Pobieranie prezentacji. Proszę czekać # Pole magnetyczne Pole magnetyczne wytwarza pole sił. Siła działa pomiędzy 2 magnesami bez ich bezpośredniego kontaktu (tak jak pole elektryczne). Pole.

## Prezentacja na temat: "Pole magnetyczne Pole magnetyczne wytwarza pole sił. Siła działa pomiędzy 2 magnesami bez ich bezpośredniego kontaktu (tak jak pole elektryczne). Pole."— Zapis prezentacji:

Pole magnetyczne Pole magnetyczne wytwarza pole sił. Siła działa pomiędzy 2 magnesami bez ich bezpośredniego kontaktu (tak jak pole elektryczne). Pole magnetyczne widać poprzez linie przepływu lub linie sił. Wskazują one kierunek działania siły i przebiegają od bieguna N do S. Linie sił pola zmieniają swój przebieg w otoczeniu innych materiałów magnetycznych takich jak np. żelazo, nikiel, kobalt zwanych ogólnie ferromagnetykami

Elektromagnetyzm I Pole magnetyczne Produkowane przez przewodnik N S I
Zasada prawej ręki: Połóż zamkniętą dłoń na przewodniku liniowym tak aby kciuk pokazywał kierunek przepływu prądu. Zagięte palce pokazują kierunek linii pola magnetycznego Cewka (uzwojenie). Połóż zamkniętą dłoń tak aby palce wskazywały kierunek przepływu prądu. Kciuk wskazuje teraz kierunek pola magnetycznego. I Pole magnetyczne Produkowane przez przewodnik Most applications of magnetism involve magnetic effects due to electric currents. When a current, I, flows in the conductor it creates a magnetic field that is concentric about the conductor, uniform along its length, and whose strength is directly proportional to I. If the conductor is wound into a coil, the fields of its individual turns combine, producing an electromagnet with the resultant field as shown in the diagram. Provided no ferromagnetic material is present, the strength of the coil’s field is also proportional to its current. If the coil is wound on a ferromagnetic core, almost all the flux is confined to the core, although a small (called stray or leakage flux) returns through the surrounding air. However, the core flux is no longer proportional to current because of the presence of the ferromagnetic material. N S I Pole magnetyczne produkowane Przez cewkę (solenoid)

Natężenie pola & Strumień pola
Strumień, F = Ilość linii pola przebiegających przez powierzchnię A. Jednostką jest Wb Natężenie pola, B = F/A w Teslach (T) lub Wb/m2 W zamkniętym obwodzie magnetycznym szeregowym F jest stałe. W połączeniu równoległym natomiast F1 = F2 + F3 . Magnetic fields are described in terms of flux, F, and flux density, B. e.g. In diagram above, A1 = .015 m2, A2 = .04 m2, B1 = 100 mT. Calculate B2. F = 100 x10-3 x .015 = 1.5 mWb; therefore, B2 = 1.5 x 10-3 / .04 = 37.5 mT. The earth’s magnetic field is approx. 50 mT near the earth’s surface; the field of a large generator or motor is on the order of 1 or 2 T. Magnetic Circuits Magnetic circuits found in motors, generators, computer disk drives, etc., are magnetic structures to guide and shape magnetic fields by providing a well-defined path for the flux. For magnetic circuits with air gaps, fringing occurs, causing a decrease in flux density in the gap. Many practical magnetic circuits (such as transformers) use thin sheets of lamination which would result in a reduced effective cross-sectional area. Series magnetic circuits have sections of different materials (e.g. sheet steel,and air gap). For these circuits, F is the same in all sections but B in each section depends on the effective cross-sectional area. A circuit may also have elements in parallel. At each junction, the sum of fluxes entering is equal to the sum leaving. This is the counterpart of KCL.

Krzywe B-H & Pętla histerezy
B = mH = mrmoH, mo = 4px10-7 H/m B Namagnesowanie resztkowe a B b a H Nasycenie Properties of Magnetic Materials Ferromagnetic materials have magnetic fields in tiny regions called domains. A nonmagnetized specimen can be magnetized by making its domain fields line up. As I through the coil increases, H increases causing more domains to align themselves in the direction of the field. If the field is strong enough, almost all domain fields line up and the material is said to be in saturation at point a. This results in the dc or normal magnetization curve. Hysteresis If I is now reduced to zero, the material still retains some magnetism called residual magnetism. If I is reversed, saturation in the reversed direction will be reached at point c. By reversing I again at point d, the curve returns to point a resulting in a hysteresis loop. Clearly, B depends not only on I (or H), it also depends on the circuit’s past history. B-H curves are the average of the two arms of the hysteresis loop. To demagnetize a ferromagnetic specimen, it is necessary to successively decrease its hysteresis loop to zero by placing it inside a coil that is driven by a variable ac source and gradually decreasing I to zero (e.g. as used by service personnel to “degauss” TV tubes). d c H Krzywa namagnesowania Pętla histerezy

