text stringlengths 1 7.76k | source stringlengths 17 81 |
|---|---|
We begin with the charge time constant: τcharge = RC τcharge = 20 kΩ220 nF τcharge = 4.4 ms Steady-state will be reached in 5 times 4.4 milliseconds, or 22 milliseconds. The capacitor is initially uncharged, so VC(0) = 0 volts. As the capacitor will have reached steady-state in 22 milliseconds, VC(50 ms) = 12 volts. The maximum charging current will occur at t = 0 when all of the 12 volt source drops across the 20 kΩ resistor, or 600 μamps, flowing left to right. At 50 milliseconds the switch is thrown to position 2. The 12 volt source and 20 kΩ resistor are no longer engaged. At this point the capacitor has 12 volts across it, positive to negative, top to bottom. As the capacitor voltage cannot change instantaneously, the capacitor now acts as a voltage source and discharges through the 120 kΩ resistor. Note that the discharge current is flowing counterclockwise, the opposite of the charging current. The discharge time constant is: τdischarge = RC τdischarge = 120kΩ220nF τdischarge = 26.4 ms The capacitor will fully discharge down to 0 volts in 5 time constants, or some 132 milliseconds after the switch is thrown to position 2. Thus steady- state occurs at t = 182 milliseconds. The maximum discharge current occurs the instant the switch is thrown to position 2 when all of the capacitor's 12 volts drops across the 120 kΩ resistor, yielding 100 μamps, flowing top to bottom. Clearly, at t = 90 milliseconds the capacitor is in the discharge phase. The capacitor's voltage and current during the discharge phase follow the solid blue curve of Figure 8.25. The elapsed time for discharge is 90 milliseconds minus 50 milliseconds, or 40 milliseconds net. We can use a slight variation on Equation 8.14 to find the capacitor voltage at this time. V C(t) = E ϵ −t τ V C (40 ms) = 12 V−ϵ −40ms 26.4 ms V C (40 ms) ≈2.637V The shape of the capacitor's voltage will appear somewhat like a rounded pulse, rising with a curve and then falling back to zero with a complementary curve (the red and then blue curves of Figure 8.25). 279 | DCElectricalCircuitAnalysis_Page_279_Chunk1201 |
Basic single resistor-capacitor circuits prove to be fairly easy to solve given a little practice, but what if a more complex circuit is used? In this situation the section feeding the capacitor may be simplified using Thévenin's theorem to determine the effective source voltage and charging resistance. The circuit then reverts back to a simple RC network which may be solved directly. If power is interrupted before the capacitor is fully charged, the equations presented previously may be used to determine the precise voltage(s) and current(s) reached. The capacitor will then behave as a voltage source and begin to discharge, its voltage curve following the blue plot line of Figure 8.25, with its maximum voltage being what the capacitor charged to, not the associated driving voltage. The following example and simulations address these issues. Example 8.8 For this example we shall revisit the circuit of Example 8.5. The circuit is redrawn in Figure 8.30 for convenience. Assume the capacitor is initially uncharged. Determine the charging time constant, the amount of time after the switch is closed before the circuit reaches steady-state, and the capacitor voltage at t = 0, 100 milliseconds, and 200 milliseconds. At 200 milliseconds, the switch is opened. Determine how long it takes for the capacitor to fully discharge and the voltage across the 6 kΩ resistor at t = 275 milliseconds (i.e., 75 milliseconds after the switch is opened). The first step is to determine the Thévenin equivalent driving the capacitor. If we remove the capacitor and determine the open circuit voltage at those points, we see that it is just a voltage divider between the 24 volt source, the 6 kΩ resistor and the 1 kΩ resistor (the 3 kΩ resistor has no current through it and thus produces no voltage drop). This works out to 20.57 volts. The Thévenin resistance will be 3 kΩ in series with 1 kΩ || 6 kΩ, or roughly 3.857 kΩ. The equivalent charging circuit is drawn in Figure 8.31. We can now determine the charging time constant: τcharge = RC τcharge = 3.857 kΩ10μ F τcharge = 38.57 ms Steady-state will be reached in 5 time constants, or 192.8 ms. Thus, we know that VC(0) = 0 volts and VC(200 ms) = 20.57 volts. For the capacitor voltage at 100 milliseconds, we simply use the charge equation. 280 Figure 8.30 Circuit for Example 8.8. Figure 8.31 Thévenin equivalent for the circuit of Figure 8.30 driving the capacitor. | DCElectricalCircuitAnalysis_Page_280_Chunk1202 |
V C(t)= E (1−ϵ −t τ) V C (100ms) = 20.57V(1−ϵ −100 ms 38.57ms) V C (100ms) ≈19.03V For the discharge phase, we need to determine the time constant. With the voltage source removed, the capacitor will discharge through the now series combination of the 3 kΩ resistor and 6 kΩ resistor. τdischarge = RC τdischarge = 9k Ω10μ F τdischarge = 90 ms Steady-state will be reached 450 milliseconds later at t = 650 milliseconds. To find V6k at t = 275 milliseconds, we can find the voltage across the capacitor and then perform a voltage divider between the 6 kΩ and 3 kΩ resistors. Remembering that t = 275 milliseconds is 75 milliseconds into the discharge phase, we have: V C(t) = E ϵ −t τ V C (75 ms) = 20.57V(1−ϵ −75ms 90ms) V C (75 ms) ≈8.94 V Finally, this voltage splits between the 6 kΩ and 3 kΩ resistors. Using the voltage divider rule, we find: V 6k = V C Rx Rx+Ry V 6k = 8.94 V 6 kΩ 6 kΩ+3kΩ V 6k = 5.96 V Computer Simulation Computer Simulation In order to verify the analysis of Example 8.8, the circuit of Figure 8.30 is entered into a simulator, as shown in Figure 8.32. In place of a DC source, a pulse generator is used to mimic the on-off nature of the switch. This starts starts at zero volts and then immediately jumps up to 24 volts. It stays at this level for 200 milliseconds before returning back to zero. This is sufficient time to check whether or not the capacitor voltage has reached steady-state (predicted to take 192.8 milliseconds). 281 | DCElectricalCircuitAnalysis_Page_281_Chunk1203 |
A transient analysis is run on this circuit, plotting the capacitor voltage (i.e., the difference between the node 2 and node 3 voltages). The result is shown in Figure 8.33. This plot confirms nicely the charge phase of the capacitor. After approximately 200 milliseconds, the voltage has leveled out at just over 20 volts, precisely as predicted. What about the discharge phase? In order to investigate that portion, the simulation circuit is modified. The pulse voltage source is disconnected from the remainder of the circuit, just as it would be if the switch in Figure 8.30 had been opened again. Also, the capacitor is modified to have an initial voltage of 20.57 volts, the precise value at had reached after it attained steady-state. A second transient analysis is run, 282 Figure 8.32 Circuit of Figure 8.30 in a simulator. Figure 8.33 Simulation results for the charge phase of the circuit of Figure 8.30. | DCElectricalCircuitAnalysis_Page_282_Chunk1204 |
again plotting the capacitor voltage. The results of this simulation are shown in Figure 8.34. It is worth noting that the time axis is relative to the switch being opened, not the original timing. That is, the horizontal origin of 0 milliseconds corresponds to t = 200 milliseconds. Note that the time required to reach the new steady-state value of zero volts has stretched out to some 450 milliseconds after the switch is opened, precisely as predicted. At this point, a fair question to ask is, “Couldn't we leave the pulse source in place in order to investigate the discharge phase?” Although the pulse source does go back down to zero volts at t = 200 milliseconds, that's not the same as opening the switch back in Figure 8.30. If the source is still connected but producing zero volts, it becomes part of the Thévenin equivalent. As such, the 1 kΩ resistor is back in the circuit, producing a discharge time constant identical to the charge time constant. To prove the point, the simulation is run again. The pulse remains at 24 volts for 200 milliseconds and then jumps down to zero, as before. The simulation time is extended out to 400 milliseconds total, enough to see both the charge and discharge phases. Further, instead of just plotting the capacitor voltage; the voltages of the source (blue trace, node 1), the 6 kΩ resistor (green trace, node 2) and the 3 kΩ resistor (red trace, node 5) are plotted. The results are shown in Figure 8.35. 283 Figure 8.34 Simulation results for the discharge phase of the circuit of Figure 8.30. | DCElectricalCircuitAnalysis_Page_283_Chunk1205 |
The first item to note is that steady-state appears to be reached in just under 200 milliseconds for both the charge and discharge phases, as expected. Second, note that the node 2 voltage (green) minus the node 3 voltage (red) starts at zero and winds up at a little over 20 volts at 200 milliseconds. This is, of course, the capacitor voltage, but what is interesting here is that this plot shows how the voltage across the 3 kΩ resistor shrinks as the capacitor voltage grows, the sum equaling the node 2 voltage. This makes perfect sense because, as the capacitor voltage increases, the current through it must be decreasing, and as this same current is flowing through the 3 kΩ resistor, the resistor's voltage must also be decreasing due to Ohm's law. The other interesting part of this plot is what happens at 200 milliseconds when the source goes back to zero. Note that the node 3 voltage immediately jumps to a negative value. This is because the voltage across the capacitor cannot change instantaneously. It must still have 20.57 volts across it the instant the source goes back to zero. In this situation, because the source is essentially a short, the capacitor winds up in series with the 3 kΩ resistor and the parallel combination of the 1 kΩ and 6 kΩ resistors, or about 857 Ω. Calculating the voltage divider between the 3 kΩ and 857 Ω resistors with 20.57 volt source shows the 3 kΩ resistor receiving approximately 16 volts. Further, the discharge current will be flowing out of the capacitor in a counterclockwise direction, meaning it flows from ground up through the 3 kΩ resistor. Thus, we expect node 3 to be at approximately −16 volts, which is precisely what the plot indicates. How cool is that? What if we didn't wait for steady-state? How would these plots change? Essentially, the trajectories of the curves would not change. After all, how would the circuit “know” that the switch would open early or the pulse would flip prematurely? 284 Figure 8.35 Simulation results using pulse voltage source for discharge phase. | DCElectricalCircuitAnalysis_Page_284_Chunk1206 |
What happens is that the curves are followed to the point in time where the circuit is interrupted. From there, the next phase occurs with the present voltages as the starting points. To test this, the simulation is run yet again, but this time the source pulse width is shortened to just 50 milliseconds, well short of steady-state. The results of the simulation are shown in Figure 8.36. Comparing the plot of Figure 8.36 to that of Figure 8.35 shows that the two are identical up to 50 milliseconds. At that point, the input pulse returns to zero and the capacitor begins to discharge. The shapes and timings of the node 2 and node 3 voltages are the same as they were in Figure 8.35, however, the amplitudes are reduced. This is because the capacitor did not have time to reach the steady-state voltage of 20.57 volts. In fact, it only reaches about 14.94 volts using Equation 8.12. Applying the voltage divider on this potential as before shows that the 3 kΩ resistor should jump to approximately −11.6 volts, which is confirmed by the simulation. 8.5 Summary 8.5 Summary The capacitor is a device that is used to store electric charge. Capacitance, C, is measured in farads, F. The idealized device consists of two conductive plates separated by some distance, that space being filled by an insulating dielectric. Capacitance is directly proportional to the plate area and the dielectric's permittivity, and inversely proportional to the plate distance. Another important characteristic of the dielectric is its breakdown strength. Along with the plate spacing, this will 285 Figure 8.36 Simulation results for interrupted discharge phase using pulse voltage source. | DCElectricalCircuitAnalysis_Page_285_Chunk1207 |
establish the capacitor's voltage rating. The permittivity will also help to determine the capacitor's volumetric efficiency, a measure of how much capacitance can be achieved within a given volume. Non-ideal parameters include the ESR, or equivalent series resistance, which is ideally zero; and the effective parallel leakage resistance, ideally infinity. Absolute accuracy, temperature stability and similar parameters round out the distinguishing features of one kind of capacitor against another. When placed in parallel, capacitors add in the same manner as resistors in series. When placed in series, capacitors behave like resistors in parallel. Perhaps the most important operational characteristic regarding capacitors is that voltage across a capacitor cannot change instantaneously. It will take some finite amount of time before the charge on the capacitor builds, leading to a predictable rise in voltage across it. Because of this, for DC circuits capacitors initially behave like shorts, but after sufficient time has passed, they behave like opens. The amount of time required to reach steady-state is five time constants, where one time constant is defined as the product of the circuit's effective resistance and its capacitance. The current charge curve is of the shape ε−t. The current starts at a maximum and eventually approaches zero as time passes. The corresponding voltage shape is of the form 1−ε−t. Here, the capacitor's voltage starts at zero and rises to some maximum value. The capacitor's voltage discharge curve effectively is swapped compared to the charge curve (e.g. voltage follows ε−t). Review Questions Review Questions 1. What are the physical characteristics of capacitors and how do they affect capacitance? 2. Define the voltage-current characteristic for capacitors. 3. What is meant by a capacitor's volumetric efficiency? 4. How do the permittivity and breakdown strength of the dielectric affect the overall capacitance and voltage rating? 5. How do capacitors combine when placed in series and how do they combine when placed in parallel? 6. Define the initial and steady-state behavior of capacitors. 7. Define time constant for an RC circuit. 8. Describe the charge and discharge characteristics of RC circuits. 286 | DCElectricalCircuitAnalysis_Page_286_Chunk1208 |
8.6 Exercises 8.6 Exercises Analysis Analysis 1. For the circuit shown in Figure 8.37, determine the effective capacitance. 2. Determine the effective capacitance of the configuration shown in Figure 8.38. 3. Given the capacitor network shown in Figure 8.39, determine the effective value. 4. Determine the effective capacitance of network shown in Figure 8.40. 287 Figure 8.37 Figure 8.40 Figure 8.38 Figure 8.39 | DCElectricalCircuitAnalysis_Page_287_Chunk1209 |
5. Determine the voltage across each capacitor for the circuit shown in Figure 8.41. 6. Determine the voltage across each capacitor for the circuit shown in Figure 8.42. 7. Determine the initial voltage across each component for the circuit shown in Figure 8.43. 8. Given the network shown in Figure 8.43, determine the steady-state voltage across each component. 288 Figure 8.41 Figure 8.42 Figure 8.43 | DCElectricalCircuitAnalysis_Page_288_Chunk1210 |
9. For the circuit shown in Figure 8.44, determine the capacitor voltage 3 microseconds after the power is switched on. 10. For the circuit shown in Figure 8.45, determine the capacitor voltage 5 seconds after the power is switched on. 11. Determine the time constant and the time required to reach steady-state for the circuit shown in Figure 8.46. 12. For the circuit shown in Figure 8.46, determine the capacitor voltage and circulating current 20 microseconds and 100 milliseconds after the switch is thrown. 13. Given the circuit shown in Figure 8.47, determine the capacitor voltage and circulating current 200 milliseconds and 10 seconds after the switch is thrown. 289 Figure 8.44 Figure 8.45 Figure 8.46 Figure 8.47 | DCElectricalCircuitAnalysis_Page_289_Chunk1211 |
14. Determine the time constant and the time required to reach steady-state for the circuit shown in Figure 8.47. 15. Determine the time constant and the time required to reach steady-state for the circuit shown in Figure 8.48, switch position 1. 16. Determine the charge and discharge time constants for the circuit shown in Figure 8.49. 17. Given the circuit shown in Figure 8.48, determine the capacitor voltage 10 milliseconds after the power is turned on. At this point, the switch is thrown to position 2. Determine how long it will take the capacitor to discharge to nearly zero volts. 18. For the circuit shown in Figure 8.49, determine the capacitor voltage 400 microseconds after the power is turned on. At this point, the switch is thrown to position 2. Determine how long it will take the capacitor to discharge to nearly zero volts. 19. Determine the time constant and the time required to reach steady-state for the circuit shown in Figure 8.50, switch position 1. 290 Figure 8.48 Figure 8.49 Figure 8.50 | DCElectricalCircuitAnalysis_Page_290_Chunk1212 |
20. Given the circuit shown in Figure 8.50, determine the capacitor voltage 12 milliseconds after the power is turned on. At this point, the switch is thrown to position 2. Determine how long it will take the capacitor to discharge to nearly zero volts. Design Design 21. Given the circuit of Figure 8.46, determine a new resistor value such that steady-state is reached in 2 milliseconds. 22. Given the circuit of Figure 8.47, determine a new resistor value such that steady-state is reached in 5 seconds. 23. When an audio amplifier is turned on, the power surge can cause an audible pop from the loudspeaker. To prevent this, amplifiers often connect to the loudspeaker via a relay. The relay is energized to connect the loudspeaker once the output has settled, typically a few seconds after power is applied. This delay may be created via an RC network. Suppose the driving circuit is 5 volts and the relay trips at 4 volts. Further, the associated charging resistance is 10 kΩ. Determine the capacitance required to achieve a 2 second delay time. Challenge Challenge 24. Determine the time constant and the time required to reach steady-state for the circuit shown in Figure 8.51. 25. For the circuit shown in Figure 8.51, determine the capacitor voltage 1 second after the power is turned on. At this point, the switch is thrown to position 2. Determine how long it will take the capacitor to discharge to nearly zero volts. 291 Figure 8.51 | DCElectricalCircuitAnalysis_Page_291_Chunk1213 |
