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Physics 169 Kitt Peak National Observatory Tuesday, April 19, 16 Luis anchordoqui 1 field, the induced emf varies sinusoidally with time and leads to an alternating current (AC), and provides a source of AC power. The symbol for an AC voltage source is 10.1 Alternating Current (AC) Voltage Driven RLC Circuits We learned that changing magnetic flux could induce an emf according to Faraday’s law 12.1 AC Sources An example of an AC source is In Chapter 10of we learned that changing magnetic If coil rotates in presence magnetic fieldflux could induce an emf according to Faraday’s law of induction. In particular, if a coil rotates in the presence of a magnetic field, the induced emfVvaries withω time and to an alternating current ,for leads (t)ofsinusoidally = V0 sin( t)varies induced emf sinusoidally with (12.1.1) time and leads to AC (AC), and provides a source AC power. The symbol an AC voltage source is Symbol for AC voltage source ☛ where the maximum value amplitude. The voltage varies between V0 and V0 isof ancalled An example AC sourcethe is (12.1.1) as a function of = V sin(−1. ω t) , A graph of voltage −V0 since a sine function varies between +1V (t)and 0 V (t) = V sin(!t) the maximum value V is called the amplitude. 0 The voltage varies between V and time is shown in Figurewhere 12.1.1. The phase of the voltage source is φV = ω t , (the phase −V since a sine function varies between +1 and −1. A graph of voltage as a function of time is shown in Figure 12.1.1. The phase of the voltage source is φ = ω t , (the phase constant is zero in Eq. (12.1.1)). 0 0 0 V constant is zero in Eq. (12.1.1)). Figure 12.1.1 Sinusoidal voltage source The sine function is periodic in time. This means that the value of the voltage at time t will be exactly the same at a later time t ′ = t + T where T is the period. The frequency, f , defined as f = 1/ T , has the unit of inverse seconds ( s-1 ), or hertz ( Hz ). The angular frequency is defined to be ω = 2π f . Figure 12.1.1 Sinusoidal voltage source When a voltage source is connected to a RLC circuit, energy is provided to compensate the energy dissipation in the resistor, and the oscillation will no longer damp out. The oscillations of charge, current and potential difference are called driven or forced oscillations. ! = 2⇡f = 2⇡/T The sine function is periodic in time. This means that the value of the voltage at time t After an initial “transient time,” an AC current will flow in the circuit as a response to the source. The current is also sinusoidal, T is the period. The frequency, will be exactly the samedriving at avoltage later time t ′ =intthe+circuit T where 1 f , defined unit ☛ as f = 1/ T , has the unit of inverse seconds ( s-1 ), or hertz12-2( Hz ). The angular frequency is defined to be ω = 2π f . s Tuesday, April 19, 16 ⌘ Hz 2 purely resistive circuit corresponds to infinite capacitance C = ∞ and 10.2 AC circuits L = 0 .)with a source and one circuit element Purely Resistive Load We’d like to find current through resistor IR (t) = I0,R sin(!t + R) Applying Kirchhoff’s loop rule yields V (t) IR (t)R = 0 V (t) V0 sin(!t) Figure 12.2.1 A purely resistive circuit IR (t) = = = I0,R sin(!t) R R would to find the current through the resistor, VR (t) = IRWe (t)R ☛ like instantaneous voltage drop across the resistor I0,R V0,R V0,R = = R XR I R (t) = I R0 sin(ω t − φ R ) . Applying Kirchhoff’s loop rule yields XR = R ☛ resistive reactance V (t) − I R (t)R = 0 , and VR (t) are in phase with each other R = 0 ☛ IR (t) Tuesday, April 19, 16 where VR (t) = I R (t)R is the instantaneous voltage drop across 3 me dependence of the current and the voltage across the resistor is depict Time dependence of VR (t) and IR (t) Phasor diagram for resistive circuit 12.2.2(a). (a) (b) Phasor ☛ rotating vector having following properties: (i) length: the length correspondsof to the I R (tamplitude ) and VR (t ) across the resistor. 