text
stringlengths 1
7.76k
| source
stringlengths 17
81
|
|---|---|
71. Use any method or combination of methods in the circuit of Figure 5.95 to determine vad. E1 = 900°, E2 = 1200° and I = 400E−3180°. 72. Consider the 2.7 kΩ + j4 kΩ combo to be the load in Figure 5.60. Determine if this value achieves maximum load power. If not, determine a new value for the load in order to achieve maximum load power. Also determine the maximum load power. 73. In the circuit of Figure 5.96, assume the source E is 120 volts RMS at 60 Hz. Determine the value for the load, Z, that will produce maximum load power. Express Z in terms of a resistor and either an inductor or capacitor. Further, specify both the series and parallel equivalents for the load. 74. Convert the circuit of Figure 5.97 into the equivalent delta configuration. 201 Figure 5.95 Figure 5.96 Figure 5.97
|
ACElectricalCircuitAnalysis_Page_201_Chunk701
|
75. Find the Thévenin equivalent looking into nodes a and b for the circuit of Figure 5.98. I = 40°. 76. Find voltage vbc in the circuit of Figure 5.99 through the use of one or more delta-Y conversions. E = 1000°, R1 = R2 = 2 kΩ, R3 = 3 kΩ, R4 = 10 kΩ, R5 = 5 kΩ, XC1 = XC2 = −j2 kΩ. 77. Given the circuit of Figure 5.100, determine an equivalent circuit using a single voltage source. E1 = 1000°, E2 = 60180°, E3 = 4090°, E4 = 750°. 202 Figure 5.98 Figure 5.99 Figure 5.100
|
ACElectricalCircuitAnalysis_Page_202_Chunk702
|
78. As mentioned earlier, it is possible that certain AC Y and delta networks cannot be converted. Consider the circuit of Figure 5.101. Can this circuit be converted with a practical outcome? Why/why not? Simulation Simulation 79. Verify the voltage computed for problem 1 by running a transient analysis. 80. Verify the voltage computed for problem 4 by running a transient analysis. 81. Using multiple transient analysis simulations, compare the original circuit of problem 54 to its converted equivalent. Do this by connecting various components to the output terminals, trying several different impedance values and checking to see if the two circuits always produce the same voltage across this impedance. 82. Using multiple transient analysis simulations, compare the original circuit of problem 59 to its converted equivalent. Do this by connecting various components to the output terminals, trying several different impedance values and checking to see if the two circuits always produce the same voltage across this impedance. 83. Run a transient analysis to verify the design of problem 70. 84. Run a transient analysis to verify the design of problem 71. 203 Figure 5.101
|
ACElectricalCircuitAnalysis_Page_203_Chunk703
|
6 6 Nodal and Mesh Analysis Nodal and Mesh Analysis 6.0 Chapter Learning Objectives 6.0 Chapter Learning Objectives After completing this chapter, you should be able to: • Utilize nodal analysis techniques to solve for voltages in multi-source series-parallel RLC networks. • Utilize mesh analysis techniques to solve for currents in multi-source series-parallel RLC networks. • Analyze networks that use dependent voltage and/or current sources. 6.1 Introduction 6.1 Introduction This chapter presents methods for the analysis of AC circuits that employ resistors, capacitors and inductors along with any number of voltage and/or current sources. The methods of interest are nodal analysis and mesh analysis. Nodal analysis is the most general technique and can be applied to virtually any circuit. Mesh analysis is nearly as versatile and works well if only voltage sources are present. Both analysis methods generate a system of simultaneous linear equations that are used to solve the circuit for desired voltages or currents. That is, the system generates a set of values, either currents or node voltages, rather than individual currents or voltages. There are several methods that can be used to solve the simultaneous equations. These include substitution, Gauss-Jordan elimination and expansion by minors. These methods are reviewed in Appendix B and are not covered in this chapter. Instead, to focus on the circuit analysis aspects with minimal distraction, the explanations and examples will simply detail the process of examining the circuit and developing the system of equations. The specific technique employed to solve these simultaneous equations depends solely on your personal preferences. At this point in the study of AC circuits, it is particularly efficient to obtain an advanced scientific calculator that can solve the system of equations directly versus working through the solution manually. By doing so, you can spend your time more effectively; meaning, mastering the process of circuit analysis and creating the equations. Manual solution techniques, though not necessarily difficult, can be tedious, time consuming and error prone. Indeed, on larger circuits, there can be a 10:1 differential in time when using a capable calculator versus a standard scientific calculator9. If you have not already done so, you should consider obtaining a calculator that can solve simultaneous equations with complex coefficients (i.e., the complex real/imaginary quantities we have been using). Such calculators can be expensive when purchased new, such as the Texas Instruments TI-89 and Nspire models. On the used market, perfectly satisfactory older models such as the TI-85 and TI-86 can be found at considerable discount. Another model to consider is the Casio FX-9750GII, although it is not quite as powerful as some of the other units mentioned. 9 That's like doing three times as many problems in one-third the time. We don't often get these kinds of opportunities. 204
|
ACElectricalCircuitAnalysis_Page_204_Chunk704
|
Along with nodal and mesh, we shall also introduce the concept of dependent AC sources. Dependent sources do not exhibit a fixed value, but rather the current or voltage is dependent on some other current or voltage in the circuit. What makes this interesting is that this controlling current or voltage may itself be affected by the value produced by the dependent source. Dependent sources are not lab instruments, like signal or function generators. Instead, they are used to model the behavior of active electronic devices such as bipolar and field effect transistors. Mastering the analysis of circuits using dependent sources is critical to the understanding of active circuitry that use transistors and similar devices10. 6.2 Nodal Analysis 6.2 Nodal Analysis Nodal analysis can be considered a universal solution technique as there are no practical circuit configurations that it cannot handle. It does not matter if there are multiple sources or if there are complex configurations that cannot be reduced using series-parallel simplification techniques, nodal analysis can handle them all. Further, nodal analysis tends to “give us what we want”, namely, a set of node voltages for the circuit. Once the node voltages are obtained, finding any branch currents or component powers becomes an almost trivial exercise. Nodal analysis relies on the application of Kirchhoff's current law to create a series of node equations that can be solved for node voltages. These equations are based on Ohm's law and will be of the form i = v/Z, or more generally, i = (1/ZX)∙vA + (1/ZY)∙vB + (1/ZZ)∙vC... We will examine two variations on the theme; first, a general version that can be used with both voltage and current sources, and a second somewhat quicker version that can be used with circuits only driven by current sources. General Method General Method Consider the circuit shown in Figure 6.1. We begin by labeling connection nodes. We are interested in identifying current junctions, that is, places where currents can combine or split. These are also known as summing nodes and are circled in blue on the figure. We do not concern ourselves with points where just two components connect without any other connection, such as points a and c. Once the proper nodes are identified, reference current directions are assigned. The reference current directions are chosen arbitrarily and for convenience. They may be the opposite of reality. This is not a problem. If we assign directions that are reversed, we'll simply wind up with a current version of a double negative, and the computed node voltages will work out just fine. 10 For more on transistors and other semiconductors, see Semiconductor Devices: Theory and Application. Another free OER text by the author. 205
|
ACElectricalCircuitAnalysis_Page_205_Chunk705
|
One node is chosen as the reference. This is the point to which all other node voltages are measured against. Typically, the reference node is ground, although it does not have to be. We now write a current summation equation for each summing node, except for the reference node. In this circuit there is only one node where currents combine (other than ground) and that's node b. Points a and c are places where components connect, but they are not summing nodes, so we can ignore them for now. Using KCL on node b we can say: i1 + i2 = i3 Now we'll describe these currents in terms of the source and node voltages, and associated components. For example, i3 is the node b voltage divided by −jXC while i1 is the voltage across R divided by R. This voltage is va − vb. va −v b R + v c −v b jX L = v b −jX C Noting that va = E1 and vc = E2, with a little algebra this can be reduced to: ( 1 R)E1 +( 1 jX L)E2 =( 1 R + 1 jX L + 1 −jX C)v b All quantities are known except for vb and thus it is easily found with a little more algebra. If there had been more nodes, there would have been more equations and more unknowns, one for each node. As we shall see, this conductance-voltage product format turns out to be a convenient way of writing these equations. Also, note that the first two terms on the left reduce to fixed current values. For current sources, the process is similar but a bit more direct. Consider the circuit of Figure 6.2. We start as before, identifying nodes and labeling currents. We then write the current summation equations at each node (except for ground). We consider currents entering a node as positive and exiting as negative. There are two nodes of interest here, and thus, two equations each with two unknowns will be generated. 206 Figure 6.1 A simple dual voltage source circuit with the currents and nodes defined.
|
ACElectricalCircuitAnalysis_Page_206_Chunk706
|
Node a: I1 = i3 + i4 Node b: i3 + I2 = i5, and rearranging in terms of the fixed source, Node b: I2 = −i3 + i5 The currents are then described by their Ohm's law equivalents: Node a: I 1 = va −vb R2 + v a R1 Node b: I 2 = −va −vb R2 + vb jX L Expanding and collecting terms yields: Node a: I 1 =( 1 R1 + 1 R2)va −( 1 R2)v b Node b: I 2 = −( 1 R2)va +( 1 R2 + 1 jX L)vb As the impedance values and currents are known, simultaneous equation solution techniques may be used to solve for the node voltages. Once again, there are as many equations as node voltages. One practical point before continuing: It is very important that the coefficients for the various node voltage terms “line up” when the final system of equations is written out. That is, there should be a column for the va terms, a column for the vb terms, and so on. They should not be written out in random order, but rather following the style shown in Figure 6.3. This format will make it much it easier to enter the coefficients into a calculator or solve manually. Further, the set of coefficients must show diagonal symmetry. That is, if we draw a major diagonal from upper-left to lower-right (red), whatever coefficients are above-right from the diagonal should be mirrored below-left of the diagonal (blue, purple, green). If the set of values does not show diagonal symmetry, an error has been made. You must go back and recheck the original node summations. Simple as that. Even the simple 2x2 of Figure 6.2 shows this symmetry (namely, the coefficient of −1/R2 for vb in the first equation and va in the second). 207 Figure 6.2 A dual current source circuit with the currents and nodes defined. Figure 6.3 Diagonal symmetry. (1+j2) a -(-j2) a -(1) a -(-j2) b (2-j3) b -(-j3) b -(1) c -(-j3) c (5-j3) c v v v v v v v v v
|
ACElectricalCircuitAnalysis_Page_207_Chunk707
|
Example 6.1 Determine the voltage across the inductor in the circuit of Figure 6.4. Source one is 50° volts RMS and source two 290° volts RMS. Other than ground, there is only one current summing node in this circuit, and that's the junction at the top of the inductor. We will refer to this junction as node a. Following the outline of Figure 6.1, we define three currents; i1 entering from the left, i2 entering from the right, and i3 exiting down through the inductor. i1 + i2 = i3 Next, these currents are described in terms of the voltages and components. We'll number the resistors from left to right. E1 −v a R1 −jX C + E2 −v a R2 = va jX L This can be rearranged as: ( 1 R1 −jX C)E1 +( 1 R2)E2 =( 1 R1 −jX C + 1 R2 + 1 jX L)v a Populate with values: ( 1 500 −j 300Ω)50°V +( 1 400Ω)290° V =( 1 500−j 300Ω + 1 400Ω + 1 j 200Ω)v a This simplifies to: 8.575E-331° A +5E-390° A = (5.72E-3−46 °S)va Solving for the unknown, we find that va = 2.08798° volts RMS. 208 Figure 6.4 Circuit for Example 6.1.
|
ACElectricalCircuitAnalysis_Page_208_Chunk708
|
Example 6.2 In the circuit of Figure 6.5, determine va and vb. E is 200° volts peak while I is 0.10° amps peak. The system frequency is 2 kHz. There are two nodes of interest here, other than ground. This means we will generate two equations with two unknowns (va and vb). Using the standard reactance formulas, the inductive and capacitive reactances are found to be j125.7 Ω and −j159.2 Ω, respectively. If we assume the reference direction for current is from node a to node b, and that the current flow through the two center resistors is downward, the equations are: Node a: 20 0°V −v a 100Ω+ j125.7Ω = va 250Ω + v a −vb −j159.2Ω Node b: va −vb −j159.2Ω +0.10° A = vb 400Ω Expanding and collecting terms yields: Node a: 0.1245−51.5° A =( 1 250Ω + 1 100Ω+ j 125.7Ω + 1 −j 159.2Ω)v a −( 1 −j 159.2Ω)v b Node b: 0.10° A = −( 1 −j 159.2Ω)va +( 1 400Ω + 1 −j 159.2Ω)vb These are simplified, ready for manipulation (note diagonal symmetry). 0.1245−51.5° A = (8E-310.1 °S)va −(6.281E-390°S)vb 0.10°A = −(6.281E-390 °S)va +(6.76E-368.3° S)vb After solving the system of equations, we see that va = 16.240.09° volts and vb = 20.99−22.3° volts. 209 Figure 6.5 Circuit for Example 6.2.
|
ACElectricalCircuitAnalysis_Page_209_Chunk709
|
Computer Simulation Computer Simulation To verify the results of the preceding example, the circuit of Figure 6.5 is captured in a simulator as shown in Figure 6.6. A transient analysis is performed on the circuit. Node voltages 1, 3 and 4 are plotted, corresponding to the voltage source and nodes a and b, respectively, in Figure 6.7. 210 Figure 6.6 The circuit of Example 6.2 captured in a simulator. Figure 6.7 Simulation results for the circuit of Figure 6.6.
|
ACElectricalCircuitAnalysis_Page_210_Chunk710
|
The amplitudes are just as computed. Node voltage a appears to be nearly in phase with the voltage source, as expected. Node voltage b lags the source by between one-quarter to one-third of a division, or some 30 microseconds. For a 2 kHz source, this translates to around −22 degrees, verifying the calculated result. Example 6.3 For the circuit of Figure 6.8, find va and vb. The system frequency is 1 kHz. I1 = 2.50° A and I2 = 10° A. We will generate two equations with two unknowns, va and vb. The reactance formulas yield j6.28 Ω and −j15.9 Ω for the inductor and capacitor. If we assume the reference direction for current is from node a to node b, and that the current flow through the capacitor and inductor is from nodes a and b downward, the equations are: Node a: 2.50° A = va −j 15.9Ω + va −vb 10 Ω Node b: va −v b 10Ω = 10° A + vb 4Ω+ j6.28Ω Expanding and collecting terms yields (note diagonal symmetry): 2.50° A =( 1 10Ω + 1 −j 15.9Ω)va −( 1 10Ω)vb −10° A =−( 1 10Ω)v a +( 1 10Ω + 1 4Ω + j6.28Ω)vb 2.50° A = (0.11832.2 °S)va −(0.10°S)v b −10° A = −(0.10°S)v a +(0.206−33.3° S)vb The results are: va = 30.39-38.7° volts and vb = 11.37−20.8° volts. 211 Figure 6.8 Circuit for Example 6.3.
|
ACElectricalCircuitAnalysis_Page_211_Chunk711
|
Computer Simulation Computer Simulation To verify the results of the preceding example, the circuit of Figure 6.8 is captured in a simulator as shown in Figure 6.9. A transient analysis is performed on the circuit. Node voltages 1 and 2 (i.e., nodes a and b, respectively) are plotted in Figure 6.10. 212 Figure 6.9 The circuit of Example 6.3 captured in a simulator. Figure 6.10 Simulation results for the circuit of Figure 6.9.
|
ACElectricalCircuitAnalysis_Page_212_Chunk712
|
The simulation results agree nicely with the computed values in terms of both amplitude and phase. Inspection Method Inspection Method The system of equations can be obtained directly through inspection if the circuit contains current sources and no voltage sources. Let's take another look at the equations developed in the preceding example. For convenience, the circuit is reproduced in Figure 6.11 with reactance values. I1 = 2.50° A and I2 = 10° A. 2.50° A =( 1 10Ω + 1 −j 15.9Ω)va −( 1 10Ω)vb −10° A =−( 1 10Ω)v a +( 1 10Ω + 1 4Ω + j6.28Ω)vb The top equation was built around a current summation at node a while the bottom was built around a summation at node b. The first thing that might be apparent is that on the left of the equals signs are the current sources connected to these nodes. Positive means the current is entering while negative denotes an exiting current. The second thing is that, for the node of interest (node a for the top equation, node b for the bottom), the coefficients represent the items connected to that particular node. For example, in the top equation, the components connected to node a are the 10 Ω resistor and the − j15.9 Ω reactance. Likewise, in the bottom equation, the components connected to node b are the 10 Ω resistor and the 4 + j6.28 Ω impedance. The third thing is that the remaining coefficients consist of the components that are in common between the node of interest and the other node (i.e., 10 Ω connects a to b for the first equation, and also connects b to a for the second equation). These other connections always show up as negative. The reason for this should be apparent if you examine the structure of the original equations from Example 6.3. If there is no bridging element between a node and the node of interest, then that coefficient will be zero. If voltage sources exist in the circuit, source conversions can be used to obtain an equivalent circuit that uses only current sources. 213 Figure 6.11 Circuit of Example 6.3 with reactances.