Simulation Simulation 26. Perform a transient analysis to verify the time to steady-state of Figure 8.46 (problem 11). 27. Perform a transient analysis to verify the time to steady-state of Figure 8.47 (problem 14). 28. Use a transient analysis to verify the design of problem 21. 29. Use a transient analysis to verify the design of problem 22. 30. Use a transient analysis to verify the operation of the circuit shown in Figure 8.51 as specified in problem 25. You may wish to do this as two separate simulations, one for each switch position, with the second position using a pre-charged capacitor. 292 | DCElectricalCircuitAnalysis_Page_292_Chunk1214 |
Notes Notes ♫♫ ♫♫ 293 | DCElectricalCircuitAnalysis_Page_293_Chunk1215 |
9 Inductors Inductors 9.0 Chapter Learning Objectives 9.0 Chapter Learning Objectives After completing this chapter, you should be able to: • Describe the theoretical and practical aspects of inductor construction. • Describe the current-voltage characteristic behavior of inductors. • Utilize component data sheets to determine operating characteristics of inductors. • Determine the initial and steady-state equivalents of resistor-inductor networks. • Determine the initial and steady-state equivalents of resistor-capacitor-inductor networks. • Determine the transient response of basic RL networks. 9.1 Introduction 9.1 Introduction This chapter introduces our third passive device, the inductor. Inductors are fundamentally different from both resistors and capacitors in terms of their construction and their operation. Inductors do, however, share certain broad traits with capacitors. First, they are energy storage devices. In the case of the inductor, energy is stored in a magnetic field, similar to the case of the capacitor which utilizes an electric field. Further, in the ideal case, inductors do not dissipate power. Also like capacitors, when inductors are placed in DC circuits they are not ohmic, meaning that their current-voltage characteristic does not respond to Ohm's law. Instead, their current- voltage characteristic is dynamic in nature. In some respects, though, their current-voltage behavior is opposite to the way in which capacitors behave, and thus they offer their own unique performance characteristics. Inductors have a long history of use in electronic systems. In fact, one of the most common uses in the home is as an integral part of a typical loudspeaker system. They are used in modern switch-mode DC power supplies in products such as personal computers and televisions. In audio and communications systems they are used in filters and tuning circuits alongside capacitors. Like a capacitor, for any application that needs to smooth out a varying voltage, store energy or filter a signal; an inductor is a likely candidate. Unfortunately, real-world inductors generally do not behave as close to their desired ideal operation as do real- world capacitors. The secondary effects of inductor construction limit their performance; perhaps the most notable factor being their potentially large equivalent series resistance. They are also susceptible to external magnetic fields which can introduce noise and interference, degrading signal quality. For these reasons, there are areas where, given a choice, capacitors would be preferred over inductors. But this is by no means a broad condemnation and there are areas where the use of inductors is essential. Beyond this, the very concept of inductance is important in that informs designers of the practical limits of performance of their circuits. 294 | DCElectricalCircuitAnalysis_Page_294_Chunk1216 |
The concept of electromagnetic induction was first discovered by English scientist Michael Faraday in the early 19th century. He noticed that if he wrapped two wires around an iron ring and introduced a current in one of them, then a transient (short- lived) current would appear in the second coil of wire. He noticed a similar effect when he slid a bar magnet through a wire coil. Around the same time, American Joseph Henry discovered much the same independently of Faraday and performed considerable research in this area. In his honor, the unit of inductance, the henry (abbreviated H), is named after him. The symbol for inductance is L, named after physicist Heinrich Emil Lenz. Inductors are commonly referred to as coils or chokes, for reasons that will be apparent shortly. 9.2 Inductance and Inductors 9.2 Inductance and Inductors To begin, we need to examine the interrelation between electric current and magnetic fields in a conductor. When a current passes through a conductor, such as a wire, a magnetic field is created around the conductor that is proportional to the strength of the current. This is illustrated in Figure 9.1. The magnetic field can be thought of as sets of concentric rings around the conductor, although for clarity only single loops are drawn in the Figure. The number of magnetic lines in a given area is known as the magnetic flux and is given the symbol Φ (the Greek letter phi). The unit of magnetic flux is the weber, Wb, named after Wilhelm Weber, a 19th century German physicist. Magnetic flux ≡ the number of magnetic lines enclosed in a given area. (9.1) Note that the magnetic field runs the length of the conductor. The direction of the field lines follows the right hand rule: if you grasp the wire with your right hand such that your thumb is pointing in the direction of conventional current flow, then your fingers wrap in the direction of the magnetic field. This rule is illustrated in Figure 9.2. If we form the conductor into a loop, the field lines are corralled into the center of the loop. This is illustrated in Figure 9.3. In this diagram it can be seen that the lines effectively collect in the center, going into the page. 295 Figure 9.1 Magnetic field around a conductor. I Figure 9.2 Right hand rule. Image source (modified) | DCElectricalCircuitAnalysis_Page_295_Chunk1217 |
The enhancement effect can be magnified by adding more loops in tandem. This is known as a solenoid and is shown in Figure 9.4. It is the most basic form of an inductor. The concentrating effect of the magnetic field is shown in Figure 9.5. In this figure, the coil is shown from the side, as a cross-section of the individual loops. The dots inside the conductors indicate that the current is flowing toward you, out of the page; while the the crosses indicate that the current is flowing into the page. The lines of flux exit out the right, loop around, and reenter from the left. Due to limited space, the entire loop for each line is not drawn, and it is important to remember that magnetic flux lines do not end, but always create a loop. Further, although it is shown as a plane here, this field is three dimensional, with lines looping back into the page as well as in front of it. This is how electromagnets can be created20. The north pole is the exiting end (right side) of Figure 9.5 while the south pole is the entering end (left side). 20 Truly one of the coolest inventions of all time: a magnet with an on/off switch. 296 Figure 9.3 Magnetic field around a loop. Figure 9.4 Solenoid. Figure 9.5 Magnetic field in a solenoid. Image source. | DCElectricalCircuitAnalysis_Page_296_Chunk1218 |
If the current changes, there will be a commensurate change in the magnetic field. Further, this change in the field will induce a current in the conductor that creates a magnetic field that opposes the original change in the field. This is known as Lenz's law. Alternately, it can be stated that the induced current caused by a changing magnetic field will oppose the change in the original current that created that change in the original magnetic field. At this point we can offer a proper definition of the weber: 1 weber ≡ the magnetic flux that, acting on a single loop of a conductor, produces a potential of 1 volt if the flux is reduced to zero at an even rate over 1 second. (9.2) In magnetic circuits we are also interested in the magnetic flux density which is the magnetic flux per unit area. The symbol for flux density is B and has units of teslas (T), named after Nicola Tesla, the Serbo-Croatian-American engineer and inventor. One tesla is defined as one weber per square meter. 1 tesla ≡ 1 weber / meter2 (9.3) To provide a reference, the magnetic flux density of the Earth near the equator is approximately 31 μT, while the value of the voice coil gap in a loudspeaker is around 1 to 2 T, with medical MRI scanners being a little higher still. Finally, we come to the definition of inductance and its unit, the henry: Inductance is a measure of the tendency of a conductor to oppose a change in the current flowing through it. (9.4) 1 henry ≡ 1 weber / 1 amp (9.5) Unsurprisingly, the energy stored in the magnetic field of an inductor is proportional to the inductance. It is also proportional to the square of the current through the inductor. W =1 2 L I 2 (9.6) Where W is the energy in joules, L is the inductance in henries, I is the current in amps. 297 | DCElectricalCircuitAnalysis_Page_297_Chunk1219 |
An inductor in its simplest form consists of a series of wire loops. These might be wound around an iron core, although a non-ferrous core might also be used. For a simple single layer inductor, such as the one drawn in Figure 9.6, the inductance is described by the following formula: L=μ A N 2 l (9.7) Where L is the inductance in henries, μ is the permeability of the core material, A is the cross sectional area of the coil, N is the number of coils or turns, l is the length of the coil. Inductors may also be wound using multiple layers or around a toroidal core, and these designs utilize alternate formulas. Inductor Styles and Packaging Inductor Styles and Packaging Equation 9.7 indicates that, in order to achieve high inductance, we would like a core with high permeability, permeability being a measure of how easy it is to establish magnetic flux in said material. Substances such as iron or ferrite have a much greater permeability than air and are used commonly for cores. They do have a disadvantage in that they will saturate sooner than an air core, and this can lead to distortion. Another approach is to pack as many turns as is possible within a given length. One way to do this is to minimize the thickness of the insulation around the wire21. This can be achieved by using a thin enamel coating instead of the typical plastic insulation. A second method is to use a very fine wire. This leads to two problems, namely an undesirable increase in equivalent series resistance (known colloquially as coil resistance or Rcoil), and limited current carrying capacity. All of these effects have to be balanced in order to achieve the best performance for a given application. Commercial inductors range in value from a fraction of a nanohenry for small surface mount “chip” inductors up to several henries. Some devices exhibit large internal inductances even though they are not specifically used as inductors. One common example is a transformer. Another example is an electric guitar or bass pickup, such as the one shown in Figure 9.7 with its cover removed. Units such as this may be constructed of several thousand turns of very fine wire (typically AWG 41 to 44) and achieve inductances in excess of one henry. 21 The wound wire must be insulated otherwise each loop will short to the loops next to it and we'd be left with a tube instead of a series of loops. 298 Figure 9.6 Simple air-core inductor dimensions. Figure 9.7 Electric bass guitar pickup. | DCElectricalCircuitAnalysis_Page_298_Chunk1220 |
A variety of inductors is shown in Figure 9.9, all of which are of the through-hole type (surface mount inductors do not appear considerably different from their surface mount resistor and capacitor cousins). The two units toward the left are molded inductors and use a standard color code, similar to the ones used for resistors and capacitors. The unit at the top (yellow) is a high current inductor that features low Rcoil. The three inductors in the center use obvious ferrite cores, two wound on straight cores and the third wound on a toroidal core. The unit to the right uses a high permeability material at the very top and is wrapped in a plastic sheath for protection. Variable inductors are also a possibility and can be made by using a ferrite core that can be slid within the coils, effectively changing the permeability of the core (part ferrite, part air). The schematic symbols for inductors are shown in Figure 9.10. The standard symbol is at the top. The variable inductor symbol is in the middle and is a two- lead device, somewhat reminiscent of the symbol for a rheostat. At the bottom is the symbol for an inductor with an iron, ferrite, or similar high permeability core. In general, like resistors, single inductors are not polarized and cannot be inserted into a circuit backwards. There are, however, special applications where multiple coils can be wound on a common core, and for these, the polarity of their interconnection can make a difference. Inductor Data Sheet Inductor Data Sheet A portion of an inductor data sheet is shown in Figure 9.12. This page lists the available sizes of this particular model, each with corresponding quantities. We can see this model is available in inductance values ranging from 1 μH up to 100 mH. Tolerance of the smaller values is ±10% while values at and above 33 μH are at ±5%. Q is the quality factor and is particularly important in AC circuits (higher being better), along with its associated frequency, fQ. Continuing across we find IR. This is the maximum rated current. For the smaller values, we find they can withstand in excess of 2 amps while the larger units can withstand only tens of milliamps. 299 Figure 9.9 A collection of inductors. Figure 9.10 Inductor schematic symbols (top-bottom): standard, variable, iron/ferrite core. | DCElectricalCircuitAnalysis_Page_299_Chunk1221 |
Finally, we come to Rmax. This is also known as Rcoil. It represents the equivalent series resistance of the inductor. In general, smaller is better. For this model, it ranges from a fraction of an ohm to a few hundred ohms. This trend is typical of inductors; all else being equal, the larger the inductance, the larger the value of the associated series resistance. In many circuits, the value of Rcoil cannot be ignored. 300 Figure 9.12 Inductor data sheet. Courtesy of TDK | DCElectricalCircuitAnalysis_Page_300_Chunk1222 |
Inductors in Series and in Parallel Inductors in Series and in Parallel Suppose we take two identical inductors and place them in series. This effectively doubles both the length and the number of loops. From Equation 9.7 we can see that doubling both the number of loops and the length would double the inductance. This is because N2 goes up by a factor of four which is then halved be the increased length. Consequently, inductors in series add values just like resistors in series. By extension, inductors in parallel behave like resistors in parallel. The equivalent of parallel inductors can be found by using either the product-sum rule or by taking the reciprocal of the sum of their reciprocals. Example 9.1 Find the equivalent inductance of the network shown in Figure 9.11. The 6 mH and 12 mH inductors are in parallel. The equivalent value of the pair is: Lparallel = L2 L3 L2+L3 Lparallel = 6 mH12 mH 6 mH+12 mH Lparallel = 4mH This combination is in series with the 5 mH inductor. Therefore the total equivalent inductance is 4 mH + 5 mH, or 9 mH. Current-Voltage Relationship Current-Voltage Relationship The fundamental current-voltage relationship of the inductor is the mirror image of that of the capacitor: v=L di dt (9.8) This states that the voltage across the inductor is a function of how quickly the current is changing. If the current is not changing (i.e., in steady-state), then the voltage across the inductor is zero. In this case, the inductor behaves like a short, or more accurately, like its Rcoil value. In contrast, during a rapid initial current change, the inductor voltage can be large, and thus the inductor behaves like an open. 301 Figure 9.11 Circuit for Example 9.1. | DCElectricalCircuitAnalysis_Page_301_Chunk1223 |
If we rearrange Equation 9.8 and solve for the rate of change of current, we find that: di dt = v L (9.9) Thus if an inductor is fed by a constant voltage source, the current will rise at a constant rate equal to v/L. For example, considering the circuit in Figure 9.11, we see a voltage source feeding a single inductor. If we were to plot the inductor's current over time, we would see something like the graph of Figure 9.12. As time progresses, the current through the inductor increases, flowing from top to bottom. With a theoretically perfect inductor and source, this would continue as long as the circuit was energized. In reality, this line would either begin to deflect horizontally as the source reached its limits, or the inductor would fail once its maximum current or power handling was reached. The slope of this line is dictated by the size of the applied voltage source and the inductance. Example 9.2 Determine the rate of change of current through the inductor in the circuit of Figure 9.13. Also determine the inductor's current 10 microseconds after power is switched on. From Equation 9.9, the rate of change of current is: di dt = v L di dt = 10V 50 mH di dt = 200A per second In other words, for every second, the current rises another 200 amps. Thus, after just 10 microseconds it will have risen to 200 A/s times 10 μs, or 2 mA. 302 I t Figure 9.11 Inductor with voltage source. Figure 9.12 Inductor current versus time. Figure 9.13 Circuit for Example 9.2. | DCElectricalCircuitAnalysis_Page_302_Chunk1224 |