12.2.2 (a) Time dependence (ii) angular speed: the vector rotates counterclockwise with angular speed ! m for(iii) theprojection: resistivethe circuit. projection of vector along vertical axis corresponds to value (b) P t ehavior of I R (t ) and VR (t ) can also be represented with a phasor diagram Tuesday, April 19, 16 4 in Figure 12.2.2(b). A phasor is a rotating vector having the following proper of the alternating quantity at time Average value of current over one period 1 hIR (t)i = T Z T 0 Z 1 IR (t) dt = T T 0 I0,R I0,R sin(!t)dt = T Z T sin(2⇡t/T ) dt = 0 0 Average of the square of the current is non-vanishing 2 hIR (t)i 1 = T Z T 0 2 IR (t) 1 dt = T Z T 0 2 I 0,R 2 I0,R sin2 (!t)dt = T Z T sin2 (2⇡t/T ) dt = 0 1 2 I0,R 2 q I0,R 2 hIR (t)i = p root-mean-square (rms) current ☛ Irms = 2 q V0,R 2 rms voltage ☛ Vrms = hVR (t)i = p 2 rms voltage supplied to domestic wall outlets in US ☛ Vrms = 120 V @ 60 Hz It is convenient to define: Power dissipated in the resistor ☛ 2 PR (t) = IR (t)VR (t) = IR (t)R Average power over one period 2 1 V 2 2 2 hPR (t)i = hIR (t)Ri = I0,R R = Irms R = Irms Vrms = rms 2 R Tuesday, April 19, 16 5 shall see below, a purely inductive circuit corresponds to infinite capacitan zero resistance R = 0 . Purely Inductive Load We’d like to find current in the circuit IL (t) = I0,L sin(!t L) Applying Kirchhoff’s loop rule yields V (t) VL (t) = V (t) dIL Figure 12.2.3 A purely inductive circuit L =0 dt dIL V (t)We would V0,Llike to find the current in the circuit, = = sin(!t) dt L L I L (t) = I L0 sin(ω t − φ L ) . IL (t) = Z Z V0,L V0,L V0,L dILApplying = sin(!t)dt = cos(!t) = ⇡/2) yi the modified Kirchhoff’s rule for inductors, thesin(!t circuit equation L !L !L V0,L V0,L I0,L = = !L XL XL = !L ☛ inductive reactance L = +⇡/2 ☛ phase constant Tuesday, April 19, 16 dI L V (t ) − VL (t ) = V (t ) − L =0. dt 6 (1 φ L = π / 2 ; it reaches its maximum value one quarter of a cycle later than VL (t ) . 2 The current lags voltage by π / 2 in a purely inductive circuit oltage plots and the corresponding phasor diagram are shown w. The word “lag” means that the plot of I (t ) is shifted to the right of the plot of V (t ) in L L Figure 12.2.4 (a), whereas in the phasor diagram the phasor I L (t) is “behind” the phasor for VL (t) as they rotate counterclockwise in Figure 12.2.4(b). 12.2.3 Purely Capacitive Load Consider now a purely capacitive circuit with a capacitor connected to an AC generator with AC source voltage given by V (t) = V0 sin(ω t) . In the purely capacitive case, both resistance R and inductance L are zero. The circuit diagram is shown in Figure 12.2.5. (a) (b) Figure 12.2.4 (a) Time dependence of I L (t ) and VL (t ) across the inductor. (b) Pha Time for dependence of I L (t ) and VL (t ) across the inductor. (b) diagram the inductive circuit. ductive circuit. As can be seen from Tuesday, April 19, 16 Figure 12.2.5 Athe purely capacitive I L (circuit t ) is out of phase with VL7(t ) the figures, current Purely Capacity Load We’d like to find current in circuit IC (t) = I0,C sin(!t C) Again ☛ Kirchhoff’s loop rule yields V (t) VC (t) = V (t) Charge on capacitor Q(t) =0 C Figure 12.2.5 A purely capacitive circuit Q(t) = CV (t) = CVC (t) = CV0,C sin(!t) We would like to find the current in the circuit, Current dQ IC (t) = = !CV0,C cos(!t) = !CV0,C sin(!t + ⇡/2) dt V0,C I0,C = !CV0,C = XC 1 XC = ☛ capacitance reactance !C ☛ phase constant C = ⇡/2 Tuesday, April 19, 16 8 VC (t ) . The current leads the voltage by π /2 in a capacitive circuit The word “lead” means that the plot of I C (t) is shifted to the left of the plot of VC (t) in Figure 12.2.6 (a), whereas in the phasor diagram the phasor I C (t) is “ahead” the phasor for VC (t) as they rotate counterclockwise in Figure 12.2.6(b). 12.3 The RLC Series Circuit Consider now the driven series RLC circuit with V (t) = V0 sin(ω t + φ ) shown in Figure 12.3.1. (b) (a) gure 12.2.6 (a) Time dependence of I C (t ) and VC (t ) across the capacitor. (b) agram for the capacitive circuit. Tuesday, April 19, 16 Figure 12.3.1 Driven series RLC Circuit 9 RLC circuit series V (t)what = V0 sin( I = + dQ now / dt , the Differentiate Eq. (12.3.3), using Consider anddriven divide through by L ,with yields isω t + φ 12.3.1. called aSingle second order damped driven differential equation, 10.3 Loop RLClinear AC Circuit We’d like to find current in 2circuit I(t) = I0 sin(!t ωV0 d I R dI I + + = cos(ω t + φ ) . 