Zasada przekaźników Armatura Przekaźnik jest elektromagnetycznie włączanym przełącznikiem Namagnesowanie cewki przyciąga ramię Ruch ramienia zamyka lub otwiera kontakt 2/3 z 1 2 1 3 4 Sprężyna Cewka 5 Schemat podstawowy NC NO Cewka 2 1 3 4 5 Symbol

Napięcie Indukowane Prawo Faraday’a: W obwodzie o N uzwojeniach
indukuje się napięcie jeżeli strumień magnetyczny przepływający przez obwód ulega zmianie. Wielkość tego napięcia zależy od szybkości zmian strumienia: (volts) Electromagnetic Induction Consider a magnet moving toward a coil (pg. 476). The induced voltage is governed by Faraday’s and Lenz’s laws. Induced Voltage & Induction The induced voltage in an inductor coil is called a counter emf or back voltage. This voltage opposes any changes in current and it prevents the current from changing abruptly. By Faraday’s law, the induced voltage is proportional to the rate of change of the flux linkage, NF. e.g. A voltage of 15 V is induced in a 300-turn coil. What is the rate of change of the flux through the coil? Rate of change of flux = 15 / 300 = .05 Wb/s Prawo Lenz’a : Polaryzacja napięcia jest taka że przeciwdziała zmianom strumienia, tzn. jeżeli np.. strumień maleje to pole magnetyczne wywołane prądem indukowanym zwiększa ten strumień

Samoindukcja Napięcie indukowane w cewce: (V) Indukcyjność cewki dla
Napięcie na cewce Indukcyjność cewki dla warunku l/d > 10 wyrażamy: In an air-core inductor, not all flux lines pass through all windings; hence, it would be difficult to determine flux linkages through the coil. But flux is proportional to current. Therefore, it is more useful to convert the rate of change of flux, dF/dt, in the equation for induced voltage to the rate of change of current, di/dt. e.g. If the current through a 2-mH inductor changes steadily from 50 mA to 100 mA in 50 ms, what is the induced voltage? di/dt = 50 mA / 50 ms = 1 kA/s; therefore, vL = 2 x10-3 x 1000 = 2 V. The accuracy of the equation for inductance of a coil breaks down for small l /d ratios (if l /d > 10, the error is < 4%). Improved formulas may be found in design handbooks. e.g. Determine the inductance of a .25 m long air-core coil with a radius of .05 m and 25 turns. L = 4p x 10-7 x 252 x p x .052 / .25 = 24.7 mH. To provide greater inductance in smaller spaces, iron cores are used. To keep the core flux below saturation, an air gap may be used. If the gap is wide enough to dominate, then L is approximately: L = moN2Ag / lg , where subscript g refers to the air gap. (H)  = przenikalność Cewka

Indukcyjności połączone szeregowo i równolegle
Dla N indukcyjności szeregowych: LT = L1 + L LN Dla N indukcyjności równoległych: Poł. szeregowe Formulas for inductances in series and parallel are very similar to those for resistors. e.g. Determine the total inductance for the circuit on the right. L = 1 / (1/5 +1/ (4 +1)) = 2.5 mH Inductance & Steady State The voltage across an inductor with constant dc current is zero because there is no change of current. Thus, an ideal inductor looks like a short circuit in steady state dc. Energy Stored by an Inductor Energy is stored in the magnetic field of the inductor. An ideal inductor has no resistance; therefore, no power loss. The energy stored is given by: W = Li2 / 2 (J) where i = instantaneous current. At steady state, W = LI2 / 2, where I is the steady state current. This energy remains stored in the field as long as the current flows. It is returned to the circuit when the current goes to zero. Poł. równoległe Energia zmagazynowana W = 1/2 LI2

Chwilowe i ustalone stany natężenia i napięcia w cewce
vR = E(1-e-t/t) Stan chwilowy Stan ustalony Statement of continuity of current for inductance: current through an inductor cannot change instantaneously, but must be continuous at all values of time. An inductor with zero initial current looks like an open circuit at the instant of switching. e.g. For the circuit above, E = 10 V, R = 100 W, L = 5 mH. Determine: i, and vL after 20 ms. t = 5 x 10-3 / 100 = 50 ms; therefore, i = 10/100 (1-e-20/50) = mA ; and vL = Ee-20/50 = 6.7 V. For all practical purposes, inductive transients, like capacitive transients, last for about 5 time constants. One can also use the universal time constant graphs to estimate the voltage rise and fall of the inductive circuits. vL = Ee-t/t i = (E/R)(1-e-t/t) t = L/R

Stany chwilowe i ustalone w cewce
Stan chwilowy (przejściowy) W momencie włączenia cewki w obwód napięcie indukowane na cewce, vL = -E a więc iL = 0. Zatem obwód z cewką wygląda jak obwód otwarty. Następnie napięcie na cewce opada wykładniczo a prąd płynący w cewce (indukcyjności) wzrasta odpowiednio do wartości maksymalnej. Ten stan przejściowy trwa około 5. Potem stan się ustala Stan ustalony vL = 0, i iL = E/R. Cewka wygląda jak zwarcie.