Equation 9.8 is key to understanding the behavior of inductors. As noted previously, if an inductor is driven by a fixed voltage source and ignoring Rcoil, the current through it rises at the constant rate of v/L. This change in current through the inductor is not limitless. An instantaneous change requires that di/dt is infinite, and thus, the voltage driving the inductor would also have to be infinite, which is a clear impossibility. Therefore we can state a particularly important characteristic of capacitors: The current through an inductor cannot change instantaneously. (9.10) This observation will be central to analyzing the operation of inductors in DC circuits. 9.3 Initial and Steady-State Analysis of RL Circuits 9.3 Initial and Steady-State Analysis of RL Circuits When analyzing resistor-inductor circuits, remember that current through an inductor cannot change instantaneously as this would require an infinite voltage source. When a circuit is first energized, the current through the inductor will still be zero, which is characteristic of opens. Once at steady-state, the current has leveled out and therefore the voltage across the inductor will approach zero, which is characteristic of shorts. Thus, we can state the general behavior of inductors at the beginning and ending of the charge cycle: For DC analysis, initially inductors appear as opens. (9.11) At steady-state, inductors appear as shorts. (9.12) This is the opposite of what was seen with capacitors. For example, in the circuit of Figure 9.14, initially L is open, leaving us with R1 and R2 in series with the source, E. At steady-state, L shorts out, leaving R1 in series with the parallel combination of R2 and R3. All practical inductors will exhibit some internal resistance, so it is often best to think of an inductor as an ideal inductance with a small resistance (Rcoil) in series with it. 303 Figure 9.14 Basic RL circuit. | DCElectricalCircuitAnalysis_Page_303_Chunk1225 |
Example 9.3 Assuming the initial current through the inductor is zero in the circuit of Figure 9.15, determine the voltage across the 2 kΩ resistor when power is applied and after the circuit has reached steady-state. Draw each of the equivalent circuits. First, we'll redraw the circuit for the initial-state equivalent. To do so, open the inductor. The new equivalent is shown in Figure 9.16. By opening the inductor, the 6 kΩ resistor has been removed from the circuit and sees no voltage. What we are left with is a voltage divider between the source and the 1 kΩ and 2 kΩ resistors. Using the voltage divider rule, V 2k = E Rx Rx+R y V 2k = 25 V 2k Ω 2k Ω+1 kΩ V 2k ≈16.67V For steady-state, we redraw using a short in place of the inductor, as shown in Figure 9.17. Here we have another voltage divider, this time between the 1 kΩ resistor and the parallel combination of 2 kΩ and 6 kΩ, or 1.5 kΩ. V 2k = E Rx Rx+R y V 2k = 25 V 1.5kΩ 1.5kΩ+1kΩ V 2k = 15V 304 Figure 9.15 Circuit for Example 9.3. Figure 9.16 Initial-state equivalent of the circuit of Figure 9.15. Figure 9.17 Steady-state equivalent of the circuit of Figure 9.15. | DCElectricalCircuitAnalysis_Page_304_Chunk1226 |
Computer Simulation Computer Simulation To verify the results of Example 9.3, the circuit is entered into a simulator as shown in Figure 9.18. A DC operating point analysis is run and the results are shown in Figure 9.19. The steady-state potential at node 2 corresponds to the voltage across the 2 kΩ resistor and agrees with the theoretical calculation of 15 volts. Note that node 3 is also 15 volts, indicating that the steady-state voltage across the inductor is zero, meaning it is behaving as a short, exactly as expected. 305 Figure 9.18 Circuit of Figure 9.15 in a simulator. Figure 9.19 Simulation results for the circuit of Figure 9.15. | DCElectricalCircuitAnalysis_Page_305_Chunk1227 |
9.4 Initial and Steady-State Analysis of RLC Circuits 9.4 Initial and Steady-State Analysis of RLC Circuits When analyzing resistor-inductor-capacitor circuits, remember that capacitor voltage cannot change instantaneously, thus, initially, capacitors behave as a short circuit. Once the capacitor has been charged and is in a steady-state condition, it behaves like an open. This is opposite of the inductor. As we have seen, initially an inductor behaves like an open, but once steady-state is reached, it behaves like a short. For example, in the circuit of Figure 9.20, initially L is open and C is a short, leaving us with R1 and R2 in series with the source, E. At steady-state, L shorts out both C and R2, leaving all of E to drop across R1. For improved accuracy, replace the inductor with an ideal inductance in series with the corresponding Rcoil value. Similarly, practical capacitors can be thought of as an ideal capacitance in parallel with a very large (leakage) resistance. Example 9.4 Assuming the initial current through the inductor is zero and the capacitor is uncharged in the circuit of Figure 9.21, determine the current through the 2 kΩ resistor when power is applied and after the circuit has reached steady-state. Draw each of the equivalent circuits. For the initial-state equivalent we open the inductor and short the capacitor. The new equivalent is shown in Figure 9.22. The shorted capacitor removes everything to its right from the circuit. All that's left is the source and the 2 kΩ resistor. We can find the current through the 2 kΩ resistor using Ohm's law. 306 Figure 9.20 Basic RLC circuit. Figure 9.21 Circuit for Example 9.4. Figure 9.22 Initial-state equivalent of the circuit of Figure 9.21. | DCElectricalCircuitAnalysis_Page_306_Chunk1228 |
I 2k = E R I 2k = 14 V 2k Ω I 2k = 7 mA Steady-state is redrawn in Figure 9.23, using a short in place of the inductor, and an open for the capacitor. We are left with a resistance of 2 kΩ in series with the parallel combination of 1 kΩ and 4 kΩ, or 2.8 kΩ in total. I2k = E R I 2k = 14 V 2.8k Ω I 2k = 5mA 9.5 Transient Response of RL Circuits 9.5 Transient Response of RL Circuits The transient response of RL circuits is nearly the mirror image of that for RC circuits. To appreciate this, consider the circuit of Figure 9.24. Again, the key to this analysis is to remember that inductor current cannot change instantaneously. When power is first applied, the circulating current must remain at zero. Therefore no voltage drop is produced across the resistor, and by KVL, the voltage across the inductor must equal the source, E. This establishes the initial rate of change of current via Equation 9.9 (di/dt = E/L) and is represented by the dashed red line in the graph of Figure 9.25. As the current starts to increase, the voltage drop across the resistor begins to increase. This reduces the voltage available for the inductor, thus slowing the rate of change of current. This is depicted by the solid red 307 Figure 9.23 Steady-state equivalent of the circuit of Figure 9.21. Figure 9.24 RL circuit for transient response analysis. | DCElectricalCircuitAnalysis_Page_307_Chunk1229 |
curve on the graph. Meanwhile, the solid blue curve represents the decreasing inductor voltage. Thus, in the RL circuit, the inductor's voltage curve echoes the RC circuit's current curve (or resistor voltage curve), and the RL current curve echoes the RC circuit's capacitor voltage curve. The curves presented in Figure 9.25 are identical to those presented in Chapter 8 when we discussed capacitors. They are reproduced here for your convenience. As noted before, the rate of current change versus time is equal to v/L, and therefore in this case, E/L. If the initial rate of change were to continue unabated, the maximum (steady-state) current, E/R, would be reached in L/R seconds22. Therefore the time constant for an RL circuit is: τ = L R (9.13) Once again, five constants will achieve steady-state. 22 Which is to say, E/L amps per second times L/R seconds yields E/R amps. 308 Normalized Charge/Discharge Curves Percent of Maximum 0 20 40 60 80 100 Time Constants 0 1 2 3 4 5 Figure 9.25 Normalized charge and discharge curves. | DCElectricalCircuitAnalysis_Page_308_Chunk1230 |
Following the prior work on capacitors, the relevant equations for the RL circuit can be shown to be23: V L(t) = Eϵ −t τ (9.14) V R(t) = E(1 −ϵ −t τ) (9.15) I (t)= E R (1 −ϵ −t τ) (9.16) Where VL(t) is the inductor voltage at time t, VR(t) is the resistor voltage at time t, I(t) is the current at time t, E is the source voltage, R is the series resistance, t is the time of interest, τ is the time constant, ε (also written e) is the base of natural logarithms, approximately 2.718. Time for an example. Example 9.5 Given the circuit of Figure 9.26, assume the switch is closed at time t = 0. Determine the charging time constant, the amount of time after the switch is closed before the circuit reaches steady-state, and the inductor voltage and current at t = 0, t = 2 microseconds and t = 1 millisecond. Assume the inductor is initially uncharged. First, the time constant: τ = L R τ = 400μ H 150Ω τ ≈2.667μs Steady-state will be reached in five time constants, or approximately 13.33 microseconds. Thus we know that VL(0) = 9 volts and VL(1 ms) = 0 volts. Because the inductor is an open initially, IL(0) = 0 amps. At IL(1 ms), the circuit is in steady-state and the inductor acts like a short. Therefore, all of the 9 volt source drops across the 150 Ω resistor, for 60 mA. 23 See also Appendix C. 309 Figure 9.26 Circuit for Example 9.5. | DCElectricalCircuitAnalysis_Page_309_Chunk1231 |
To find VL(2 μs) we simply solve Equation 9.14. V L(t)= E ϵ−t τ V L(2μ s) = 9V ϵ − 2μs 2.667μ s V L(2μ s) ≈4.251V This value can also be determined graphically from Figure 9.25. The time of 2 microseconds represents 75% of a time constant. Find this value on the horizontal axis and then track straight up to the solid blue curve that represents the charging inductor voltage. The point of intersection is right around 47% of the maximum value on the vertical axis. The maximum value here is the source voltage of 9 volts. Therefore the inductor will have reached approximately 47% of 9 volts, or just over 4.2 volts. The current can be found in a similar manner using Equation 9.16. I L(t)= E R (1 −ϵ −t τ) I L(2μs)= 9V 150Ω(1 −ϵ − 2μs 2.667μ s) I L(2μs)= 60 mA(1 −ϵ −0.75) I L(2μs)= 31.66mA Computer Simulation Computer Simulation To verify our analysis, the circuit of Figure 9.26 is entered into a simulator, as shown in Figure 9.27. In order to reflect the notion of a time-varying circuit with a switch, the 9 volt DC voltage source has been replaced with a rectangular pulse voltage source. This source starts at 0 volts and then immediately steps up to 9 volts. It stays at this level for 20 microseconds before dropping back to 0 volts. 310 Figure 9.27 The circuit of Figure 9.26 in a simulator. | DCElectricalCircuitAnalysis_Page_310_Chunk1232 |
The results of a transient analysis are shown in Figure 9.28. The waveform shown tracks the inductor's voltage at node 2 with respect to ground. We can see that the voltage starts at 9 volts as expected. It then falls back to zero and is at steady-state in less than 15 microseconds, just as predicted. At 20 microseconds, the pulse source returns to zero volts. At this instant, the current through the inductor must still be the steady-state current of 60 milliamps. This current will still be flowing in a clockwise direction, thus it will produce a 9 volt drop across the 150 Ω resistor with a + to − polarity from left to right. This effectively places node 2 negative with respect to ground. The result is that the polarity of the inductor's voltage flips, with the inductor now acting as a short-lived source. We see this on the transient analysis as a negative 9 volt spike. The discharge time constant is identical to the charge constant, and thus we see the inductor's voltage fall back to zero in the same amount of time. Example 9.6 Given the circuit of Figure 9.29, find VL at t = 1 microsecond after the circuit is energized. Assume the inductor is initially uncharged. First, the time constant: τ = L R τ = 6 mH 15 kΩ τ = 400ns 311 Figure 9.28 Simulation results for the circuit of Figure 9.26. Figure 9.29 Circuit for Example 9.6. | DCElectricalCircuitAnalysis_Page_311_Chunk1233 |
Steady-state will be reached in five time constants, or 2 microseconds, at which point the inductor voltage will be zero as it will be behaving as a short. In contrast, as the inductor is initially an open (IL(0) = 0), all of the current from the source will flow into the 15 kΩ resistor, producing 30 volts across this parallel network. Therefore, we can state that VL(0) = 30 volts and VL(2 μs) = 0 volts, defining the extremes. In order to find VL(1 μs), we can use Equation 9.14. V L(t)= E ϵ−t τ V L(1μs) = 30 V ϵ −1μs 0.4μ s V L(1μs) ≈2.463V There are a few different ways to crosscheck this value. For starters, we can determine the inductor current using a slight modification of Equation 9.16 (the current source value is used in place of E/R as the equation effectively requires the maximum or steady-state current). I L(t) = I (1 −ϵ −t τ) I L(1μs) = 2mA(1 −ϵ −1μs 0.4μ s) I L(1μs) = 1.836mA The inductor voltage of 2.463 volts would also have to appear across the parallel 15 kΩ resistor. This produces 2.463 V / 15 kΩ, or 0.164 mA of resistor current. By KCL, the remainder of the 2 mA source current must be flowing down through the inductor. This yields a net inductor current of 2 mA − 0.164 mA, or 1.836 mA, verifying our prior result. Example 9.6 also reinforces the concept of the time constant being inversely proportional to the resistance, rather than directly proportional as in the RC case. In the circuit of Figure 9.29, it should be obvious that the larger the resistance value, the larger the resulting initial-state voltage. From Equation 9.9 it can be seen that if the voltage across the inductor is increased, then the initial rate of change of current with respect to time will increase, and that implies a shorter time constant. For more complex circuits, Thévenin's theorem may be used in order to determine the effective source voltage and charging resistance. As we saw with RC circuits, it's also possible that the discharge resistance may be considerably different from the charging resistance. In such a case, the charge and discharge curves can be highly asymmetric in both time and amplitude. Generally, the larger the discharge resistance is when compared to the charge resistance, the larger in voltage and the shorter in time the discharge spike will be (think in terms of the area under the curve staying constant). 312 | DCElectricalCircuitAnalysis_Page_312_Chunk1234 |
Example 9.7 Assume the initial current through the inductor is zero in Figure 9.30. Determine the time constant. Also, determine the inductor voltage and the voltage across the 6 kΩ resistor 200 nanoseconds after the switch is closed. This circuit is based on the circuit presented in Figure 9.15 as used in Example 9.3. In that analysis it was discovered that the steady- state voltage for the 6 kΩ and 2 kΩ resistors was 15 volts, the pair being in parallel. Further, the initial voltage across the 2 kΩ resistor and the inductor was 16.67 volts and for the 6 kΩ resistor, 0 volts. For the Thévenin circuit, the open circuit voltage at the inductor would be the potential across the 2 kΩ resistor, which is obtained from a voltage divider between it and the 1 kΩ resistor, or 16.67 volts. The equivalent resistance is obtained by shorting the voltage source which leaves the 1 kΩ and 2 kΩ resistors in parallel, which is then in series with the 6 kΩ resistor, yielding approximately 6.667 kΩ. The equivalent is shown in Figure 9.31. Immediately, we can see that the initial-state voltage across the inductor is confirmed in the equivalent circuit. We can now determine the time constant. τ = L R τ = 1mH 6.667k Ω τ = 150ns Steady-state will be reached in 750 nanoseconds. To find the inductor voltage we can use Equation 9.14. V L(t)= E ϵ−t τ V L(200ns)= 16.67V ϵ −200 ns 150 ns V L(200ns) ≈4.39 V 313 Figure 9.30 Circuit for Example 9.7. Figure 9.31 Thévenin equivalent of the circuit of Figure 9.30 driving the inductor. | DCElectricalCircuitAnalysis_Page_313_Chunk1235 |
Referring back to the original circuit, in order to determine the voltage across the 6 kΩ resistor we can find the current through it and use Ohm's law. The current would be the same as the inductor's current as the two are in series. Thus, Equation 9.16 would do the trick. I L(t) = E R (1 −ϵ −t τ) I L(200ns) = 16.67V 6.667k Ω(1 −ϵ −200 ns 150 ns) I L(200ns)= 2.5mA(1 −ϵ−1.333) I L(200ns) ≈1.841mA And finally, V 6k(200 ns) = I R V 6k(200 ns) = 1.841 mA6k Ω V 6k(200 ns) ≈11.05 V Further, the voltage across the 2 kΩ resistor must be the sum, or approximately 15.44 volts. Computer Simulation Computer Simulation The results of Example 9.7 are crosschecked in a simulator. Once again the circuit is built using a pulse generator, as shown in Figure 9.32. A transient analysis is run out to 1 microsecond which is modestly into steady-state. Node voltages 2 and 3 are plotted, as shown in Figure 9.33. The initial voltage across the 2 kΩ resistor (node 2) is as predicted, approximately 16.7 volts, and falls to 15 volts at steady-state, approximately 750 nanoseconds later. The voltage across the 6 kΩ resistor (node 2) starts at zero volts and also winds up at 15 volts in steady- state, just as predicted. Further, note that the predicted voltages across the 2 kΩ and 6 kΩ resistors at 200 nanoseconds are verified. 314 Figure 9.32 Circuit of Figure 9.30 in a simulator. | DCElectricalCircuitAnalysis_Page_314_Chunk1236 |
The voltage across the inductor is node 2 minus node 3. This differential is plotted separately in Figure 9.34. The expected initial, steady-state and 200 nanosecond voltages are as predicted. 315 Figure 9.33 Transient analysis simulation of the circuit of Figure 9.30. Figure 9.34 Simulation of the inductor voltage versus time for the circuit of Figure 9.30. | DCElectricalCircuitAnalysis_Page_315_Chunk1237 |