|
ACElectricalCircuitAnalysis_Page_213_Chunk713
|
The huge advantage of the inspection method is that it cuts out a time consuming and error prone section of the process, namely converting the original KCL summations into a set of simplified equations with coefficients for each unknown. The inspection method generates the equations directly. To further speed the process, it can be useful to turn each impedance value into a corresponding admittance value before creating the equations. In this way, the reciprocals are computed once for each item rather than multiple times in multiple equations. Finally, remember that the resulting set of equations must exhibit diagonal symmetry, as shown back in Figure 6.3. The inspection method is summarized as follows: 1. Verify that the circuit uses only current sources and no voltage sources. If voltage sources exist, they must be converted to current sources before proceeding. 2. Find all of the current summing nodes and number (or letter) them. Also decide on the reference node (usually ground). 3. To generate an equation, locate the first node. This is the node of interest and the next few steps will be associated with it. 4. Sum the current sources feeding the node of interest. Entering is deemed positive while exiting is deemed negative. The sum is placed on one side of the equals sign. 5. Next, find all of the impedances connected to the node of interest and write them as a sum of admittances on the other side of the equals sign, the group being multiplied by this node's voltage (e.g., v1). That makes one term. 6. Now for the other terms. Find all of the admittances that are connected between the node of interest and the next node (e.g., node 2). Sum these together and multiply the group by this other node's voltage (e.g., v2). Subtract that product from the equation built so far. Repeat this process until all of the other nodes have been examined (except ground). If there are no common impedances between the node of interest and the other node, use zero for the coefficient of that node's voltage. Once all other nodes are considered, this equation is finished. 7. Go to the next node and treat this as the new node of interest. 8. Repeat steps 4 through 7 until all nodes have been treated as the node of interest. Each iteration creates a new equation. There will be as many equations as there are nodes, less the reference node. Check for diagonal symmetry and solve. The inspection method is best observed in action, and is used in the following example. 214
|
ACElectricalCircuitAnalysis_Page_214_Chunk714
|
Example 6.4 Write the system of equations for the circuit of Figure 6.12. I1 = 100° A and I2 = 490° A. We begin at node a, the first node of interest. Find all of the current sources connected to this node. All we have is I1. It is exiting, and thus negative. −10 0°A = ... Now find all of the items connected to this node and create a sum of admittances. −10 0°A =( 1 4Ω + 1 2Ω + 1 8Ω)v a... Include the terms that are common between node a and node b. This is negative. −10 0°A =( 1 4Ω + 1 2Ω + 1 8Ω)v a −( 1 2Ω)vb... And finally, include the terms common between node a and node c. This is also negative. −10 0°A =( 1 4Ω + 1 2Ω + 1 8Ω)v a −( 1 2Ω)vb −( 1 8Ω)vc The first equation is done. We now make node b the node of interest and repeat the process. 215 Figure 6.12 Circuit for Example 6.4.
|
ACElectricalCircuitAnalysis_Page_215_Chunk715
|
Find all of the current sources connected to this node. There are none. 0 = ... Find all of the items connected to this node and create a sum of admittances. 0 = ...+( 1 1Ω + 1 2Ω + 1 −j5Ω)v b... Include the terms that are common between node a and node b. This is negative and goes into the lead (a before b) to keep everything nicely lined up. 0 =−( 1 2Ω)v a +( 1 1Ω + 1 2Ω + 1 −j5Ω)v b... Now include the terms common between node b and node c. This is also negative and is inserted at the tail (c after b). 0 =−( 1 2Ω)v a +( 1 1Ω + 1 2Ω + 1 −j5Ω)v b −( 1 1Ω)vc The second equation is finished. We now make node c the node of interest and repeat the process for the final time. Find all of the current sources connected to node c. We have both I1 and I2 entering. 100° A+4 90 °A = ... This current is equivalent to 10.7721.8°. Now find all of the items connected to this node and create a sum of admittances. 10.7721.8°A = ...( 1 1Ω + 1 8Ω + 1 j10Ω)v c Include the terms that are common between node c and node a. This is negative and goes into the lead (a before c). 10.7721.8°A =−( 1 8Ω)va... +( 1 1Ω + 1 8Ω + 1 j10 Ω)v c Now include the terms common between node b and node c. This is also negative and is inserted in the middle (b before c). 216
|
ACElectricalCircuitAnalysis_Page_216_Chunk716
|
10.7721.8°A =−( 1 8Ω)va −( 1 1Ω)v b +( 1 1Ω + 1 8Ω + 1 j10 Ω)vc The third and final equation is finished. The completed set of equations is: −100° A =( 1 4Ω + 1 2Ω + 1 8Ω)va −( 1 2Ω)vb −( 1 8Ω)vc 0 =−( 1 2Ω)va +( 1 1Ω + 1 2Ω + 1 −j5Ω)vb −( 1 1Ω)v c 10.7721.8° A =−( 1 8Ω)va −( 1 1Ω)vb +( 1 1Ω + 1 8Ω + 1 j 10Ω)vc Note that the set exhibits diagonal symmetry and that all coefficient groups are negative except for those along the major diagonal. Consequently, the coefficient groups may now be simplified to obtain single coefficients for the unknowns, and the equations are ready for solution. The results are: va = 10.972.8° volts, vb = 23.634.7° volts and vc = 31.237.3° volts. These values can be crosschecked by using them to find the currents through each component, and then verifying KCL for each node. Although this example may appear to be somewhat long winded, with a little practice the process will become second nature. At that point, the set of equations can be created quickly and with little possibility of error, even for large circuits with many nodes. Using Source Conversions Using Source Conversions As mentioned previously, given circuits with voltage sources, it may be easier to convert them to current sources and then apply the inspection technique rather than using the general approach outlined initially. There is one trap to watch out for when using source conversions: the voltage across or current through a converted component will most likely not be the same as the voltage or current in the original circuit. This is because the location of the converted component will have changed. For example, the circuit of Figure 6.5 (Example 6.2) could be solved using the inspection method of nodal analysis by converting the voltage source and its associated impedance of the 100 Ω resistor in series with the 10 mH inductor into a current source. Although the associated impedance still connects to the converted source, the other end no longer connects to node a. Rather, it would connect to ground. Therefore, the voltage drop across this impedance in the converted circuit is not likely to equal the voltage drop seen across it in the original circuit (the only way they would be equal is if the voltage source E turned out to be 0). 217
|
ACElectricalCircuitAnalysis_Page_217_Chunk717
|
Supernode Supernode From time to time you may see a circuit utilizing an ideal voltage source like the one shown in Figure 6.13. That is, this voltage source does not have a series impedance associated with it. Without that impedance, it becomes impossible to create an expression for the current passing through the source using the general method, and impossible to convert the voltage source into a current source in order to use the inspection method. There are a few of ways out of this quandary. The first way is to recognize that all realistic sources have some internal impedance, so we simply add a very small resistor in series with the source so that a source conversion is possible. Of course, not just any resistor will do. In order to maintain accuracy, the newly added resistance has to be much smaller than any surrounding resistances or reactances. A reduction by two orders of magnitude generally yields a variation smaller than that produced by component tolerances in all but high precision circuits and will usually do the trick. Still smaller values will further increase accuracy. Another way out is to use a supernode. A supernode is, in effect, the combination of two nodes. It relies on a simple observation. If we examine the circuit of Figure 6.13, the path of the voltage source produces identical currents flowing into and out of nodes a and b. As a consequence, if we treat the two nodes as one big node, then when we write a KCL summation, these two terms will cancel. To see just how this works, refer to Figure 6.14. In this version we have replaced the voltage source with its ideal internal impedance; a short. We have also labeled the two nodes of interest, a and b, and labeled the currents, drawn with convenient reference directions. The specific choice of direction will not matter, just use whatever scheme seems appropriate. Due to the shorted voltage source, nodes a and b are now the same node. Take a look at the currents entering and exiting this combined “super” node. On the left side (formerly node a) we see a constant current Ix entering while i1, i2 and i3 are exiting. On the right side (formerly node b) we see the Iy entering along with i1 and i2, and 218 Figure 6.13 Bridging voltage source without an internal impedance. Figure 6.14 Circuit modified for supernode analysis.
|
ACElectricalCircuitAnalysis_Page_218_Chunk718
|
exiting we see i4. At this point we'll create an expression where all of currents entering the super node are on the left side of the equals sign and all of the exiting currents are on the right: Σiin = Σiout I x+ I y+i1+i2 = i1+i2+i3+i4 This can be simplified to: I x+ I y = i3 +i4 Writing this in terms of Ohm's law we have: I x+I y = 1 −j X C va + 1 j X L vb We also know that va − vb = E from the original circuit. We know this because the reference polarity of the source is + toward the a node and − toward the b node. Therefore it must be va − vb and not vb − va. Assuming all sources and components are known, that makes two equations with two unknowns, solvable using simultaneous equation techniques. This is illustrated in the following example. Example 6.5 Find va and vb for the circuit of Figure 6.15. E = 160° volts, Ix = 0.10° amps and Iy = 0.2590° amps. The circuit is redrawn in Figure 6.16 with nodes and currents labeled. We short the 16 volt source and write a current summation at the a:b supernode: Σiin = Σiout 0.10°A+0.2590° A+i1+i2 = i1+i2+i3+i4 This can be simplified to: 0.269368.2°A = i3 +i4 Writing this in terms of Ohm's law we have: 0.269368.2°A = 1 −j100Ωva + 1 j500Ω vb 0.269368.2°A = j10 mSv a −j2 mSvb 219 Figure 6.15 Circuit for Example 6.5.
|
ACElectricalCircuitAnalysis_Page_219_Chunk719
|
We also know that va − vb = 160° volts. Therefore vb = va − 160° volts. Substituting this into the prior equation yields: 0.269368.2°A = j10 mSv a −j2 mS(va −160°V) 0.269368.2°A = j10 mSv a −j2 mSva +32E-3 90 °A 0.239467.9° A = j8 mSva va = 29.92−22.1° V We know that vb is 160° volts below va, and thus after subtracting, we find vb = 16.24−43.8° volts. To verify, we will perform a KCL summation at each node. For node a, assuming i1 exits as drawn: i1 = 0.10° A − v a −j100Ω −v a−v b 200Ω i1 = 0.10°A−29.92−22.1° V −j100Ω −29.92−22.1° V−16.24−43.8° V 200Ω i1 = 0.292−108°A Doing likewise for node b, and assuming i1 enters as drawn: i1 =−0.2590° A+ vb j500Ω −v a−vb 20 Ω i1 =−0.2590° A+16.24−43.8 ° V j500Ω −29.92−22.1° V−16.24−43.8° V 200Ω i1 = 0.292−110° A Other than the small deviation due to accumulated rounding, these currents match. That means that the current through the voltage source is verified to be the same at both terminals, as it must be. An alternative to the basic supernode technique is to recognize that the two nodes on either side of the voltage source are effectively locked together by the source voltage. That is, if one of the node voltages is found, then the other may be determined by adding or subtracting the source voltage to or from the known node voltage, depending on the reference polarity. This idea is exploited by simply describing one node voltage in terms of the other at the outset. This will reduce the total number of unknowns by one and reduce the system of equations by one. The technique is illustrated in the example following. 220 Figure 6.16 Circuit modified for supernode analysis.
|
ACElectricalCircuitAnalysis_Page_220_Chunk720
|
Example 6.6 Find node voltages va, vb and vc for the circuit of Figure 6.17. The sources are: E = 200° volts and I = 245° amps. Once again we have a situation of a voltage source lacking a series impedance which makes a source conversion impossible. Without having to short it and thus treating nodes a and c as an explicit supernode, we can take an alternate route. We begin by noting that the currents entering and exiting the voltage source must be identical. The circuit is redrawn in Figure 6.18 with the nodes and convenient current directions labeled. The circuit also uses equivalent conductances and susceptances in place of the original resistances and reactances in order to speed the process of simplifying the equations. Unlike the basic supernode technique, this time the voltage source is left in. The key observation is that vc = va − 200° V. In other words, vc is locked to va and if we find one of them, we can determine the other. Therefore, instead of writing three equations using three unknowns, we shall instead refer to node c in reference to node a. In other words, wherever we need vc we instead shall write va − 200° V. Thus, this three node circuit will only need two equations. We begin at node a and apply KCL as usual. Σiin = Σiout i1+i3 = i2 This is expanded using Ohm's law and we solve for i1: i1 = i2 −i3 i1 = j0.5Sv a −0.25S(vb −v a) i1 = (0.25 +j0.5)Sva −0.25Sv b On to node b: I = i3+i4 245 °A = 0.25S(v b −va) +0.1S(vb −v c) 245 °A = 0.25S(v b −va) +0.1S(vb −(v a −20 0° V)) 245 °A = 0.25S(v b −va) +0.1S(vb −v a +200° V) 245 °A = 0.25S(v b −va) +0.1S(vb −v a) +20°A 1.531112.5° A =−0.35Sv a +0.35Sv b 221 Figure 6.17 Circuit for Example 6.6. Figure 6.18 Circuit of Example 6.6 with currents labeled and using conductances.
|
ACElectricalCircuitAnalysis_Page_221_Chunk721
|
And finally node c: i4 = i1 +i5 i1 = i4 −i5 i1 = 0.1S(vb −v c)−(−j0.2S)vc i1 = 0.1S(vb −(v a−20 0° V)) +j 0.2S(va −20 0°V) i1 = 0.1S(vb −v a +20 0°V) +j0.2S(va −200° V) i1 = 0.1S(vb −v a) +j 0.2Sva +(2 −j 4)A i1 = (−0.1 +j0.2)Sv a +0.1Svb +4.472 −63.4°A The equations for nodes a and c both equal i1, thus they equal each other. (0.25+j 0.5)Sv a −0.25Svb = (−0.1 +j 0.2)Sv a +0.1Svb +4.472−63.4° A 4.472−63.4 °A = (0.35 +j 0.3)Sv a −0.35Svb The final equations are: 4.472−63.4° A = (0.35 +j0.3)Sva −0.35Svb 1.531112.5°A =−0.35Sv a +0.35Svb The solution is va = 9.823151.3 − ° volts and vb = 10.31176.2 − ° volts. As vc is 200° volts less than va, then vc = 29170.6 − ° volts. KCL summations at each of the three nodes will verify these values. 6.3 Mesh Analysis 6.3 Mesh Analysis Mesh analysis is similar to nodal analysis in that it can handle complex multi-source circuits. In some ways it is the mirror image of nodal analysis. While nodal analysis uses Kirchhoff's current law to create a series of current summations at various nodes, mesh analysis uses Kirchhoff's voltage law to create a series of loop equations that can be solved for mesh currents. The current through any particular component may be a mesh current or a combination of mesh currents. Of course, once those currents are found, it is a short hop to find any desired voltage. Mesh does have one limitation that nodal doesn't: Mesh analysis requires that the circuit be planar. That is, the circuit must be able to be drawn on a flat surface without any wires crossing each other. Another way of looking at it is that planar circuits can be drawn to appear as a series of boxes butting up against each other. To get a visceral idea of this notion, grab a piece of paper and place four dots on it. Try to draw a line from each dot to every other dot but without crossing any lines. After a few tries, you should be successful. Now try it with five dots. You can't do it unless you “draw in the air” and hop over other lines. Obviously, that's possible with real circuits because they're 3D. Therefore there are circuits that cannot be solved using mesh. 222
|
ACElectricalCircuitAnalysis_Page_222_Chunk722
|
Consider the circuit of Figure 6.19. This circuit has two voltage sources and cannot be simplified further, although it can be solved using either superposition or nodal analysis. For mesh analysis, we begin by designating a set of current loops. These loops should be minimal in size and together cover all components at least once. By convention, the loops are drawn with a clockwise reference direction. There is nothing magical about them being clockwise, it is just a matter of consistency. The annotated version of the circuit is redrawn in Figure 6.20. Here we have two loop currents, i1 and i2. Note that all components exist in at least one loop (and sometimes in more than one loop, like capacitor C). Depending on circuit values, one or more of these loop directions may in fact be opposite of reality. This is not a problem. If the true reference direction is opposite, then the currents will show up as negative values, and thus we know that the real reference direction is counterclockwise. Just remember that a positive result means a clockwise direction and negative indicates counterclockwise. We begin by writing KVL equations for each loop. Loop 1: E1 = voltage across R + voltage across XC Loop 2: −E2 = voltage across XC + voltage across XL Note that E2 is negative as i2 is drawn flowing out of its negative terminal. Now expand the voltage terms using Ohm's law. The resistor and inductor each see a single current, i1 and i2, respectively. The capacitor experiences both currents. From the perspective of loop 1, i2 is flowing in the opposite direction. Thus, the net current is i1 − i2. From the perspective of loop 2, i1 is flowing in the opposing direction and thus the net current is i2 − i1. The reference voltage polarities reinforce this notion. Loop 1: E1 = i1 R + (i1 − i2)(−jXC) Loop 2: −E2 = i2 (jXL) + (i2 − i1)(−jXC) Multiplying out and collecting terms yields: Loop 1: E1 = (R −jXC) i1 − (−jXC) i2 Loop 2: −E2 = − (−jXC) i1 + (jXL −jXC) i2 223 Figure 6.19 A simple two-source circuit. Figure 6.20 Circuit with mesh loops and voltage polarities drawn.