Jak uniknąć nagłych zmian napięcia
Nagłe rozłączanie lub włączanie prądu w obwodzie z indukcyjnością (np. silniki, generatory) może spowodować nagłe zmiany napięcia nawet do wielu kV!. Może to spowodować wyładowanie na przełączniku! Indukcyjność wygląda jak źródło prądowe w momencie włączania. Aby uniknąć gwałtownych zmian indukcyjnych należy w obwodzie umieścić rezystor lub diodę.

Vo= Io(R1+ R2) - Interrupting Current in an Inductive Circuit Abruptly breaking the current through a large inductor (such as a motor or generator field coil) can create voltage spikes up to several thousand volts that will cause a flashover across the switch. Even moderate sized inductances in electronic systems can create enough voltage to cause damage if protective circuitry is not used. Flashovers can be prevented by using a discharge resistor or a diode. The Inductor at Switching An inductor with an initial current looks like a current source at the instant of switching. De-energizing Transients The basic principle of minimizing the inductive kick is to use the circuit shown above with R2 much smaller than R1. In the equations, if the current has reached steady state before the switch is open, the initial current, Io = E/R1. e.g. For the circuit above, E = 50 V, R1 = 500 W, R2 = 100 W. Assuming steady state before the switch is opened, calculate Vo at the instant the switch is opened. Io = 50/500 = .1 A; therefore, Vo = -.1x ( ) = 60 V. Note that the initial voltage is > E. The larger the value of the R2 / R1 ratio, the greater the initial voltage spike, and the longer the time constant , t’. vL= -Voe-t/t’ i = Ioe-t/t’ t’= L/(R1+R2)

Transformatory z rdzeniem
ep/es = Np /Ns = N pin=ep ip = pout = es is ip / is = 1/N Ep / Es = N = Is / Ip Pin = Pout Zp= N2 ZL Transformator idealny A transformer is a magnetically coupled circuit, i.e. a circuit in which the magnetic field produced by time-varying current in one circuit induces voltage in another. Transformers are basically either iron-core or air-core type. The diagram shows a basic iron-core transformer where an alternating current in the primary winding establishes a flux which links the secondary winding. Cores for transformers are made from laminated and insulated sheet steel to keep eddy current losses low. The iron increases the coupling between coils by providing an easy path for magnetic flux. Because iron has high power loss due to hysteresis and eddy currents at high frequencies, iron cores are usable only at power and audio frequencies. For high-frequency applications, air-core and ferrite-core types are used. When the windings are wound as shown, the primary and secondary voltages are in phase and they are represented by the dot convention. If most of the flux produced by one coil links the other, the coils are said to be tightly coupled. Iron-core transformers are characterized by their turns ratios, N, while air-core transformers are characterized by self- and mutual inductances. The equations above are for an ideal transformer. A step-up transformer is one in which Es>Ep, while a step-down transformer is the reverse (N>1). e.g. Given: N = 5, Es = 10 V , Z = 8 W. Determine: Ep, Ip, Zp, Pin. Ep = 5x10 = 50 V; Ip = (10/8)/5 = 0.25 A; Zp = 52 x 8 = 200 W; Pin = 50 x .25 = 12.5 W. Transformers are rated in terms of voltage and apparent power (rather than in terms of real power). Thus a transformer rated 2400/120 V, 48 kVA, has a current rating of 48000/2400 = 20 A on the primary side and 48000/120 = 400 A on the secondary side. It can handle a 48 kVA load, regardless of power factor. Schemat transformatora

Auto-transformatory i transformatory wielostopniowe
Podnoszący Zmienny Autotransformatory a2 = N1/N2 Z2’ = a22 Z2 Transformer Applications Typical applications are: power supply; power distribution; isolation; and impedance matching. Autotransformers An autotransformer has a single winding serving as both the primary and the secondary as shown in the diagrams above. Thus, the secondary is not electrically isolated from the primary, but it is smaller and cheaper than conventional transformers for the same rated load kVA. It can either be a step-up or a step-down transformer depending on how Vin and Vout are connected. Transformers with Multiple Secondaries For a transformer with multiple secondaries, each secondary voltage is governed by the appropriate turns ratio, i.e. E1/E2 = N1/N2, etc. Loads are reflected in parallel as shown in Fig. (d), i.e. Zp = Z2’ //Z3’. a3 = N1/N3 Z3’ = a32 Z3 Obw. zastępczy dwa uzw. wtórne