One very important observation is that if an RL circuit is abruptly altered or opened, very large voltage spikes may occur. This is due to the fact that inductor current cannot change instantaneously. If the circuit is opened, the open represents a very large resistance. Ohm's law indicates that the inductor current times this very large resistance may produce a very large voltage across the new open. In fact, the potential may be sufficient to cause a spark or arc. Note that because the current cannot change instantaneously (both magnitude and direction), the inductor now behaves as a voltage source of very high magnitude and with reverse polarity. This phenomenon is used to create the ignition spark in internal combustion engines. In short, the ignition coil is charged, creating some current flow. The circuit is then interrupted leaving just the coil in series with the spark plug, the spark plug being little more than a precisely sized gap between two electrodes. This results in a large voltage being developed across the spark plug gap, typically in the neighborhood of 20,000 volts, which is sufficient to create a small arc (i.e., the spark) that then ignites the air-fuel mixture in the piston. An example of producing a discharge voltage spike that is considerably larger than the source voltage can be illustrated with the circuit of Figure 9.35. We shall assume that the inductor is initially uncharged when power is applied, and that the switch is in position 1. In this case, the circuit consists of just the 12 volt source, the 2.2 kΩ resistor, and the inductor. The circuit reaches steady-state in roughly 227 nanoseconds. At that point the inductor behaves as a short, leaving the full 12 volt source to drop across the 2.2 kΩ resistor. This produces a clockwise current of approximately 5.455 mA. If we now move the switch to position 2, this current must be maintained because the current through the inductor cannot change instantaneously. The new discharge resistance is now the series combination of the two resistors, or 49.2 kΩ. Ohm's law and KVL dictate that the resulting inductor voltage must be 49.2 kΩ times 5.455 mA, or just beyond −268 volts. This potential is negative because the clockwise current is flowing up through the 47 kΩ resistor, producing a drop + to − from ground up. The increased resistance also shortens the time constant and now steady-state will be reached in only 10.1 nanoseconds. Thus, we see a much larger magnitude spike (over twenty times that of the source voltage) with a much shorter time duration (less than one-twentieth of the charge time). 316 Figure 9.35 Circuit illustrating large discharge voltage spike. | DCElectricalCircuitAnalysis_Page_316_Chunk1238 |
Depending on the kind of switch that is used, things can be even more extreme than what has been just described. Switches come in two basic varieties: make-before- break and break-before-make. The former makes contact with the second position before it breaks contact with the first position, while the latter does the opposite. The behavior just described assumes that a make-before-break switch is being used. In contrast, if a break-before-make switch is used, we'd see a drastically different result. Let us pick up back at steady-state with 5.455 mA flowing through the inductor. We now throw the switch to position 2. This new switch breaks contact with the wire leading back to voltage source prior to it making contact with the 47 kΩ resistor. For a short instant of time the switch is not making contact with anything and the resulting resistance in the loop is determined by the air gap between the switch contacts. Even if this was a mere 10 MΩ, the resulting potential would be over 50,000 volts. This will almost certainly create a spark and we will have inadvertently recreated the spark plug scenario. Hopefully, there are no combustible gases nearby. 9.6 Summary 9.6 Summary The inductor is a device that stores energy in the form of a magnetic field. Inductance, L, is measured in henries, H. The idealized device consists of several loops or coils of wire. These may or may not be wrapped around a magnetic core material. Inductance is directly proportional to the permeability of the core material and the cross sectional area of the loops, and inversely proportional to the length of the coil. It is also proportional to the square of the number of loops. The most important non-ideal parameter is Rcoil, or equivalent coil resistance. Ideally, this value is zero. Absolute accuracy, temperature stability and similar parameters round out the distinguishing features of one kind of inductor against another. When placed in series, inductors add in the same manner as resistors in series. When placed in parallel, inductors combine like resistors in parallel. Perhaps the most important operational characteristic regarding inductors is that current through an inductor cannot change instantaneously. It will take some finite amount of time before the magnetic field reacts, leading to a predictable rise in current through it. Because of this, for DC circuits inductors initially behave like opens, but after sufficient time has passed, they behave like shorts. The amount of time required to reach steady-state is five time constants, where one time constant is defined as the inductance divided by the circuit's effective resistance. The current charge curve is of the shape 1−ε−t. The inductor's current starts at zero and rises to some maximum value. The corresponding voltage shape is of the form ε−t. Here, the voltage starts at a maximum and eventually approaches zero as time passes. 317 | DCElectricalCircuitAnalysis_Page_317_Chunk1239 |
Review Questions Review Questions 1. What are the physical characteristics of inductors and how do they affect inductance? 2. Define the voltage-current characteristic for inductors. 3. What is Rcoil? 4. How do inductors combine when placed in series? How do they combine when placed in parallel? 5. Define the initial and steady-state behavior of inductors. 6. Define time constant for an RL circuit. 7. How do the charge and discharge characteristics of RL circuits compare to those of RC circuits? 9.7 Exercises 9.7 Exercises Analysis Analysis 1. For the circuit shown in Figure 9.36, determine the effective inductance. 2. Determine the effective inductance of the network shown in Figure 9.37. 3. Given the inductor network shown in Figure 9.38, determine the effective value. 318 Figure 9.36 Figure 9.37 Figure 9.38 | DCElectricalCircuitAnalysis_Page_318_Chunk1240 |
4. Determine the effective inductance of network shown in Figure 9.39. 5. Determine the initial voltage across each component for the circuit shown in Figure 9.40. 6. Given the network shown in Figure 9.40, determine the steady-state voltage across each component. 7. Given the network shown in Figure 9.41, determine the steady-state voltage across each component. 8. Determine the initial voltage across each component for the circuit shown in Figure 9.41. 9. Given the network shown in Figure 9.42, determine the steady-state voltage across each component. 319 Figure 9.39 Figure 9.40 Figure 9.41 Figure 9.42 | DCElectricalCircuitAnalysis_Page_319_Chunk1241 |
10. Determine the initial voltage across each component for the circuit shown in Figure 9.42. 11. Determine the initial voltage across each component for the circuit shown in Figure 9.43. 12. Given the network shown in Figure 9.43, determine the steady-state voltage across each component. 13. Given the network shown in Figure 9.44, determine the steady-state voltage across each component. 14. Determine the initial voltage across each component for the circuit shown in Figure 9.44. 15. For the circuit shown in Figure 9.45, determine the inductor current 20 microseconds after the power is switched on. Assume this is an ideal inductor with no internal resistance. 320 Figure 9.43 Figure 9.44 Figure 9.45 | DCElectricalCircuitAnalysis_Page_320_Chunk1242 |
16. For the circuit shown in Figure 9.46, determine the inductor current 100 milliseconds after the power is switched on. Assume this is an ideal inductor with no internal resistance. 17. Determine the time constant and the time required to reach steady-state for the circuit shown in Figure 9.47. 18. Determine the time constant and the time required to reach steady-state for the circuit shown in Figure 9.48. 19. Given the circuit shown in Figure 9.37, determine the inductor voltage and circulating current 200 milliseconds and 2 seconds after the switch is thrown. How does this change if we include 5 Ω of internal resistance to the inductor? 20. For the circuit shown in Figure 9.48, determine the inductor voltage and circulating current 50 nanoseconds and 100 milliseconds after the switch is thrown. 321 Figure 9.46 Figure 9.47 Figure 9.48 | DCElectricalCircuitAnalysis_Page_321_Chunk1243 |
21. Determine the time constant and the time required to reach steady-state for the circuit shown in Figure 9.49, switch position 1. 22. Determine the charge and discharge time constants for the circuit shown in Figure 9.50. 23. Given the circuit shown in Figure 9.49, determine the inductor current 1 millisecond after the power is turned on. At this point, the switch is thrown to position 2. Determine the inductor voltage the instant of switch contact (assume ideal switch). 24. For the circuit shown in Figure 9.50, determine the inductor current 400 microseconds after the power is turned on. At this point, the switch is thrown to position 2. Determine the inductor voltage the instant of switch contact (assume ideal switch). 25. Determine the time constant and the time required to reach steady-state for the circuit shown in Figure 9.51, switch position 1. 322 Figure 9.49 Figure 9.50 Figure 9.51 | DCElectricalCircuitAnalysis_Page_322_Chunk1244 |
26. Given the circuit shown in Figure 9.51, determine both the inductor current and voltage 10 milliseconds after the power is turned on. At this point, the switch is thrown to position 2. Determine how long it will take the inductor to discharge to nearly zero amps. Assume ideal switch. Design Design 27. For Figure 9.47, determine a new resistor value such that steady-state is reached in 2 milliseconds. 28. For Figure 9.48, determine a new inductor value such that steady-state is reached in 5 microseconds. Challenge Challenge 29. Determine the time constant and the time required to reach steady-state for the circuit shown in Figure 9.52. 30. For the circuit shown in Figure 9.52, determine the inductor voltage 400 microseconds after the power is turned on. At this point, the switch is thrown to position 2. Determine how long it will take the inductor to discharge to nearly zero amps. Assume ideal switch. Simulation Simulation 31. Use a transient analysis to verify the time constant and time to steady-state of Figure 9.48 (problem 18). 32. Use a transient analysis to verify the inductor voltage waveform of Figure 9.47 as specified in problem 19. 323 Figure 9.52 | DCElectricalCircuitAnalysis_Page_323_Chunk1245 |
10 10 Magnetic Circuits and Transformers Magnetic Circuits and Transformers 10.0 Chapter Learning Objectives 10.0 Chapter Learning Objectives After completing this chapter, you should be able to: • Describe the basic quantities of a magnetic circuit including magnetic flux, flux density and magnetizing force. • Outline a BH curve and define aspects including hysteresis, coercivity and retentivity. • Analyze basic magnetic circuits using BH curves. • Describe the uses and operation of transformers. • Analyze basic transformer circuits. 10.1 Introduction 10.1 Introduction This chapter extends the material presented in the chapter on inductors and introduces the concept of magnetic circuits. Basic applications are also illustrated. Here we shall take a closer look at the parameters of magnetic circuits such as permeability and flux density, and introduce new parameters including reluctance and magnetizing force. We shall also examine a key graphic that helps define the magnetic properties of materials, namely the BH curve. In the process we shall define new terms including coercivity and retentivity, and introduce the concept of hysteresis. As we shall see, there are certain parallels between magnetic circuits and electrical circuits. For example, there is an “Ohm's law for magnetic circuits”, sometimes referred to as Hopkinson's law. There is also a version of Kirchhoff's voltage law dealing with the magnetic equivalents of voltage rises and drops. Perhaps the two most common applications of magnetic circuits for the new student involve the use of transformers and relays, both of which will be examined in this chapter. Other common applications include dynamic loudspeakers and microphones, magnetic resonance imaging (i.e., medical MRI), pick-up cartridges for vinyl phonographs and the various applications made available by magnetic Hall effect sensors such as advanced anti-lock braking systems, timing control systems for internal combustion engines, and other instances requiring position and motion sensing. These magnetic devices tend to be much more reliable and capable than simple devices such as mechanical switches. This does not mean that magnetic systems are in some way “taking over”, indeed, there are some applications where they have become obsolete. One example is the use of output transformers to couple a home audio power amplifier to a loudspeaker. At one time, these were an integral part of every design, however, today they generally are not used. With this in mind, it is perhaps best to think of magnetic circuits as another item in the engineer's or technician's toolkit to help solve practical problems. 324 | DCElectricalCircuitAnalysis_Page_324_Chunk1246 |
10.2 Electromagnetic Induction 10.2 Electromagnetic Induction Perhaps the most important observation regarding magnetic systems is Faraday's law of electromagnetic induction. Briefly, it states: If a conductor is cut by changing magnetic lines of force, a voltage will be induced in the conductor. (10.1) More specifically, e =−d Φ dt (10.2) Where e is the induced voltage, dΦ/dt is the rate of change of magnetic flux with respect to time. Therefore, the larger the flux and the more quickly it fluctuates, the greater the induced voltage. What is important here is a relative change with respect to the conductor and field. This can be accomplished in two basic ways: by a magnetic field that is itself fluctuating around a fixed conductor, or by a conductor moving through a magnetic field. The concept of electromagnetic induction is a key element of a variety of transducers. Simply put, A transducer is a device that transforms energy from one type to another. (10.3) Although many people don't think of them as such, electric motors and generators are perfect examples of transducers. An electric motor transforms electrical energy into mechanical energy and a generator does the precise opposite, transforming mechanical energy into electrical energy. More commonly, the word transducer is associated with devices such as loudspeakers and microphones. A loudspeaker transforms its electrical input into an acoustic pressure wave (sound) whereas the microphone performs the complementary function of transforming an acoustic pressure wave into an electrical signal. While there are different ways of constructing these devices, the most common designs are the dynamic loudspeaker and dynamic microphone, both of which rely on the principle of electromagnetic induction. The designs of the two devices are strikingly similar, mostly varying in size and parts optimization. Let's take a closer look at how they operate. 325 | DCElectricalCircuitAnalysis_Page_325_Chunk1247 |
Dynamic Loudspeakers and Microphones Dynamic Loudspeakers and Microphones Regardless of their output capabilities and frequency range, all dynamic loudspeakers share a common set of elements. Corresponding elements can be found in dynamic microphones. Indeed, the devices are so similar that in some applications, small loudspeakers might pull double duty and switch over to a microphone mode of operation. A cutaway view of a loudspeaker designed to reproduce low frequencies is shown in Figure 10.1. The heart of the system is a voice coil (H) which is a coil of tightly wound magnet wire. This sits inside of a magnetic field that is created by a powerful permanent magnet (E). The voice coil is connected to a diaphragm (F) that is is usually made of paper or plastic. This assembly is connected to the frame by the spider (D) at the top of the voice coil, and a suspension at the end of the diaphragm (B). The voice coil and diaphragm can move back and forth relative to the frame, like a piston. To understand the operation, recall that passing a current through a coil of wire creates a magnetic field around the coil, as illustrated in Figure 10.2. This magnetic field is the same as is created by a normal magnet. In this case the north pole is on the left side (i.e., lines of force exiting). Consequently, this coil will interact with a permanent magnet just like any other magnet would. In the case of the loudspeaker, the voice coil is fed a current from the amplifier that echoes the music or voice signal. This current creates a magnetic field around the voice coil which interacts with the field of the permanent magnet. As the direction of the current changes, the poles of the voice coil's field reverse. Thus, sometimes the voice coil is pushed outward and sometimes it is pulled inward. Further, the larger the current, the greater the field it creates, and the greater the push or pull. 326 Figure 10.1 Cutaway view of a dynamic loudspeaker. A. Frame B. Suspension C. Lead wire D. Spider E. Magnet F. Diaphragm G. Voice coil former H. Voice coil I. Dust cap Image courtesy of Audio Technology. Figure 10.2 Magnetic field around a coil. Image ©, courtesy of HyperPhysics | DCElectricalCircuitAnalysis_Page_326_Chunk1248 |
This results in a back-and-forth motion of the diaphragm that echoes the shape of the waveform being fed from the amplifier. As the diaphragm pushes against the air, it sets up a pressure wave, and we have sound. A dynamic microphone runs the sequence in reverse. First, the diaphragm will move back and forth in accordance to an applied sound wave. This causes the voice coil to move back and forth within the field of the permanent magnet. We now have a coil of wire being cut by magnetic lines of force, and by Faraday's law of induction, this means that a voltage will be induced in the coil. As long as the diaphragm can respond to the subtle changes in the acoustic pressure wave, then the resulting induced voltage should be of high fidelity. Clearly, to do this with great accuracy, the diaphragm/voice coil/suspension assembly will need to very light and nimble, meaning that the components of the microphone will tend to be much smaller than those of a loudspeaker. Electric Guitar Pickup Electric Guitar Pickup A set of pickups for an electric guitar is shown in Figure 10.3. Although it might be tempting to think that the pickups are just microphones, they are not. If you have any doubt of this statement, just try screaming into a guitar pickup and listen to what comes out24. The operation of the guitar pickup does, however, rely on Faraday's law. As discussed in the prior chapter covering inductance, an electric guitar pickup consists of thousands of turns of very fine wire around a permanent magnet. The pickup is placed immediately below the guitar strings. In this position, the metal guitar strings (generally various combinations of steel, nickel and/or cobalt), are within the field generated by the magnet. When a string is plucked, it vibrates in a complex pattern that depends on the note it is tuned to and the overtones that are present. This pattern distorts the magnetic field because the strings have a much higher permeability than the air around them. We now have a changing magnetic field, in the middle of which is large coil of wire. Faraday's law states that a voltage must be induced in this coil, and that it will follow the motions of the strings. This induced voltage is then fed to an amplifier, hopefully, set to 11. The astute observer might ask, “Why are there multiple pickups?” Perhaps surprisingly, it is not to get a larger, and thus louder, signal. It has to do with the timbre, or tone quality, of the sound. When a guitar string vibrates, its motion can be thought of as containing the myriad motions of the fundamental pitch and all of the overtones that go with it. Towards the bridge, where the strings are attached, the lowest pitched elements produce little motion, and thus their strength in the overall signal is reduced. The result is that a pickup placed closer to the bridge sounds “thin” or “cutting” while one placed further up sounds “full” or “thick”. 24 A little string resonance if you're lucky, and some irritated band-mates if you're not. 327 Figure 10.3 Pickups on an electric guitar. | DCElectricalCircuitAnalysis_Page_327_Chunk1249 |
