H - Fall 0 Electroagnetic Oscillations and Alternating urrent ectures 8-9 hapter 3 (Halliday/esnick/Walker, Fundaentals of hysics 8 th edition)
hapter 3 Electroagnetic Oscillations and Alternating urrent In this chapter we will cover the following topics: -Electroagnetic oscillations in an circuit -Alternating current (A) circuits with capacitors -esonance in circuits -ower in A circuits -ransforers, A power transission
Oscillations he circuit shown in the figure consists of a capacitor and an inductor. We give the capacitor an initial charge Q and then observe what happens. he capacitor will discharge through the inductor, resulting in a tiedependent current i. We will show that the charge q on the capacitor plates as well as the current i in the inductor oscillate with constant aplitude at an angular frequency. he total energy U in the circuit is the su of the energy stored in the electric field q i of the capacitor and the agnetic field of the inductor: U UE UB. du he total energy of the circuit does not change with tie. hus 0. dt du q dq di dq di d q dq i 0. i q 0 dt dt dt dt dt dt dt 3
qt () Qco s t dq dq q 0 ( ) q 0 eq. dt dt his is a hoogeneous, second order, linear differential equation that we have encountered previously. We used it to describe the siple haronic oscillator (HO). d x x 0 ( eq. ) dt With solution: xt ( ) Xcos( t) If we copare eq. with eq. we find that the solution to the differential equation that describes the circuit (eq. ) is: qt ( ) Qcos t where, and is the phase angle. dq he current i Qsin t. dt 4
he energy stored in the electric field of the capacitor: q Q UE cos t he energy stored in the agnetic field of the inductor: i Q Q UB sin t t he total energy U U U : sin E Q Q U cos t sin t he total energy is constant; energy is conserved. B Q 3 he energy of the electric field has a axiu value of at t 0,,,,... Q 3 5 he energy of the agnetic field has a axiu value of at t,,,... 4 4 4 Note : When U is axiu, U is zero, and vice versa. E B 5
t 0 t /8 3 t /4 4 3 /8 t 5 5 t / 4 3 6 6 5 /8 t 3 /4 t 7 /8 t 7 8 7 8 6
dq di dq dq dq Daped Oscillations in an ircuit If we add a resistor in an circuit (see figure) we ust odify the energy equation, because now energy is du being dissipated on the resistor: i. dt q i du q dq di U UE UB i i dt dt dt i q 0. his is the sae equation as that dt dt dt dt dt d x dx of the daped haronics oscillator: b kx0, which has the solution: dt dt x t x e t bt / () cos. he angular freque For the daped circuit the solution is: ncy k b 4. t/ q() t Qe cos t. he angular frequency. 4 7
Q qt () / Qe t qt () / () t qt Qe cos t 4 Q / Qe t he equations above describe a haronic oscillator with an exponentially decaying aplitude Qe 4 t/. he angular frequency of the daped oscillator is always saller than the angular frequency of the undaped oscillator. If the ter we can use the approxiation. 4 8
Alternating urrent A battery for which the ef is constant generates a current that has a constant direction. his type of current is known as " direct current " or " dc. " sin t In hapter 30 we encountered a different type of source (see figure) whose ef is: NABsint sin t, where NAB, A is the area of the generator windings, N is the nuber of the windings, is the angular frequency of the rotation of the windings, and B is the agnetic field. his type of generator is known as " alternating current" or " ac" because the ef as well as the current change direction with a frequency f. In the U.. f 60 Hz. Alost all coercial electrical power used today is ac even though the analysis of ac circuits is ore coplicated than that of dc circuits. he reasons why ac power was adapted will be discussed at the end of this chapter. 9
hree iple ircuits Our objective is to analyze the circuit shown in the figure ( circuit). he discussion will be greatly siplified if we exaine what happens if we connect each of the three eleents (,, and ) separately to an ac generator. A onvention Fro now on we will use the standard notation for ac circuit analysis. owercase letters will be used to indicate the instantaneous values of ac quantities. Uppercase letters will be used to indicate the constant aplitudes of ac quantities. Exaple: he capacitor charge in an circuit was written as q Qcos t. he sybol q is used for the instantaneous value of the charge. he sybol Q is used for the constant aplitude of q. 0
V I A esistive oad In fig. a we show an ac generator connected to a resistor. Fro K we have: i 0 i sin t. he current aplitude is I. he voltage v across is equal to sin t. he voltage aplitude is equal to. he relation between the voltage and current aplitudes is V I. In fig. b we plot the resistor current i and the resistor voltage v as functions of ti e t. Both quantities reach their axiu values at the sae tie. We say that voltage and current are in phase.