Rzeczywiste transformatory rdzeniowe
Straty w “rzeczywistych” transformatorach : Prąd upływu Rezystancja zwojów (grzanie zwojów) Prądy wirowe w zwojach (grzanie rdzenia) Prąd magnetyzacji oraz pole rozproszone Straty prowadzą do zmniejszenia wydajności: h = (Pout / Pin) x 100% = (EsIs/EpIp) x 100% ; Pin = Pout + Pzwojów + Prdzenia Effects not included in the idealized transformers are: Leakage flux - a small amount of the flux passes outside the core and through the air at each winding. The effect can be modeled by inductance Lp and Ls. Winding resistance - Rp and Rs cause a slight power loss (or copper loss) and hence a reduction in efficiency as well as a voltage drop. Core or iron loss - due to eddy currents and hysteresis. Eddy currents can be reduced by using thin laminated steel core (for power and audio transformers) or powered iron (ferrite cores). Hysteresis is minimized by using special grain-oriented transformer steel. The effect of core loss can be modeled as a resistor, Rc in parallel with the primary winding (pg. 935). Magnetizing current - current is required to magnetize the core. Stray capacitances - exist between various parts of the transformer. Transformer Efficiency Efficiency is the ratio of the output power to input power. For a transformer, losses are due to I2R losses in the windings (called copper losses) and losses in the core. Large power transformers have about 98 to 99% efficiency while smaller transformers are around 95% or better.

Transformatory bezrdzeniowe
W transformatory bezrdzeniowe lub w cewkach połączone w obwodzie Indukują się napięcia na skutek indukcji wzajemnych i samoindukcji. Te napięcia nie są określone przez stosunek uzwojeń (jak dla i.transform) Dodaje się: LT+ = L1 + L2 + 2M Lub odejmuje: LT- = L1+ L2 - 2M When a coupled circuit does not have an iron core, only a portion of the flux produced by one coil links another and the coils are said to be loosely coupled. This circuit cannot be characterized by turns ratio; rather, it is characterized by self- and mutual inductances. Air-core and ferrite-core transformers, and general inductive circuit coupling fall into this category. In the diagrams above, the fluxes in the coils are additive (as indicated by the dots), hence the sign in front of M is positive. When the fluxes oppose, the sign in front of M would be negative. The degree of coupling between coils is known as the coefficient of coupling, k which ranges in value between 0 and 1. It is the ratio of the flux that links the companion coil to the total flux produced by the energized coil. e.g. Given: L1 = 5 mH, L2 = 8 mH, k = Determine: LT if their fluxes are opposing. M = .25 (5x10-3x8x10-3)1/2 = 1.58 mH; therefore, LT- = x1.58 =9.84 mH. The effect of unwanted mutual inductance can be minimized by physically separating coils, shielding them, or orienting their axes at right angles. M = wzajemna indukcyjność k = wsp. sprzężenia

Napięcia zmienne sinusoidalne
In dc circuits, voltage polarities and current directions do not change. In contrast, ac voltage polarities and current directions constantly vary with time. The voltage at the wall outlet is a sinusoidal ac waveform because its amplitude varies sinusoidally with respect to time. During one half of a cycle, the voltage polarity is positive, while during the next half cycle, it is negative. Since the waveform repeats itself, it is called periodic. For a resistive circuit, the current waveform is identical in shape to the voltage waveform and its amplitude is given by i = e/R. The direction of the current flow alternates repeatedly. 1 okres Obwód AC Przebiegi sinusoidalne

Generowanie napięcie AC
Obrót Obrót Pozycja cewki Generowane Napięcie The ac generator is based on the principle that when a conductor is moving through a magnetic field, a voltage is induced whose magnitude is proportional to the rate at which flux lines are cut and whose polarity depends on the direction in which the conductor is moved through the field. In the diagram, a coil is rotated through a magnetic field and voltage is induced in the coil. Not shown are slip rings and brushes that connect the coil to the load (pg. 528). When the coil is in the 0o position, sides a and b are moving parallel to the magnetic field. Since they are not cutting the flux, the induced voltage and current are zero. As the coil leaves the 0o position, it starts to cut across flux lines and voltage starts to increase. Voltage reaches a maximum when the coil arrives at its 90o position because sides a and b cut the flux perpendicularly. As the coil passes 90o, the voltage starts to decrease, reaching zero at the 180o position. Beyond 180o, the polarity of the induced voltage reverses relative to the 0o position. The maximum negative voltage value is reached at 270o and it starts to decrease, reaching zero at 360o or 0o. The cycle then repeats. The resulting voltage waveform can be represented as: e = Em sin a. Electronic Signal Generators AC waveforms are frequently generated electronically and they are not limited to sinusoidal ac. The general-purpose signal generator used in the lab can produce a variety of variable-frequency waveforms. Cewka e = Em sin a