Bicycle Computer Bicycle Computer Our third and final illustrative example is that of a bicycle computer. These handy little devices consist of a small sensor system on the front wheel which is connected to a display unit mounted on the handlebars. Typically, these units will display current speed, average speed, elapsed time, distance covered, and other attributes of interest. The sensing apparatus of a typical bicycle computer is shown in Figure 10.4. This apparatus consists of two parts: a permanent magnet mounted to one of the wheel spokes, and a weatherproof sensing unit mounted to the front fork. The principle of operation is fairly straightforward: each time the while rotates, the magnet attached to the spoke swings by the sensor. The moving magnetic field creates a short-lived voltage spike in the sensor, an example shown in Figure 10.5. The computer records these spikes over time. Knowing the circumference of the wheel, a simple multiplication yields the accumulated distance traversed. Given internal clocking circuitry, the time recorded between the pulses can be turned into a velocity. Given these data, other attributes such as average or maximum speed are easily obtained with further computation. Using the oscilloscope trace shown in Figure 10.5 as an example, we can see that the time between the pulses is about 3.5 divisions, with each division being 200 milliseconds in length. This gives one rotation every 0.7 seconds, or roughly 5140 revolutions per hour. The sensor was mounted on a bike with 700 by 25 mm tires which yields a circumference of about 2.1 meters. Thus, the velocity would be 5140 revolutions per hour at 2.1 meters per revolution, or about 10.8 km/h (≈ 6.73 MPH).25 25 In reality, the bike was mounted on a maintenance stand so its velocity was, in fact, zero. Running alongside a bike with an oscilloscope and probes is no easy feat, especially when the scope needs AC power. 328 Figure 10.4 Bicycle computer sensor and magnet. Figure 10.5 Sensor signal feeding bicycle computer. | DCElectricalCircuitAnalysis_Page_328_Chunk1250 |
10.3 Magnetic Circuits 10.3 Magnetic Circuits Magnetic circuits include applications such as transformers and relays. A very simple magnetic circuit is shown in Figure 10.6. First, it consists of a magnetic core. The core may be comprised of a single material such as sheet steel but can also use multiple sections and air gap(s). Around the core is at least one set of turns of wire, i.e., a coil formed around the core. Multiple sets of turns are used for transformers (in the simplest case, one for the primary and another for the secondary). As we have seen already, passing current through the windings generates a magnetic flux, Φ, in the core. As this flux is constrained within the cross sectional area of the core, A, we can derive a flux density, B. B = Φ A (10.4) Where B is the magnetic flux density in teslas, Φ is the magnetic flux in webers, A is the area in square meters. Recall from Chapter 9 that one tesla is defined as one weber per square meter. An alternate unit that is sometimes used is gauss (cgs system of units), named after Carl Friedrich Gauss, the German mathematician and scientist. 1 tesla = 10,000 gauss (10.5) Example 10.1 A magnetic flux of 6E−5 webers exists in a core whose cross section has dimensions of 1 centimeter by 2 centimeters. Determine the flux density in teslas. First, convert the dimensions to meters to find the area. There are 100 centimeters to the meter. A = width×height A = 0.01 m×0.02 m A = 2E-4 m 2 B = Φ A B = 6E-5Wb 2E-4 m 2 B = 0.3T 329 Figure 10.6 Simple magnetic circuit. | DCElectricalCircuitAnalysis_Page_329_Chunk1251 |
Ohm's Law for Magnetic Circuits (Hopkinson's or Rowland's Law) Ohm's Law for Magnetic Circuits (Hopkinson's or Rowland's Law) There is a common parallel drawn between magnetic circuits and electrical circuits, namely Hopkinson's law (Rowland's law). For electrical circuits, Ohm's law states: V = I R In like manner, for magnetic circuits, we have: F = Φ R (10.6) Where F is the magnetomotive force (or MMF) in amp-turns, Φ is the magnetic flux in webers, R is the reluctance of the material in amp-turns/weber. The magnetomotive force compares to a source voltage or electromotive force (EMF), the magnetic flux is likened to the flow of current, and the reluctance stands in for resistance (i.e., on the one hand we have a material that resists the flow of current, and on the other, a material in which there is “reluctance” to establish magnetic flux). Further, magnetomotive force is the product of the current flowing through a coil and the number of loops or turns in the coil: F = N I (10.7) Where F is the magnetomotive force in amp-turns, N is the number of loops or turns in the coil, I is the current in the coil in amps. The equation for reluctance has a nice parallel with the equation for resistance (Equation 2.11 from Chapter 2): R = ρl A R = l μ A (10.8) Where R is the reluctance in amp-turns/weber, l is the length of the material in meters, A is the cross sectional area of the material in square meters, μ is the permeability of the material in henries/meter. 330 | DCElectricalCircuitAnalysis_Page_330_Chunk1252 |
Given the characteristics of the coil and the path length of the magnetic circuit, the magnetic flux gives rise to a magnetizing force, H. H = N I l (10.9) Where H is the magnetizing force in amp-turns/meter, N is the number of turns or loops in the coil, I is the coil current in amps, l is the length of the magnetic path in meters. Equation 10.8 reveals that ferromagnetic materials (i.e., materials that have high permeability, such as steel) produce a low reluctance. The practical problem here is that μ, unlike the resistivity ρ of resistors, is not a constant for these sorts of materials. It can vary considerably, as seen in the general diagram presented in Figure 10.7. As a result, it is impractical to find reluctance in same manner that we find resistance. All is not lost, though. The flux density and corresponding magnetizing force for any given material are related by the following equation: B = μ H (10.10) Where B is the flux density in teslas, μ is the permeability of the material in henries/meter, H is the magnetizing force in amp-turns/meter. Once again, the tricky bit here is the permeability of the core material. For air, we can use the permeability of free space, μ0. μ0 = 4π × 10−7 H/m ≈ 1.257×10−6 H/m (10.11) For other materials, such as sheet steel or cast steel, we shall take another path; namely an empirically derived curve that plots flux density B against magnetizing force H. Such graphs generically are referred to as “BH curves”. An example is shown in Figure 10.8. Clearly, this curve is not a nice straight line, or even an obvious, predictable function. The immediately apparent attributes are the initial steep rise which is followed by a flattening of the curve. This flattening corresponds to saturation of the magnetic material. In contrast, a plot for air would reveal a straight line with a very shallow slope. As we shall see, being able to achieve a high flux density for a given magnetizing force will result in an effective and efficient magnetic circuit. So while air has the positive attribute of not saturating, the resulting flux density is low, usually leading to lower performance. 331 B H Figure 10.7 Typical permeability curve for a high permeability material. H μ Figure 10.8 Generic BH curve. | DCElectricalCircuitAnalysis_Page_331_Chunk1253 |
The The BH BH Curve Curve The process of generating a BH curve is as follows. First, we create a core of the material to be investigated. A coil of wire is then wrapped around this core. An example is shown in Figure 10.9. Here we have a basic toroid with a coil of N turns. We begin with the system at rest and not energized. A small current, I, is applied to the coil. This produces a magnetizing force via Equation 10.9. There will be a corresponding flux density, as per Equation 10.10. The current is then increased. This produces an increase in magnetizing force and a corresponding change in flux density. The current is increased further, until the curve flattens, indicating that saturation has been reached. This trajectory is shown as the dashed line in Figure 10.10, starting at point a with point b indicating saturation. The current is then reduced. This causes a reduction in magnetizing force, but while the flux density decreases, it does not trace back perfectly along the original trajectory. Instead, the curve is displaced above the original. Eventually, the current will be reduced to zero. This corresponds to point c on the graph of Figure 10.10. At this point, even though there is no current in the coil, there is still flux in the core. The resulting flux density is referred to as retentivity and is a measure of residual magnetism. This phenomenon is what makes it possible to magnetize materials. If we now reverse the direction of the coil current and begin to increase its magnitude, the flux density will continue to drop. At point d, it will have reached 332 Figure 10.9 Toroidal core with coil. I N turns B H -B -H b a c d e f g Retentivity (residual magnetism) Coercivity (coercive force) Saturation Figure 10.10 Generation of BH curve. | DCElectricalCircuitAnalysis_Page_332_Chunk1254 |
zero. As we have effectively coerced the flux back to zero, we call the magnetizing force required to do this the coercivity or coercive force. As the current magnitude increases, the flux density also increases but with opposite sign. Eventually, saturation is reached again at point e. Once again, if the current's magnitude is decreased, the magnitude of the flux density will also decrease but will not perfectly trace back along the original trajectory. This time it will follow a lower path. When the current is reduced to zero at point f, a mirror retentivity is evident. Further positive increases in current show a mirror coercivity at point g. Finally, as current is increased to maximum, we again reach saturation at point b. If the current is cycled in this manner again, the process is repeated, and the outer path indicated by the arrows is taken again. Thus, a specific value of magnetizing force can give rise to different values of flux density: it depends on the recent history of the material. This effect is known as hysteresis and is found in other areas as well. Effectively, published BH curves follow the middle of the hysteresis loop. An example of BH curves for three different core materials is shown in Figure 10.1126. Curve A is for sheet steel (which is common in transformers), curve B is for cast steel, and curve C is for cast iron. We shall make use of these in upcoming examples. Curves for other materials are also available. 26 Curves based on https://en.wikipedia.org/wiki/Saturation_(magnetic); Steinmetz (1917), Theory and Calculation of Electric Circuits; and Boylestad (2010), Introductory Circuit Analysis. 333 Figure 10.11 BH curves for: A. Sheet steel B. Cast steel C. Cast iron B (T) H (At/m) 1.0 1.5 0.5 100 200 300 400 500 A B C 600 700 | DCElectricalCircuitAnalysis_Page_333_Chunk1255 |
Example 10.2 Assume the toroid of Figure 10.9 is made of cast steel, has a 500 turn coil, a cross section of 2 cm by 2 cm, and an average path length of 50 cm. Determine the flux in webers if a current of 0.3 amps feeds the coil. We shall use Equation 10.9 to find the magnetizing force and from the BH curve, find the flux density. H = N I l H = 500turns×0.3 A 0.5 m H = 300At/m (amp-turns per meter) Cast steel corresponds to curve B (green) in Figure 10.10. A decent estimate for the flux density is 0.52 teslas. This is the corresponding flux density. In order to find the flux, we need to find the area of the core. A = width×height A = 0.02 m×0.02 m A = 4E-4m2 Φ = B×A Φ = 0.52 T×4E-4 m 2 Φ = 2.08E-4 Wb KVL for Magnetic Circuits KVL for Magnetic Circuits A cursory examination of Equations 10.7 and 10.9 shows that: F = H l = N I (10.12) Continuing the Ohm's law analogy, the amp-turns product of the coil, NI, is analogous to a voltage rise. Further, the Hl product is analogous to a voltage drop. If we then extend the analogy to include the concept of Kirchhoff's voltage law, it should come as no surprise that the sum of NI “rises” must equal the sum of the Hl “drops”. In the circuit of Figure 10.9, there is a single “rise” and a single “drop”. Thus, the magnetic circuit is analogous to the minimal electric circuit shown in Figure 10.12. F is the magnetomotive force, NI, while R is the reluctance of the toroidal core. This reluctance will experience a “drop” of Hl. The core could consist of two or more different materials, creating the equivalent of a series circuit. In this case, a table such as the one found in Figure 10.13 can be 334 Figure 10.12 Electrical circuit analogy for the magnetic system of Figure 10.9. | DCElectricalCircuitAnalysis_Page_334_Chunk1256 |
used to aid in computation. In this table, each section of the core gets its own row. The table is divided into two sides (note the thick separating line in the center). In general, we will be working through problems where we know the data on the left side and need to find something on the right side, or vice versa. The bridge between the two sides (i.e., traversing the thick line) is a BH curve for the material of that particular section. The exception to this rule is if the section is an air gap. In that case we can use Definition 10.11 which shows that for air, H ≈ 7.958E5 B or alternately, B ≈ 1.257E−6 H. Section Flux Φ (Wb) Area A (m2) Flux Density B (T) Magnetizing Force H (At/m) Length l (m) “Drop” Hl (At) 1 2 Time for a few illustrative examples. We shall consider a simple system like that of Figure 10.9, a two section core, a core with an air gap, and a core with two coils. Example 10.3 Assume the core of Figure 10.14 is made of sheet steel, has a 200 turn coil, a cross section of 1 cm by 1 cm, and an average path length of 12 cm. Determine the coil current required to achieve a flux of 1E−4 webers. The analogous circuit consists of a single source and reluctance, like that of Figure 10.12. Thus, N I = H sheetl sheet We shall begin by filling in the portions of the table that can be addressed directly, such as the path length, area and flux. Don't forget to convert the centimeter values into meters. Section Flux Φ (Wb) Area A (m2) Flux Density B (T) Magnetizing Force H (At/m) Length l (m) “Drop” Hl (At) Sheet Steel 1E−4 1E−4 0.12 Now we determine the flux density to complete the left side of the table. B = Φ A B = 1E-4 Wb 1E-4 m 2 B = 1T 335 Figure 10.13 Table format used for magnetic circuit analysis. Figure 10.14 Magnetic system for Example 10.3. N turns I | DCElectricalCircuitAnalysis_Page_335_Chunk1257 |
Section Flux Φ (Wb) Area A (m2) Flux Density B (T) Magnetizing Force H (At/m) Length l (m) “Drop” Hl (At) Sheet Steel 1E−4 1E−4 1 0.12 We use the BH curve of Figure 10.11 to find H from this flux density. From the blue curve (A) the value is about 190 amp-turns per meter. Section Flux Φ (Wb) Area A (m2) Flux Density B (T) Magnetizing Force H (At/m) Length l (m) “Drop” Hl (At) Sheet Steel 1E−4 1E−4 1 190 0.12 At this point we find the Hl “drop” with a simple multiplication. Section Flux Φ (Wb) Area A (m2) Flux Density B (T) Magnetizing Force H (At/m) Length l (m) “Drop” Hl (At) Sheet Steel 1E−4 1E−4 1 190 0.12 22.8 This “drop” is 22.8 amp-turns and the coil has 200 turns. Therefore the required current is: I = H l N I = 22.8At 200 t I = 114mA Next, let's consider a core with two sections. Example 10.4 Given the magnetic system shown in Figure 10.15, assume section A is made of sheet steel and section B is made of cast steel. Each part has a cross section of 2 cm by 2 cm. The path length of A is 12 cm and the path length of B is 4 cm. If the coil has 50 turns, determine the coil current required to achieve a flux of 2E−4 webers. The analogous circuit consists of a single source and two reluctances. This is shown in Figure 10.16. 336 Figure 10.15 Magnetic system for Example 10.4. N turns I A B | DCElectricalCircuitAnalysis_Page_336_Chunk1258 |
Therefore: N I = H sheetl sheet+H castlcast For our table there will be two rows, one for sheet steel and the second for cast steel. The first row will require using curve A (sheet steel) from Figure 10.11 while the second row will require using curve B (cast steel). The flux, like the current in a series loop, will be the same in both sections. From here we can fill out a series of other values in the table to arrive at: Section Flux Φ (Wb) Area A (m2) Flux Density B (T) Magnetizing Force H (At/m) Length l (m) “Drop” Hl (At) Sheet Steel 2E−4 4E−4 0.5 0.12 Cast Steel 2E−4 4E−4 0.5 0.04 The BH curves are used to transition to the right side, and we arrive at: Section Flux Φ (Wb) Area A (m2) Flux Density B (T) Magnetizing Force H (At/m) Length l (m) “Drop” Hl (At) Sheet Steel 2E−4 4E−4 0.5 70 0.12 Cast Steel 2E−4 4E−4 0.5 290 0.04 And now we fill in the Hl “drops”: Section Flux Φ (Wb) Area A (m2) Flux Density B (T) Magnetizing Force H (At/m) Length l (m) “Drop” Hl (At) Sheet Steel 2E−4 4E−4 0.5 70 0.12 8.4 Cast Steel 2E−4 4E−4 0.5 290 0.04 11.6 337 Figure 10.16 Analogous electrical circuit for the system of Figure 10.15. | DCElectricalCircuitAnalysis_Page_337_Chunk1259 |