A convenient ethod for the representation of ac quantities is that of phasors. he resistor voltage v and the resistor current i are represented by rotating vectors known as phasors using the following conventions:. hasors rotate in the counterclockwise direction with angular speed.. he length of each phasor is proportional to the ac quantity aplitude. 3. he projection of the phasor on the vertical axis gives the instantaneous value of the ac quantity. 4. he rotation angle for each phasor is equal to the phase of the ac quantity ( t in this exaple).
Vrs Average ower for tdt () sin t 0 sin 0 tdt 0 sin tdt We define the "root ean square" (rs) value of V as follows: Vrs Vrs. he equation looks the sae as in the dc case. his power appears as heat on. 3
A apacitive oad In fig. a we show an ac generator connected to a capacitor. q Fro K we have: 0 q sint O dq i cost sin t90 dt he voltage aplitude V is equal to. V he current aplitude is I V. / he quantity X / is known as the capacitive reactance. In fig. b we plot the capacitor current and the capacitor voltage v as functions of tie t. he current leads the voltage by a quarter of a period. he voltage and current are out of phase by 90. i X X 4
Average ower for 0 VI sintcos t X sincos sin 0 0 X sin t () t dt = sintdt 0 X Note : A capacitor does not dissipate any power on the average. In soe parts ofthe cycle it absorbs energy fro the ac generator, but at the rest of the cycle it gives the energy back so that on the average no power is used! 5
An Inductive oad In fig. a we show an ac generator connected to an inductor. di di Fro K we have: 0 sint dt dt i di sintdt cost sin t90 he voltage aplitude V is equal to. X V he current aplitude is I. he quantity X is known as the O inductive reactance. X In fig. b we plot the inductor current i and the inductor voltage v as functions of tie t. he current lags behind the voltage by a quarter of a period. he voltage and current are out of phase by 90. 6
0 Average ower for ower VI sintcos t X sincos sin 0 0 X sin t () t dt = sintdt 0 X Note : An inductor does not dissipate any power on the average. In soe parts of the cycle it absorbs energy fro the ac generator, but at the rest of the cycle it gives the energy back so that on the average no power is used! 7
UMMAY ircuit Eleent Average ower eactance hase of urrent Voltage Aplitude esistor urrent is in phase with the voltage V I apacitor 0 X urrent leads voltage by a quarter of a period V I X I Inductor 0 X urrent lags behind voltage by a quarter of a period V I X I 8
i I t sin he eries ircuit An ac generator with ef sint is connected to an in-series cobination of a resistor, a capacitor, and an inductor, as shown in the figure. he phasor for the ac generator is given in fig. c. he current in this circuit is described by the equation: i I sin t. he current i is coon for the resistor, the capacitor, and the inductor. he phasor for the current is shown in fig. a. In fig. b we show the phasors for the voltage v across, the voltage v across, and the voltage v across. he voltage v is in phase with the current i. he voltage v lags behind the current i by 90. he voltage v leads ahead of the current i by 90. 9
A B i I t sin Kirchhoff's loop rule (K) for the circuit: v v v. his equation O is represented in phasor for in fig. d. Because V and V have opposite directions we cobine the two in a single phasor V X X of the circuit. V Z X X he current aplitude I. Fro triangle OABwe have: V V V I IX IX I X X I. he denoinator is known as the " ipedance" Z I. Z Z X X. I Z 0
i I t sin X A B Z X X O X X tan X V V IX IX X X Fro triangle OAB we have: tan. V I We distinguish the following three cases depending on the relative values of X and X.. X X 0 he current phasor lags behind the generator phasor. he circuit is ore inductive than capacitive.. X X 0 he current phasor leads ahead of the generator phasor. he circuit is ore capacitive than inductive. 3. X X 0 he current phasor and the generator phasor are in phase.
