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6/15/2011 Chapter 31 A horizontal power line carries a current of 5000 A from south to north. Earth's magnetic eld (60 µT) is directed towards north and inclined down-ward 50 degrees to the horizontal. Find the magnitude and direction of the magnetic force on 100 m of the line due to the Earth's field. (1)23 N, west (2) 23 N, east (3) 30N, west (4) 30N, east (5) none of these Electromagnetic Oscillations and Alternating Current In this chapter we will cover the following topics: -Electromagnetic oscillations in an LC circuit -Alternating current (AC) circuits with capacitors -Resonance in RCL circuits -Power in AC-circuits -Transformers, AC power transmission I 50 N (31 - 1) B Suppose this page is perpendicular to a uniform magnetic field and the magnetic flux through it is 5Wb. If the page is turned by 30◦ around an edge the flux through it will be: A car travels northward at 75 km/h along a straight road in a region where Earth’s magnetic field has a vertical component of 0.50 × 10−4 T. The emf induced between the left and right side, separated by 1.7m, is: A. 2.5Wb B. 4.3Wb C. 5Wb D. 5.8Wb E. 10Wb A. 0 B. 1.8mV C. 3.6mV D. 6.4mV E. 13mV LC Oscillations L C d 2q 1 d 2q 1 + q = 0→ 2 + q = 0 (eqs.1) 2 dt C dt LC C This is a homogeneous, second order, linear differential equation L L The circuit shown in the figure consists of a capacitor C and an inductor L. We give the capacitor an initial chanrge Q and then abserve what happens. The capacitor which we have encountered previously. We used it to describe will discharge through the inductor resulting in a time the simple harmonic oscillator (SHO) d 2x + ω 2 x = 0 (eqs.2) dt 2 with solution: x(t ) = X cos(ωt + φ ) q (t ) = Q cos (ωt + φ ) dependent current i. We will show that the charge q on the capacitor plates as well as the current i 1 LC The total energy U in the circuit is the sum of the energy stored in the electric field in the inductor oscillate with constant amplitude at an angular frequency ω = of the capacitor and the magnetic field of the inductor. U = U E + U B = q 2 Li 2 + . 2C 2 The total energy of the circuit does not change with time. Thus dU =0 dt dU q dq di dq di d 2 q d 2q 1 = + Li = 0. i = → = 2 →L 2 + q=0 dt C dt dt dt dt dt dt C (31 - 2) ω= 1 LC If we compare eqs.1 with eqs.2 we find that the solution to the differential equation that describes the LC-circuit (eqs.1) is: q(t ) = Q cos (ωt + φ ) The current i = where ω= 1 , and φ is the phase angle. LC dq = −ωQ sin (ωt + φ ) dt (31 - 3) 1 6/15/2011 2 L 2 4 t = 3T / 8 2 q Q = cos 2 (ωt + φ ) 2C 2C The energy stored in the magnetic field of the inductor UE = 3 t =T /4 t =T /8 The energy stored in the electric field of the capacitor C 8 6 t = 7T / 8 1 Q2 T 3T at t = 0, , T , ,... 2C 2 2 Q2 T 3T 5T The energy of the magnetic field has a maximum value of at t = , , ,... 2C 4 4 4 Note : When U E is maximum U B is zero, and vice versa (31 - 4) t = 5T / 8 3 7 5 The energy of the electric field has a maximum value of 7 2 4 6 t = 3T / 4 8 (31 - 5) q (t ) Damped oscillations in an RCL circuit dU = −i 2 R dt 2 2 q Li dU q dq di + U = UE +UB = → = + Li = −i 2 R 2C 2 dt C dt dt t =T /2 t =0 Q2 Q2 2 2 U = cos ( ωt + φ ) + sin ( ωt + φ ) = 2C 2C The total energy is constant; energy is conserved If we add a resistor in an RL cicuit (see figure) we must modify the energy equation because now energy is 5 1 Li 2 Lω 2Q 2 Q2 = sin 2 ( ωt + φ ) = sin 2 ( ωt + φ ) UB = 2 2 2C The total energy U = U E + U B q (t ) = Qe− Rt / 2 L cos (ω ′t + φ ) Qe− Rt / 2 L Q q (t ) ω′ = being dissipated on the resistor. 