Using our KVL analogy, the total “drop” is 8.4 At + 11.6 At, or 20 amp- turns. The coil was specified as having 50 turns. Therefore: I = H l N I = 20 At 50 t I = 400mA Notice that even though the cast steel section is shorter than the sheet steel section, it produces a larger “drop”; just like a larger resistor in an electrical circuit. This requires a larger current in the coil than if the entire core had been made of sheet steel. Finally, it should be apparent that the current demand can be reduced if the coil has more turns. There is a practical limit here because all of the turns have to fit within the opening of the core. Once that space is full, the only way to increase the number of turns is to use finer gauge wire but doing so increases wire resistance and power loss, and also lowers maximum current capacity. Dealing With Air Gaps Dealing With Air Gaps Some systems include an air gap in the core. If there is only a single material for the remainder, this will create a system like that depicted in Figure 10.17. The main observation here is that the permeability of air is far below that of ferromagnetic materials and thus the gap will exhibit a relatively large reluctance compared to its path length. The obvious question then is, why would we use a gap? One possibility is an electro-magnetic relay, the internals of which are illustrated in Figure 10.18. 338 Figure 10.17 Analogous electrical circuit for a system with an air gap. Figure 10.18 Internals of an electrical relay. | DCElectricalCircuitAnalysis_Page_338_Chunk1260 |
To create a relay, we place one portion of the core on a hinge which can be held open with a small spring. This creates the air gap (to the immediate left of the core in the Figure). If we apply a sufficiently large current, the resulting magnetic flux will be enough to overcome the spring tension and close the second piece onto the first (i.e., magnetic attraction). Once this happens, the gap disappears, reducing the reluctance around the loop. The system will stay closed until the coil current is turned off and the restoring spring pulls apart the two sections again. We add insulated metal contacts to the moving section and what we end up with is a heavy duty switch that is “thrown” by a controlling current rather than a mechanical lever moved by a human. It is worth repeating that for air gaps, instead of a BH curve we can use Definition 10.11 which shows that for air, H ≈ 7.958E5 B or alternately, B ≈ 1.257E−6 H. Example 10.5 Given the magnetic system shown in Figure 10.19, assume the core is made of sheet steel. The cross section throughout is 3E−4 m2. The path length of the main core is 8 cm and the path length of the gap is 1 mm. How many turns will the coil need in order for a coil current of 400 mA to achieve a flux of 1.2E−4 webers? The analogous circuit consists of a single source and two reluctances, one for the sheet steel core and a second for the air gap. This is shown in Figure 10.17. Further, the analogous KVL relation is: N I = H sheetl sheet+H gapl gap For this table there will be two rows, one for the sheet steel core and one for the air gap. As we have seen, the flux will be the same in both sections. From there we can fill out some of the other values in the table resulting in: Section Flux Φ (Wb) Area A (m2) Flux Density B (T) Magnetizing Force H (At/m) Length l (m) “Drop” Hl (At) Sheet Steel 1.2E−4 3E−4 0.4 8E−2 Gap 1.2E−4 3E−4 0.4 1E−3 339 Figure 10.19 Magnetic system for Example 10.5. N turns I gap | DCElectricalCircuitAnalysis_Page_339_Chunk1261 |
The BH curve for sheet steel and Definition 10.11 for the air gap are used to transition to the right side, yielding: Section Flux Φ (Wb) Area A (m2) Flux Density B (T) Magnetizing Force H (At/m) Length l (m) “Drop” Hl (At) Sheet Steel 1.2E−4 3E−4 0.4 63 8E−2 Gap 1.2E−4 3E−4 0.4 3.183E5 1E−3 The Hl “drops” are filled in: Section Flux Φ (Wb) Area A (m2) Flux Density B (T) Magnetizing Force H (At/m) Length l (m) “Drop” Hl (At) Sheet Steel 1.2E−4 3E−4 0.4 62 8E−2 5.0 Gap 1.2E−4 3E−4 0.4 3.183E5 1E−3 318.3 Using our KVL analogy, the total “drop” is 5.0 At + 318.3 At, or 323.3 amp- turns. The coil current was specified as being 400 mA. Therefore: N = H l I N = 323.3At 400mA N ≈808turns And finally, an example using two coils. Example 10.6 Given the magnetic system shown in Figure 10.20, assume the core is made of sheet steel. The cross section throughout is 3E−4 m2 and the length of the core is 20 cm. Coil one consists of 2000 turns and coil two consists of 500 turns. If a current of 120 mA into coil one achieves a flux of 2.4E−4 webers, determine the current of coil 2. The analogous circuit consists of two voltage sources and a single reluctance for the sheet steel core. This is illustrated in Figure 10.21. 340 Figure 10.20 Magnetic system for Example 10.6. N1 I1 N2 I2 | DCElectricalCircuitAnalysis_Page_340_Chunk1262 |
The analogous KVL relation is: N 1 I1 −N 2 I 2 = H sheetl sheet As we are seeking I2, we can rearrange this equation into a more useful form: N 2 I 2 = N1 I 1 −H sheetl sheet Only one row will be required for this table. We fill out the obvious values in the table resulting in: Section Flux Φ (Wb) Area A (m2) Flux Density B (T) Magnetizing Force H (At/m) Length l (m) “Drop” Hl (At) Sheet Steel 2.4E−4 3E−4 0.8 0.2 The BH curve for sheet steel is used to transition to the right side: Section Flux Φ (Wb) Area A (m2) Flux Density B (T) Magnetizing Force H (At/m) Length l (m) “Drop” Hl (At) Sheet Steel 2.4E−4 3E−4 0.8 100 0.2 The Hl “drop” is computed: Section Flux Φ (Wb) Area A (m2) Flux Density B (T) Magnetizing Force H (At/m) Length l (m) “Drop” Hl (At) Sheet Steel 2.4E−4 3E−4 0.8 100 0.2 20 Using our KVL analogy, we find: N 2 I 2 = N1 I 1 −H sheetl sheet N 2 I 2 = 2000turns×120mA −20 At N 2 I 2 = 240At −20 At N 2 I 2 = 220At Coil two was specified as having having 500 turns, therefore its current is: I 2 = 220At 500turns I 2 = 440mA 341 Figure 10.21 Analogous electrical circuit for the system of Figure 10.20. | DCElectricalCircuitAnalysis_Page_341_Chunk1263 |
An item of key importance is that the preceding example must make use of an AC current in order to function as described. A DC current will not produce the predicted results. The reason for this goes back to Definition 10.1 and Equation 10.2: Unless the flux is changing relative to the conducting coil, no voltage will be induced in a coil. So, while a DC input current will create a flux in the core, that flux will be static and unchanging. Consequently, the second coil will not produce an output. In contrast, an AC current smoothly varies in amplitude and polarity. This creates a smoothly varying magnetic flux which in turn allows a voltage to be induced in the second coil. Assuming for a moment that an AC current was used in the preceding example, notice that the current is nearly four times higher in coil two than in coil one. If we had been able to reduce the core “drop” to zero, perhaps by reducing the core's reluctance to a negligible value, then the current would have been increased precisely by a factor four (to 480 mA versus 440 mA). This is the same as the ratio of the number of turns of coil one to the number of turns of coil two, and is known as the turns ratio. It is a key parameter describing transformers, which just happens to be the subject of the next section. 10.4 Transformers 10.4 Transformers Transformers have been in wide use in the electrical industry for over a century. Beginning with Michael Faraday and working through a sequence of scientists and inventors, the fundamentals of modern transformer design had been established by the late 1800s. Transformer have three basic uses: • Voltage scaling. • Isolation. • Impedance matching. Transformers range from very small units used in communications systems up to large power distribution units many times the size of an adult human. A couple of transformers typical of the type used in consumer electronics devices is shown in Figure 10.22. These devices are designed to handle standard residential AC input voltages (120 volts in North America, 220 to 240 volts in much of the remainder of the planet) and feeding circuits that demand perhaps 10 to 30 watts of power. The unit on the left is designed to be mounted directly onto a printed circuit board while the unit on the right is designed for chassis mount. 342 | DCElectricalCircuitAnalysis_Page_342_Chunk1264 |
In contrast, Figure 10.23 shows pole mounted transformers used to reduce typical North American residential distribution lines (in the vicinity of 15 kV) down to home voltage (120 volts). These transformers are capable of delivering 1000 or more times as much power as the transformers of Figure 10.22. The transformers in these figures by no means present the extremes. Transformers are available that are considerably smaller or larger than the ones pictured here. Transformers exploit magnetic circuit theory for their operation, as illustrated previously in Example 10.6. In its most basic form, the device has two coils or windings; one for the input side (called the primary winding) and one for the output side (the secondary winding). It is possible to have multiple input and/or output windings. Such is the case for the transformer on the left side of Figure 10.22; it has two input primary coils and two secondary coils. This allows for a variety of interconnection options, as we shall see. A typical schematic symbol is shown in Figure 10.24. This symbol is appropriate for the magnetic system of Figure 10.20 which featured two coils. The vertical lines in the symbol represent the magnetic core. Voltage Scaling and Isolation Voltage Scaling and Isolation A particularly important transformer parameter is the ratio of the number of turns on the primary coil (NP) to the number of turns on the secondary coil (NS). This is denoted N, and referred to as the turns ratio. N = N P N S (10.13) 343 Figure 10.22 Transformers suitable for consumer electronics devices. The unit on the left features four separate coils for enhanced flexibility. Figure 10.23 A cluster of three residential power transformers. | DCElectricalCircuitAnalysis_Page_343_Chunk1265 |
As the two windings are on the same core and ideally see the same flux, the voltage ratio of the primary to secondary will be equal to N for an ideal lossless transformer. Thus, if N=10 and the primary voltage (VP) is 60 volts, the secondary voltage (VS) will be 6 volts. Thus: V S = V P N (10.14) The primary and secondary appear as “NI” potentials in the magnetic circuit, therefore, the current will change inversely with N. For example, if N=10 and the primary current is 1 amp, the secondary current will be 10 amps. Thus: I S = N I P (10.15) It is important to note that the product of primary current and voltage must equal the product of secondary current and voltage for an ideal transformer. Such a device does not dissipate power but rather transforms it (hence the name). V S I S = V P I P (10.16) Real world devices will exhibit some loss due to finite winding wire resistance, non- ideal behavior of the magnetic circuit and the like. Instead of a power dissipation rating, power transformers are given a VA (volt-amps) rating, typically thought of as secondary voltage times maximum secondary current. Transformers like the ones pictured in Figure 10.22 have ratings in the range of 10 to 30 VA. Generally, the higher the VA rating, the larger and heavier the transformer. On the other hand, applications using signal transformers in communications systems tend to focus on the turns ratio and not on the VA rating as the associated voltages and currents tend to be small. Their goal generally is to increase the amplitude of a signal voltage. Also, because there is no direct electrical connection between the primary and secondary, the transformer may be used solely for isolation. This is a safety feature. If N=1, then there is no change in the voltage being presented to the device, ideally. Example 10.7 A microphone feeds a small transformer which is used as the first stage of a preamplifier. The transformer has a primary winding of 50 turns and a secondary winding of 250 turns. Assuming an ideal transformer, determine the output voltage of the transformer if the signal from the microphone is 35 microvolts. As this is an ideal transformer, the turns ratio indicates how much the input voltage is scaled. First we use Equation 10.13 to determine the turns ratio. 344 Figure 10.24 Schematic symbol for a simple transformer. | DCElectricalCircuitAnalysis_Page_344_Chunk1266 |
N = N P N S N = 50 250 N = 0.2 Now we use Equation 10.14 to determine the secondary voltage. V S = V P N V S = 35μ V 0.2 V S = 175μV Power applications will make use of N to scale the voltage in order to create a more appropriate or efficient level. If the voltage is brought down (N>1), it is referred to as a step-down transformer. If N<1, it is referred to as a step-up transformer. For example, a consumer electronics product may require a modest DC voltage for its operation, say 15 or 20 volts. To achieve this, the standard North American residential “wall” voltage of 120 volts may be reduced using a step-down transformer with N=5. This would result in a secondary voltage of 24 volts that could then be rectified, filtered, and regulated to the desired DC level. Conversely, a step-up transformer could be used to create a much higher secondary potential (for example, in long distance power transmission to reduce I2R losses). For some applications, split secondaries or multiple secondaries may be used. A center tap is relatively common and is used to divide the secondary into two equal voltage sections (e.g., a 24 volt center-tapped secondary would behave as two 12 volt secondaries in series). The schematic symbol for a transformer with a center- tapped secondary is shown in Figure 10.25. Multiple secondaries are also common and are more flexible than a center-tapped secondary. Multiple secondaries can be combined in series to increase the secondary voltage or combined in parallel to increase secondary current capacity (i.e, load current). Example 10.8 The power transformer shown in Figure 10.26 has turns ratio of 10. If the primary is connected to a 120 volt source and the secondary is connected in series with a 32 Ω load resistor, determine the current through the load resistor. Assume this is an ideal transformer. The turns ratio indicates secondary voltage via Equation 10.4. 345 Figure 10.25 Schematic symbol for a transformer with a center- tapped secondary. Figure 10.26 Circuit for Example 10.8. | DCElectricalCircuitAnalysis_Page_345_Chunk1267 |
V S = V P N V S = 120 V 10 V S = 12V Ohm's law can be used to find the secondary current. I S = V S RL I S = 12V 32Ω I S = 375mA As mentioned earlier, transformers work with AC voltages and currents. Unlike the consistent polarity of DC sources, AC (literally, Alternating Current) flips back and forth in voltage polarity and current direction. In order to determine the phase relationship between the primary and secondary, transformers use “dot notation”. The dot indicates the positive polarity of voltage. This is illustrated in the circuit of Figure 10.27. On the primary side, the current flows into the dot and establishes the positive reference. In this case, the primary is seen as the load for the source, E. On the secondary side, current flows out of the dot and also notates the positive reference because the secondary is seen as the source for the load, R. Impedance Matching The third use of the transformer is impedance matching. Impedance is an AC phenomenon that consists of both resistance and reactance (an ohmic value associated with capacitors and inductors). The symbol for impedance is Z. Although impedance is measured in ohms, it differs from ordinary resistance in that it is a vector quantity and has a phase angle. For our present purposes, we can ignore this distinction.27 27 The concept of impedance is covered in great detail in the companion text, AC Electrical Circuit Analysis, another free OER title. 346 Figure 10.27 Transformer dot convention. | DCElectricalCircuitAnalysis_Page_346_Chunk1268 |
Because both the voltage and current are being scaled, the source's “view” of the load changes. The impedance seen by the source can be defined via Ohm's law as the voltage it produces divided by its output current. Z P = V P I P (10.17) From Equation 10.14, we know that VP = VS ∙ N. Substituting this into Equation 10.17 yields: Z P = V S N I P (10.18) From Equation 10.15, we know that IP = IS / N. Substituting this into Equation 10.18 yields: Z P = V S N 2 I S (10.18) By definition, Z S = V S I S (10.19) Combining Equations 10.18 and 10.19 leaves us with: Z P = N 2 Z S (10.20) Thus, the impedance seen by the source is the secondary impedance times the square of the turns ratio. This is called the reflected impedance. Example 10.9 Assume that the transformer shown in Figure 10.27 has turns ratio of 5. If the primary is connected to an 20 volt source and the secondary is connected to an 8 Ω load resistor, determine the primary current and the reflected impedance. Assume this is an ideal transformer. First we'll use Equation 10.20 to find the reflected impedance seen in the primary. 347 | DCElectricalCircuitAnalysis_Page_347_Chunk1269 |
Z P = N 2 Z S Z P = 5 2×8Ω Z P = 200Ω Ohm's law can be used to find the primary current. I P = V P RP I P = 20V 200Ω I P = 100mA Here is where impedance matching comes in. In the preceding example, if the source, E, had a large internal resistance compared to the load resistance, R, and there was no transformer, there would be considerable loss in the system. For example, if the source impedance was 10 Ω and it was directly connected to the 8 Ω load without the transformer, the resulting voltage divider would cause a great deal of internal power dissipation in the source leading to inefficiency. On the other hand, in the given circuit with transformer, the source sees a reflected impedance of 200 Ω. The resulting voltage divider between 200 Ω and the 10 Ω internal resistance of the source will cause very little loss, yielding a much more efficient transfer. 10.5 Summary 10.5 Summary This chapter has extended the material presented in Chapter 9 on inductance into the realm of magnetic circuits. We began with the concept of electromagnetic induction, which in simple terms states that if a conductor experiences changing magnetic lines of force, then a voltage will be induced in that conductor. This concept is exploited by an array of different devices including electric motors, generators, relays, loudspeakers, microphones, and other transducers and sensors. For example, dynamic loudspeakers and microphones both make use of a coil of wire suspended in a fixed magnetic field. In the case of the loudspeaker, the signal fed into the coil creates a changing magnetic field that interacts with the permanent magnet's field and causes an attached diaphragm to vibrate against the air, which in turn creates sound. A dynamic microphone works in an opposite manner. Sound waves cause a diaphragm to vibrate back and forth. This diaphragm is attached to a coil, and the motion of the coil within the permanent magnet's field causes a voltage to be induced in the coil. This voltage can then be amplified by electronic circuitry. Magnetic circuits are constructed by creating one or more coils of wire wrapped around a core that is made of a ferromagnetic material such as steel. A key part of analyzing magnetic circuits is Hopkinson's law (also called Rowland's law), which is the magnetic circuit version of Ohm's law. In this analogy, magnetic flux is likened 348 | DCElectricalCircuitAnalysis_Page_348_Chunk1270 |