Parametry impulsu sinusoidalnego
Okres f = 1/T Okres trwania cyklu, T. Epk= Em; Ep-p= 2Em; Eave= 0; Erms= 0.707Em e1 = Em sin w t ; e2 = Em sin (w t - q) gdzie w = 2pf w radianach/s. Moc efektywna lub średnia, Pavg = Irms2R or Erms2/R The number of cycles/sec of a waveform is defined as frequency. The period of a waveform is measured between any two corresponding points. For 60 Hz, T = ms. If the sinewave, e, is riding on top of a dc voltage of E volts, then, v = E + e; Epk = E + Em ; Eave = E and Erms must be computed by calculus or other numerical methods (also for the case when the waveform is nonsinusoidal). Effective Values An effective (or rms) value is an equivalent dc value: it tells us how many volts or amps of dc that a time varying waveform is equal to in terms of its ability to produce average power: Pavg = Irms2 R or Ipk2 R/2. e.g. A sinusoidal source has a peak-to-peak amplitude of 40 V and a period of 4 ms. Calculate: Erms, f and the voltage when t = 5 ms. Erms = x 40/2 = V; f = 1/4ms = 250 Hz; e = 20 sin (2 x 180o x 250 x 5 x 10-3) = 20 V e.g. Given i = 100 sin (1000 t + 50o) mA, compute the voltage across a 500 W resistor when t = 2 ms and the average power dissipated. What is the frequency of the current waveform? vR = 500 x100 x10-3 sin (1000x2x10-3x180o /p+50o) = V; Pavg = (100 x 10-3)2 500 /2 = 2.5 ; f = 1000 / (2 x p) = Hz.

Liczby zespolone. Interpretacja
w Vm v p 2p a wt a Vm Obracamy wektor v(t) = Vm sin a = Vm sin wt Obrót wektora o kąt  powoduje iż jego rzut Na oś pionową zmienia się jak funkcja sinus

Napięcie i natężenie zmienne
Im v(t) q + wt Vm q i(t) Im wyprzedza Vm (lub i wyprzedza v) o fazę qo Zależność od czasu: v(t) = Vm sin w t; i (t)= Im sin (w t + q) Napięcie i natężenie mają tą samą częstotliwość !

Natężenie i napięcie zmienne (c.d)
j i(t) + wt q Vm q Im v(t) Im opóźnia się względem Vm (lub i opóźnia się wzgl. v) o qo Zależność od czasu: v(t) = Vm sin w t; i = Im sin (w t - q) Napięcie i natężenie ma tą samą częstotliwość.

Pojęcie liczby zespolonej
Liczba zespolona jest w postaci Z = r + jx, gdzie r i x są częściami rzeczywistymi i urojonymi: tzn. j = (-1), a więc j2 = -1. W układzie polarnym , Z = |Z| e jq, gdzie |Z| = (r2 + x2) 1/2, i q = tan-1 (x/r). Transfromacje pomiędzy układem polarnym i kartezjańskim: r = |Z| cos q ; x = |Z| sin q. Czyli: Z=|Z|(cos q + j sin q) Dodawanie, mnożenie liczb zespolonych: Z1 + Z2 = (r1 + r2) + j (x1 + x2); Z1Z2 = |Z1||Z2| e j( q1+ q2 ) Z1/Z2 = |Z1|/|Z2| e j( q1 - q2 ) . The number r is called the real part of Z and x is called the imaginary part of Z. Z can be represented geometrically either in the rectangular form or in the polar form as a point on a two-dimensional complex plane. (pg. 576) e.g. Express 30 - j100 in the polar form. Answer: / -73.3o. e.g. (70 + j50) x (20 - j40) = j1800 or / -27.9o. e.g. 1 / (35 + j86) = j or / o. The conjugate of Z is Z* = r - jx. ZZ* = r2 + x2.

Połączenie szeregowe AC
XL= jwL +j Z L C R = R Z = R1 + jXL = |Z|/q1 ejq1 Diagram fazowy XC= j/wC Ohm’s Law for AC Circuits: VR = IRR; VL = ILZL ; VC = ICZC . Note that IL lags VL by 90o, and ZL = jXL (i.e. +90o in the complex plane). Similarly, IC leads VC by 90o , and ZC = -jXC (i.e. -90o in the complex plane). AC Series Circuits For an RL or RC circuit, Z = R + jX = |Z|/q, where |Z| = (R2 + X2)1/2 and q = +tan-1 (X/R). For an RLC circuit, ZT = R + j (XL -XC). When XL = XC , the circuit is said to be in resonance and ZT = R. In a series circuit with Z1, Z2, , Zn, then ZT = Z1 + Z Zn. The power dissipated in a reactive circuit is P = |I|2R. From the impedance triangle, R = |Z|cos q. Therefore, P = |I|2|Z| cos q = |V||I| cos q = (|V|2/|Z|) cos q. Also, the power factor, Fp = cos q = |R|/|Z|. e.g. Determine Z at 400 Hz for a 300 W resistor in series with an L = 5 mH and a C = 10 mF. Z = j(2xpx400x5x /(2xpx400x.1x10-6)) = j 27.2 = /-5.2o. e.g. A voltage source E = 45 V /30o is applied to a circuit with Z = j 75. Determine I and P. Z = 125 /36.9o; I = 45 /30o/(100 + j75) = .36 A /-6.9o; P = .362 x 100 or .362 x 125 x cos 36.9o = W. Z = R2 - jXC = |Z|/q2 + ejq2