2 Qe− Rt / 2 L 2 dq di d q d q dq 1 → = 2 → L 2 + R + q = 0 This is the same equation as that dt dt dt dt dt C d 2x dx of the damped harmonics oscillator: m 2 + b + kx = 0 which has the solution: dt dt i= −Q x(t ) = xme −bt / 2 m cos (ω ′t + φ ) The angular frequency ω ′ = k b2 − m 4m 2 For the damped RCL circuit the solution is: q(t ) = Qe − Rt / 2 L cos ( ω ′t + φ ) The angular frequency ω ′ = The equations above describe a harmonic oscillator with an exponetially decaying amplitude Qe − Rt / 2 L . The angular frequency of the damped oscillator 1 R2 1 − of the is always smaller than the angular frequency ω = LC 4 L2 LC 2 1 R undamped oscillator. If the term we can use the approximation ω ′ ≈ ω 4 L2 LC ω′ = 1 R2 − LC 4 L2 (31 - 7) (31 - 6) Three Simple Circuits Our objective is to analyze the circuit shown in the figure (RCL circuit). The discussion will be greatly simplified if we examine what happens if we connect each of the three elements (R, C , and L) separately Alternating Current A battery for which the emf is constant generates a current that has a constant direction. This type of current is known as "direct current " or "dc" E = Em sin ωt 1 R2 − LC 4 L2 In chapter 30 we encountered a different type to an ac generator. of sourse (see figure) whose emf is: A convention E = ω NAB sin ωt = Em sin ωt where Em = ω NAB, A is the area of the generator From now on we will use the standard notation for ac circuit analysis. Lower case letters will be used to indicate the windings, N is the number of the windings, ω is the angular frequency of the rotation of the windings, and B is the magnetic field. This type of generator L C is known as "alternating current" or "ac" because the emf as well as the current change direction with a frequency f = 2πω. In the US f = 60 Hz. will be used to indicate the constant amplitudes of ac quantities. Example: The capacitor charge in an LC circuit was written as: q = Q cos ( ωt + φ ) Almost all commercial electrical power used today is ac even though the analysis of ac circuits is more complicated than that of dc circuits. The symbol q is used for the instantaneous value of the charge The reasons why ac power was adapted will be discussed at the end of this The symbol Q is used for the constant amplitude of q chapter. (31 - 8) instantaneous values of ac quantities. Upper case letters (31 - 9) 2 6/15/2011 VR = I R R A resistive load In fig.a we show an ac generator connected to a resistor R From KLR we have: E − iR R = 0 → iR = E Em = sin ωt R R Em R across R is equal to Em sin ωt The current amplitude I R = The voltage vR The voltage amplitude is equal to Em The relation between the voltage and current amplitudes is: VR = I R R The resistor voltage vR and the resistor current iR are represented In fig.b we plot the resistor current iR and the resistor voltage vR as function of time t. Both quantities reach their maximum values at the same time. We say that voltage and current are in phase. (31 - 10) 2 2 ⟨ P⟩ = E Em Vrms ≡ m 2R 2 ⟨ P⟩ = 2 ⟨ P⟩ = 2 1 P (t ) dt T ∫0 P= Em sin 2 ωt → R 2. The length of each phasor is proportional to tha ac quantity amplitude 3. The projection of the phasor on the vertical axis gives the instantaneous value of the ac quantity. 4. The rotation angle for each phasor is equal to the phase of the ac quantity (ωt in this example) (31 - 11) Em 1 sin 2 ωtdt R T ∫0 T 2 → ⟨ P⟩ = In fig.a we show an ac generator connected to a capacitor C XC q From KLR we have: E − C = 0 → C qC = EC = EmC sin ωt T 1 1 sin 2 ωtdt = 2 T ∫0 by rotating vectors known as phasors using the following conventions: 1. Phasors rotate in the counterclockwise direction with angular speed ω A capacitive load Average Power for R T A convenient method for the representation of ac quantities is that of phasors Em 2R dqC = EmCω cos ωtdt = sin ( ωt + 90° ) dt The voltage amplitude VC equal to Em ω O iC = 1 XC = ωC VC 1/ ωC The quantity X C = 1 / ωC is known as the The current amplitude IC = ωCVC = capacitive reactance We define the "root mean square" (rms) value of V as follows: In fig.b we plot the capacitor current iC and the capacitor 2 Em V2 → ⟨ P⟩ = rms The equation looks the same R 2 as in the DC case. This power appears as heat on R voltage vC as function of time t. The current leads the Vrms ≡ (31 - 12) An inductive load Average Power for C PC = 0 2 E P = VC I C = m sin ωt cos ωt XC 2 2sin θ cos θ = sin 2θ T ⟨ P⟩ = → P= 2 voltage by a quarter of a period. The voltage and (31 - 13) current are out of phase by 90 °. Em sin 2ωt 2XC T E 1 1 P (t ) dt = m sin 2ωtdt = 0 T ∫0 2 X C T ∫0 Note : A capacitor does not dissipate any power