to current flow, magnetic reluctance stands in for resistance, and magnetomotive force is analogous to electromotive force (i.e., voltage). Further, the magnetomotive force is the product of the number of turns in a coil and the current through said coil, or NI. When current is passed through the coil, a magnetizing force, H, and an associated magnetic flux are created. Given this, the core material will have a corresponding flux density, B. The linkage between B and H is found through a BH curve for that particular material. BH curves are non-linear for ferromagnetic materials. They also exhibit hysteresis, meaning that the recent operational history of the material plays a role in its current state. Every element of the core has a certain reluctance and thus exhibits an analogous “voltage drop”. These “drops” must add up to the NI “rises” of the coil(s), in a KVL analogy. Transformers use one or more primary coils to transform an input voltage into one more output voltages appearing on secondary coils. The ratio between the number of turns of the primary coil to the number of turns of the secondary coil is called the turns ratio, N. The secondary voltage is equal to the primary voltage divided by N while the secondary current is equal to the primary current times N. If N is greater than one, we have a step-down transformer (reducing the voltage), and if N is less than one, we have a step-up transformer (increasing the voltage). Ideally, transformers do not dissipate power. Thus, they do not have a power rating, but instead have a VA (volt-amps) rating to indicate their capacity. Transformers can also be used for impedance matching to increase the efficiency of a system. The impedance seen by the source on the primary is equal to the secondary impedance times the square of the turns ratio. This is called the reflected impedance. Review Questions Review Questions 1. Describe the concept of electromagnetic induction. 2. Describe Ohm's law for magnetic circuits (Hopkinson's or Rowland's law). 3. Describe hysteresis. 4. Define residual magnetism and coercivity. 5. Outline the operation of a relay. 6. What are some of the uses of transformers? 7. What is turns ratio? How does it affect voltage and current in the primary and secondary? 8. What is the difference between a step-up and a step-down transformer? 9. What is reflected impedance? 349 | DCElectricalCircuitAnalysis_Page_349_Chunk1271 |
10.6 Exercises 10.6 Exercises Analysis Analysis 1. In the magnetic circuit shown in Figure 10.1, assume the cross section is 1 cm by 1 cm with a path length of 8 cm. The entire core is made of sheet steel and there are 100 turns on the winding. Determine the current to establish a flux of 8E−5 webers. 2. Repeat problem 1 using cast steel for the core. 3. Given the core shown in Figure 10.1, assume the cross section is 2 cm by 2 cm with a path length of 10 cm. The entire core is made of cast steel and there are 200 turns on the winding. Determine the current to establish a flux of 4E−4 webers. 4. Repeat problem 3 using sheet steel for the core. 350 N turns I Figure 10.1 | DCElectricalCircuitAnalysis_Page_350_Chunk1272 |
5. In the magnetic circuit shown in Figure 10.1, assume the cross section is 1 cm by 1 cm with a path length of 8 cm. The entire core is made of sheet steel. Determine the number of turns required to establish a flux of 8E−5 webers given a current of 50 mA. 6. Given the core shown in Figure 10.1, assume the cross section is 2 cm by 2 cm with a path length of 10 cm. The entire core is made of sheet steel Determine the number of turns required to establish a flux of 4E−4 webers given a current of 200 mA. 7. In the magnetic circuit shown in Figure 10.2, assume the cross section is 1 cm by 1 cm. Section A is sheet steel with a path length of 6 cm. Section B is cast steel with a length of 2 cm. There are 500 turns on the winding. Determine the current to establish a flux of 5E−5 webers. 8. In the magnetic circuit shown in Figure 10.2, assume the cross section is 1 cm by 1 cm. Section A is sheet steel with a path length of 6 cm. Section B is cast steel with a length of 2 cm. Determine the number of turns on the coil to establish a flux of 6E−5 webers with a current of 50 mA. 9. A transformer is shown in Figure 10.3, assume the cross section is 5 cm by 5 cm. The core is sheet steel with a path length of 20 cm. N1 is 500 turns and N2 is 200 turns. Determine the secondary current (I2) if a primary current (I1) of 1 amp establishes a flux of 1.5E−3 webers. 10. Given the same conditions of problem 9, alter the secondary turns (N2) so that the secondary current (I2) is 3 amps given the original primary current of 1 amp. 351 Figure 10.2 N turns I A B Figure 10.3 N1 I1 N2 I2 | DCElectricalCircuitAnalysis_Page_351_Chunk1273 |
11. In general, how would the performance noted in problem 9 change if cast steel was substituted for sheet steel? 12. Given the results of problems 9 through 11, what does the ratio of N1 to N2 represent in terms of idealized performance, and what steps should be taken to make the transformer operate as close to ideal as possible? 13. Given the magnetic circuit shown in Figure 10.4, assume the cross section is 1 cm by 2 cm with a path length of 6 cm. The entire core is made of sheet steel with the exception of a 1 mm air gap. Determine the current required to establish a flux of 8E−5 webers if N = 1000 turns. 14. Using the data given in problem 13, determine the number of turns required to establish the same flux when the input current is 200 mA. 15. An ideal transformer has a 6:1 voltage step-down ratio. If the primary is driven by 24 VAC and the load is 100 Ω, determine the load voltage and current, and the primary side current. 16. A 120 VAC transformer is specified as having a 36 volt center-tapped secondary. If each side of the secondary is connected to its own 50 Ω load, determine the load currents and the primary side current. 17. An ideal transformer has a 12:1 voltage step-down ratio. If the secondary is connected to a 10 Ω load, what impedance is seen from the primary side? 18. An ideal transformer has a 1:5 voltage step-up ratio. If the secondary is connected to a 2 Ω load, what impedance is seen from the primary side? Design Design 19. A step-down transformer with N=8 has a 15 volt RMS secondary which is connected to a load with an effective value of 5 Ω. Determine the minimum acceptable VA rating of the transformer. 352 N turns I gap Figure 10.4 | DCElectricalCircuitAnalysis_Page_352_Chunk1274 |
20. A step-up transformer with N=.5 is driven from a 120 VAC source. The secondary is connected to a load with an effective value of 150 Ω. Determine the minimum acceptable VA rating of the transformer. Challenge Challenge 21. A transformer specified as having a 120 VAC primary with an 18 volt secondary is accidentally connected backwards, with its secondary connected to the source and its primary connected to a 16 Ω load. Determine the load current in both the normal and reversed connections. Also determine the required transformer VA rating for both connections. 22. Consider the distributed public address system for an airport as shown in Figure 10.5. It consists of an audio power amplifier with a nominal 70 volt RMS output that is connected to four remote loudspeakers, each separate from the others and some 150 meters away from the amplifier. Each loudspeaker assembly includes a 10:1 voltage step-down transformer that feeds the loudspeaker impedance of 8 Ω (resistive) off its secondary. These four lines are fed in parallel by the amplifier. Determine the power delivered to each loudspeaker and the total current delivered by the power amplifier. Assume the transformers are ideal and ignore any cable resistance. 23. Continuing with the preceding problem, assume that the wiring connecting each transformer back to the amplifier is AWG 22. Determine the power lost in each of the 150 meter long sections of dual cable. Further, suppose the system is reconfigured without the transformers and the output of the amplifier is lowered to 7 volts RMS to compensate. Determine the power lost in each of the cable feeds under the new configuration. 353 Figure 10.5 | DCElectricalCircuitAnalysis_Page_353_Chunk1275 |
Simulation Simulation 24. Use a transient analysis to verify the results of problem 15. 25. Use a transient analysis to verify the results of problem 16. 354 | DCElectricalCircuitAnalysis_Page_354_Chunk1276 |
Appendix A Appendix A Standard Component Sizes Standard Component Sizes Passive components (resistors, capacitors and inductors) are available in standard sizes. The tables below are for resistors. The same digits are used in subsequent decades up to at least 1 Meg ohm (higher decades are not shown). Capacitors and inductors are generally not available in as many standard values as are resistors. Capacitors below 10 nF (.01 μF) are usually available at the 5% standard digits while larger capacitances tend to be available at the 20% standards. 5% and 10% standard values, EIA E24 and EIA E12 5% and 10% standard values, EIA E24 and EIA E12 10% values (EIA E12) are bold 20% values (seldom used) are every fourth value starting from 10 (i.e., every other 10% value) 10 11 12 13 15 16 18 20 22 24 27 30 33 36 39 43 47 51 56 62 68 75 82 91 1% and 2% standard values, EIA E96 and EIA E48 1% and 2% standard values, EIA E96 and EIA E48 2% values (EIA E48) are bold 10.0 10.2 10.5 10.7 11.0 11.3 11.5 11.8 12.1 12.4 12.7 13.0 13.3 13.7 14.0 14.3 14.7 15.0 15.4 15.8 16.2 16.5 16.9 17.4 17.8 18.2 18.7 19.1 19.6 20.0 20.5 21.0 21.5 22.1 22.6 23.2 23.7 24.3 24.9 25.5 26.1 26.7 27.4 28.0 28.7 29.4 30.1 30.9 31.6 32.4 33.2 34.0 34.8 35.7 36.5 37.4 38.3 39.2 40.2 41.2 42.2 43.2 44.2 45.3 46.4 47.5 48.7 49.9 51.1 52.3 53.6 54.9 56.2 57.6 59.0 60.4 61.9 63.4 64.9 66.5 68.1 69.8 71.5 73.2 75.0 76.8 78.7 80.6 82.5 84.5 86.6 88.7 90.9 93.1 95.3 97.6 355 | DCElectricalCircuitAnalysis_Page_355_Chunk1277 |
Appendix B Appendix B Methods of Solution of Linear Simultaneous Equations Methods of Solution of Linear Simultaneous Equations Some circuit analysis methods, such as nodal analysis and mesh analysis, yield a set of linear simultaneous equations. There will be as many equations as there are unknowns. For example, a particular circuit might yield three equations with three unknown currents (often referred to as a “3 by 3” for the matrix it creates). There are several techniques that may be used to solve this system of equations. The methods include graphical, substitution, Guass-Jordan elimination and determinants (determinants may be solved via Cramer's Rule/Sarrus’ Rule or via expansion by minors). Graphical Graphical Graphical solutions involve plotting the individual equations on graph paper. The location of where the lines cross is the solution to the system (i.e., values that satisfy all of the equations). This technique will not be discussed further because it is only practical for two unknowns. It would be very difficult to draw something like a four dimensional graph for four equations with four unknowns! Substitution Substitution The idea here is to write one of the equations in terms of one of the unknowns and then substitute this back into one of the other equations resulting in a simplified version. This process is iterated for as many unknowns as the system includes. Take for example the following 2x2: 10 = 20I1 + 8I2 2 = 8I1 + 4I2 Solve the second equation for I2. 2 = 8I1 + 4I2 4I2 = 2 − 8I1 I2 = 0.5 − 2I1 Substitute this back into the first equation and expand/simplify/solve. 10 = 20I1 + 8I2 10 = 20I1 + 8(.5 − 2I1) 10 = 20I1 + 4 − 16I1 10 = 4I1 + 4 I1 = 1.5 Finally, substitute this value back into one of the two original equations to determine I2. 356 | DCElectricalCircuitAnalysis_Page_356_Chunk1278 |
2 = 8I1 + 4I2 2 = 8(1.5) + 4I2 2 = 12 + 4I2 4I2 = −10 I2 = −2.5 For a 3x3, this process is iterated as follows: Equation 2 would be solved for I3 and this would be substituted back into equation 1 yielding a new equation (let’s call it A) with only I1 and I2 terms. Similarly, Equation 3 would be solved for I3 and this would be substituted back into equation 2 yielding a new equation (let’s call it B) with only I1 and I2 terms. Equations A and B now make a 2x2 with I1 and I2 as the unknowns and can be solved as outlined above. This would yield values for I1 and I2 which could then be substituted into one of the three original equations to obtain I3. While the substitution method is perfectly valid for an arbitrarily sized system, it proves cumbersome as the system gets larger. Gauss-Jordan Elimination Gauss-Jordan Elimination In some respects, Gauss-Jordan is similar to substitution but it tends to involve less overhead for larger systems and thus is generally preferred. This method involves multiplying one equation by a constant such that when it is subtracted from another equation, one of the unknown terms disappears. The process is then iterated for as many unknowns exist in the system. Using the same example from before: 10 = 20I1 + 8I2 2 = 8I1 + 4I2 Multiply the second equation by the ratio of the coefficients for I2 (8/4 = 2). 2 = 8I1 + 4I2 4 = 16I1 + 8I2 Subtract this new equation from the first equation. The I2 terms will cancel leaving just I1. 10 = 20I1 + 8I2 4 = 16I1 + 8I2 6 = 4I1 I1 = 1.5 Substitute this result back into one of the original equations to obtain I2. For a 3x3, iterate as follows: Using equations 1 and 2, multiply equation 2 by the ratio of the coefficients for I3. Subtract this equation from equation 1 to generate a new equation (let’s call it equation A) that only has I1 and I2 as unknowns. Using equations 2 and 3, multiply equation 3 by the ratio of the coefficients for I3. Subtract this equation from equation 2 to generate a new equation (let’s call it equation B) that only has I1 and I2 as unknowns. Equations A and B now make a 2x2 with I1 and I2 as the unknowns and can be solved as outlined previously. This would yield values for I1 and I2 which could then be substituted into one of the three original equations to obtain I3. Like the substitution method, Gauss-Jordan grows rapidly as the system size increases. The process tends to be formulaic though, and thus easier to handle. 357 | DCElectricalCircuitAnalysis_Page_357_Chunk1279 |
Determinants Determinants Determinants revolve around the concept of a matrix which itself is little more than an ordered collection of coefficients and/or constants. It is imperative that the unknowns be in the same order in each equation (i.e., I1 ascending to IX left to right) A simple coefficient matrix for the original 2x2 example is: 20 8 8 4 The resultant value (properly referred to as the determinant) for a 2x2 matrix such as this may be solved using Sarrus’ Rule: Simply multiply the two values along the upper right-lower left diagonal and then subtract that product from the product of the two values found along on the upper left-lower right diagonal. In this example that’s: 20*4 − 8*8 = 16 A solution involves dividing one determinant by another determinant (Cramer's Rule). That is, each matrix is solved for its resultant value and then these two values are divided to determine the final answer. One of these matrices will be the coefficient matrix just discussed. This will be placed in the denominator. The numerator matrix is a modified version of the basic coefficient matrix. It is created by replacing one column of coefficients with the constant values from the original system of equations. For example, the numerator matrix used to find I1 would replace the first column (the I1 coefficients 20 and 8) with the constants 10 and 2: 10 8 2 4 The resultant value is 40 − 16 or 24. Similarly, the numerator matrix for I2 would replace the I2 coefficients in the second column (8 and 4) with the constants 10 and 2: 20 10 8 2 The resultant value is 40 − 80 or −40. To find any particular unknown, simply divide the modified matrix by the basic coefficient matrix. 10 8 2 4 I1 = -------- 20 8 8 4 24 I1 = ----- 16 358 | DCElectricalCircuitAnalysis_Page_358_Chunk1280 |
I1 = 1.5 In like fashion I2 is found: 20 10 8 2 I2 = -------- 20 8 8 4 −40 I2 = ----- 16 I2 = −2.5 Sarrus’ Rule may also be used with a 3x3 matrix. This is achieved by extending the matrix. Fourth and fifth columns are added to the right of the 3x3 matrix by simply making copies of the first two columns. This creates three right to left diagonals with three values each and three left to right diagonals with three values each. The three values along each diagonal are multiplied together. The three right to left products are then subtracted from the sum of the three left to right products yielding a single resultant value (the determinant). To create the modified numerator matrix, replace the coefficient column of interest with the constant terms and then replicate columns one and two. For example, given these three equations: 10 = 20I1 + 8I2 + 3I3 2 = 8I1 + 4I2 + 5I3 7 = 3I1 + 5I2 + 6I3 The basic coefficient matrix (i.e., denominator) is: 20 8 3 8 4 5 3 5 6 The extended matrix is: 20 8 3 20 8 8 4 5 8 4 3 5 6 3 5 The result is: 20*4*6 + 8*5*3 + 3*8*5 − 3*4*3 − 20*5*5 − 8*8*6 Sarrus’ Rule does not work beyond 3x3. 359 | DCElectricalCircuitAnalysis_Page_359_Chunk1281 |