|
ACElectricalCircuitAnalysis_Page_223_Chunk723
|
As the component values and source voltages are known, we have two equations with two unknowns. These can be solved for i1 and i2 using the simultaneous equation solution techniques of your choice. Example 6.7 For the circuit of Figure 6.21, determine vb and vc. The sources are: E1 = 90° volts and E2 = 12−90° volts. We begin by labeling our loops, as shown in Figure 6.22. Each loop will generate an equation based on a KVL summation around that loop. We will number the components from left to right, as usual. For loop 1: E1 −E2 = ( jX L1+R1)i1+R2(i1 −i2) E1 −E2 = (R1+R2+ jX L1)i1 −R2i2 90°V −12−90 °V = (20Ω+80Ω+ j 20Ω)i1 −20Ωi2 1553.1° V = (100Ω+ j 20Ω)i1 −20Ωi2 Repeat for loop 2: 224 Figure 6.21 Circuit for Example 6.7. Figure 6.22 Circuit of Figure 6.21 with current loops drawn.
|
ACElectricalCircuitAnalysis_Page_224_Chunk724
|
E2 = (−jX C+ jX L2)i2+R2(i2 −i1) E 2 =−R2i1+(R2 −jX C+ jX L2)i2 12−90 °V = −20Ωi1 +(20Ω −j75+ j 50Ω)i2 12−90 °V = −20Ωi1 +(20Ω −j 25Ω)i2 The two loop equations are: 1553.1°V = (100Ω+ j 20Ω)i1 −20Ωi2 12−90 °V = −20Ωi1 +(20Ω−j 25Ω)i2 The equations show diagonal symmetry. The currents are i1 = 0.178519.9° amps and i2 = 0.3529−21.4° amps. To find the voltages vb and vc, we just need to apply Ohm's law. The voltage vc is the potential across the j50 Ω inductor. v c = i2× jX L2 v c = 0.3529−21.4° A× j50Ω v c=17.6468.6 ° V The potential vb is found similarly. vb = i2×(−jX C+jX L2) vb = 0.3529−21.4 °×(−j 75Ω +j50 Ω) vb=8.823−111.4° V For verification, we can also find vb by subtracting the voltage developed across the series inductor/resistor pair from the first source. vb = E1 −i1×(R1+jX L1) vb = 90° V −0.178519.9° A×(80 +j20 Ω) vb=8.823−111.4° V Example 6.8 In the circuit of Figure 6.23, find vb. E = 100° volts peak at a frequency of 10 kHz. The circuit is a bridge network. Even though it has only a single voltage source, basic series-parallel techniques will not work here. Nodal analysis can also work here as can delta-Y conversion, however, mesh is an excellent choice for this layout. 225
|
ACElectricalCircuitAnalysis_Page_225_Chunk725
|
We start by finding the reactance values. Using the standard reactance formulas we find that XL = j628.3 Ω and XC = −j318.3 Ω. After substituting these into the original circuit and defining the loops, we have Figure 6.24. We have three loops with three unknown currents, and therefore three equations. We'll number the resistors from left to right. For loop 1: E = vC+v R1 E =−jX C(i1 −i2)+R1(i1 −i3) E = (R1 −jX C)i1 −(−jX C)i2 −R1i3 100° V = (1k Ω−j 318.3Ω)i1+j318.3Ωi2 −1k Ωi3 For loop 2: 0 = vC+v L+vR2 0 =−jX C(i2 −i1)+ jX L(i2 −i3)+ jX Li3 0 =−(−jX C)i1+(R2+ X L −jX C)i2 −R2i3 0 =−(−j 318.3Ω)i1+(500Ω+ j628.3 −j318.3Ω)i2 −500Ωi3 0 = j318.3Ωi1+(500Ω+ j 310Ω)i2 −500Ωi3 226 Figure 6.23 Circuit for Example 6.8. Figure 6.24 Circuit of Figure 6.23 with reactance values and current loops drawn.
|
ACElectricalCircuitAnalysis_Page_226_Chunk726
|
For loop 3: 0 = v R1+v R2+v R3 0 = R1(i3 −i1)+R2(i3 −i2)+R3i3 0 =−R1i1 −R2i2+(R1+R2+R3)i3 0 =−1 kΩi1 −500Ωi2+(1k Ω+500Ω+2k Ω)i3 0 =−1 kΩi1 −500Ωi2+3.5k Ωi3 The final set of equations is: 100° V = (1k Ω −j318.3Ω)i1+ j318.3Ωi2 −1 kΩi3 0 = j318.3Ω i1+(500Ω+ j 310Ω)i2 −500Ωi3 0 =−1 kΩi1 −500Ωi2+3.5k Ωi3 The system of equations has diagonal symmetry. The results are: i1 = 10.24E-316° amps, i2 = 6.754E-3−85.7° amps and i3 = 2.888E-3−3.13° amps. For vb, this is just the voltage across the 1 kΩ resistor. Note that a pair of meshing currents (i1 and i3) are flowing through that resistor, so we must determine the net value of current. If we assume the reference polarity for vb is positive, that coincides with the direction of i1, and thus the net current must be i1 − i3. The result is inet = 7.57E−323.2° amps. Therefore, vb = 7.5723.2° volts. Computer Simulation Computer Simulation Figure 6.25 shows the bridge circuit of Example 6.8 captured in a simulator. This will be used to verify the computed result. 227 Figure 6.25 Circuit of Figure 6.23 in a simulator.
|
ACElectricalCircuitAnalysis_Page_227_Chunk727
|
A transient analysis is run on the circuit, plotting node 2 which corresponds to vb, and node 1, the input voltage, which is handy for phase reference. Examining the plot, we can see that the node 2 voltage is just above 7.5 volts peak, as calculated. Further, this waveform leads the the input waveform by just over a quarter of a division. As this plot shows four divisions per cycle, each division is 90 degrees. This indicates a leading or positive phase shift in the low 20 degree range, and that corroborates nicely the computed value of 23.2 degrees. Inspection Method Inspection Method Like nodal analysis, it is possible that the system of equations can be obtained directly through inspection. This is true only if the circuit contains no current sources. Look at the final set of equations derived in Example 6.8 from Figure 6.24. A clear pattern will emerge. To generate an equation for a given loop, simply focus on that loop and ask the following questions: What is the total source voltage in this loop? This yields the voltage constant on the left side of the equals sign. Next, sum the resistance and reactance values in the loop under inspection. This yields the coefficient for that current term. For the other current coefficients, sum the resistances and reactances that are in common between the loop under inspection and the other loops (e.g., for loop 1, XC is in common with loop 2). These values will always be negative (an exception arises with a “double negative”, as seen with the 228 Figure 6.26 Transient analysis for the circuit of Figure 6.25.
|
ACElectricalCircuitAnalysis_Page_228_Chunk728
|
capacitor). As usual, the set of equations produced must exhibit diagonal symmetry. While it is possible to extend this technique to include current sources, usually it is easier and less error-prone to convert the current sources into voltage sources. Then the process can continue with the direct inspection method outlined above. Finally, it is important to remember that the number of loops determines the number of equations to be solved. This method will be illustrated in the example following. Example 6.9 For the circuit of Figure 6.27, find vb and the current through the 15 Ω resistor. E1 = 100° volts peak and E2 = 2090° volts peak. We identify and label three loops, as shown in Figure 6.28. This circuit utilizes only voltage sources and no current sources. Therefore, we can apply the inspection method without extra effort. We start at loop 1 and sum all of the voltage sources. The sole source is E1. 100° V = ... Now we sum all of the resistances and reactances in this loop. This is the coefficient for the first current term. 100° V = (20Ω+6Ω+j8Ω)i1... 229 Figure 6.27 Circuit for Example 6.9. Figure 6.28 Circuit of Figure 6.27 with current loops drawn.
|
ACElectricalCircuitAnalysis_Page_229_Chunk729
|
We continue the process by determining the components that are in common between this loop and the next loop. Remember, this coefficient is negative. 100° V = (20Ω+6Ω+j8Ω)i1 −(6Ω+j 8Ω)i2... We repeat the process by determining the common components with the next loop. This coefficient is negative. In this situation, no components are in common between loops 1 and 3. We shall leave in placeholder with a coefficient of zero, just as a reminder that we didn't forget anything and also to ensure that the coefficients in the final set of equations line up nicely. 100° V = (26Ω+ j 8Ω)i1 −(6Ω+ j8Ω)i2 −0i3 We now move to loop 2 and repeat the sequence of steps. There is only one source, E2, and it shows up negative as mesh current i2 is flowing out of its negative terminal. The result is: −2090° V =−(6Ω+ j8Ω)i1+(6Ω+10Ω+ j 8 −j 14Ω)i2 −10Ωi3 −2090° V =−(6Ω+ j8Ω)i1+(16Ω −j6Ω)i2 −10Ωi3 And finally the third loop. Here the second source shows up as positive. 20 90°V =−0i1−10Ωi2 −(10Ω+15Ω−j 12Ω)i3 20 90°V =−0i1−10Ωi2 −(25Ω −j 12Ω)i3 The completed system of equations is: 100° V = (26 Ω+ j8Ω)i1 −(6Ω+ j8Ω)i2 −0i3 −2090° V =−(6Ω+ j8Ω)i1+(16Ω −j6Ω)i2 −10Ωi3 20 90 °V =−0i1−10Ωi2 −(25Ω −j 12Ω)i3 The system has diagonal symmetry. The resulting currents are: i1 = 0.6131−15.5° amps, i2 = 0.6687−49.2° amps and i3 = 0.561199.3° amps. The current through the 15 Ω resistor is i3, so that much is done. Regarding vb, it can be found using Ohm's law as vb is the series connection of the 6 Ω resistor and j8 Ω inductor times the current through them. This current is the pair of meshing currents i1 and i2. Assuming the reference polarity for vb is positive, that is the direction of i1, and thus the net current must be i1 − i2. The result is 0.37565.8° amps. Therefore, vb = 3.75119° volts. To crosscheck this, we can subtract the voltage across the 20 Ω resistor from E1. That's 100° volts minus 20 Ω times 0.6131−15.5° amps, or 3.75119° volts, as expected. 230
|
ACElectricalCircuitAnalysis_Page_230_Chunk730
|
Supermesh Supermesh On occasion you may find a current source which has no associated internal impedance, such as the one in the circuit of Figure 6.29. This is similar to the situation discussed previously with nodal analysis where a voltage source does not have a specified internal impedance. As with nodal, there are two ways of solving this predicament. The first technique is to add a very large impedance in parallel with the current source and then perform a source conversion on the pair so that the inspection method of mesh can be used. The larger the value of this impedance, the greater the accuracy. As a general rule it should be at least a couple of orders of magnitude larger than any surrounding impedance, and preferably larger. The second technique is to use supermesh. A supermesh is a larger mesh loop than contains other mesh loops inside of it. Refer to the circuit shown in Figure 6.29. In the center we have a current source, Is, which lacks an associated internal impedance. Two traditional mesh loops, i1 and i2, are labeled as usual. The problem here is that we cannot use an Ohm's law-based iZ voltage drop for vb. We have no way to express this as the voltage across Is is an unknown. On the other hand, what we do know is that Is must equal the combination of the original mesh currents i1 and i2. That is, from the perspective of the first loop, Is = i2 − i1. Remember, one or both of the mesh currents could be negative, and thus rotating counterclockwise. At this point we invoke the idea of a supermesh loop. First, we replace the problematic current source with its ideal internal impedance, an open. Second, a supermesh loop is drawn which encompasses the original two loops. This is shown in Figure 6.30. The supermesh loop is shown in red and labeled. 231 Figure 6.29 Circuit for supermesh. Figure 6.30 Supermesh labeled.
|
ACElectricalCircuitAnalysis_Page_231_Chunk731
|
We now perform a KVL summation around the supermesh loop, similar to what we have done in prior work. The difference this time around is that we need to recognize that components each see one of the original mesh currents; namely i1 or i2 here. We do not solve for a supermesh current. Instead, we just use the supermesh loop to define the KVL summation. The summation follows: Σv rises = Σ vdrops E1 = v R+v XL+E2 The voltage drops across the resistor and inductor can be expanded using Ohm's law, using the original mesh current associated with each component. E1 −E 2 = i1 R +i2 jX L Also, by inspection, I s = i2 −i1 or i2 = i1 +I s We now have two equations with two unknowns and can solve for i1 and i2. This procedure is illustrated in the following example. Example 6.10 Find vb for the circuit of Figure 6.31. E1 = 200° volts, E1 = 1890° volts and IS = 10E−30° amps. First, we label the loops, as shown in Figure 6.32. Now we perform a KVL summation around the supermesh loop. Σv rises = Σ vdrops 20 0°V = vR+v XL+1890° V Expand using Ohm's law and rearrange: 20 0°V −1890° V = 1kΩi1+j 400Ωi2 26.9 −42° V = 1kΩi1+j 400Ωi2 232 Figure 6.31 Circuit for Example 6.10.
|
ACElectricalCircuitAnalysis_Page_232_Chunk732
|
By inspection we can see that: 10E-30° A = i2 −i1 or i2 = i1 +10E-30° A We can substitute this expression into the prior supermesh expression and solve for i1: 26.9 −42° V = 1kΩi1+ j 400Ωi2 26.9 −42° V = 1kΩi1+ j 400Ω(i1+10E-30° A) 26.9 −42° V = 1kΩi1+ j 400Ωi1+490° V 29.7−44.7°V = (1 kΩ+ j 400Ω)i1 i1 ≈27.6E-3−69.5° A Thus, i2 = 27.6E−3−69.5° amps + 10E−30° amps, or 32.5E−352.8 − ° amps. To determine vb we simply subtract the drop across the 1 kΩ resistor from E1: vb = 20 0°V −i11k Ω vb = 20 0°V −27.6E-3−69.5° A1 kΩ vb ≈27.8568.2°V As a crosscheck, we could also add the voltage across the inductor to E2: vb = 1890° V +i2 j 400Ω vb = 1890° V +32.5E-3−52.8° A j400Ω vb ≈27.8568.2° V Comparison of Nodal and Mesh Comparison of Nodal and Mesh Having covered both nodal and mesh in some detail, it is fair to look at the two techniques to gauge their strengths and weaknesses. Compared to nodal analysis, mesh analysis has the advantage of dealing with impedances rather than admittances when writing the system of equations. Further, the mesh inspection method works with voltage sources, which tends to be convenient for many circuits, while the nodal inspection method requires current sources. On the down side, the resulting set of mesh currents requires further processing in order to find either branch currents or node voltages. In contrast, nodal analysis produces node voltages directly with no further processing. Mesh also has the disadvantage of being limited to planar circuits while there is no such limit to nodal. Ultimately, instead of thinking in terms of which technique is “better” overall, it is more efficient to use the proper tool for the job at hand. For example, if a circuit is populated with voltage sources, mesh might be the more efficient route, especially if specific currents are desired. On the other 233 Figure 6.32 Circuit of Figure 6.31 with supermesh labeled.
