1 −j338.6Ω Ztotal = 4.32889.9°Ω In rectangular form this is 10.4E−3 + j4.328 Ω. The result is inductive, the opposite of what we saw at 10 kHz. Using the inductive reactance formula, it can be shown that at 1 kHz this parallel network has the same impedance as a 10.4 milliohm resistor in series with a 689 μH inductor. 3.4 Parallel Circuit Analysis 3.4 Parallel Circuit Analysis 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 equal 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← (3.6) A similar situation occurs in AC parallel circuits to that seen AC series circuits; namely that it appears that KCL (like KVL) is “broken”. That is, a simplistic summing of the magnitudes of the currents might not balance. Once again, this is because the summation must be a vector summation, paying attention to the phase angles of each separate current. It is possible to drive a parallel circuit with multiple current sources. These sources will add in much the same way that voltage sources in series add, that is, polarity and phase must be considered. Ordinarily, voltage sources with differing values are not placed in parallel as this violates the basic rule of parallel circuits (voltage being the same across all components). The current divider rule remains valid for AC parallel circuits. Given two components, Z1 and Z2, and a current feeding them, IT, the current through one of the components will equal the total current times the ratio of the opposite component over the sum of the impedance of the pair. i1 = iT Z 2 Z 1+Z 2 (3.7) 85
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This rule is convenient in that the parallel equivalent impedance need not be computed, but remember, it is valid only when there are just two components involved. When analyzing a parallel circuit, if it is being driven by a voltage source, then this same voltage must appear across each of the individual components. Ohm's law can then be used to determine the individual currents. According to KCL, the total current exiting the source must be equal to the sum of these individual currents. For example, in the circuit shown in Figure 3.4, the voltage E must appear across both R and L. Therefore, the currents must be iL = E / XL and iR = E / R, and itotal = iL + iR. itotal can also be found be determining the parallel equivalent impedance of R and XL, and then dividing this into E. This technique can also be used in reverse in order to determine a resistance or reactance value that will produce a given total current: dividing the source by the current yields the equivalent parallel impedance. As one of the two is already known, the known component can be used to determine the value of the unknown component via Equation 3.4. Example 3.3 Determine the currents in the circuit shown in Figure 3.4 if the source is 100º volts peak, XL = j2 kΩ and R = 1 kΩ The same voltage must appear across all elements in a parallel connection. In this case that's 100º volts peak. The two branch currents are found via Ohm's law: iL = v Z iL = 100° V j2 kΩ iL = 5E-3−90° A iR = v Z iR = 100° V 1k Ω iR = 10E-30°A The source current is the sum of the these two currents, or 11.18E−3−26.6º amps. This can be verified by determining the equivalent parallel impedance and then applying Ohm's law: 86 Figure 3.4 A simple parallel RL circuit.
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ZT = Z 1×Z 2 Z 1+Z 2 ZT = 1k Ω× j 2k Ω 1k Ω+j 2k Ω ZT = 89426.6°Ω As a side note, in rectangular form this is 800 + j400, meaning that this parallel combination is equivalent to a series combination of an 800 ohm resistance and a 400 ohm inductive reactance. Continuing with Ohm's law, we have: iSource = v Z T iSource = 100° V 89426.6°Ω iSource = 11.18E-3−26.6° A A phasor diagram of the currents is shown in Figure 3.5, below. Current Measurement in the Laboratory Current Measurement in the Laboratory At this point, a practical question arises; how do we verify these currents in the laboratory? After all, the preeminent measurement tool is the oscilloscope, and these are designed to measure voltage, not current. While it is possible to obtain current measurement probes, a simple method to measure current is to use a small current sensing resistor with a standard oscilloscope setup. We have seen already that the voltage across a resistor must be in phase with the current through it. Therefore, 87 Figure 3.5 Phasor diagram of currents in the circuit of Figure 3.4.