Expansion by Minors Expansion by Minors Expansion by Minors is another method that may be used to generate a determinant solution. This involves breaking the matrix into a series of smaller matrices (minors) that are combined using row-column coefficients. The position of these coefficients will also indicate whether the determinant of any particular sub-matrix is added or subtracted to the total. The first step is to establish a single row or column from which to derive the coefficients. This can be any horizontal row or vertical column (no diagonals). Each element of the chosen row or column determines the associated minor (essentially, that which is left over). Consider the 3x3 used previously: 20 8 3 8 4 5 3 5 6 Choosing the top row yields coefficients of 20, 8 and 3. For each of these, blot out its row and column and see what is left. This leaves three 2x2 matrices, one for each coefficient. Multiply each coefficient by the determinant of its 2x2 matrix. To determine whether this result is added or subtracted to the others, the sign may be found using the following map for the coefficients: + − + − etc. − + − + etc. + − + − etc. The origin in the upper left is positive and the signs continually alternate across from it and down from it. The result using the top row for the coefficients is found thus (the 2x2 matrices are bold red for clarity): 4 5 8 5 8 4 20 * 5 6 − 8 * 3 6 + 3 * 3 5 If the second column was used instead (8, 4, 5), the result is found like so: 8 5 20 3 20 3 −8 * 3 6 + 4 * 3 6 − 5 * 8 5 In closing, whichever method is used, always look for null coefficient terms (that is, places in the equations and matrices where the coefficients are zero). Smart use of these can considerably simplify the computations as there are few mathematical operations easier than multiplying by zero. For example, if a particular row of a matrix contains a few zeros, that would be a good candidate for the coefficient row when using expansion by minors because some 2x2 minors need not be computed (they will just be multiplied by the zero coefficient). 360 | DCElectricalCircuitAnalysis_Page_360_Chunk1282 |
Appendix C Appendix C Equation Proofs Equation Proofs Maximum Power Transfer: Refining the Maximizing Value of P = R/(R2+2R+1) While the algebraic and graphing technique explored in Chapter 6 leads to a proper answer, it is incomplete. For the ultimate value we can use a little differential calculus. Examining the curve of the function indicates a single peak, and thus we may find the corresponding value by taking the first derivative of the function, setting it to zero (i.e., find the point where the slope goes to zero), and solving for R. There are three basic techniques for determining the derivative of the equation above: using quotient rule, using product rule, and using chain rule. The quotient rule is perhaps the most obvious route although not necessarily the most efficient. The quotient rule states that the derivative of a quotient is 2) ( ) (' ) ( ) (' ) ( ) ( )' ( x g x g x f x f x g x g x f f(x) is R and therefore f '(x) is 1. g(x) is R2+2R+1 and therefore g′(x) is 2R+2. Note that for simplicity, g(x) may also be written as (R+1)2. Plugging these values into the quotient formula yields 4 2 )1 ( ) 2 2 ( 1 )1 ( R R R R dR dP This may be simplified to 4 2 )1 ( 1 R R dR dP For dP/dR to be zero, R must equal 1, meaning that maximum power is developed in the load when the load resistance matches the resistance of the driving source. QED 361 | DCElectricalCircuitAnalysis_Page_361_Chunk1283 |
Exponential Charge and Discharge Equations Utilizing a simple series RC circuit, such as depicted in the accompanying figure, we may write an expression for the natural discharge of the capacitor using KCL. We shall assume that the capacitor has some initial starting voltage, vC(0) = v0. Looking at node A, the total current entering and leaving this node must be 0. Expressing this using the fundamental current/voltage relationships for capacitors and resistors gives us: C dvC dt + v 0 R = 0 Applying a little algebra: dvC dt + v 0 RC = 0 dvC v 0 = −1 RC dt Remember, the voltage at t = 0 is v0 and at time t is vC(t). Integrating will allow us to obtain an expression for vC(t). Using x and y as variables of integration yields: ∫ v0 vC(t) dx x =−1 RC∫ 0 t dy Using the rule regarding natural logs, this can be written as: ln( vC(t) v 0 ) =−1 RC t And solving for vC(t) we discover: vC(t) = v0 ϵ −t RC Finally, given that RC = τ and that the starting voltage for the capacitor normally would be its fully charged value; that of the associated voltage source E, we find: vC(t) = Eϵ−t τ (discharge phase) The equation above also describes the shape of the current (and hence the resistor voltage) during the charge phase, and therefore to satisfy KVL during this phase we find: vC(t) = E(1 −ϵ −t τ) (charge phase) 362 | DCElectricalCircuitAnalysis_Page_362_Chunk1284 |
Appendix D Appendix D Answers to Selected Odd-Numbered Problems Answers to Selected Odd-Numbered Problems 1 Fundamentals 1 Fundamentals 1. 14.54, 30060, 76.90, 0.0008475 3. 5.1002, 1020.8, 1.0054, 0.000045781 5. 2.361E1, 1.2E4, 7.632E3, 5.09E−3 7. 4.156E1, 9.54000E5, 8.4035E1, 1.632E−4 9. 12E3 (or 12 k), 470, 6.5, 1.98E−3 (or 1.98 m) 11. 33.2, 313.6, 43E3 (or 43 k), 76E−6 (or 76 μ) 13. 3.6E6, 16.74E3, 1.202E3, 217.3E3 15. 0.205, 1.21E−3, 301E3, 3.2E12 17. 482, 22.42, approx. 6.09E3 (6.08997E3), −4.14E6 19. 15.8, 4.734E3, 7E−3 21. 320E6, 667E3, 307E−6 23. 2 kg·m 25. 3 km/s 363 | DCElectricalCircuitAnalysis_Page_363_Chunk1285 |
2 Basic Quantities 2 Basic Quantities 1. 33.858 volts to 34.542 volts 3. 2 volt, 20 volt and 200 volt scales 5. 6.81 volts to 7.19 volts 7. 1.602E−7 coulombs 9. 1.248E20 electrons 11. 0.4 amps 13. 0.5 coulombs 15. 0.2 volts 17. 20 joules 19. 1492 watts 21. 24 watts 23. 82.9% 25. 125 watts 27. 3 Ah 29. 1200 hours 31. 160 ohms 33. 8 ohms 35. 1200 ohms 37. 3.2 ohm-cm 39. yes, +2.26% 41. yes, −0.2% 43. 200 k, 68 k, 2.7 k 45. 3.3 M, 91, 390 47. 160 k to 240 k, 54.4 k to 81.6 k, 2.43 k to 2.97 k 49. 2.97 M to 3.63 M, 81.9 to 100.1, 370.5 to 409.5 51. yellow-violet-black-gold, red-red-orange-gold, orange-white-yellow-gold, brown-red-orange-gold, violet-green-brown-gold. 364 | DCElectricalCircuitAnalysis_Page_364_Chunk1286 |
3 Series Resistive Circuits 3 Series Resistive Circuits 1. 120 mA 3. 1.44 W 5. 15 V 7. 33 V 9. 20.6 k 11. 40 mA 13. 320 mW (200), 160 mW (100), 480 mW (source) 15. 6 V (2 k), 12 V (4 k), −6 V 17. 0.25 mA 19. 15 mA, + − left to right on 200, + − top to bottom on 100, + − right to left on 500, + − top to bottom on source. 21. 180 mW 23. 2.5 V (50), 0.5 V (10), 3 V (60), Vc = 3 V, Vac = 3 V, Va = 6 V 25. 200 mA, + − left to right on 25, + − top to bottom on 10, + − right to left on 5, + − top to bottom on source. 27. 2 V (400), 1 V (200), Vb = 10 V, Vc = 1 V, Vac = 11 V 29. 3.6 V (2 k), 14.4 V (8 k), Vb = 9.6 V, Vc = −14.4 V, Vac = 20.4 V 31. 35 mA, + − left to right on 400, + − right to left on 200, + − top to bottom on 12 V source, + − bottom to top on 9 V source. 33. 1.8 mA, + − left to right on 2 k, + − right to left on 8 k, + − top to bottom on sources. 35. Vb = 14.4 V, Vc = 9 V, Vac = 6 V 37. 1 V, it's lower 39. 20 V 41. 40 V (20), 100 V (50) 43. 12 V (4), 30 V (10), 15 V (5) 45. 20 V at a, 0 V at b, 10 V halfway 47. Change the 100 to 120 49. 5.5 V 51. R1 = 6 k, R2 = 1.5 k, R3 = 500 53. 19.2 55. Yes, 10 k 365 | DCElectricalCircuitAnalysis_Page_365_Chunk1287 |
4 Parallel Resistive Circuits 4 Parallel Resistive Circuits 1. 80 Ω 3. 14.3 Ω 5. 4 mA down 7. 120 mA (200), 480 mA (50), 600 mA (source) 9. 219.5 mA (82), 264.7 mA (68), 484.2 mA (source) 11. No. It is already the smallest current by an order of magnitude and this makes it smaller. 13. 333.3 μA (each 36 k), 250 μA (each 48 k) 15. 0.1915 mA (47 k), 1.765 mA (5.1 k), 5 mA (1.8 k) 17. 1.6 A (10), 0.4 A (40), 16 V 19. 50 mA (200), 25 mA (each 400), 10 V 21. 145.5 mA (3 k), 218.2 μA (2 k), 36.37 μA (12 k), 0.4364 V 23. 2.571 mA (each 25 k), 0.8571 mA (75 k), higher because there is one less path for current flow. 25. 5.714 mA (5 k), 2.857 mA (each 10 k), 28.57 mA (1 k), 28.57 V 27. The voltage is larger but now negative, so it becomes more negative. 29. 11.25 mA (2 k), 3.75 mA (6 k), 5 mA (4.5 k) 31. 1.714 k 33. 6.667 mA 35. 857 Ω 37. 3 k 366 | DCElectricalCircuitAnalysis_Page_366_Chunk1288 |
5 Series-Parallel Resistive Circuits 5 Series-Parallel Resistive Circuits 1. 10 k with 30 k 3. None 5. 200 with 300 7. None 9. 12.5 k 11. 23.3 k 13. 106 Ω 15. 3 k 17. Va = 2 V, Vb = 1.34 V, Vab = 0.662 V 19. 0.4 mA (12 k), 0.4 mA (3 k), 0.8 mA (7.5 k) 21. Va = −18 V, Vb = −9.35 V, Vab = −8.65 V 23. 903 μA (3.3 k), 602 μA (10 k), 301 μA (20 k) 25. Va = 100 V, Vb = 59.5 V, Vab = 40.5 V 27. 15 mA (1 k), 3.75 mA (2.2 k), 3.75 mA (1.8 k) 29. Vb = 2.87 V, Vc = 5.1 V, Vcb = 2.23 V 31. 4 mA (2 k), 0.702 mA (5.1 k), 4.702 mA (source) 33. Va = 10 V, Vb = 5.263 V 35. 0.2222 mA (30 k), 0.3333 mA (36 k), 0.8889 mA (source) 37. Yes, because they also have the same voltage. 39. Va = −22.4 V, Vb = −36 V, Vab = 13.6 V 41. 191.5 μA (47 k), 448.9 μA (5.1 k), 1.72 mA (3.9 k) 43. Vb = 10.53 V, Vc = 7.643 V, Vd = 5.945 V 45. 133.3 μA (left), 66.67 μA (middle), 33.33 μA (right) 47. Va = 22 V, Vb = 16 V, Vab = 6 V 49. 375 mA (100 k), 125 mA (60 k), 125 mA (240 k) 51. Va = 4.91 V, Vb = 3.27 V, Vab = 1.64 V 53. 50 mA (200), 25 mA (400), 25 mA (100) 55. Va = −1.2 V, Vb = −3.2 V, Vab = 2 V 57. 1.2 mA (9 k), 0.6 mA (82 k) 59. Yes, because they also have the same voltage. 61. Va = 3.54 V, Vb = 35.4 V, Vab = −31.86 V 63. 878 μA (1 k), 878 μA (2.2 k), 122 μA (18 k) 65. Va = −5.29 V, Vb = −3.31 V, Vab = −1.98 V 67. 4 mA (left), 2 mA (middle), 1 mA (right) 69. 133.3 Ω 71. 110 Ω 73. 2.55 k 75. 1 A 77. 4 mA 367 | DCElectricalCircuitAnalysis_Page_367_Chunk1289 |
6 Analysis Theorems and Techniques 6 Analysis Theorems and Techniques 1. 4.26 mA in parallel with 4.7 k 3. 4.56 mA in parallel with 5.7 k 5. 26.4 V in series with 2.2 k 7. −40 V in series with 800 Ω 9. −18 V in series with 9 k 11. 2.55 mA left to right 13. 49.5 V 15. 8.18 V 17. 2.55 mA 19. 4.2 V 21. 1.25 mA 23. 0.7 V 25. 8.09 mA 27. 42.7 V 29. 0.696 mA 31. Yes, it's only current sources and resistors in parallel. 33. 1.17 mA 35. −15.65 V 37. −11.6 V 39. 4.03 mA (up) 41. 38.7 mV 43. 1.19 mA left to right 45. 7.2 V with 6.2 k 47. 2 mA with 3 k 49. 5.71 V with 1.71 k 51. 1.538 A with 13 Ω 53. 66.7 μA with 3 k 55. 1.92 mW. Yes, match Rthev. 57. 4.27 μW. Yes, match Rthev. 59. 3 k, 3 mW 61. 4 k, 4.44 μW 63. Replace 10 V/2 k with 5 mA||2 k, and 24 V/10 k with 2.4 mA||10 k 65. Replace 2 mA||20 k with 40 V + 20 k, and 0.4 mA|| 3 k with 1.2 V + 3 k 368 | DCElectricalCircuitAnalysis_Page_368_Chunk1290 |
7 Nodal & Mesh Analysis, Dependent Sources 7 Nodal & Mesh Analysis, Dependent Sources 1. Loop ordering is left to right 9 = (1 k + 3 k)I1 − (3 k)I2 −12 = − (3 k)I1 + (3 k + 2 k)I2 3. 0.818 mA left to right 5. 3.14 V 7. 50 = (200 + 40)I1 − (40)I2 −120 = − (40)I1 + (40 + 75)I2 9. 1.03 A right to left 11. −2.3 V 13. First, combine the 4 k||12 k = 3 k −34 = (2 k + 10 k)I1 − (10 k)I2 24 = − (10 k)I1 + (10 k + 3 k)I2 15. 0.696 mA (up) 17. 12.27 V 19. 60 = (100 + 200)I1 − (200)I2 − (0) I3 − (0) I4 0 = − (200)I1 + (200 + 300 + 400)I2 − (400) I3 − (0) I4 0 = − (0)I1 − (400)I2 + (400 + 500 + 600) I3 − (600) I4 −150 = − (0)I1 − (0)I2 − (600) I3 + (600 + 700) I4 21. 47.7 mA right to left 23. −9.34 V 25. Loop 1 at top, then left to right. 30 = (15 k + 10 k + 9 k + 3 k)I1 − (3 k)I2 − (9 k) I3 − (10 k) I4 25 = − (3 k)I1 + (2 k + 3 k + 4 k)I2 − (4 k) I3 − (0) I4 10 = − (9 k)I1 − (4 k)I2 + (4 k + 9 k + 5 k) I3 − (5 k) I4 −10 = − (10 k)I1 − (0)I2 − (5 k) I3 + (5 k + 10 k + 7 k) I4 27. 0.86 mA left to right 29. −3.26 V 369 | DCElectricalCircuitAnalysis_Page_369_Chunk1291 |
31. Loop ordering: left, top, bottom. 24 = (2 k + 8 k)I1 − (2 k)I2 − (8 k) I3 0 = − (2 k)I1 + (2 k + 22 k + 800)I2 − (22 k) I3 0 = − (8 k)I1 − (22 k)I2 + (8 k + 22 k + 400) I3 33. 19.8 mA (down) 35. −22.4 V 37. 20 = (6.8 k + 10 k)I1 − (10 k)I2 − (0 k) I3 0 = − (10 k)I1 + (10 k + 8.5 k + 30 k)I2 − (30 k) I3 0 = − (0)I1 − (30 k)I2 + (30 k + 20 k + 70 k) I3 39. 0.34 mA left to right 41. 49.5 V 43. Current source converts to 10 V + 20 k. 8 = (5 k + 4 k)I1 − (4 k)I2 −10 = − (4 k)I1 + (4 k + 20 k)I2 45. 0.76 mA left to right 47. −2.73 V 49. −6 mA = (1/24 k + 1/12 k + 1/9 k)Va − (1/9 k)Vb −2 mA = − (1/9 k)Va + (1/9 k + 1/25 k)Vb 51. 4.03 mA (up) 53. −774 mV 55. Combine 15 k + 25 k = 40 k 2 mA = (1/20 k + 1/40 k + 1/5 k)Va − (1/5 k)Vb 0.4 mA = − (1/5 k)Va + (1/5 k + 1/3 k)Vb 57. 0.538 mA 59. 324 mV 61. (300 − 20)mA = (1/40 + 1/10 + 1/100)Va − (1/10)Vb − (1/40)Vc 10 mA = − (1/10)Va + (1/10 + 1/50 + 1/250)Vb − (1/250)Vc (−300 − 50)mA = − (1/40)Va − (1/250)Vb + (1/250 + 1/25 + 1/40)Vc 63. 177 mV 65. 5.94 V 370 | DCElectricalCircuitAnalysis_Page_370_Chunk1292 |
67. This is deceptively simple. First combine the parallel 36 k resistors to a single 18 k. There is only one node of concern, node a. The 4 mA source enters this node. Define an exiting current, I1, down through the 18 k. I1 = Va/18 k. Define an entering current from the voltage source, I2. I2 = (12 − Va)/30 k. From KCL, 4 mA + I2 = I1. Substitute the current expressions into the KCL equation, expand and simplify. This results in a single node equation: 4.4 mA = (1/30 k + 1/18 k) Va. 69. 1.25 mA left to right 71. 4.2 V 73. 240 V 75. 720 V 77. 441 mV 79. 6 V 8 Capacitors 8 Capacitors 1. 5 μF 3. 2 μF 5. 8 V across 10 μF, 4 V across 20 μF 7. 2 k gets 5.38 V, 5 μF gets 0 V, 4 k and 3 k get 4.62 V 9. 23. 1.5 V 11. 20 μs, 100 μs 13. 13.2 V, 24 V 15. 1 ms, 5 ms 17. 10 V, 15 ms 19. 3 ms, 15 ms 21. Replace 100 Ω with 2 k 23. 125 μF 371 | DCElectricalCircuitAnalysis_Page_371_Chunk1293 |
9 Inductors 9 Inductors 1. 6 mH 3. 0.75 mH 5. 2 k gets 3.33 V, 5 μF gets 0 V, 4 k and 3 mH get 6.67 V 7. 2 k gets 3.33 V, 3 mH gets 0 V, 4 k and 5 μF get 6.67 V 9. 2 k gets 5.38 V, 5 mH gets 0 V, 4 k and 3 k get 4.62 V 11. 2 k, 10 k and 20 μF get 0 V, 8 k and 12 mH get 20 V 13. 33 k gets 15 V, all other components get 0 V 15. 0.2 A 17. 20 ms, 100 ms 19. 0 V and 480 mA for both times. With 5 Ω, 2 V and 400 mA for both times. 21. 40 ns, 200 ns 23. 3 mA, −60 V 25. 833 ns, 4.17 μs 27. Replace 25 Ω with 250 Ω. 10 Magnetic Circuits and Transformers 10 Magnetic Circuits and Transformers 1. 80 mA 3. 95 mA 5. 160 turns 7. 20 mA 9. 2.42 A 11. Cast steel causes greater core losses so the secondary current would be reduced. 13. 324 mA 15. vload = 4 volts, iload = 40 mA, iprimary = 6.67 mA 17. 1440 Ω 19. 45 VA 372 | DCElectricalCircuitAnalysis_Page_372_Chunk1294 |
Appendix E Appendix E Base Units Base Units In Chapters 1 and 2, a variety of measurement units were introduced and defined, along with the metric system. It is useful to understand that all of the units used are built from three foundational units. These units describe the conceptual building blocks of the system. They are mass (kilograms, kg), length (meters, m) and time (seconds, s). They are the foundation of the MKS system. Every other unit used, such as watts, volts or amps, can be built from these three fundamental elements. For example, it is not arbitrary that ohms times amps yields volts. In Chapter 2 it was noted that power equals current times voltage, and that this can be derived from the associated units. Current was defined as the rate of charge movement per unit time, and one amp equals one coulomb per second. Similarly, voltage was defined as the energy required to move a certain charge, and one volt equals one joule per coulomb. If we multiply the units together we get: P = I×V P = C s × J C P = J s Joules per second is energy per unit time and that is the very definition of power, with one watt being equal to one joule per second. But this brings up the question, “Then what is a joule in terms of the three fundamental units?” Energy can be defined as a force working over a specified distance. One joule is defined as the force of one newton working over a distance of one meter. This creates a new question, “Then what is a newton?” Newton's second law of motion states that force is equal to mass times acceleration. Acceleration, in turn, is the rate of change of velocity, or distance per unit time, per unit time. For these units, we find: force = mass×acceleration force = kg× m s2 force = kg m s 2 energy = force×distance energy = kg m s 2 ×m energy = kg m 2 s 2 Thus, we verify that one joule is equal to one kilogram-meter2 per second2. If we apply this back to the definition for power, we find that one watt is equal to one kilogram-meter2 per second3. 373 | DCElectricalCircuitAnalysis_Page_373_Chunk1295 |
Appendix F Appendix F “Could I do one more immediately?” – Bill Bruford 374 | DCElectricalCircuitAnalysis_Page_374_Chunk1296 |
ELECTROMAGNETICS VOLUME 1 | Electromagnetics_Vol1_Page_2_Chunk1297 |
Publication of this book was made possible in part by the Virginia Tech University Libraries’ Open Education Faculty Initiative Grant program: http://guides.lib.vt.edu/oer/grants | Electromagnetics_Vol1_Page_3_Chunk1298 |
VT Publishing Blacksburg, Virginia ELECTROMAGNETICS STEVEN W. ELLINGSON VOLUME 1 | Electromagnetics_Vol1_Page_4_Chunk1299 |
Copyright © 2018, Steven W. Ellingson This textbook is licensed with a Creative Commons Attribution Share-Alike 4.0 license https://creativecommons.org/licenses/by-sa/4.0. You are free to copy, share, adapt, remix, transform and build upon the material for any purpose, even commercially as long as you follow the terms of the license https://creativecommons.org/licenses/by-sa/4.0/legalcode. You must: Attribute — You must give appropriate credit, provide a link to the license, and indicate if changes were made. You may do so in any reasonable manner, but not in any way that suggests the licensor endorses you or your use. Suggested citation: Ellingson, Steven W. (2018) Electromagnetics, Vol. 1. Blacksburg, VA: VT Publishing. https://doi.org/10.21061/electromagnetics-vol-1 Licensed with CC BY-SA 4.0 https://creativecommons.org/licenses/by-sa/4.0 ShareAlike — If you remix, transform, or build upon the material, you must distribute your contributions under the same license as the original. You may not: Add any additional restrictions — You may not apply legal terms or technological measures that legally restrict others from doing anything the license permits. This work is published by VT Publishing, a division of University Libraries at Virginia Tech, 560 Drillfield Drive, Blacksburg, VA 24061, USA publishing@vt.edu. The print version of this book is printed in the United States of America. Publication Cataloging Information Ellingson, Steven W., author Electromagnetics (Volume 1) / Steven W. Ellingson Pages cm ISBN 978-0-9979201-8-5 (print) ISBN 978-0-9979201-9-2 (ebook) DOI: https://doi.org/10.21061/electromagnetics-vol-1 1. Electromagnetism. 2. Electromagnetic theory. I. Title QC760.E445 2018 621.3 Cover Design: Robert Browder Cover Image: © Michelle Yost. Total Internal Reflection https://flic.kr/p/dWAhx5 is licensed with a Creative Commons Attribution-ShareAlike 2.0 license https://creativecommons.org/licenses/by-sa/2.0/ (cropped by Robert Browder) iv | Electromagnetics_Vol1_Page_5_Chunk1300 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.