on the average. In some parts of the cycle it absorbes energy from the ac generator but at the rest of the cycle it gives the energy back so that on the average no power is used! In fig.a we show an ac generator connected to an inductor L di di E E From KLR we have: E − L L = 0 → L = = m sin ωt dt dt L L E E E iL = ∫ diL = ∫ m sin ωtdt = − m cos ωtdt = m sin (ωt − 90° ) L ωL ωL The voltage amplitude VL equal to Em XL V The current amplitude I L = L ωL ω The quantity X L = ω L is known as the O inductive reactance X L = Lω In fig.b we plot the inductor current iL and the inductor voltage vL as function of time t. The current lags behind the voltage by a quarter of a period. The voltage and current are out of phase by 90 °. (31 - 14) (31 - 15) 3 6/15/2011 PL = 0 Average Power for L SUMMARY 2 Power P = VL I L = − Em sin ωt cos ωt XL 2 2sin θ cos θ = sin 2θ → P=− T ⟨ P⟩ ≡ 2 Em sin 2ωt 2X L T E 1 1 P (t ) dt = − m sin 2ωtdt = 0 T ∫0 2 X L T ∫0 Circuit element Average Power Resistor R ⟨ PR ⟩ = Reactance R Note : A inductor does not dissipate any power Capacitor on the average. In some parts of the cycle it absorbes energy from the ac generator but at the rest of the cycle C ⟨ PC ⟩ = 0 XC = Inductor L ⟨ PL ⟩ = 0 X L = ωL it gives the energy back so that on the average no power is used! Voltage amplitude Current is in phase with the voltage 2 Em 2R Phase of current 1 ωC VR = I R R Current leads voltage by a quarter of a period VC = I C X C = Current lags behind voltage by a quarter of a period VL = I L X L = I Lω L (31 - 16) IC ωC (31 - 17) (31 - 18) A i = I sin (ωt − φ ) B Z = R2 + ( X L − X C ) O The series RCL circuit Em Z Kirchhoff's loop rule (KLR) for the RCL circuit: E = vR + vC + vL . This equation An ac generator with emf E = Em sin ωt is connected to is represented in phasor form in fig.d. Because VL and VC have opposite directions an in series combination of a resistor R, a capacitor C we combine the two in a single phasor VL − VC . From triangle OAB we have: and an inductor L, as shown in the figure. The phasor Em2 = VR2 + (VL − VC ) = ( IR ) + ( IX L − IX C ) = I 2 R 2 + ( X L − X C ) → Em I= The denominator is known as the "impedance" Z 2 R2 + ( X L − XC ) 2 for the ac generator is given in fig.c. The current in i = I sin (ωt − φ ) I= 2 this circuit is described by the equation: i = I sin ( ωt − φ ) The current i is common for the resistor, the capacitor and the inductor The phasor for the current is shown in fig.a. In fig.c we show the phasors for the voltage vR across R, the voltage vC across C , and the voltage vL across L. of the circuit. I= The voltage vR is in phase with the current i. The voltage vC lags behind 2 2 Z = R 2 + ( X L − X C ) → The current amplitude I = 2 Em Z Em 1 R + ωL − ωC 2 2 the current i by 90°. The voltage vL leads ahead of the current i by 90°. 2 (31 - 19) i = I sin (ωt − φ ) (31 - 20) 1 XC = ωC A B Z = R2 + ( X L − X C ) tan φ = O 2 X L − XC R X L = ωL 1. Fig.a and b: X L > X C → φ > 0 V −V IX − IX C X L − X C From triangle OAB we have: tan φ = L C = L = VR IR R The current phasor lags behind the generator phasor. The circuit is more We distinguish the following three cases depending on the relative values inductive than capacitive of X L and X L . 1. X L > X C → φ > 0 The current phasor lags behind the generator phasor. The circuit is more inductive than capacitive 2. X C > X L → φ < 0 The current phasor leads ahead of the generator phasor The circuit is more capacitive than inductive 3. X C = X L → φ = 0 The current phasor and the generator phasor are in phase 2. Fig.c and d: X C > X L → φ < 0 The current phasor leads ahead of the generator phasor. The circuit is more capacitive than inductive 3. Fig.e and f: X C = X L → φ = 0 The current phasor and the generator phasor are (31 - 21) in phase 4 6/15/2011 ω= 1 LC I res = 2 Pavg = I rms R Resonance Em R (31 - 23) Power in an RCL ciruit the angular frequency ω of the ac generator can be varied continuously. The current amplitude We already have seen that the average power used by a capacitor and an inductor is equal to zero. The power on the average is consumed by the