|
ACElectricalCircuitAnalysis_Page_233_Chunk733
|
hand, if you need to find voltages in a circuit that contains numerous current sources, nodal would be more effective. 6.4 Dependent Sources 6.4 Dependent Sources A dependent source is a current or voltage source whose value is not fixed. Instead, the value depends on some other circuit current or voltage. The general form for the value of a dependent source is N = kM where M and N are currents and/or voltages and k is the proportionality constant. For example, the value of a dependent voltage source may be a function of a current, so instead of the source being equal to, say, 10 volts, it could be equal to twenty times the current passing through a particular resistor, or v = 20i. There are four possible dependent sources: They are the voltage-controlled voltage source (VCVS), the voltage-controlled current source (VCCS), the current-controlled voltage source (CCVS), and the current-controlled current source (CCCS). The source and control parameters are the same for both the VCVS and the CCCS so k has no units, although it may be given as volts/volt and amps/amp, respectively. For the VCCS and CCVS, k has units of amps/volt and volts/amp, respectively. These are referred to as the transresistance and transconductance of the sources with units of ohms and siemens. The schematic symbols for dependent or controlled sources are usually drawn using a diamond instead of a circle. Also, for simulators, there will be a secondary connection for the controlling current or voltage. Examples of voltage-controlled and current-controlled sources are shown in Figure 6.33. On each of these symbols, the control element is shown to the left of the source. The control portion can be thought of as a connection for a voltmeter or ammeter which senses the control parameter. These sensing connections are not always drawn on a schematic. Instead, the source simply may be labeled as a function, as in v = 0.02 iX where iX is the controlling current. These simpler controlled sources are shown in Figure 6.34 and are typical in electronic schematics and texts. In some cases, these sources are drawn with a circle instead of a diamond. Also, the sine wave shape shown here is often omitted from the inside of symbol, however, current sources are always drawn with an arrow pointing in the reference direction and voltages sources always include the reference polarity. 234 Figure 6.33 Dependent source symbols as used in common simulators (left to right): VCVS, CCVS, VCCS, and CCCS. Figure 6.34 Generic symbols for dependent voltage source (left) and current source (right).
|
ACElectricalCircuitAnalysis_Page_234_Chunk734
|
Dependent sources are not “off-the-shelf” items in the same way that a battery or signal generator are. Rather, dependent sources are used to model the behavior of more complex devices. For example, a bipolar junction transistor commonly is modeled as a CCCS while a field effect transistor may be modeled as a VCCS. Similarly, many amplifier circuits are modeled as VCVS systems. Solutions for circuits using dependent sources follow along the lines of those established for independent sources (i.e., the application of Ohm's law, KVL, KCL, etc.), however, the sources are now dependent on the remainder of the circuit which tends to complicate the analysis. In general, there are two possible circuit configurations for dependent sources: isolated and coupled. An example of the isolated form is shown in Figure 6.35. In this example, the dependent source (center, a CCVS) does not interact with the sub-circuit on the left driven by the independent source E. Thus it can be analyzed as two separate circuits as shown in Figure 6.36. Solutions for this form are relatively straightforward. The control value for the dependent source can be computed directly using standard techniques. Then this value is substituted into the dependent source and the analysis continues as normal. Sometimes it is convenient if the solution for a particular voltage or current is defined in terms of the control parameter rather than as a specific value (e.g., the current through a particular component might be 75 i1 instead of just 1 milliamp). The second type of circuit (coupled) is somewhat more complex in that the dependent source can affect the parameter that controls the dependent source. An example is shown in Figure 6.37. 235 Figure 6.35 Dependent source circuit: isolated. Figure 6.36 Dependent source circuit: isolated treated as two circuits.
|
ACElectricalCircuitAnalysis_Page_235_Chunk735
|
In this example it should be obvious that the voltage from the dependent source can affect the voltage at node a, and it is this very voltage that defines ix, which in turn sets up the value of the dependent source. As far as analysis is concerned, either mesh or nodal can be used. The dependent source(s) will contribute terms that include the controlling parameter(s) so some additional effort will be required. To illustrate the technique, consider the circuit of Figure 6.38. We shall use the general method of nodal analysis. We begin by defining current directions. Assume that the currents through R1 and C are flowing into node a, the current through R2 is flowing out of node a, and the current through L is flowing out of node b. We shall number the branch currents to reflect the associated components, from left to right. The resulting KCL equations are: Σiin = Σiout Node a: i1 +i3 = i2 Nodeb: k va = i3 +i4 The currents are then described by their Ohm's law equivalents: Node a: E−v a R1 + v b−va −jX C = va R2 Node b: k v a = v b−va R2 + vb jX L Expanding terms yields: Node a: E R1 −va R1 + vb −jX C − va −jX C = v a R2 Nodeb: k va = vb R2 −va R2 + v b jX L 236 Figure 6.37 Dependent source circuit: coupled. Figure 6.38 Circuit with a voltage- controlled current source.
|
ACElectricalCircuitAnalysis_Page_236_Chunk736
|
Collecting terms and simplifying yields: Node a : E R1 =( 1 R1 + 1 R2 + 1 −jX C)va − 1 −jX C vb Nodeb: 0 =−(k + 1 R2)va +( 1 R2 + 1 jX L)vb At this point, the component values and independent source value would be inserted into the equations and the system solved. Finally, referring back to the prior chapter, it is possible to perform source conversions on dependent sources, within limits. The new source will remain a dependent source (e.g., VCVS to VCCS). This process is not applicable if the control parameter directly involves the internal impedance (i.e., is its voltage or current). Example 6.11 For the circuit shown in Figure 6.39, determine vc if the source is 10° volt peak. This is an example of the isolated or uncoupled dependent source. The value of the dependent current source is 30 times the value of the current labeled ix, which is the current flowing through the −j2 kΩ capacitive reactance. We can find this current first and then determine the resulting value of the dependent source. The analysis will require nothing beyond basic series- parallel techniques. We will number the components from left to right. The current ix is found via Ohm's law: ix = E −jX C1 ix = 10° V −j 2 kΩ ix = 0.5E-390 °A 237 Figure 6.39 Circuit for Example 6.11.
|
ACElectricalCircuitAnalysis_Page_237_Chunk737
|
The dependent source is 30 times this value, or 15E−390° amps. Given the reference direction of this source, the current is flowing upwards through the 65 kΩ resistor and parallel −j50 kΩ capacitor. This establishes vc as negative. Multiplying ix by the parallel impedance yields the desired voltage. Z RC = R3× jX C2 R3 −jX C2 Z RC = 65k Ω×(−j50 k Ω) 65 kΩ−j50 k Ω Z RC = 39.6E3−52.4 °Ω v c =−ix×Z RC v c =−15E-390°Ω×39.6E3−52.4°Ω v c = 594−142.4°V The next example features a coupled configuration solved using nodal analysis. Example 6.12 In the circuit of Figure 6.40, determine va. E =200° volts peak at 50 kHz. In this circuit we have a current controlled voltage source, or CCVS. The proper unit for the constant of 1000 is ohms (volts over amps). First, we need to determine the reactance value for the inductor. X L = 2π f L X L = 2π50 kHz 4mH X L = j1257Ω There is only one node of interest here (a) so we will only need one KCL equation. The sole unknown is va. Let's assume that the reference directions of the currents flowing through the 1 kΩ and 3 kΩ resistors are entering node a. We'll call these i1 and i2, respectively. The exiting current is ix. 238 Figure 6.40 Circuit for Example 6.12.
|
ACElectricalCircuitAnalysis_Page_238_Chunk738
|
The KCL summation is: Σiin = Σiout i1+i2 = i x This is expanded using Ohm's law (in several steps, for clarity). ix = i1+i2 va R2 = E −v a R1+ jX L + 1000(Ω)i x−va R3 va 2k Ω = 20 0°V−v a 1k Ω+ j1257Ω + 1000(Ω)i x−va 3k Ω 20 0°V 1k Ω+ j1257Ω = va 1 kΩ+ j1257Ω + v a 2 kΩ + va 3k Ω −1000(Ω)i x 3k Ω 12.45−51.5° A = v a 1k Ω+ j1257Ω + va 2k Ω + va 3 kΩ −1000(Ω)va 2 kΩ×3 kΩ 12.45−51.5°A = v a 1k Ω+ j1257Ω + va 2k Ω + v a 3 kΩ −v a 6k Ω 12.45−51.5° A =( 1 1 kΩ+ j 1257Ω + 1 2 kΩ + 1 3k Ω − 1 6 kΩ)va 12.45−51.5° A = 1.161E-3−24.8°S va va = 10.7−26.7° V Computer Simulation Computer Simulation For verification, the dependent source circuit of Example 6.12 is entered into a simulator as shown in Figure 6.41. The results are shown in Figure 6.42. 239 Figure 6.41 Circuit of Figure 6.40 in a simulator.
|
ACElectricalCircuitAnalysis_Page_239_Chunk739
|
Note the connection to sense the current ix. It is inserted just like an ammeter. As mentioned previously, the constant for the dependent source is a transresistance and has units of ohms. A transient analysis is run the circuit, plotting the independent source, E, as node 1 (blue), and va as node 3 (red). Both the amplitude and lagging phase shift line up nicely with the computed result. The voltage of the dependent source is also plotted as node 5 (green). Verifying this potential is left as an exercise. 6.5 Summary 6.5 Summary Nodal analysis can be used to solve virtually any complex multi-source AC electrical circuit. It is based on KCL, writing expressions involving each node in the circuit. A system of equations results, there being as many equations as there are nodes in the circuit, minus the reference node (typically taken as ground). The set of equations will exhibit diagonal symmetry, which can be used as a crosscheck before setting out to solve them. The solution will be a complete set of node voltages. From these, any branch current may be determined as needed. There are two different methods of creating the system of equations. The first method is deemed the general method and will work for a mix of current sources and voltage sources. Individual currents are defined based on the node voltages and any known current sources. KCL is then applied at each node, followed by simplification 240 Figure 6.42 Transient analysis for the circuit of Figure 6.41.
|
ACElectricalCircuitAnalysis_Page_240_Chunk740
|
and combination of terms to arrive at the end equations. The second approach is referred to as the inspection method. If the circuit contains only current sources (or if the voltage sources are converted to equivalent current sources), this method allows direct generation of the system of equations without the need for simplification and thus is less prone to error. Mesh analysis can be used to solve any planar complex multi-source AC electrical circuit. In some respects it is the mirror of nodal analysis. It is based on KVL, writing expressions involving each closed loop in the circuit. The loops are minimally sized and the set of loops must cover every component in the circuit. A system of equations results, there being as many equations as there are loops. As with nodal analysis, the set of equations will exhibit diagonal symmetry. The solution will be a complete set of mesh currents. From these, any branch current and node voltage may be determined. Like nodal, mesh offers two different methods of creating the system of equations. The general method will work for a mix of current sources and voltage sources. Individual loops are defined based on the meshing currents passing through each component. KVL is then applied around each loop, followed by simplification and combination of terms to arrive at the end equations. In contrast, if the circuit contains only voltage sources (or if the current sources are converted), then the inspection method may be used. This method allows direct generation of the system of equations and is faster and less error prone. Dependent sources are current or voltage sources whose value depends on the current or voltage developed in some other part of the circuit. There are four types: current controlled current source (CCCS), current controlled voltage source (CCVS), voltage controlled current source (VCCS) and voltage controlled voltage source (VCVS). These sources are used commonly to model the characteristics of active devices such bipolar and field effect transistors. Techniques for solution tend to be a bit more involved than when using constant sources, however, nodal analysis in particular tends to work well. Review Questions Review Questions 1. Describe the practical differences between nodal analysis and mesh analysis. 2. What is diagonal symmetry? Of what use is it? 3. What are the differences between the general method and the inspection method of nodal analysis? 4. What are the differences between the general method and the inspection method of mesh analysis? 5. What is a supernode? 6. What is a supermesh? 7. Describe the concept of dependent sources and how they differ from independent or constant sources. 241
|
ACElectricalCircuitAnalysis_Page_241_Chunk741
|
6.6 Exercises 6.6 Exercises Analysis Analysis (All source values are in amps or volts unless specified otherwise) 1. Given the circuit in Figure 6.43, use nodal analysis to determine vc. I1 = 30°, I2 = 0.90°. 2. Use nodal analysis to find the current through the 120 Ω resistor in the circuit of Figure 6.44. I1 = 0.590°, I2 = 1.60°. 3. Use nodal analysis to find the current through the 43 Ω resistor in the circuit of Figure 6.44. The sources are in phase. 4. Given the circuit in Figure 6.44, use nodal analysis to determine vb. The sources are in phase. 5. Given the circuit in Figure 6.45, determine vc. I1 = 30°, I2 = 20°. 242 Figure 6.43 Figure 6.45 Figure 6.44
|
ACElectricalCircuitAnalysis_Page_242_Chunk742
|
6. Use nodal analysis to find the current through the j45 Ω inductor in the circuit of Figure 6.45. I1 = 20°, I2 = 1.560°. 7. Use nodal analysis to find the current through the 4 Ω resistor in the circuit of Figure 6.46. I1 = 145°, I2 = 245°. 8. Given the circuit in Figure 6.46, use nodal analysis to determine vc. I1 = 630°, I2 = 40°. 9. Given the circuit in Figure 6.47, use nodal analysis to determine vac. I1 = 100°, I2 = 60°. 10. Use nodal analysis to find the current through the j8 Ω inductor in the circuit of Figure 6.47. I1 = 30°, I2 = 530°. 11. Use nodal analysis to find the current through the 22 Ω resistor in the circuit of Figure 6.48. I1 = 800E−30°, I2 = 2.50°, I3 = 220°. 243 Figure 6.46 Figure 6.47 Figure 6.48
|
ACElectricalCircuitAnalysis_Page_243_Chunk743
|
12. Given the circuit in Figure 6.48, use nodal analysis to determine vc. I1 = 490°, I2 = 10120°, I3 = 50°. 13. Given the circuit in Figure 6.49, use nodal analysis to determine vc. I1 = 3E−30°, I2 = 10E−30°, I3 = 2E−30°. 14. Use nodal analysis to find the current through the −j2 kΩ capacitor in the circuit of Figure 6.49. I1 = 1E−30°, I2 = 5E−30°, I3 = 6E−3−90°. 15. Use nodal analysis to find the current through the 3.3 kΩ resistor in the circuit of Figure 6.50. E = 360°, I = 4E−3−120°. 16. Given the circuit in Figure 6.50, write the node equations and determine vc. E = 180°, I = 7.5E−3−30°. 17. Given the circuit in Figure 6.51, use nodal analysis to determine vc. E = 40180°, I = 20E−30°. 244 Figure 6.49 Figure 6.50 Figure 6.51
|
ACElectricalCircuitAnalysis_Page_244_Chunk744
|
18. Use nodal analysis to find the current through the 2.2 kΩ resistor in Figure 6.51. E = 2400°, I = 100E−30°. 19. Use nodal analysis to find vbc in the circuit of Figure 6.52. 20. Use nodal analysis to find the current through the 2.7 kΩ resistor in the circuit of Figure 6.53. 21. Given the circuit in Figure 6.54, use nodal analysis to determine vba. E1 = 10°, E2 = 20°. 245 Figure 6.52 Figure 6.53 Figure 6.54
|
ACElectricalCircuitAnalysis_Page_245_Chunk745
|
22. Given the circuit in Figure 6.55, use nodal analysis to determine vad. E1 = 90°, E2 = 540°. 23. Use nodal analysis to find vcb in the circuit of Figure 6.56. E1 = 10−180°, E2 = 250°. 24. Given the circuit in Figure 6.57, use nodal analysis to determine vbc. E = 200°, R1 = 10 kΩ, R2 = 30 kΩ, R3 = 1 kΩ, XC = −j15 kΩ, XL = j20 kΩ. 246 Figure 6.55 Figure 6.56 Figure 6.57
|
