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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. Th... | 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 the... | 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 Ω10... | 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 vo... | 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... | 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 cours... | 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 ... | 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 ... | 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 cap... | 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 ... | 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 ... | 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 sho... | 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 ne... | 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 ... | 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 d... | 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... | 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 indi... | 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 cau... | 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 i... | 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 ... | 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... | 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... | 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... | 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 ... | 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... | 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 agree... | 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 i... | 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... | 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 Figur... | 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 sourc... | 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 b... | 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 microsecond... | 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 ... | 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.... | 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 ... | 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 ver... | 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 ti... | 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 o... | 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 a... | 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 th... | 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... | 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... | 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 ... | 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.... | 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... | 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 specifical... | 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 ... | 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 accorda... | 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, elaps... | 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 a... | 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... | 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 p... | 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 ener... | 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 mag... | 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 f... | 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. T... | 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 ... | 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... | 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 e... | 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 ... | 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 i... | 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 Dens... | 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 ... | 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 ... | 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 second... | 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... | 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 polari... | 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.... | 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 ... | 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 fo... | 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... | 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 secti... | 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... | 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... | 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 ind... | 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 par... | 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 ... | 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 ori... | 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 t... | 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 t... | 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 curv... | 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 cu... | 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, 1.98... | 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. 16... | 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... | 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 (eac... | 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 ... | 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 2... | 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... | 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 ... | 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 K... | 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 ... | 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. The... | 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 ... | Electromagnetics_Vol1_Page_5_Chunk1300 |
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