Wzory dla obwodów szeregowych AC
Prawo Ohma: UR= IR; UL= jXL; UC = -jXC Obwód RLC: ZT = R+jX, gdzie X = XL - XC |ZT| = (R2+X2)1/2 ; q = tan-1 (X/R) R = ZT cos q; X = ZT sin q kiedy XL = XC, układ jest w rezonansie napięciowym a ZT = R, tzn. obwód ma impedancje całkowitą=rezystancji UL=-UC IT=UR/R Zależy od częstości : 1/wC=wL w=1/(LC)1/2 w=2pf

Rezonans w obw. szeregowym
W rezonansie: XL= XC ZT = R; Imax = E/R UL = UC = QsE, gdzie Qs = 2pfRL/R=1/(2pfRRC) Szerokość połówkowa rezonansu w częstotliwości: BW = f2 - f1 = fR/Qs (Hz) = R/(2pL). For a series RLC circuit, ZT = R + j(XL - XC), where R represents the sum of the generator resistance, RG , the inductor coil resistance, RL , and the series resistance, RS. A plot of ZT , and I (= E/ZT) versus f are shown above. At the resonant frequency, fR , XL = XC , ZT is minimum (note: purely resistive), and I is maximum. The quality factor of the resonant circuit is defined as Qs = (reactive power)/(average power) = XL/R = VL/E = (L/C)1/2/R. Since Qs is generally significantly greater than 1, VL and VC can be quite high at the resonant frequency. e.g. A 12 V ac source is applied to a series circuit consisting of a 100 W resistor, a 0.5 mH inductor, and a 1 nF capacitor. Calculate: fR , I, VL, and the circuit’s bandwidth at resonance. fR = 1/(2xpx(.5x10-3x1x10-9)1/2) = kHz; I = 12/100 = .12 A; VL = .12x2xpx225.1x103x.5x10-3 = 84.9 V; BW = 100/(2xpx.5x10-3) = 31.8 kHz. Note: The higher the value of Qs , the narrower the circuit’s bandwidth, i.e., it is more selective. Parallel resonant circuits have the same formula for fR , but other characteristics are opposite to those for the series resonant circuits. For example, at resonance, ZT is maximum while the current, IT , going into the parallel LC circuit is minimum (see pg of text). Resonant circuits are widely used in RF oscillator, filter and RF amplifier circuits.

Prawo napięć Kirchhoff’a
Prawo napięć Kirchhoff’a . Zasada dzielnika napięcia dla obwodów szeregowych Całkowita impedancja: ZT = Z1 + Z ZX ZN Prawo Kirch. napięciowe: Suma spadków i wzrostów napięć zespolonych w oczku wynosi zero . E - U1 - U UX UN = 0. Zasada dzielnika napięcia: UX = EZX / ZT Ohm’s law, KVL, and the voltage divider rule for ac circuits are similar to those for dc circuits except resistances are replaced by impedances and voltages and currents are phasor quantities. e.g. A voltage source, e = 90 sin 2500t, is applied to a 250 W resistor in series with an 80 mH inductor. Find ZT, VR and VL. ZT = j 2500x80x10-3 = j200 W or W/38.7o; VR = 90x.707x250/ /38.7o = 49.7V/-38.7o ; VL = 90x.707x200 /90o/ /38.7o = 39.7 V/51.3o. e.g. A voltage source, E = 50 V /30o, is applied across two series impedances, Z1 = j200 W and Z2 = 340 W/50o. Determine: V1 and V2. Draw the voltage phasor diagram. ZT = j /50o = j60.46 W; V1 = (100 - j200)x50/30o/ZT = 34.5 V/-44.2o ; V2 = 340/50ox50/30o/ZT = 52.4 V/69.3o. j V2 E Voltage phasor diagram: V1 -j

Obwody R, L, C (połączenie równoległe)
j E R C L I T g I C E g + I R I L e Diagram fazowy g i Phasor diagram L i For a purely resistive circuit, current and voltage are in phase. For a purely inductive circuit, current lags voltage by 90o. Reactance, XL = 2pfL, represents the opposition that inductance presents to current for an ac source. XL increases proportionally with frequency. For a purely capacitive circuit, current leads voltage by 90o. Reactance, XC = 1/(2pfC), is inversely proportional to frequency. The opposition that a circuit element presents to current in the phasor domain is defined as its impedance. Z = V/I = V/I /q or |Z| /q. e.g. A 20 V, 60 Hz source is applied across a 100 mH inductor. What is the current flowing through it? XL = 2xpx60x100x10-3 = 37.3 W; therefore, IL = 20/37.7 = .53 A. e.g. Given vC = 40 sin (2400t + 20o) V across a 5 mF capacitor, determine iC. XC = 1/(2400x5x10-6) = 83.3 W; therefore, iC = (40/83.3) sin (2400t +20o + 90o) = .48 sin (2400t + 110o) A. e.g. The current through a 10 mF capacitor is 30 mA with a frequency of 60 Hz. What is the voltage across the capacitor? XC = 1/(2xp x60x10x10-6) = W; therefore, V = 30x10-3x265.3 = 7.96 V. C Eg = IRZR = ILZL = ICZC 2p t ZR = R p i R ZL = jXL = jwL = wL ej90 ZC = -jXC = -j/ (wC)=(1/ wC) e -j90 Przebiegi Waveforms w = 2p f