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whatever phase angle we see on a resistor's voltage, its current phase angle must be the same. The idea is to simply insert a resistor into the branch of which we wish to measure the current. As long as the resistance is much smaller than the impedance seen in the rest of that branch, it will have little impact on the precise value of the current. We can then place a probe across this sensing resistor to read its voltage. Ohm's law is then be used to determine the current's magnitude, while the phase angle is obtained directly from the oscilloscope (again, because the sensing resistor's voltage must be in phase with its current, and thus the voltage phase angle is equal to the current phase angle). This technique will be illustrated in the simulation portion of the next example problem. Example 3.4 Determine the branch currents in the circuit of Figure 3.6. The first step is to determine the capacitive reactance. X C =−j 1 2 π f C X C =−j 1 2 π500Hz 250nF X C ≈−j1273Ω Both the resistor and capacitor will see 20 volts peak from the source. Their currents can be determined via Ohm's law: iC = v X C iC = 20 V 1273−90°Ω iC ≈15.71E-390° A iR = v R iR = 20 V 2200°Ω iR ≈90.91E-30° A The source current is the sum of these two currents, or 92.26E−39.8° amps. The source current may also be determined by dividing the source voltage by the equivalent parallel impedance, as follows. 88 Figure 3.6 Circuit for Example 3.4.
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ZT = Z 1×Z 2 Z 1+Z 2 ZT = 220Ω×(−j1273Ω) 220Ω−j 1273k Ω ZT = 216.8−9.8°Ω iSource = v Z T iSource = 20 0°V 216.8−9.8°Ω iSource = 92.26E-39.8° A A time domain plot of the currents is illustrated in Figure 3.7 along with a phasor plot in Figure 3.8. Note that the source current is close in both amplitude and phase to the resistor current. By comparison, the capacitor current is considerably smaller and with an obvious leading phase shift. 89 Time Domain Graph Current (milliamps) -100.0 -50.0 0.0 50.0 100.0 Time (milliseconds) 0 0.5 1 1.5 2 isource ic ir Figure 3.7 Current waveforms for the circuit of Figure 3.6. Figure 3.8 Phasor diagram for the circuit of Figure 3.6.
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Computer Simulation Computer Simulation The circuit of Figure 3.6 is captured in a simulator as shown in Figure 3.9. Individual 2 ohm resistors are used to sense the currents in the resistor and capacitor branches. These sensing resistors are inserted directly above ground for convenience of measure. In this way a differential measurement is not needed. The value of a sensing resistor needs to be considerably smaller than the impedance of the branch in which it is inserted. Two orders of magnitude (i.e., less than 1 %) would be a good place to start. While even smaller values will increase accuracy, from a practical standpoint in the lab, such tiny resistor values would produce extremely small voltages that would be difficult to measure with great accuracy. 90 Figure 3.9 Circuit of Figure 3.6 in a simulator with added current sensing resistors. Figure 3.10 Simulation results for the circuit of Figure 3.9.
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A basic transient analysis is then performed as shown in Figure 3.10. The results of the simulation are in tight agreement with the plot of Figure 3.7. Note that the plotted resistor and capacitor “currents” are, in fact, the voltages across the associated sensing resistors (nodes 2 and 3) divided by 2 (ohms), in direct application of Ohm's law. Not only does the simulation verify the results computed earlier, it also validates the concept of using small current sensing resistors in the laboratory. Example 3.5 Determine the the branch currents in the circuit of Figure 3.11. Use the source voltage as the reference (0° ). Although this circuit is drawn a little differently than the prior examples, it remains a simple parallel circuit with just two nodes. Therefore, each element sees a 2 volt peak potential. Ohm's law will suffice to find the three component currents. These are then added to find the source current. Their reference directions are right-to-left given the source's polarity. iC = v X C iC = 20° V 12 −90°Ω iC ≈0.166790° A iL = v X L iL = 20° V 890 °Ω iL = 0.25 −90 °A iR = v R iR = 20°V 100°Ω iR = 0.20°A The sum of these three currents is approximately 0.2167−22.6° amps. This value can be verified by finding the combined parallel impedance. Using Equation 3.4 it can be shown that this impedance is 9.23122.6° Ω. Thus, 91 Figure 3.11 Circuit for Example 3.5.
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iSource = v Z iSource = 20°V 9.23122.6°Ω iSource ≈0.2167−22.6° A A phasor diagram showing the vector summation of the currents is illustrated in Figure 3.12. Note that iC and iL are in perfect opposition as expected. Subtracting the magnitude of iC from iL leaves us with the precise vertical component of the source current. 92 Figure 3.12 Current phasor diagram for the circuit of Figure 3.11.