resistor. in the circuit is given by the equation: Em I= 1 R2 + ω L − ωC 2 The current amplitude This occurs when ω = The instantaneous power P = i 2 R = I sin (ωt − φ ) R 2 T has a maximum when the term ω L − The average power Pavg = 1 =0 ωC Em R A plot of the current amplitude I as function of ω is shown in the lower figure. This plot is known as a "resonance curve" (31 - 22) Transmission lines Εrms =735 kV , I rms = 500 A (31 - 24) The transformer is a device that can change the voltage amplitude of any ac signal. It consists of two coils with different number R = 220Ω of turns wound around a common iron core. l = 1000 km Power Station Energy Transmission Requirements The resistance of the power line R = 2 Heating of power lines Pheat = I rms R ρl . R is fixed (220 Ω in our example) A This parameter is also fixed (31 - 25) The transformer home T2 T1 The coil on which we apply the voltage to be changed is called the "primary" and it has N P turns. The transformer output appears on the second coils which is known as the "secondary" and has N S turns. The role of the iron core is to insure that the magnetic field lines from one coil also pass through the second. We assume that if voltage equal to VP is applied across the primary then a voltage VS appears ( 55 MW in our example) Power transmitted Ptrans = Erms I rms power consumed by the circuit is maximum when φ = 0 Step-down transformer 110 V Step-up transformer 1 Pdt T ∫0 1 T I 2R 2 2 Pavg = I 2 R ∫ sin (ωt − φ ) dt = = I rms R 2 T 0 E R Pavg = I rms RI rms = I rms R rms = I rmsErms = I rms Erms cos φ Z Z The term cosφ in the equation above is known as the "power factor" of the circuit. The average 1 LC The equation above is the condition for resonance. When its is satisfied I res = (368 MW in our example) In our example Pheat is almost 15 % of Ptrans and is acceptable To keep Pheat we must keep I rms as low as possible. The only way to accomplish this is by increasing Erms . In our example Erms = 735 kV. To do that we need a device that can change the amplitude of any ac voltage (either increase or decrease) VS V = P NS NP Pavg = I rms Erms cos φ In the RCL circuit shown in the figure assume that dΦ P dB = − NP A (eqs.1) dt dt d ΦS dB Φ S = N S BA → VS = − = −NS A (eqs.2) dt dt If we divide equation 2 by equation 1 we get: Φ P = N P BA → VP = − dB − NS A VS dt = N S → VS = VP = VP − N A dB NP NS NP P dt on the secondary coil. We also assume that the magnetic field through both coils is equal to B and that the iron core has cross sectional area A. The magnetic flux dΦP dB = −NP A (eqs.1) dt dt dΦS dB = −NS A The flux through the secondary Φ S = N S BA → VS = − dt dt through the primary Φ P = N P BA → VP = − IS IP VS V = P NS NP (eqs.2) IS NS = I P NP We have that: VS V = P NS NP → VS N P = VP N S (eqs.1) If we close switch S in the figure we have in addition to the primary current I P a current I S in the secondary coil. We assume that the transformer is "ideal" The voltage on the secondary VS = VP NS NP If N S > N P → NS > 1 → VS > VP We have what is known a "step up" transformer NP If N S < N P → NS < 1 → VS < VP We have what is known a "step down" transformer NP Both types of transformers are used in the transport of electric power over large distances. (31 - 26) i.e. it suffers no losses due to heating then we have: VP I P = VS I S (eqs.2) VP I P V I If we divide eqs.2 with eqs.1 we get: = S S → I P N P = I S NS VP N S VS N P IS = NP IP NS In a step-up transformer (N S > N P ) we have that I S < I P In a step-down transformer (N S < N P ) we have that I S > I P (31 - 27) 5 6/15/2011 Hitt A generator supplies 100 V to the primary coil of a transformer. The primary has 50 turns and the secondary has 500 turns. The secondary voltage is: A. 1000 V B. 500 V C. 250 V D. 100 V E. 10V hitt The main reason that alternating current replaced direct current for general use is: A. ac generators do not need slip rings B. ac voltages may be conveniently transformed C. electric clocks do not work on dc D. a given ac current does not heat a power line as much as the same dc current E. ac minimizes magnetic effects 6