ACElectricalCircuitAnalysis_Page_246_Chunk746
|
25. Given the circuit in Figure 6.58, write the mesh loop equations and determine vb. 26. Use mesh analysis to find the current through the 2.7 kΩ resistor in the circuit of Figure 6.58. 27. Use mesh analysis to find the current through the 75 Ω resistor in the circuit of Figure 6.52. 28. Given the circuit in Figure 6.52, write the mesh loop equations and determine vc. 29. Given the circuit in Figure 6.53, write the mesh loop equations and determine vb. 30. Use mesh analysis to find the current through the 1.8 kΩ resistor in the circuit of Figure 6.53. 31. Use mesh analysis to find the current through the j200 Ω inductor in Figure 6.54. E1 = 10°, E2 = 20°. 32. Given the circuit in Figure 6.54, write the mesh loop equations and determine vb. Consider using parallel simplification first. E1 = 36−90°, E2 = 24−90°. 33. Given the circuit in Figure 6.55, use mesh analysis to determine vcd. E1 = 0.10°, E2 = 0.50°. 34. Use mesh analysis to find the current through the 600 Ω resistor in the circuit of Figure 6.55. E1 = 90°, E2 = 540°. 247 Figure 6.58
|
ACElectricalCircuitAnalysis_Page_247_Chunk747
|
35. Use mesh analysis to find the current through the −j200 Ω capacitor in the circuit of Figure 6.59. E1 = 180°, E2 = 1290°. 36. Given the circuit in Figure 6.59, use mesh analysis to determine vac. E1 = 10°, E2 = 500E−30°. 37. Given the circuit in Figure 6.56, use mesh analysis to determine vc. E1 = 10−180°, E2 = 250°. 38. Use mesh analysis to find the current through the 22 kΩ resistor in the circuit of Figure 6.56. E1 = 240°, E2 = 360°. 39. Use mesh analysis to find the current through the j300 Ω inductor in Figure 6.60. E1 = 10°, E2 = 1090°. 40. Given the circuit in Figure 6.60, use mesh analysis to determine va. E1 = 1000°, E2 = 900°. 41. Given the circuit in Figure 6.57, use mesh analysis to determine vbc. E = 100°, R1 = 1 kΩ, R2 = 2 kΩ, R3 = 3 kΩ, XC = −j4 kΩ, XL = j8 kΩ. 42. Use mesh analysis to find the current through resistor R3 in the circuit of Figure 6.57. E = 200°, R1 = 10 kΩ, R2 = 30 kΩ, R3 = 1 kΩ, XC = −j15 kΩ, XL = j20 kΩ. 248 Figure 6.59 Figure 6.60
|
ACElectricalCircuitAnalysis_Page_248_Chunk748
|
43. Use mesh analysis to find the current through resistor R3 in Figure 6.61. E = 600°, R1 = 1 kΩ, R2 = 2 kΩ, R3 = 3 kΩ, XC = −j10 kΩ, XL = j20 kΩ. 44. Given the circuit in Figure 6.61, use mesh analysis to determine vbc. E = 12090°, R1 = 100 kΩ, R2 = 20 kΩ, R3 = 10 kΩ, XC = −j5 kΩ, XL = j20 kΩ. 45. Given the circuit in Figure 6.62, use mesh analysis to determine vb. Consider using source conversion. E = 120°, I = 10E−30°. 46. Use mesh analysis to find the current through the 3 Ω resistor in the circuit in Figure 6.62. Consider using source conversion. E = 1590°, I = 10E−30°. 47. Use mesh analysis to find the current through the 2.2 kΩ resistor in the circuit in Figure 6.63. E = 3.30°, I = 2.1E−30°. 48. Given the circuit in Figure 6.63, use mesh analysis to determine vb. E = 100°, I = 30E−390°. 249 Figure 6.61 Figure 6.62 Figure 6.63
|
ACElectricalCircuitAnalysis_Page_249_Chunk749
|
49. Given the circuit in Figure 6.64, use nodal analysis to determine vab. 50. Use nodal analysis to find the current through the 100 mH inductor in the circuit of Figure 6.64. 51. Use nodal analysis to find the current through the 330 Ω resistor in the circuit of Figure 6.65. 52. Given the circuit in Figure 6.65, write the node equations and determine vb. 53. Given the circuit in Figure 6.61, use nodal analysis to determine vbc. E = 1200°, R1 = 1 kΩ, R2 = 2 kΩ, R3 = 3 kΩ, XC = −j10 kΩ, XL = j20 kΩ. 54. Determine the current through the 10 kΩ resistor in the circuit of Figure 6.66 if I1 = 10E−3−90°. 55. Determine vb in the circuit of Figure 6.66 if the source I1 = 20E−30°. 250 Figure 6.64 Figure 6.65 Figure 6.66
|
ACElectricalCircuitAnalysis_Page_250_Chunk750
|
56. Determine vc in the circuit of Figure 6.67 if the source E = 3120°. 57. Determine the current through the 5 kΩ resistor in the circuit of Figure 6.67 if E = 100°. 58. In the circuit of Figure 6.68, determine the capacitor current if the source E = 120°. 59. In the circuit of Figure 6.68, determine vc if the source E = 890°. 60. In the circuit of Figure 6.69, determine vb if the source E = 12−90°. 61. In the circuit of Figure 6.69, determine the current flowing into the 1 kΩ resistor if the source E = 60°. 251 Figure 6.67 Figure 6.68 Figure 6.69
|
ACElectricalCircuitAnalysis_Page_251_Chunk751
|
62. In the circuit of Figure 6.70, determine the current flowing into the 600 Ω resistor if I1 = 1E−3180°. 63. Determine va and vb in the circuit of Figure 6.70 if the source I1 = 2E−30°. 64. Determine va in the circuit of Figure 6.71 if the source E = 20°. 65. Given the circuit in Figure 6.71, determine the current flowing through the 1 kΩ resistor. Assume that E = 1545°. 66. Given the circuit in Figure 6.72, determine the current flowing through the 3 kΩ resistor if the source E = 2533°. 67. Given the circuit in Figure 6.72, determine vab. Assume the source E = 15−112°. 252 Figure 6.70 Figure 6.71 Figure 6.72
|
ACElectricalCircuitAnalysis_Page_252_Chunk752
|
68. In the circuit of Figure 6.73, determine vd. 69. Given the circuit in Figure 6.73, determine the current flowing through the 1 kΩ resistor. 70. Given the circuit in Figure 6.74, determine the current flowing through the 100 Ω resistor. 71. Determine vd in the circuit of Figure 6.74. 72. Determine vab in the circuit of Figure 6.75. E = 100° 253 Figure 6.73 Figure 6.74 Figure 6.75
|
ACElectricalCircuitAnalysis_Page_253_Chunk753
|
Challenge Challenge 73. Given the circuit in Figure 6.76, write the node equations. E1 = 500°, E2 = 35120°, I = 500E−390°. 74. Given the circuit in Figure 6.76, use either mesh or nodal analysis to determine ved. E1 = 90°, E2 = 120°, I = 50E−30°. 75. Given the circuit in Figure 6.77, use mesh analysis to determine vfc. E1 = 120°, E2 = 480°, E3 = 3670°. 254 Figure 6.76 Figure 6.77
|
ACElectricalCircuitAnalysis_Page_254_Chunk754
|
76. Find voltage vbc in the circuit of Figure 6.78 using either mesh or nodal analysis. E = 1000°, R1 = R2 = 2 kΩ, R3 = 3 kΩ, R4 = 10 kΩ, R5 = 5 kΩ, XC1 = XC2 = −j2 kΩ. 77. Given the circuit in Figure 6.79, use nodal analysis to find vac. I1 = 8E−30°, I2 = 12E−30°, E = 500°. 255 Figure 6.78 Figure 6.79
|
ACElectricalCircuitAnalysis_Page_255_Chunk755
|
78. Given the circuit in Figure 6.80, use nodal analysis to determine vad. I1 = 0.10°, I2 = 0.20°, I3 = 0.30°. 79. Given the circuit in Figure 6.81, determine vad. E1 = 150°, E2 = 60°, I = 100E−30°. 80. Given the circuit in Figure 6.82, determine vad. E1 = 220°, E2 = −100°, I = 2E−30°. 256 Figure 6.80 Figure 6.81 Figure 6.82
|
ACElectricalCircuitAnalysis_Page_256_Chunk756
|
81. Given the circuit in Figure 6.83, determine vab. I1 = 1.20°, I2 = 2120°, E = 750°. 82. Given the circuit in Figure 6.84, determine vad. I1 = 0.80°, I2 = 0.2180°, I3 = 0.10°, E = 150°. Simulation Simulation 83. Perform a transient analysis simulation on the circuit of problem 25 (Figure 6.58) to verify the results for vb. 84. Investigate the variation of vb due to frequency in problem 25 (Figure 6.58) by performing an AC simulation. Run the simulation from 10 Hz up to 100 kHz. 85. Investigate the variation of vb due to component tolerance in problem 25 (Figure 6.58) by performing a Monte Carlo simulation. Apply a 10% tolerance to the resistors and capacitor. 86. Perform a transient analysis simulation on the circuit of problem 28 (Figure 6.52) to verify the results for vc. 257 Figure 6.83 Figure 6.84
|
ACElectricalCircuitAnalysis_Page_257_Chunk757
|
87. Investigate the variation of vb due to frequency in problem 28 (Figure 6.52) by performing an AC simulation. Run the simulation from 1 Hz up to 10 kHz. 88. Investigate the variation of vb due to component tolerance in problem 28 (Figure 6.52) by performing a Monte Carlo simulation. Apply a 10% tolerance to the resistors and capacitors. 258
|
ACElectricalCircuitAnalysis_Page_258_Chunk758
|
Notes Notes ♫♫ ♫♫ 259
|
ACElectricalCircuitAnalysis_Page_259_Chunk759
|
7 7 AC Power AC Power After completing this chapter, you should be able to: • Describe current, voltage and power relationships in AC RLC networks. • Plot and make use of the power triangle to determine real, apparent and reactive power components in an AC power system. • Compute the power factor of an RLC network. • Determine necessary components for basic power factor correction. • Perform basic power calculations for systems involving power factor and efficiency. 7.1 Introduction 7.1 Introduction This chapter introduces the concept of power and power waveforms in AC systems. This is an important part of AC circuit analysis and turns out to have striking differences compared to the DC counterpart. While it remains true that power is the product of current and voltage, a naive application of that definition can lead to erroneous conclusions for the AC case. In Chapter One, RMS (i.e., root-mean-square) values were defined and explained. As a general rule, RMS values are used for power calculations, not peak or peak-to-peak values. Further, while complex non-sinusoidal waveshapes are a decided possibility in electronic systems, we shall limit ourselves here to sinusoids. One of the tools we shall use is the power triangle. This is a simple trigonometric device designed to illustrate the power relations between resistive and reactive components in a complex impedance. One of its parameters is the power factor, PF. As we shall see, ordinarily we like the power factor to be unity as this implies best use of the available current. It turns out that this is not the case in many systems. As a consequence, we shall also investigate a simple means of compensating or shifting the power factor back to unity. This is known as power factor correction. As part of our discussion involving power factor, we shall examine typical applications such as motors. Here we shall consider the power factor of a motor along with its efficiency. The efficiency is defined as the useful output power relative to the supplied power and is always less than 100%. Finally, we shall consider basic power factor correction for this application. One practical item to remember here is that, while the instantaneous power changes over time due to the sinusoidal cycling of the voltage and current, what matters to most electrical and electronic devices is the production of internal heat. Devices such as resistors, transistors and so forth, have mass, and thus exhibit a thermal time constant. That time constant tends to be much longer than the period of the wave. The effect is an averaging of the power waveform. In other words, components do not heat up and cool down instantaneously, any more than a hot fry pan would drop to room temperature the moment it was removed from its burner. 260
|
ACElectricalCircuitAnalysis_Page_260_Chunk760
|
7.2 Power Waveforms 7.2 Power Waveforms Computation of power in AC systems is somewhat more involved than the DC case due to the phase between the current and voltage. It has been stated in prior work that power dissipation is characteristic of resistors, and that ideal inductors and capacitors do not dissipate power. We shall show precisely why this is the case by examining three distinct cases for AC circuits: purely resistive, purely reactive and complex impedance. Resistive Load Resistive Load First, consider the case of the purely resistive load, that is, a load with a phase angle of 0 degrees. To determine the power, we simply multiply the voltage by the current. Recall that the basic expression for a sine wave voltage without a DC offset is: v(t)=V sin(2π f t +θ) Where v(t) is the voltage at some time t, V is the peak value, f is the frequency, θ is the phase shift. We know that the current and voltage are always in phase for a resistor, and thus θ is zero degrees. Thus, the expression for a sinusoidal current is similar, using I in place of V for the peak current. We multiply the current and voltage together to arrive at an expression for power:11 P(t)=v(t)×i(t) P(t)=V sin(2π ft)×I sin(2π ft) P(t)=VI( 1 2 −1 2 cos(2π2 ft)) P(t)= VI 2 −VI 2 cos(2π2 ft) (7.1) The final expression is made of two parts; the first portion which is fixed (not a function of time) and the second portion which consists of a negative cosine wave at twice the original frequency. This can be visualized as a time shifted sine wave that is riding on a DC level which is equal to the peak value of the new sinusoid. This is shown in Figure 7.1 using current and voltage peaks normalized to unity. In this figure, the current waveform (green) is drawn just slightly above its true value so that it may be seen easily next to the otherwise identical red voltage waveform. 11 A useful trigonometric identity here is (sin x)2 = ½ – ½ cos 2x 261
|
ACElectricalCircuitAnalysis_Page_261_Chunk761
|
The power product is shown in blue. Unless the frequency is ridiculously low, the resistor's heating will respond to the average value of this waveform thanks to the device's thermal time constant. Due to the fact that sinusoids are symmetrical around zero, the effective power dissipation averaged over time will be the offset value, or VI/2. For example, a one volt peak source delivering a current of one amp peak, as shown here, should generate VI/2, or 0.5 watts. This crosschecks nicely with the RMS calculation of roughly 0.707 volts RMS times 0.707 amps RMS also yielding 0.5 watts. Reactive Load Reactive Load The situation is considerably different if the load is purely reactive. For a load consisting of just an inductor, the voltage leads the current by 90 degrees. This is equivalent to a cosine wave. Once again, we multiply the voltage by the current to arrive at an expression for power: P(t)=V cos2π ft×I sin 2π ft P(t)=VI( 1 2 sin 2π 2 ft) P(t)= VI 2 sin 2π2 ft (7.2) 262 Figure 7.1 Waveforms for a resistive load (the current is shifted slightly positive to ease viewing). AC Power - Resistive -1.0 -0.8 -0.6 -0.4 -0.2 0.0 0.2 0.4 0.6 0.8 1.0 1.2 Time 0 0.2 0.4 0.6 0.8 1 Voltage Current Power
|
ACElectricalCircuitAnalysis_Page_262_Chunk762
|
Note that this expression does not contain a constant term and only contains a time- varying term. Consequently, without an offset, there is no net power dissipation. The result is shown in Figure 7.2. Here power is being alternately generated and dissipated (i.e., positive values indicate dissipation while negative values indicate generation). In this respect, the reactive element can be thought of as alternately storing and releasing energy in the manner of an ideal spring being compressed and then released. Complex Impedance Load Complex Impedance Load Finally, we come to the case of a complex load, part resistive and part reactive. Given some phase angle, θ, we have: P(t)=V sin 2π ft×I sin(2π ft+θ) P(t)=VI( 1 2 cosθ −1 2 cos(2π 2 ft+θ)) P(t)= VI 2 cosθ −VI 2 cos(2π 2 ft+θ) (7.3) This expression contains both a constant term and a time varying term, like the case for the purely resistive load shown in Equation 7.1. There is, however, an important distinction. The constant term is multiplied by the cosine of the impedance angle, a 263 Figure 7.2 Waveforms for a purely reactive load. AC Power - Reactive -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 Voltage Current Power
|
ACElectricalCircuitAnalysis_Page_263_Chunk763
|
value whose magnitude ranges from 0 up to 1. Therefore, unless θ is zero, the offset will not equal the peak value of the sinusoidal portion. This is a particularly important point which shall be amplified in a moment. Example waveforms using θ = 45° are shown graphically in Figure 7.3. The power waveform dips slightly below zero but is not symmetrical around the time axis. Consequently, there is some power dissipation but not as much as in the purely resistive case. In short, the long term power average is now a function of the phase angle, θ. As cosine θ may range between 0 and 1, the power for the complex impedance case will never be more than that of the purely resistive version. Indeed, we can see that Equation 7.3 is the general case. If the load is purely resistive then θ is zero, and Equation 7.3 reduces to Equation 7.1. Similarly, if the load is purely reactive then θ is ±90 degrees, and Equation 7.3 reduces to Equation 7.2. While this analysis used an inductive load, the same can be said regarding the capacitive case (simply swap the labels for the current and voltage waveforms). Finally, in the equations above, V and I are peak values. If RMS values are used, there is no need to divide VI by 2. 264 AC Power - Complex -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 Voltage Current Power Figure 7.3 Waveforms for a complex impedance load.