Rezonans w obw. równoległym
C L I T g W rezonansie: XL= XC (YL=1/XL= YC=1/XC) YT = Ymin= 1/R; Imin = EYmin I L = IC = QsImin, gdzie dobroć Qs Qs = R/(2pfRL)=2pfRRC Szerokość połówkowa rezonansu w częstotliwości: BW = f2 - f1 = fR/Qs (Hz) = 1/(2pRC). IT I min For a series RLC circuit, ZT = R + j(XL - XC), where R represents the sum of the generator resistance, RG , the inductor coil resistance, RL , and the series resistance, RS. A plot of ZT , and I (= E/ZT) versus f are shown above. At the resonant frequency, fR , XL = XC , ZT is minimum (note: purely resistive), and I is maximum. The quality factor of the resonant circuit is defined as Qs = (reactive power)/(average power) = XL/R = VL/E = (L/C)1/2/R. Since Qs is generally significantly greater than 1, VL and VC can be quite high at the resonant frequency. e.g. A 12 V ac source is applied to a series circuit consisting of a 100 W resistor, a 0.5 mH inductor, and a 1 nF capacitor. Calculate: fR , I, VL, and the circuit’s bandwidth at resonance. fR = 1/(2xpx(.5x10-3x1x10-9)1/2) = kHz; I = 12/100 = .12 A; VL = .12x2xpx225.1x103x.5x10-3 = 84.9 V; BW = 100/(2xpx.5x10-3) = 31.8 kHz. Note: The higher the value of Qs , the narrower the circuit’s bandwidth, i.e., it is more selective. Parallel resonant circuits have the same formula for fR , but other characteristics are opposite to those for the series resonant circuits. For example, at resonance, ZT is maximum while the current, IT , going into the parallel LC circuit is minimum (see pg of text). Resonant circuits are widely used in RF oscillator, filter and RF amplifier circuits. 0.7 I min Y T 1/R f f f f 1 R 2

Obwody równoległe AC Całkowita admitancja (Y=1/Z): YT = Y1 + Y YN = 1/ZT Prawo prądów Kirchoffa: Suma prądów zespolonych wchodzących i wychodzących z węzła jest zero, tzn: IT - I1 - I IN = 0. Dzielnik prądu: IX = ZTIT /ZX lub YXIT /YT The admittance Y of any impedance is defined as a vector quantity which is the reciprocal of Z and it has units of siemens (S). In particular, YR = G/0o ; YL = -jBL or BL/-90o ; and YC = jBC or BC/90o, where B is known as the susceptance. The admittance diagram shows the different Y’s in the complex plane. The total admittance YT is the vector sum of the admittances of the network. For two impedances in parallel, ZT = Z1Z2 /(Z1 + Z2). e.g. A 1kW resistor is in parallel with a 500 W capacitive reactance. Find ZT . YT = 1/ /(-j500) = j0.002 S or 2.24 mS /63.4o; therefore, ZT = 1/YT = j400 W or W /-63.4o. Current divider rule for two branches in parallel : I1 = Z2IT/(Z1 + Z2). e.g. A voltage source, E = 25 V /0o, is applied to a circuit with a 100 W resistor, a 250 W inductive reactance, and a 180 W capacitive reactance, all in parallel. Determine: IT , and IC. ZT = 1/(1/ /(j250) + 1/(-j180)) = j15.19 W or 98.8 W/-8.8o; therefore, IT = 25/ZT = .25 A/8.8o; IC = .25/8.8ox 98.8/-8.8o/(-j180) = .14 A/90o. e.g. An impedance, Z = 60 + j75 W is in series with the parallel impedance circuit above. Find ZT. ZT = 60 + j j15.19 = j59.81 W or W/20.8o.

Wzory dla obwodów równoległych AC
YR = G ej0 , G=1/R; YL = -jBL , BL =1/wL, YL=Bl e-i YC = jBC , BC =wC, YC = BC ej90 gdzie G = konduktancja, oraz B = susceptancja = 1/X Całkowita impedancja 2 impedancje równoległe:

Moc w obwodach AC Dla obwodu z rezystancją, u oraz i są w fazie.
UmIm Pmaks i p(t) UmIm Pśred + + 2 R v e _ _ t ½T i T v Dla obwodu z rezystancją, u oraz i są w fazie. Średnia wartość mocy (lub rms) rzeczywistej lub czynnej, P = URIR = 1/2 UmIm (W), gdzie UR and IR są wartościami rms , a Um i Im są wartościami w maksimum.