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If a parallel circuit is driven by a current source, as shown in Figure 3.13, there are two basic methods of solving for the component currents. The first method is to use the current divider rule. If desired, the component voltage can then be found using Ohm's law. An alternate method involves finding the parallel equivalent impedance first and then using Ohm's law to determine the voltage (remember, being a parallel circuit, there is only one common voltage). Given the voltage, Ohm's law can be used to find the current through one component. To find the current through the other, either Ohm's law can be applied a second time or KCL may be used; subtracting the current through the first component from the source current. If there are more than two components, usually the second method would be the most efficient course of action. It is worth noting that both methods described above will yield the correct answers. One is not “more correct” than the other. We can consider each of these as a separate solution path; that is, a method of arriving at the desired end point. In general, the more complex the circuit, the more solution paths there will be. This is good because one path may be more obvious to you than another. It also allows you a means of crosschecking your work. Example 3.6 Determine the currents in the circuit of Figure 3.13 if the source is 300 milliamps peak at 10 kHz, R is 750 Ω and L is 15 mH. The first step is to determine the inductive reactance. X L = j 2π f L X L = j 2π10 kHz15 mH X L≈j 942.4Ω The current divider rule can be used to find the current through the resistor. iR = iSource X L R+X L iR = 0.30° A 942.490 ° Ω 7500°Ω +942.490°Ω iR = 0.234738.5 °A KVL can be used to find the remaining current through the inductor as the source current must equal the sum of the inductor and resistor currents. Therefore: 93 Figure 3.13 Parallel network driven by a current source.
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iL = iSource−i R iL = 0.30° A −0.234738.5° A iL = 0.1868−51.5°A This value could also have been obtained by applying the current divider rule a second time: iL = iSource R R+X L iL = 0.30° A 7500°Ω 7500°Ω+942.490 °Ω iL = 0.1868−51.5°A A time domain plot of the current waveforms is shown in Figure 3.14. From the plot it is apparent that the inductor and resistor currents do indeed add up to the source current. Further, the current through the inductor is lagging, as expected. 94 Figure 3.14 Current waveforms for the circuit of Example 3.6. Time Domain Graph Current (amps) -0.3 -0.2 -0.1 0.0 0.1 0.2 0.3 Time (microseconds) 0 20 40 60 80 100 isource il ir
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Computer Simulation Computer Simulation The circuit of Example 3.6 is captured in a simulator as shown in Figure 3.15. A pair of 1 ohm current sensing resistors have been added. By using 1 ohm, the sensing voltage will have the same magnitude as the current, with no scaling required. A transient analysis is performed, plotting the voltages at nodes 2 and 3 along with their sum (the source current). The results are shown in Figure 3.16. The plot is delayed for a millisecond in order to avoid the initial power-up transient. The results are in full agreement with the plot of Figure 3.14. 95 Figure 3.15 Circuit of Example 3.6 in a simulator. Figure 3.16 Results of the transient response simulation for the circuit of Example 3.6.
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Example 3.7 Determine the branch currents in the circuit of Figure 3.17. Use the current source as the time reference (i.e., 0°). Instead of using repeated current dividers here, we can instead find the equivalent impedance and then apply Ohm's law to determine the system voltage. Each branch current may be obtained by dividing this voltage by the impedance of that branch; a quick application of Ohm's law. The system impedance is computed through Equation 3.4 as follows: Z total = 1 1 Z 1 + 1 Z 2 + 1 Z 3 Ztotal = 1 1 −j12Ω + 1 j8Ω + 1 10Ω Ztotal = 9.23122.6°Ω The system voltage is found using Ohm's law. As the reference direction of the current source is flowing down into ground, the branch currents will be flowing up. This makes their voltage reference negative with respect to ground (i.e., + to − from bottom to top). We can deal with this by describing the current source as negative (i.e., flowing out of the top node). The source can be written as either −10° or as 1180°, it makes no difference. v = i source×Z total v =−10° A×9.23122.6° v ≈−9.23122.6 °V This potential may also be written as 9.231−157.4° volts (remember, an inversion is the same as a 180 degree phase shift). And now for the branch currents: 96 Figure 3.17 Circuit for Example 3.7.
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iC = v X C iC = 9.231−157.4° V 12−90° Ω iC ≈0.7692−67.4 °A iL = v X L iL = 9.231−157.40° V 890°Ω iL = 1.154112.6° A iR = v R iR = 9.231−157.4°V 100°Ω iR = 0.9231−157.4°A The sum of these three currents is approximately 1180° amps (i.e., −10°), the value of the current source, as expected. A phasor diagram of the currents is plotted in Figure 3.18. 97 Figure 3.18 Current phasor diagram for the circuit of Figure 3.16.