|
ACElectricalCircuitAnalysis_Page_264_Chunk764
|
At this point we can see that resistors dissipate true power but that reactive components do not. This raises a practical problem, namely, what to call the current- voltage product for purely reactive or complex loads. That is, we can't lump together the current-voltage values for an inductor with those of a resistor any more than we would simply add the magnitudes of resistance and reactance. The practical solution is that we refer to the “power” in reactive components as reactive power. Reactive power uses the symbol Q. Further, the units are not watts, but volt-amps reactive, or more commonly, VAR12. Continuing, for a complex impedance we refer to apparent power. It uses the symbol S and has units of volt-amps, abbreviated VA. It is called apparent power because it appears to be the power if you naively multiply the value obtained from a voltmeter by the value obtained from an ammeter. Those devices would not account for the phase angle between the voltage and current, unlike a proper power meter, and their product would not be the true power. The various power terms are summarized in Figure 7.4. Quantity Symbol Unit, Abbreviation Power P watts, W Apparent Power S volt-amps, VA Reactive Power Q volt-amps reactive, VAR A few examples are in order to help solidify these concepts. Example 7.1 Determine the power dissipated by the resistor in the circuit of Figure 7.5. Also find the apparent power drawn by the circuit and the reactive power of the inductor. The source frequency is 1 kHz. The first item is to find the reactance of the inductor. X L = 2π f L X L = 2π1 kHz1mH X L = j6.283Ω There are several ways to find power. In a series loop like this, the most direct is to use the i2R forms. The source current can be found via Ohm's law. As power calculations utilize RMS values, first find the RMS value of the source voltage. 12 For the plural form, some sources use “VARs” while others use “VAR”. We shall use the latter. 265 Figure 7.4 Symbols and units for power quantities. Figure 7.5 Circuit for Example 7.1.
|
ACElectricalCircuitAnalysis_Page_265_Chunk765
|
v RMS = v peak √2 v RMS = 10V √2 v RMS ≈7.07V i = v Z i = 7.07 V 10+j 6.283Ω i = 0.5986−32.1° A For the power calculations, we shall only use the magnitudes of the voltage and current. Here, the symbol “| |” refers to just the magnitude of the reactance or impedance. P = i2R P = (0.5986A) 210Ω P ≈3.58 W Q = i 2| X | Q = (0.5986 A) 26.283Ω Q ≈2.25 VAR, inductive S = i2|Z | S = (0.5986 A) 2|10+j 6.283Ω| S ≈4.23 VA, inductive Computer Simulation Computer Simulation The circuit of Figure 7.5 is captured in a simulator as shown in Figure 7.6. Three different transient analysis simulations are run. 266 Figure 7.6 The circuit of Figure 7.5 in a simulator.
|
ACElectricalCircuitAnalysis_Page_266_Chunk766
|
The first simulation plots the circulating current (green), the resistor voltage (red) and their product (the power, in blue). This is shown in Figure 7.7. We can see that current and voltage are perfectly in phase, as expected. Also, the power waveform ranges from zero up to about 7 watts. The average of this is approximately half of the peak-to-peak, or about 3.5 watts, just as calculated. The power value can also be computed from voltage times current as a crosscheck. The voltage across the resistor can be found via the voltage rule, and its magnitude is approximately 5.986 volts RMS. Multiplying this by the RMS current will also yield 3.58 watts. It is instructive to compare these curves to those generated in Figure 7.1 for the general resistive case. The current and voltage values in Figure 7.1 were normalized to unity so they do not appear to be identical to those of Figure 7.7, however, the important part is that the phase relationships are the same along with the position of the power waveform. In both cases the power waveform ranges from a minimum of zero up to some maximum value. Consequently, its average value must be half of its peak-to-peak value. 267 Figure 7.7 Transient analysis for the resistor of Figure 7.5.
|
ACElectricalCircuitAnalysis_Page_267_Chunk767
|
A second set of plots is generated for the inductor. This is shown in Figure 7.8. Again, compare this set against the curves seen in Figure 7.2 for the general reactive case. We can see that the current (green) is 90 degrees out of phase with the inductor's voltage (red) and lagging, as expected. More importantly, we see that the power waveform (blue) is centered around zero. The full cycle average of this is zero, meaning that no power is dissipated. But how does this square with the 2.25 VAR reactive power that was calculated for the inductor? A close look at the power plot shows that that value corresponds to the maximum value of the reactive power waveform (i.e., half of its peak-to-peak value). Finally, a third set of curves are created for the circuit as a whole. In other words, now we're treating the series combination of the inductor and resistor as the load. The results are illustrated in Figure 7.9. The computed impedance phase angle was lagging at 32.1 degrees. We can see this same shift between the voltage (red) and current (green) waveforms. The interesting bit here is the offset and amplitude of the power waveform (blue). The waveform has a peak-to-peak value of about 8.5 VA. Once again, the computed value for apparent power, S, works out to one-half of the plotted peak-to-peak value. This will be the case for P, Q and S. Further, it turns out that if we find the full cycle average of this waveform, those small negative peaks would subtract from the total area and reduce the value. The result would be the true power of 3.58 watts. We'll look at this more closely following another example. 268 Figure 7.8 Transient analysis for the inductor of Figure 7.5.
|
ACElectricalCircuitAnalysis_Page_268_Chunk768
|
Example 7.2 Determine the power dissipated by the resistor in the circuit of Figure 7.10. Also find the apparent power drawn by the circuit and the reactive power of the capacitor. The source frequency is 1 kHz. The first item is to find the reactance of the capacitor. X C = 1 2π f C X C = 1 2π1kHz 1μ F X C =−j15.92Ω Unlike the the previous example, we shall use the v2/R forms for power as an alternative. The RMS value of the source voltage is 7.07 volts. First, find vR. v R = E R R−jX C v R = 7.07V 10Ω 10Ω −j15.92Ω v R = 3.76157.9° V 269 Figure 7.9 Transient analysis for the resistor and inductor together in Figure 7.5. Figure 7.10 Circuit for Example 7.2.
|
ACElectricalCircuitAnalysis_Page_269_Chunk769
|
The capacitor voltage is found via KVL: vC = E −vR vC = 7.070° −3.76157.9 °V vC = 5.99−32.1° V For the powers, we just use the magnitude of the voltage. P = v 2 R P = (3.761V) 2 10Ω P ≈1.414W Q = i2 | X | Q = (5.99 V) 2 15.92Ω Q ≈2.25 VAR, capacitive S = E 2 |Z | S = (7.07 V) 2 |10Ω −j15.92Ω| S ≈2.66 VA, capacitive Computer Simulation Computer Simulation In order to verify the results, the circuit of Figure 7.10 is captured in a simulator as shown in Figure 7.11. Once again, three different transient analysis simulations are run; one each for the resistor, the capacitor, and the pair together. 270 Figure 7.11 The circuit of Figure 7.10 in a simulator.
|
ACElectricalCircuitAnalysis_Page_270_Chunk770
|
In Figure 7.12 we see the results of a transient analysis run on the resistor. We can see that the voltage and current are in phase. Also, the power waveform swings from zero up to around 2.8 watts or so. This corresponds to an average value of just under 1.5 watts, and this agrees nicely with the computed result. Figure 7.13 illustrates the results from a transient analysis run on the capacitor. As expected, the current is leading the voltage by 90 degrees. We can also see that the power waveform is swinging symmetrically around zero, meaning that there is no net power dissipation. The peak value is just under 2.5 VAR, aligning nicely with the calculated value. Finally, in Figure 7.14 we have the results of a transient analysis using both the capacitor and the resistor as the load. The current waveform is still leading the voltage, but by less than 90 degrees. In fact, it leads by about 2/3rds of a division, some 60° or so, which conforms nicely to the expected impedance angle of −57.9° (i.e. from 10 − j15.92 Ω). The power waveform sits slightly below the horizontal axis indicating it is neither true power nor reactive power, but a combination. The peak-to-peak value is somewhat over 5 units, indicating an apparent power of just over 2.5 VA, again, just as expected. 271 Figure 7.12 Transient analysis for the resistor of Figure 7.11.
|
ACElectricalCircuitAnalysis_Page_271_Chunk771
|
272 Figure 7.13 Transient analysis for the capacitor of Figure 7.11. Figure 7.14 Transient analysis for the resistor and capacitor together in Figure 7.11.
|
ACElectricalCircuitAnalysis_Page_272_Chunk772
|
7.3 Power Triangle 7.3 Power Triangle The prior section revealed that the phase angle between the current and voltage cannot be ignored when computing power. For example, if a 120 volt RMS source delivers 2 amps of current, it appears that it delivers 240 watts. This is only true if the load is purely resistive. For a complex load, the true power is somewhat less. In fact, as we've just seen, if the load is purely reactive, there will be no true load power at all. Although plotting the current, voltage and power waveforms is instructive, it can be somewhat cumbersome. Instead, we use a power triangle as shown in Figure 7.15. The horizontal axis represents true power, P, in watts. The vertical axis represents reactive power, Q, in VAR. The vector combination of P and Q results in the apparent power, S, in VA. Remember, the apparent power is the product of the magnitudes of the current and voltage. This is what the power “appears to be” based on simple current and voltage measurements from a DMM, versus a proper power meter. In the resistive case, there is no reactive power and thus S and P are the same. Consequently, the S vector collapses onto the P vector. In a purely reactive case, there is no true power and S and Q are the same; both vectors identical and vertical. For the complex case, S is the vector sum of P and Q. This simple graphic nicely encapsulates the relationship between the three vectors. Further, given any two of the four parts (three vector magnitudes and θ) and with just a little trigonometry, the other two parts may be found. For example, knowing the real and reactive powers, the apparent power can be found via the Pythagorean theorem. Similarly, if the apparent power and angle are known, the real and reactive powers may be found using sine and cosine. Remember, apparent power can be found from the product of the RMS voltage and current magnitudes for any complex impedance, and θ is the same as the impedance angle (i.e., the voltage angle minus the current angle). For your convenience, some useful power triangle relationships are listed following. 273 Figure 7.15 The power triangle. Real Power Reactive Power Real Power, P Reactive Power, Q Apparent Power, S θ
|
ACElectricalCircuitAnalysis_Page_273_Chunk773
|
S = v RMS×iRMS (7.4) S = √P 2+Q 2 (7.5) θ = tan −1 Q P (i.e., the impedance angle) (7.6) P = S cos θ (7.7) P = √S 2−Q 2 (7.8) Q = S sinθ (7.9) Q = √S 2−P 2 (7.10) Power Factor Power Factor As we are often interested in the true power, it is worth noting that a rearrangement of Equation 7.7 shows that the ratio of true power to apparent power is the cosine of the impedance angle, P/S = cos θ. This is known as the power factor and is abbreviated PF. Thus, PF = cos θ. Knowing the phase angle and the apparent power, true power can be calculated. If PF is positive it is said to be a lagging power factor. This is the case for inductive loads where the current is lagging the voltage. In contrast, a capacitive load results in a negative or leading PF. Recall that for capacitors, current leads the voltage. The sign is only used to indicate leading or lagging and will be useful when we examine power factor correction, shortly. For example, if a 100 volt RMS source delivers 1 amp for an apparent power of 100 VA and the phase angle is −30°, PF is cos(−30°) or 0.866 leading and the true power is P = 100 cos(−30°) = 86.6 watts. PF = P S = cosθ (positive is lagging and inductive) (7.11) We'll illustrate the use of the power triangle and the power factor in the next thrilling and action-packed example.13 Example 7.3 Find S, P and Q in the circuit of Figure 7.16. E = 120 volts RMS. The source frequency is 60 Hz. The first step is to determine the inductive reactance. X L = 2π f L X L = 2π60 Hz150 mH X L = j56.55Ω 13 Hyperbole seems to work for the movie industry, anyway. 274 Figure 7.16 Circuit for Example 7.3.
|
ACElectricalCircuitAnalysis_Page_274_Chunk774
|
From here we can determine the system impedance which will, in turn, allow us to determine the source current. Z = R× jX L R + jX L Z = 160Ω×( j56.55Ω) 160Ω+ j56.55Ω Z = 53.370.5°Ω isource = esource Z i source = 1200° V 53.370.5°Ω isource = 2.252−70.5° A The apparent power is the product of the magnitudes of circuit voltage and current. S = E×isource S = 120V×2.252A S = 270.1VA, inductive P = S cos θ P = 270.1VAcos 70.5° P = 90 W Q = S sinθ Q = 270.1VAsin 70.5° Q = 254.7VAR, inductive As a crosscheck, the true power can also be determined by squaring the RMS voltage and then dividing by the 160 Ω resistance. Similarly, dividing the squared voltage by XL will generate Q. P = v R 2 R P = 120V 2 160Ω P = 90 W The power triangle for this circuit is shown in Figure 7.17. 275 Figure 7.17 Power triangle for the circuit of Figure 7.16. 90 W 254.7 VAR 270.1 VA 70.5°
|
ACElectricalCircuitAnalysis_Page_275_Chunk775
|
Power Factor Correction Power Factor Correction One issue with a reactive load is that the current is higher than it needs to be in order to achieve a certain true load power. This is wasteful and would require larger conductors. To alleviate these issues, an opposite reactance can be added to the load such that the resulting load is purely resistive. This can be realized by determining the original Q value and then adding a sufficient reactance to produce an additional Q of the opposite sign resulting in cancellation. From there, it is short step to determine the required impedance. Then, knowing the frequency, the required capacitance or inductance can then be found using the appropriate reactance formula. This is illustrated in the following example. Example 7.4 Find circuit PF, S, P and Q for Figure 7.18. E = 200° volts peak at a frequency of 10 kHz. Also find an appropriate component which when placed from node a to ground brings PF to unity. The inductive reactance of 1 mH at 10 kHz is j62.83 Ω. This is in parallel with the 100 Ω resistor, which is then in series with the 20 Ω resistor. Z = 20Ω + (100Ω|| j62.83Ω) Z = 20Ω + (28.3Ω+ j 45Ω) Z = 66 43°Ω This time we shall use peak values to illustrate the difference compared to using RMS values as in the previous example. The source current is: isource = E Z isource = 200° V 6643°Ω isource = 0.303−43° A peak To find S, multiply the magnitudes of source voltage and current. As these are peak values, multiply each by 0.707 to arrive at RMS, or just cut the answer in half (i.e., 0.7072 is 0.5). This is the apparent power of the whole circuit. S = E×isource 2 S = 20 V×0.303A 2 S = 3.03 VA, inductive 276 Figure 7.18 Circuit for Example 7.4.
|
ACElectricalCircuitAnalysis_Page_276_Chunk776
|
P = S cos θ P = 3.03VA cos43° P = 2.22 W To be clear, 2.22 watts is the combined power dissipation between the two resistors. Q = S sinθ Q = 3.03 VAsin43 ° Q = 2.07 VAR, inductive The power triangle for this circuit is shown in Figure 7.19. For the second part involving power factor correction, we need to add a reactive power equal in magnitude to the existing value but of the opposite sign. This means we need to add 2.07 VAR capacitive. The new power triangle is shown in Figure 7.20. The vertical components cancel, resulting in the apparent power equaling the true power with PF = 1. We can place the correction capacitor where it's convenient in physical terms, just as long as we add 2.07 VAR capacitive. We don't have a physical circuit, so that's not a consideration here. A convenient location would be to place it across the existing resistor-inductor combo. Our goal then is to first find the required reactance, and from that, determined the required capacitance. We shall do this two different ways. In the first case, we note that the capacitor appears across the only other reactive element in the circuit. Therefore, in order for them to cancel they must have the same reactance magnitude, and thus XC must equal −j62.8 Ω. But what if there were multiple reactive elements in the circuit or if it was impractical to locate the component there, for example, due to space restrictions? In that case, we would simply work the power relation backwards. For comparison, suppose we place the capacitor in the same location; from node a to ground. A voltage divider can be used to determine the present value of va. This works out to 11.4 volts RMS. Using the voltage form of the power rule: Q = v X 2 X X = v X 2 Q X = 11.4V 2 2.07VAR X = 62.8Ω Either way we derive XC, now we just solve the capacitive reactance formula to find C. 277 Figure 7.19 Power triangle for the circuit of Figure 7.18. 2.22 W 2.07 VAR 3.03 VA 43 degrees Figure 7.20 Power triangle for the circuit of Figure 7.18 with power factor correction. 2.22 W 2.07 VAR Inductive 2.07 VAR Capacitive After, S=P Before
|
ACElectricalCircuitAnalysis_Page_277_Chunk777
|
C = 1 2π f X C C = 1 2π10 kHz62.8Ω C = 253.3nF The resulting circuit is shown in Figure 7.21. Computer Simulation Computer Simulation It is useful to see the reduction in current demand caused by using power factor correction. To so so, the circuit of Figure 7.21 is captured in a simulator as shown in Figure 7.22. We will run two transient analyses to find the source current. The first version will be run using the original circuit configuration without the capacitor. The second run will include the capacitor. To plot the currents, we will make use of Ohm's law. First we obtain the voltages at nodes 1 and 2. Then, using the simulator's post processor, we subtract v1 from v2 which yields the voltage across the 20 Ω resistor. This quantity is then divided by 20 Ω to arrive at the input current. This is similar to the current sense resistor technique used in earlier chapters. 278 Figure 7.21 The circuit of Figure 7.18 with power factor correction. Figure 7.22 The circuit of Figure 7.21 in a simulator.