Moc na cewce i UI p(t) + i u + v e L _ t _ ¼T T u wyprzedza i o 90o
W pierwszej ćwiartce, p = ui jest dodatnie, a więc moc wpływa na cewkę. Energia magaz. Energia wydziel. Energia magazyn. Energia wydziel. W czasie 2 ćwiartki, p jest ujemne i moc zmagazynowana w L jest zwalniana z powrotem do obwodu

Moc na kondensatorze i UI p(t) u + + i e C v _ _ t ¼T T
u opóźnia się o 90o -UI Energia wydziel. Energia wydziel. W czasie 1 ćwiartki, p = ui jest dodatnie, a więc Moc magazynowana jest w C. Energia magaz. Energia magazy. W czasie 2 ćwiartki, p jest ujemne a energia zmagazynowana na C jest oddawana do obwodu.

Moc bierna Dla obwodów czysto pojemnościowych lub indukcyjnościowych (cewka), średnia moc uśrednionia po jednym okresie wynosi zero; tzn. nie odbiera się i nie traci się żadnej mocy Moc bierna płynąca do L i C: Indukcyjna, QL = ULIL = IL2XL = UL2/XL Pojemnościowa, QC = UCIC = IC2XC = UC2/XC Przez konwencję moc bierną na pojemności definiuje się ujemnie.

Moc w złożonych systemach
Dla obwodów składających się z k rezystorów, m indukcyjności, i n pojemności połączonych, szeregowo, lub równolegle moc czynną i bierną można obliczyć : Całkowita moc czynna, PT = P1 + P Pk Całkowita moc bierna, QT = QLT - QCT, gdzie QLT = QL1 + QL QLm, pochodzi od L a QCT = QC1 + QC QCn od C

Trójkąt mocy dla obwodów RLC
I |S| = EI _ + QT = QL-QC VR + + q UL E _ P = IUR _ _ + UC Moc zespolona, S = P + QT= EI* = I2Z = E2/Z = |S|e jq (VA) gdzie Z = R + j (XL - XC) Moc czynna, P = |S| cos q lub EI cos q (W) Moc bierna, QT = |S| sin q lub EI sin q (VAR) The power that appears to flow into a complex load is the apparent power, S = VI = I2Z = V2/Z. Most electrical equipment are rated in kVA or MVA. The relationship between the apparent, real and reactive power for an inductive circuit is given by the power triangle above. For a capacitive circuit, it will be similar except QL is replaced by -QC (pg. 617). For an RLC circuit, the reactive power, Q will be the resultant of QL and -QC and S = P + Q =VI*, where I* is the conjugate of current I. e.g. A 120 V ac source is applied to a circuit with series-parallel loads as follow: S1 = 35 + j 47 VA, S2 = 59 - j93 VA, and Q3 = 26 VAR (ind.). Draw the power triangle and determine IT. PT=35+59=94 W; QT= = -20 VAR; ST = 94 -j20 = 96.1 / -12o VA; IT = 96.1/120 = .8 A. Note: ST must be added vectorially. From the power triangle, it is obvious that P = S cos q or VI cos q, and Q = S sin q. cos q is defined as the power factor. The power factor angle q = cos-1 (P/S). Since I lags V in an RL circuit, its power factor is lagging while for an RC circuit, the power factor is leading. Electrical apparatus are rated in VA rather than W in order to take into account the extra current required for any reactive load.

Czynnik Mocy Czynnik mocy, Fp = cos q = P/S
Faza czynnika mocy, q = cos-1 (P/S) W obwodach RL, czynnik mocy się opóźnia ponieważ natężenie się opóźnia , w czynnik mocy wyprzedza. Urządzenia elektryczne pracujace w AC mają moc wyrażana w VA a nie w W aby uwzględnić extra prąd potrzebny do obciążeń typu indukcyjnego.

Poprawka ze względu na czynnik mocy
Poprawiony czynnik mocy Poprawka na czynnik mocy is używana jest poprzez dodanie do obwodu reaktancji odwrotnego typu do danej w obwodzie. W większości elektrowni gdzie obciążenie są typu indukcyjnego (cewki), dodaje się pojemności równolegle aby zredukować prąd ze źródła i obniżyć moc bierną (rezonans prądów). An utility company would impose a penalty on industrial customers if their power factor drops below a prescribed value because the load current would be higher which necessitates the use of larger conductors, larger transformers, etc. e.g. A factory using a 440 V, 60 Hz motor draws 100 kW with a 0.75 lagging power factor. Determine the reactive power and the value of capacitance required to raise the power factor (a) to .85, and (b) to 1.0. q = cos = 41.4o ; therefore, QL = 100 k tan 41.4o = 88.2 kVAR (ind.); (a) q’ = cos = 31.79o; therefore, QT = 100k tan 31.79o = kVAR = QL + QC. Thus, QC = kVAR, and XC = 4402/26.22 k = 7.39 W Therefore, C = 1/(2xpx60x7.39) = 359 mF. (b) To get PF =1.0, QC = QL; therefore, XC = 4402/88.2 k = W Hence, C = 1/(2px60x2.195) = 1208 mF. AC power measurements are done with wattmeters and varmeters (read pg ).

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