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Now for the sneaky bit. In case you didn't notice, the components in this circuit are identical to those of Figure 3.11 with the exception of the source. As a consequence, the phase angles and magnitudes of the branch currents through the three components will remain unchanged relative to each other. Due to the change in both magnitude and direction of the source, the entire phasor plot has been rotated clockwise by 154.7 degrees and all of the magnitudes have been lengthened by the ratio of the voltages (9.231 / 2). This is perhaps easiest to see by focusing on the vector for isource in Figure 3.12. Rotate that vector clockwise, along with every other vector, until isource lines up with the negative real axis. The result is the plot of Figure 3.18, albeit with newly lengthened magnitudes. If a parallel circuit contains multiple current sources, they may be added together (a vector sum, of course) to generate an equivalent single current source. The analysis then proceeds as before. This is demonstrated in the following example. Example 3.8 For the circuit of Figure 3.19, determine the currents through the capacitor, inductor and resistor, and also determine the system voltage. i1 is 0.50° amps and i2 is 345° amps. The first step here is to determine the effective current driving the circuit. If we treat the top node as positive, i2 is entering (positive) while i1 is exiting (negative). The combination is: itotal = i1 +i2 itotal = −0.50° A +345° A itotal = 2.6752.6° A This current has the same reference direction as source two. A phasor diagram of the process is illustrated in Figure 3.20 to help visualize the vector addition. 98 Figure 3.19 Circuit for Example 3.8.
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The next step is to find the combined impedance so that we can find the applied voltage. Ztotal = 1 1 Z1 + 1 Z 2 + 1 Z 3 Ztotal = 1 1 −j100 Ω + 1 j50Ω + 1 40 Ω Ztotal = 37.1421.8°Ω v = i source×Z total v = 2.6752.6° A×37.1421.8° v ≈99.1674.4° V We now apply Ohm's law to each of the components to determine the branch currents. iC = v X C iC = 99.1674.4° V 100−90° Ω iC ≈0.9916164.4° A iL = v X L iL = 99.1674.4 °V 50 90° Ω iL = 1.983−15.6° A 99 Figure 3.20 Phasor diagram of the combined current sources for the circuit for Example 3.8.
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iR = v R iR = 99.1674.4° V 400°Ω iR = 2.47974.4 °A The sum of these three currents is approximately 2.6752.6° amps. This balances the net entering current from the two sources, verifying KVL. 3.5 Summary 3.5 Summary In this chapter we have examined parallel circuits using either a single voltage source, or one or more AC current sources, along with two or more resistors, capacitors and inductors. The defining characteristic of a parallel configuration is that all components are connected to just two nodes and that all elements in this configuration see the same voltage. If multiple AC current sources are present, they may be combined into a single equivalent current source via vector addition. Unlike the purely resistive case, the equivalent impedance of a group of parallel RLC components will not always be smaller than the smallest resistance or reactance in the group due to cancellations between the inductance and capacitance. As voltage is identical for all components, then the currents through the capacitors must be 180 degrees out of phase with the currents through the inductors. Thus it is quite possible that the current through one of the reactance branches could be larger than the current supplied by the source. In general, the effective impedance is found by summing the individual conductances and susceptances to find the total admittance of the group, and then taking the reciprocal of this value. The product-sum rule may be used so long as the angles are taken into account (i.e., a vector process). Kirchhoff's current law (KCL) states that the sum of currents entering a node must equal the sum of currents leaving that node. This remains true for AC circuit analysis, however, it must be remembered to always use a vector summation. A simple summation of current magnitudes will not achieve proper results. Individual branch currents in a circuit driven by a voltage source may be determined by using Ohm's law: simply divide the source voltage by the individual resistance and reactance values. These branch currents must sum to the total current delivered by the voltage source thanks to KCL. If the parallel network is driven by current sources, the individual branch currents can be found by first determining the effective parallel impedance and then using Ohm's law to find the system voltage. Once the voltage is known, Ohm's law is used again on each component to find the associated branch current. Alternately, the current divider rule may be used repeatedly on successive pairings of components. 100
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