|
ACElectricalCircuitAnalysis_Page_278_Chunk778
|
The resulting current for the original circuit is shown in Figure 7.23. Note that the peak current is just over 300 milliamps, as calculated in step two of the example. The second simulation is shown in Figure 7.24, now with power factor correction. 279 Figure 7.23 Simulation results for the circuit of Figure 7.22 without power factor correction. Figure 7.24 Simulation results for the circuit of Figure 7.22 with power factor correction.
|
ACElectricalCircuitAnalysis_Page_279_Chunk779
|
The amplitude here is much reduced, in the range of 160 to 170 milliamps peak. By adding the capacitor, the two reactive currents cancel, leaving a parallel impedance of just 100 Ω. This is in series with the 20 Ω resistor. Dividing 120 Ω into the 20 volt peak source yields a peak current of approximately 167 mA, in agreement with the simulation. The cancellation of the reactive currents is also shown by the fact that the source current is no longer out of phase. In Figure 7.24 the current waveform is in phase and starts at zero, as expected. This implies a load that is the equivalent of a pure resistance. In contrast, the plot of Figure 7.23 shows a current that is lagging by around one-half division, or about 45 degrees, which unsurprisingly is the approximate value of the circuit's impedance angle. Obviously then, this combined impedance must be inductive. To sum up, power factor correction is used to lower current demand while keeping load power constant. If the load is fixed, this can be achieved through the use of a capacitor (for inductive loads) or an inductor (for capacitive loads). If the load demand is dynamic, then a more complex system is required, for example, switching through a bank of capacitors to get a value close to the precise value needed for that particular load. More examples of power factor correction are in the next section. 7.4 7.4 Power Systems Power Systems In this section we shall change our focus and consider power systems at a more functional or abstract level rather than at the component level. Specifically, we would like to consider common loads such as motors, heating elements, lighting devices and the like, and how to analyze a system consisting of a variety of different loads. For the most part, we shall stick with the basic scheme of multiple loads being supplied by a common and ideally constant voltage source with each load configured in parallel with the others. Systems do not have to be configured this way but it is typical of residential and commercial applications. Typical Loads Typical Loads Three common loads are heating elements, lighting devices and motors or compressors. The first two are fairly obvious, but the the latter are also quite common, although often hidden from view inside devices such as refrigerators, air conditioners and heat pumps. We can classify loads in terms of their typical impedance. Devices such as baseboard electrical heating units, toasters, coffee makers and electric ovens are classified as resistive heating devices. The load they present is easily approximated as a simple resistance. The same is true for incandescent lights, however, there can be a very large change in resistance between the on and off states of incandescent lights. Motors, and other devices integrating motors such as compressors, present an inductive impedance. This is due to the large coils of wire, or windings, inside the motor. Any motorized electrical device falls into this category, such as a washing machine, drill press or the fan of Figure 7.25. 280 Figure 7.25 A small boxer fan for cooling electronic components.
|
ACElectricalCircuitAnalysis_Page_280_Chunk780
|
Purely capacitive loads are not nearly as common but some everyday devices exhibit highly complex impedances that can vary widely in both amplitude and angle. A standard moving coil dynamic loudspeaker falls into this category. While you would never plug a loudspeaker directly into an AC power system14, the amplifier that drives it must be designed to meet the challenges of an impedance whose magnitude can change by a factor of ten across the audible range of frequencies. To complicate matters, while the phase typically is inductive, at some frequencies it can be purely resistive or even capacitive. Efficiency Efficiency Efficiency is defined as usable power output divided by applied power input and is denoted by the Greek letter eta, η. Normally it is expressed as a percentage and it can never be over 100%. η = Poutput Pinput (7.12) Some loads, such as typical heating elements, can be thought of as being 100% efficient, meaning that all of the electrical input is turned into useful output (in this case that's heat, although there are more effective ways of using that electrical input to generate heat, such as a heat pump15). In contrast, incandescent light bulbs turn very little of their input into the desired quantity (light) and thus suffer from low efficiency. Just a few percent of the electrical energy fed into an incandescent bulb turns into light while the vast majority turns into heat. From that perspective, incandescent lights are more efficient at producing heat than light. LED lighting, on the other hand, is perhaps an order of magnitude more efficient than incandescent lighting, meaning that we can get the same light output for a much smaller electrical power input while simultaneously producing less heat. Motors lose power in the form of mechanical losses (e.g., friction) and electrical losses (e.g., resistance of windings). Motors are rated in terms of their output power, not the power they draw from the source. For example, a motor with a 1 HP rating might be said to generate 1 HP (approximately 745.7 W) at the shaft. If the motor is 90% efficient, the electrical draw would be 745.7 W/0.9 or 829 W. This situation is further complicated by the phase angle (i.e., power factor) of the motor due to reactive elements, as noted in the previous sections. Compared to motors, home 14 That is, unless you enjoy hearing a short-lived blast of sound which is followed by a burst of flames. Plugging a nominal 8 Ω loudspeaker into a 120 VAC outlet would generate 1800 watts of power, well beyond the design limits of any common loudspeaker. 15 A heat pump moves heat from one place to another. In a properly designed system it takes less energy to move heat than it takes to generate it, and thus it's much more effective than simple resistive heating when it comes to warming the interior of a building. 281
|
ACElectricalCircuitAnalysis_Page_281_Chunk781
|
loudspeakers are particularly inefficient, typically converting only about 1% of the electrical input into usable acoustic output. Example 7.5 A certain 120 VAC RMS 60 Hz 1.2 HP motor has an efficiency of 90% and a lagging power factor of 0.85. Determine the apparent power and the current drawn from the system. Also draw the power triangle. 1 HP ≈ 745.7 watts. The motor output is 1.2 horsepower which is equivalent to: Pwatts ≈1.2 HP×745.7W/HP Pwatts ≈895W The input required to achieve this is computed with Equation 7.12. Pinput = Poutput η Pinput = 895W 0.9 Pinput = 994 W From Equations 7.10 and 7.11, the remaining powers are: S = P PF S = 994W 0.85 S = 1170VA Q = √S 2−P 2 Q = √(1170VA) 2−(994W) 2 Q = 617VAR The current is found via Equation 7.4, essentially power law: S = vRMS×i RMS iRMS = S v RMS iRMS = 1170 VA 120V iRMS = 9.75 A The power triangle is shown in Figure 7.26. 282 Figure 7.26 Power triangle for the motor described in Example 7.5. 994 W 617 VAR Inductive 1170 VA
|
ACElectricalCircuitAnalysis_Page_282_Chunk782
|
Example 7.6 For the motor described in Example 7.5, determine an appropriate component that when placed in parallel will produce unity power factor. Draw the complete power triangle for the system and determine the new current draw. In Example 7.5 the reactive power was determined to be 617 VAR and was inductive, so we'll need 617 VAR capacitive to compensate. The new power triangle is shown in Figure 7.27. The source voltage was stated to be 120 V. We can use a variation on power law to determine the required reactance. Q = v RMS 2 X C X C = v RMS 2 Q X C = 120V2 617VAR X C =−j23.34Ω Now use the capacitive reactance formula to determine the capacitance value. The line frequency was specified as 60 Hz. C = 1 2π f X C C = 1 2π60 Hz23.34Ω C = 114μ F After correction, the apparent power and real power are the same. Thus, iRMS = S v RMS iRMS = 994 VA 120V iRMS = 8.28A The current draw has been reduced by nearly 1.5 amps, a considerable savings. The new current is 85% of what it used to be. Now let's consider a larger system that contains several devices. 283 994 W 617 VAR Inductive 617 VAR Capacitive After, S=P Before Figure 7.27 Power triangle for the corrected system of Example 7.6.
|
ACElectricalCircuitAnalysis_Page_283_Chunk783
|
Example 7.7 The system shown in Figure 7.28 is supplied by a 240 VAC RMS 60 Hz source. Load 1 is 1000 watts of resistive heating elements. Load 2 is a 3 HP motor with an efficiency of 92% and a lagging power factor of 0.75. Load 3 is a capacitor bank equivalent to 75 μF. Draw the system power triangle and determine whether or not load 3 is appropriate to bring the system power factor to unity. First, determine the power triangle for the motor. Then we can add the other loads to it to create a system power triangle. The motor output is 3 HP which is equivalent to: Pwatts ≈3 HP×745.7W/HP Pwatts ≈2237W The required input power is. Pinput = Poutput η Pinput = 2237W 0.92 Pinput = 2432W The computations for apparent and reactive power follow. S = P PF S = 2432W 0.75 S = 3242VA Q = √S 2−P 2 Q = √(3242VA) 2−(2432W) 2 Q = 2144 VAR, inductive The power triangle for the motor is shown in Figure 7.29. 284 Figure 7.29 Power triangle for the motor of Example 7.7. 2432 W 2144 VAR Inductive 3242 VA Figure 7.28 System block diagram for Example 7.7.
|
ACElectricalCircuitAnalysis_Page_284_Chunk784
|
Now for the capacitor. First, we need to determine the reactance, then we can use the v2/Z form of power law to determine Q. X C = 1 2π f C X C = 1 2 π60 Hz75μ F X C =−j35.37Ω Q = v 2 X Q = (240V) 2 35.37Ω Q = 1629VAR, capacitive There are no other reactive elements in the system to consider. We have 515 fewer capacitive VAR than inductive VAR, so the correction will not be ideal. We would need to increase the total capacitance by about 24 μF to achieve a power factor of unity. The system power diagram with all of the individual parts is shown in Figure 7.30. This is then simplified to the final system power triangle as shown in Figure 7.31. The system PF is 0.989 lagging. 285 2432 W 2144 VAR Inductive 1629 VAR Capacitive 1000 W Figure 7.30 Power components for the system of Example 7.7. Figure 7.31 Final power triangle for the system of Example 7.7. 3432 W 515 VAR Inductive 3470 VA
|
ACElectricalCircuitAnalysis_Page_285_Chunk785
|
7.5 Summary 7.5 Summary Power is the product of current and voltage, however, the phase relationship between the two has a major impact on the result. The most simple and straightforward case is when the current and voltage are in phase, meaning the load is a pure resistance. In this case, the true power can be computed directly as the product of RMS current and voltage. At the other extreme, when the voltage and current are 90 degrees out of phase, as in the case of a purely capacitive or inductive load, power is alternately generated and dissipated. That is, in a reactive load, true power dissipation is zero. A mechanical analogy is the storage and release of energy in an ideal spring as it is alternately compressed and then allowed to expand. In between these two extremes, that is, when the load is a complex impedance, the true power dissipation is somewhere between zero and the resistive maximum. The power triangle is used to make visual sense of this situation. It is a right triangle comprised of three legs. The horizontal leg represents the true, or resistive, power and is denoted by the letter P. It is measured in watts. The vertical leg represents the so-called “reactive power”. It is denoted by the letter Q and has units of VAR (volt- amps-reactive). Q can be either inductive or capacitive. The third leg, the hypotenuse, is the apparent power, S. It is measured in VA (volt-amps). It is called apparent power because that is what the power appears to be if it is naively measured with a voltmeter and an ammeter, ignoring the phase difference between them. The angle between the real and apparent powers, θ (theta), is the phase angle between the voltage and current. In other words, theta is the impedance angle. Knowing theta, real power can be determined using right angle trigonometry, namely, P = S cos θ. The cosine of theta is also know as the power factor, PF. It ranges from 0 (purely reactive) to 1 (purely resistive). Positive or inductive impedance angles are said to have lagging power factor while negative or capacitive impedance angles produce leading power factor. Ideally, loads are purely resistive and have a power factor of unity. If this is not the case, then for a given voltage, a higher current is needed to create the same real power as in the purely resistive case. This is not advantageous. Power factor correction is the process of shifting the power factor back to unity for complex loads. This is done by inserting a reactance of the opposite sign to counterbalance the reactive portion of the load, for example, adding capacitive reactance to a system that is inductive. The added reactive power must have the same magnitude as the original reactive power but be of the opposite sign, resulting in cancellation. Efficiency is the measure of usable output power to applied power. Ideally, electromechanical systems such as motors would be 100% efficient, meaning that there is no power loss, but this is not a practical possibility. For example, there will always be frictional losses and power losses in wires. 286
|
ACElectricalCircuitAnalysis_Page_286_Chunk786
|
Review Questions Review Questions 1. Define apparent power, real power and reactive power. 2. Describe the power triangle. 3. What is power factor, PF? 4. Describe the difference between leading and lagging power factor. 5. What is power factor correction? 6. Give examples of resistive loads and inductive loads. 7.6 Exercises 7.6 Exercises Analysis Analysis 1. For the circuit shown in Figure 7.32, determine apparent power S, real power P, reactive power Q and power factor PF. Also, draw the power triangle. 2. For the circuit shown in Figure 7.33, determine apparent power S, real power P, reactive power Q and power factor PF. Also, draw the power triangle. 287 Figure 7.32 Figure 7.33
|
ACElectricalCircuitAnalysis_Page_287_Chunk787
|
3. For the circuit shown in Figure 7.34, determine apparent power S, real power P, reactive power Q and power factor PF. Also, draw the power triangle. The source is 120 volts. 4. For the circuit shown in Figure 7.35, determine apparent power S, real power P, reactive power Q and power factor PF. Also, draw the power triangle. The source is 120 volts. 5. For the circuit shown in Figure 7.36, determine apparent power S, real power P, reactive power Q and power factor PF. The source is 90 volts, XL = j30 Ω, R = 50 Ω. 288 Figure 7.34 Figure 7.35 Figure 7.36
|
ACElectricalCircuitAnalysis_Page_288_Chunk788
|
6. For the circuit shown in Figure 7.37, determine apparent power S, real power P, reactive power Q and power factor PF. Also, draw the power triangle. The source is 240 volts, XC = −j200 Ω, R = 75 Ω. 7. For the circuit shown in Figure 7.38, determine apparent power S, real power P, reactive power Q and power factor PF. The source is 120 volts, XL = j40 Ω, XC = −j25 Ω, R = 20 Ω. 8. For the circuit shown in Figure 7.38, determine apparent power S, real power P, reactive power Q and power factor PF. The source is 120 volts, 60 Hz. R = 80 Ω, C = 20 μF, L = 400 mH. 9. An audio power amplifier delivers a 30 volt RMS 1 kHz sine to a loudspeaker. If the loudspeaker impedance at this frequency is 745°, determine the RMS current delivered to the load and the true power. 10. An audio power amplifier delivers an 80 volt peak 35 Hz sine to a subwoofer. If the subwoofer impedance at this frequency is 4−30°, determine the peak current delivered to the load and the true power. 11. A certain load is specified as drawing 8 kVA with a lagging power factor of 0.8. Determine the real power P, and the reactive power Q. Further, if the source is 120 volts at 60 Hz, determine the effective impedance of the load in both polar and rectangular form, and the requisite resistance/inductance/capacitance values. 12. A certain load is specified as drawing 20 kVA with a leading power factor of 0.9. Determine the real power P, the reactive power Q and draw the power triangle. If the source is 240 volts at 60 Hz, determine the effective impedance of the load in both polar and rectangular form, and the requisite resistance/inductance/capacitance values. 289 Figure 7.37 Figure 7.38
|
ACElectricalCircuitAnalysis_Page_289_Chunk789
|