. Figs. a and b: X X 0 he current phasor lags behind the generator phasor. he circuit is ore inductive than capacitive.. Figs. c and d: X X 0 he current phasor leads ahead of the generator phasor. he circuit is ore capacitive than inductive. 3. Figs. e and f: X X 0 he current phasor and the generator phasor are in phase.
I res esonance In the circuit shown in the figure, assue that the angular frequency of the ac generator can be varied continuously. he current aplitude in the circuit is given by the equation I. he current aplitude has a axiu when the ter 0. his occurs when. he equation above is the condition for resonance. When it is satisfied, Ires. A plot of the current aplitude I as a function of is shown in the lower figure. his plot is known as a " resonance curve. " 3
I I cos avg rs rs avg rs ower in an ircuit We already have seen that the average power used by a capacitor and an inductor is equal to zero. he power on the average is consued by the resistor. he instantaneous power i Isin t. he average power avg dt. I avg I sin t dt Irs 0 rs avg IrsIrs Irs Irsrs Irsrs cos Z Z he ter cos in the equation above is known as the " power factor" of the circuit. he average power consued by the circuit is axiu 4 when 0. 0
I V V = 40 V V V tan=(x X )/=(88 44)/300= 0.847 = 40.3 5
ransission lines rs =735 kv, I rs = 500 A tep-up transforer tep-down transforer 0 V Hoe ower tation = 0Ω 000 k Energy ransission equireents he resistance of the power line. is fixed (0 in our exaple). A Heating of power lines I. his paraeter is also fixed (55 MW in our exaple). heat heat heat rs ower transitted I (368 MW in our exaple). trans rs rs In our exaple is alost 5 % of and is acceptable. rs trans o keep we ust keep I as low as possible. he only way to accoplish this is by increasing. In our exaple 735 kv. o do that we need a device rs that can change the aplitude of any ac voltage (either increase or decrease). rs 6
he ransforer he transforer is a device that can change the voltage aplitude of any ac signal. It consists of two coils with a different nuber of turns wound around a coon iron core. he coil on which we apply the voltage to be changed is called the " priary" and it has N turns. he transforer output appears on the second coil, which is known as the "secondary" and has N turns. he role of the iron core is to ensure that the agnetic field lines fro one coil also pass through the second. We assue that if voltage equal to V is applied across the priary then a voltage V appears on the secondary coil. We also assue that the agnetic field through both coils is equal to B and that the iron core has cross-sectional area A. he agnetic flux d db through the priary NBAV NA ( eq. ). dt dt d db he flux through the secondary NBA V NA ( eq. 7 ). dt dt
V N V N d db NBAV NA ( eq. ) dt dt d db NBA V NA ( eq. ) dt dt If we divide equation by equation we get: V V db N A N dt V V db N N A N N dt. N he voltage on the secondary V V. N N If N N V V, we have what is known as a " step up" transforer. N N If N N V V, we have what is known as a " step down" transforer. N Both types of transforers are used in the transport of electric power over large distances. 8
I I V N V N I N I N We have that: VN VN V N V N (eq. ). If we close switch in the figure we have in addition to the priary current a current I in the secondary coil. We assue that the transforer is " ideal, " i.e., it suffers no losses due to heating. hen we have: VI VI (eq. ). If we divide eq. with eq. we get: I N N I In a step-up transforer ( N N ) we have that I I. In a step-down transforer ( N VI VN VI VN I N N ) we have that I I. I N. I 9