2) L dt LC L dt (12.3.4) Kirchhoff’s loop rule yields We shall find the amplitude, I 0 , of the current, and phase constant φ which is the phase the voltage Vshift (t)between VR (t) VL source (t) and VCthe (t)current = 0 by examining the phasors associates with the three circuit elements R , L and C . VThe (t)instantaneous = VR (t) voltages + VL (t) + Veach C (t) across of the three circuit elements R , L , and C has a different amplitude and phase compared to the current, as can be seen from the phasor Figure 12.3.1 Driven series RLC Circuit Using phasor representation diagrams shown in Figure 12.3.2. ~0 = V ~0,R (a) ~0,L + V ~0,C V +V (b) (c) Figure 12.3.2 Phasor diagrams for the relationships between current and voltage in (a) 10 Tuesday, April 19, 16 q ~ 0 = |V ~ 0 | = |V ~0,R + V ~0,L + V ~0,C | = V 2 + (V0,L V0,C )2 V 0,R q p 2 + (X 2 + (I X Eq.I(12.3.2) 2 =beIwritten 2 =Using(I X ) the0 X phasor R ) representation, 0 L 0 XC ) can 0 Xas L c R current amplitude V0 I0 = p V0 = VR 0 + VL 0 + VC 0 (12.3.5) V0 q = 2 12.3.3(a). 2 see that current phasor as shown in Figure Again we X + (X X ) L C R R2 + !L I 0 1leads2 the capacitive phase voltage phasor VC 0 by π / 2 but lags the inductive voltage phasor VL!C 0 by π / 2 . The three ✓ phasors rotate ◆ counterclockwise ✓ ◆as time increases, with✓ their relative◆positions voltage XL XC 1 1 1 1 1 fixed. tan = = !L ! = tan !L XR Note that ☛ R !C V0 6= V0,R + V0,L + V0,C R !C Figure 12.3.3 (a) Phasor diagram for the series RLC circuit. (b) voltage relationship Tuesday, April 19, 16 11 The relationship between Z , X R , X L , and X C can be represented by th in Figure q12.3.4: X L − X C Impedance ☛ Z= 2 + (X XR l XC ) 2 V0 I(t) = sin(ω t) Z (12.3.13) 0 Z also depends on the angular frequency ω , as do X L and Notice that the V impedance I(t) = sin(!t) Z XC . Vector representation of the between Z , we may φ and Using Eq. (12.3.9) for theFigure phase 12.3.4 constant Eq. (12.3.12) forrelationship readilyZ , X R recover theSimple-circuit limits for simple circuit (with only series one element). A summary is provided in limits of the RLC circuit The impedance also has SI units of ohms. In terms of Z , the current (Eq Table 12.1 below: (12.3.7)) may be rewritten as Simple Circuit purely resistive purely inductive purely capacitive ⎛ X − XC ⎞ φ = tan ⎜ L X R ⎟⎠ ⎝ Z= X R 2 + ( X L − X C )2 R L C X L = ωL 1 XC = ωC R 0 ∞ 0 0 0 XR 0 L ∞ XL 0 π /2 XL 0 0 C 0 XC −π / 2 XC −1 Table 12.1 Simple-circuit limits of the series RLC circuit Tuesday, April 19, 16 12 10.4 Resonance In driven RLC series circuit ☛ amplitude of current has a maximum value: a resonance which occurs at the resonant angular frequency Because current amplitude is inversely proportional to impedance φ =0. !0 maximum occurs when A qualitative plot I of0 the amplitude of the current as a functi Z is minimum frequency for two driven RLC circuits, with different values o This occurs at angular frequency !0 such that XL = XC illustrated in Figure 12.3.5. The amplitude is larger for small resistance. 1 1 !0 L = ) !0 = p !0 C LC V0 At resonance ☛ Z = R ) I0 = R Tuesday, April 19, 16 13 Figure 12.3.5 The amplitude of the current as a function of ω in t 10.5 Power in AC Circuits In series RLC circuit ☛ instantaneous power delivered by AC generator is given by V0 V02 P (t) = I(t)V (t) = sin(!t )V0 sin(!t) = sin(!t) sin(!t Z Z V02 = [sin2 (!t) cos( ) sin(!t) cos(!t) sin( )] Z ) we have used trigonometric identity sin(!t ) = sin(!t) cos( ) cos(!t) sin( ) Time average of the power is Z 1 T V02 hP (!)i = sin2 (!t) cos( )dt T 0 Z 2 Vrms = cos = Irms Vrms cos Z power factor Tuesday, April 19, 16 ☛ cos 1 T Z T 0 V02 1 V02 sin(!t) cos(!t) sin( )dt = cos Z 2 Z R = Z 14 In Figure 12.4.1, we plot the time-averaged power as a function of th frequency ω for two driven RLC circuits, with different values of 2 hP (!)i = Irms (!)R V0 1 Irms (!) = p q 2 R2 + !L 1 2 !C = 1power R Maximum @ resonance condition ☛ cos or Z Figure 12.4.1 Average as = a function of frequency in a driven We see that P(ω ) attains the maximum value when cos φ = 1 , o resonance condition. At resonance, we have 2 hP (!0 )i = Irms Vrms = Vrms R P(ω 0 ) = I rmsVrms = Tuesday, April 19, 16 2 Vrms R . 