13. Consider the system shown in Figure 7.39. E is a standard 120 V input. If the three loads are 45 W, 60 W and 75 W incandescent light bulbs, respectively, determine the apparent power delivered to the system, the source current, the reactive power and the real power. 14. Given the system shown in Figure 7.39, determine the apparent power delivered to the system, the source current, the real power, the reactive power and the efficiency. E is 120 V. The three loads are resistive heating elements of 500 W, 1200 W and 1500 W, respectively. 15. Consider the system shown in Figure 7.40. E is 240 V. If the three loads are 200 W, 400 W and 1000 W resistive, respectively, determine the apparent power delivered to the system, the real power and the reactive power. 16. Given the system shown in Figure 7.40, determine the apparent power delivered to the system, the real power, the reactive power and the efficiency. E is 480 V. The three loads are resistive heating elements of 1500 W, 2000 W and 3500 W, respectively. 17. Consider the system shown in Figure 7.39. E is 120 V. Load 1 is 1 kW resistive, load 2 is 400 W resistive and load 3 is 600 VAR inductive. Determine the apparent power delivered to the system, the source current, the reactive power, the real power and the power factor. 18. Consider the system shown in Figure 7.39. E is 240 V. Load 1 is 2 kW resistive, load 2 is 800 W resistive and load 3 is 1200 VAR capacitive. Determine the apparent power delivered to the system, the source current, the real power, the reactive power and the power factor. 19. Given the system shown in Figure 7.39, determine the apparent power delivered to the system, the source current, the real power, the reactive power and the power factor. E is 120 V. Load 1 is 600 W of incandescent lighting, load 2 is 1200 W of heating elements and load 3 is 200 VAR capacitive. 290 Figure 7.40 Figure 7.39
|
ACElectricalCircuitAnalysis_Page_290_Chunk790
|
20. Given the system shown in Figure 7.39, determine the apparent power delivered to the system, the source current, the real power, the reactive power and the power factor. E is 60 V. Load 1 is 90 W of incandescent lighting, load 2 is 800 W of heating elements from a dryer and load 3 200 VAR inductive. 21. A 120 V 3 HP motor draws a real power of 2500 W from the source. Determine its efficiency. 22. A 120 V 12 HP motor draws a real power of 10 kW from the source. Determine its efficiency. 23. An ideal 120 V 2 HP motor draws an apparent power of 1800 W from the source. Determine its power factor. 24. An ideal 120 V 0.3 HP motor draws an apparent power of 270 W from the source. Determine its power factor. 25. A 120 V motor is rated at 0.5 HP. It has an efficiency of 78% and a lagging power factor of 0.7. Determine the apparent power drawn from the source (S), the real power (P), and the reactive power (Q) supplied. Also draw the power triangle and find the delivered current. 26. A motor is rated at 10 HP. It has an efficiency of 92% and a lagging power factor of 0.8. Determine the apparent power drawn from the source (S), the real power (P), and the reactive power (Q) supplied. Also draw the power triangle. Finally, determined the current drawn from the 120 V source. 27. Consider the system shown in Figure 7.39. E is 120 V. Load 1 is 1 kW resistive, load 2 is 400 W resistive and load 3 is a 1 HP motor that is 80% efficient and has a 0.85 lagging power factor. For the system, determine the apparent power delivered, the source current, the real power, the reactive power and the power factor. 28. Consider the system shown in Figure 7.39. E is 120 V. Load 1 is 2.5 kW resistive, load 2 is 500 VAR capacitive and load 3 is a 2 HP motor that is 85% efficient and has a 0.9 lagging power factor. For the system, determine the total power delivered, the source current, the apparent power, the real power and the power factor. 291
|
ACElectricalCircuitAnalysis_Page_291_Chunk791
|
29. For the system shown in Figure 7.41, E is 240 V. Load 1 is 1.2 kW resistive heating, load 2 is 400 W resistive lighting, load 3 is a 0.5 HP motor that is 80% efficient with a 0.7 lagging power factor, and load 4 is a 1 HP motor that is 85% efficient with a 0.8 lagging power factor. For the system, determine the apparent power delivered, the source current, the real power, the reactive power and the power factor. Design Design 30. A 120 V 60 Hz source drives a load equivalent to a 75 Ω resistor in parallel with a 25 μF capacitor. Determine the appropriate capacitance or inductance value to place across this load to produce unity power factor. 31. A 240 V 60 Hz source drives a load equivalent to a 10 Ω resistor in parallel with a 50 mH inductor. Determine the appropriate capacitance or inductance value to place across this load to produce unity power factor. 32. A load of 5030° is driven by a 120 V 60 Hz source. Determine the appropriate capacitance or inductance value to place across this load to produce unity power factor. 33. A load of 50−50° is driven by a 240 V 60 Hz source. Determine the appropriate capacitance or inductance value to place across this load to produce unity power factor. 34. A certain load is specified as drawing 8 kVA with a lagging power factor of 0.8. The source is 120 volts at 60 Hz. Determine the appropriate capacitor or inductor to place in parallel with this load to produce unity power factor. 35. A certain load is specified as drawing 20 kVA with a leading power factor of 0.9. The source is 240 volts at 60 Hz. Determine the appropriate capacitor or inductor to place in parallel with this load to produce unity power factor. 36. A 240 V 60 Hz source is connected to a load consisting of heating elements of 10 kW along with a 15 HP motor with η=90%, PF=0.85. Determine an appropriate capacitor or inductor to place in parallel to produce unity power factor. 37. A 120 V 60 Hz source is connected to a load consisting of 350 W of resistive lighting along with a 1.5 HP motor with η=70%, PF=0.75. Determine an appropriate component to place in parallel to produce unity power factor. 292 Figure 7.41
|
ACElectricalCircuitAnalysis_Page_292_Chunk792
|
Challenge Challenge 38. A power distribution system for a concert has the following specifications: Ten class D audio power amplifiers rated at 2 kW output each with 90% efficiency and unity power factor, 10 kW worth of resistive stage lighting to illuminate the musicians alongside a troupe of trained dancing kangaroos, a 3 HP motor used to continuously rotate the drum riser throughout the performance (η=80%, PF=0.75) and a 2 HP compressor which inflates and deflates a giant rubber T. rex during particularly exciting parts of the show (η=85%, PF=0.8). For the system, determine the total power delivered, the source current, the apparent power, the real power, and the power factor. Finally, make a sketch of this extravaganza with its entertainers in full regalia singing their latest tune “Maximum Volume”. Simulation Simulation 39. Verify the design of problem 28 by performing a transient analysis. The design will have been successful if the source current and voltage are in phase. 40. Verify the design of problem 29 by performing a transient analysis. The design will have been successful if the source current and voltage are in phase. 293
|
ACElectricalCircuitAnalysis_Page_293_Chunk793
|
8 8 Resonance esonance 8.0 Chapter Learning Objectives 8.0 Chapter Learning Objectives After completing this chapter, you should be able to: • Describe series resonance and plot impedance, voltage and current as a function of frequency. • Describe parallel resonance and plot impedance, voltage and current as a function of frequency. • Determine system resonant frequency, Q and bandwidth in series and parallel resonant networks. • Perform series-to-parallel transforms for practical inductors. 8.1 Introduction 8.1 Introduction Resonance can be thought of as a preferred frequency of vibration. In other words, it is a frequency at which a system operates with reduced limitation. It is exploited in a variety of areas, for example, a good mechanical resonance can be used for the construction of acoustic musical instruments. In this case, we strive to maintain the resonant oscillation in order to enhance the instrument's sustain. On the other hand, we might want to control or limit the resonance, as in an automotive suspension system. In electrical systems, resonance tends to produce either a maximum or a minimum response to current or voltage. As a result, resonant systems can be used to filter out or select specific frequencies across the spectrum. Obvious uses include tuning circuits, oscillators, filters and the like. There are two basic forms of resonance for electrical circuits: series RLC resonance and parallel RLC resonance. Series resonance tends to be the less complicated of the two. Both types share many similarities but in some respects they are mirror images of each other. Series and parallel resonant circuits both exhibit wide fluctuations in impedance magnitude and phase across the frequency spectrum. In the case of series resonance, the impedance is at a minimum at the resonance frequency. This implies a current maximum if the circuit is driven by a voltage source. In contrast, parallel resonance produces an impedance peak at resonance. This implies a voltage maximum if the circuit is driven by a current source. Much of the initial circuit analysis in this chapter concentrates on the response of the circuits across a range of frequencies rather than at some random specific frequency. Consequently, we will be making considerable use of frequency domain plots; that is, plots of impedance, voltage and current magnitude and phase as they vary across the frequency spectrum. Once that is established we can zero in on the area of resonance and determine precisely the shape of these curves in the region around the resonant frequency. Along the way we will define and utilize new circuit parameters such as Q, bandwidth, half-power frequencies and more.16 16 “Q” as used in this chapter is not to be confused with the Q used to represent reactive power, although it is related to the Q, or quality factor, of an inductor. 294
|
ACElectricalCircuitAnalysis_Page_294_Chunk794
|
8.2 Series Resonance 8.2 Series Resonance Let's begin with the simplest RLC circuit; one consisting of a single voltage source in series with a single resistor, inductor and capacitor, as shown in Figure 8.1. Of particular interest is how the total impedance varies across the frequency spectrum and what impact this has on the current and the three component voltages. The impedance as seen by the source is simply the sum of the three components, or Z = R+ jX L −j X C This can be expanded into Z= = R+ j 2π f L −j 1 2π f C The interesting part here is that the first term is not a function of frequency, the second term is directly proportional to frequency and the third is inversely proportional to frequency. Further, given that the positive and negative reactances behave oppositely, it appears that at some frequency they may cancel out, leaving just the resistance. To refine this, we expect that at low frequencies the capacitor will dominate the impedance. In other words, XC will be the largest of the three ohmic values. This means that the overall impedance will tend to mimic both the magnitude and phase of the capacitive reactance. On the other hand, at very high frequencies the inductor will tend to dominate the impedance. XL will be the largest of the three values. In this region, the combined impedance will echo that of the inductor. In short, at low frequencies the impedance magnitude will be large and the circuit will appear capacitive while at high frequencies the impedance magnitude will be large and the circuit will appear inductive. In the middle is where things get interesting. A plot of the resistance or reactance of the three elements is shown in Figure 8.2. The sum of the three is also shown (red). The frequency axis is uses a logarithmic scale to show the symmetrical nature of the combined impedance curve. 295 Figure 8.1 A series RLC circuit.
|
ACElectricalCircuitAnalysis_Page_295_Chunk795
|
The dip in the center corresponds to an impedance equal to R. At this frequency the capacitive and inductive reactances are equal in magnitude and effectively cancel each other. All that's left is the resistive component, R. This frequency is known as the resonant frequency and is denoted by f0. The series resonant frequency, f0, is the frequency at which the magnitudes of the inductive and capacitive reactances are equal. (8.1) This implies that the power factor is unity at resonance. Also, in a real world circuit R is the combination of the series resistance plus any resistance from the inductor's coil. We can derive a formula for f0 as follows. The definition declares that the magnitude of XL must equal the magnitude of XC. Therefore, we can set the capacitive and inductive reactance formulas equal to each other and then solve for the resulting frequency. X C = 1 2π f C X L = 2π f L 296 Series Resonance Impedance Plot Impedance Magnitude 0 2 4 6 8 10 0 2 4 6 8 10 Normalized Frequency 0.1 1.0 10.0 0.1 1 10 R+jXL-jXC XC XL R Figure 8.2 Series impedance variation across frequency.
|
ACElectricalCircuitAnalysis_Page_296_Chunk796
|
X L = X C 2π f 0 L = 1 2π f 0C f 0 2 = 1 (2π) 2 LC f 0 = 1 2π√LC (8.2) Note that a particular resonant frequency can be obtained through a variety of LC pairs. This, along with the value of R, will alter the specific shape of the impedance curve in terms of how narrow or broad the dip is. These components will also affect how quickly the phase response shifts from fully capacitive (−90°) to fully inductive (+90°). This shape factor is described by the parameter, Q. The tighter or more narrow the curve, the higher the Q. Given a constant voltage source, it should be no surprise that a plot of the resulting current will be an inversion of the impedance curve. This is shown in Figure 8.3. If we scale the curves such that they both have a normalized peak of unity, the difference in the shapes may be a little easier to see. This is shown in Figure 8.4. 297 Series Resonance Current Plot Relative Current 0 2 4 6 8 10 Normalized Frequency 0.1 1.0 10.0 0.1 1 10 Low Q High Q Figure 8.3 Series current versus frequency.
|
ACElectricalCircuitAnalysis_Page_297_Chunk797
|
At this point we can more precisely define Q. Specifically, the “sharpness” of the curve is related to the half-power or “−3 dB” frequencies, f1 and f2.17 These are the frequencies at which the current (assuming voltage source drive) falls off to 0.707 of the maximum value at resonance. Therefore, they represent the frequencies at which power will have fallen to one-half of the maximum value seen at resonance (recall that power varies as the square of current and that 0.707 squared is approximately 0.5). f1 is below f0 and f2 is found above. The difference between these two frequencies is called the bandwidth, BW. BW = f 2 −f 1 (8.3) Qcircuit = f 0 BW (8.4) The relationship between these variables is illustrated in Figure 8.5. The vertical axis is shown as a percentage of maximum. For a series resonant circuit driven by a voltage source, this axis is current; however, it can be voltage in the the case of a parallel resonant circuit, as we shall see. If this plot is compared to the curves in Figure 8.4, it should be apparent that for lower Q circuits, f1 and f2 spread apart, moving away from the resonant frequency, f0. Thus, for any given f0, a lower Q means a wider (larger) bandwidth. 17 Decibels are covered in detail in Chapter 10. 298 Series Resonance Current Plot Normalized Current 0 0.2 0.4 0.6 0.8 1 Normalized Frequency 0.1 1.0 10.0 0.1 1 10 Low Q High Q Figure 8.4 Normalized series current versus frequency.
|
ACElectricalCircuitAnalysis_Page_298_Chunk798
|
The resonant frequency, f0, in general is not located evenly between f1 and f2. It is, in fact, located at their geometric mean. In other words, f 0 = √f 1 f 2 (8.5) From Equation 8.5 we may derive: f 0 f 1 = f 2 f 0 (8.6) To find accurate values for f1 and f2 we can define a factor, k0. The derivation of k0 is found in Appendix C. k 0 = 1 2Qcircuit +√ 1 4Qcircuit 2 +1 (8.7) f 1 = f 0 k 0 (8.8) f 2 = f 0×k 0 (8.9) 299 Figure 8.5 Location of f1 and f2, and definition of bandwidth (BW). Resonance Frequencies Percent of Maximum 0 10 20 30 40 50 60 70 80 90 100 Normalized Frequency 0.1 1.0 10.0 0.1 1 10 f1 f0 f2 70.7% BW
|
ACElectricalCircuitAnalysis_Page_299_Chunk799
|
For higher Q circuits (Qcircuit ≥ 10), we can approximate symmetry, and thus f 1 ≈f 0 −BW 2 (8.10) f 2 ≈f 0+ BW 2 (8.11) As mentioned previously, the Q can be a function of either R or the L/C ratio. In Figures 8.6 and 8.7 we have impedance curves for the two cases. The frequency axis is normalized to f0 (i.e., f0 is unity). In Figure 8.6 we vary the resistance value to see how it affects both the magnitude and phase of the impedance across frequency. Figure 8.7 is similar except we vary the inductor/capacitor ratio. Looking first at the phase (blue, left axis), we see in both cases that high Q circuits exhibit a quick transition from a negative (capacitive) phase angle to a positive (inductive) phase angle. We also notice that the phase shift hits zero at f0, implying unity power factor. The impedance magnitude plots show a slightly different story. While it is true that the higher Q plots are sharper, they get that way through different mechanisms. In the case of the resistor, a lower Q is achieved via a larger resistance. This has the effect of blunting the tip of the curve and lowering current flow at f0 when compared to the high Q case (as seen in Figure 8.3). In contrast, reducing Q by reducing the 300 Series Resonance Impedance Variation with Q: R Value Impedance Phase (degrees) -100 -50 0 50 100 Relative Impedance Magnitude 0 2 4 6 8 10 Normalized Frequency 0.1 1.0 10.0 0.1 1 10 Dashed: Lower Q via higher R value Figure 8.6 Magnitude and phase of impedance for a variation of resistance.
|
ACElectricalCircuitAnalysis_Page_300_Chunk800
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.