15 Width of the peak 12.4.1 Width of the Peak The peak has a line width hasisa to linedefine width. One the width is to defi One way to characterizeThe thepeak width ! way = !to+characterize ! where ω ± are the values of the driving angular frequency such that the !± ☛ values of driving angular frequency halfthat its power maximum powerto athalf resonance. This power is called width at h such is equal its maximum at full resonance illustrated in Figure 12.4.2. The width Δω increases with resistance R . This is called full width at half maximum Width ! increases with resistance R Tuesday, April 19, 16 Figure 12.4.2 Width of the peak To find Δω , it is instructive to first rewrite the average power P(ω16) a To find ! it is instructive to first rewrite the average power as 1 V02 R hP (!)i = 2 R2 + !L 1 2 !C 1 V02 R! 2 = 2 ! 2 R2 + L2 (! 2 !02 )2 V02 hP (!0 )i = 2R condition for finding !± is 2 V02 R!± 1 V0 1 hP (!0 )i = hP (!± )i ) = 2 R2 + L2 (! 2 2 4R 2 !± ± after some algebra 2 ☛ (!± !02 )2 !02 )2 = (R!± /L)2 Taking square roots yields two solutions which we analyze separately Tuesday, April 19, 16 17 Case 1: Taking the positive root leads to 2 !+ !02 R!+ R =+ ) !+ = + L 2L s✓ Case 2: Taking the negative yields ! 2 !02 = R! L width at half maximum )! = ☛ R + 2L ! = !+ R 4L s✓ ◆2 R 4L + !02 ◆2 + !02 R ! = L Quality factor Qqual Tuesday, April 19, 16 !0 !0 L = = ! R 18 10.6 Transformer Transformer is device used to increase or decrease the AC voltage in a circuit Typical device consists of two coils of wire primary and secondary wound around an iron core Primary coil with N1 turns is connected to alternating voltage source V (t) Secondary coil has N2 turns and is connected to a load with resistance R2 Figure 12.5.1 A transformer The way transformers operate is based on the principle that Inalternating the primary circuit, neglecting in the Faraday’scoil law of current in primary coil the will small induce resistance an alternating emfcoil, on secondary induction implies due to their mutual inductance dΦB Tuesday, April 19, 16 19 V1 = − N1 , (12.5.1) Neglecting small resistance in coil V1 = B ☛ N1 Faraday’s law of induction implies d B dt ☛ magnetic flux through one turn of primary coil Iron core extends from primary to secondary coils Iron core serves to increase magnetic field produced by current in primary coil and ensures that nearly all magnetic flux through primary coil also passes through each turn of the secondary coil Voltage (or induced emf) across secondary coil is ☛ V2 = N2 d B dt Ideal transformer ☛ power loss due to Joule heating can be ignored so that power supplied by primary coil is completely transferred to secondary coil I 1 V1 = I 2 V2 In addition ☛ if no magnetic flux leaks out from iron core through each turn is same in both primary and secondary coils V2 N2 = V1 N1 Tuesday, April 19, 16 20 Transformation of currents in the two coils reads V2 N2 I1 = I2 = I2 V1 N1 Ratio of output voltage to input voltage is determined by turn ratio N 2 > N 1 ) V 2 > V1 N2 N1 output voltage in secondary coil is greater than input voltage in primary coil Transformer with N2 > N1 ☛ step-up transformer N 1 > N 2 ) V 1 > V2 output voltage is smaller than input Transformer with N1 > N2 ☛ step-down transformer For home safety ☛ we would like LOW emf supply Tuesday, April 19, 16 21 Why do we use high voltages? As electricity flows down a metal wire electrons carrying its energy jiggle through the metal structure That's why wires get hot when electricity flows through them (useful for electric toasters and other appliances that use heating elements) Electricity that comes from power plants is sent dow wires at extremely high voltages to save energy 2 hP (!)i = Irms R 2 For power transmission ☛ we’d like to keep Irms at minimum The higher the voltage and the lower the current ☛ the less energy is wasted Tuesday, April 19, 16 22