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Transcript

Physics 121 - Electricity and Magnetism Lecture 14 - AC Circuits, Resonance Y&F Chapter 31, Sec. 3 - 8 • • • • • • • Phasor Diagrams for Voltage and Current The Series RLC Circuit. Amplitude and Phase Relations Impedance and Phasors for Impedance Resonance Power in AC Circuits, Power Factor Examples Transformers Copyright R. Janow –Fall 2016 Current/Voltage Phases in pure R, C, and L circuit elements • Sinusoidal current i (t) = Imcos(wDt). Peak is Im • Peak voltage drops in R, L, or C loads lead/lag current by 0, p/2, -p/2 radians • Reactances (generalized resistances) are ratios of peak voltages to peak currents VR& Im in phase Resistance VR / Im R VC lags Im by p/2 Capacitive Reactance VC / Im C 1 wDC VL leads Im by p/2 Inductive Reactance VL / Im L wDL Phases of voltages in series components are referenced to the current phasor Same Phase currrent Copyright R. Janow –Fall 2016 Phasors applied to a Series LCR circuit E vR Applied EMF: E(t) Emax cos(wDt F) R Current: i(t) Imax cos(wDt) L Same current everywhere in the single branch vL • Same frequency dependance as E (t) C • Same phase for the current in E, R, L, & C, but...... • Current leads or lags E (t) by a constant phase angle F • F in interval +/- p/2 vC Phasors all rotate CCW at frequency wD • Lengths of phasors are the peak values (amplitudes) F Em Im wDt+F • The “x” components are instantaneous values wDt Apply Kirchhoff Loop rule to series LRC: E(t) vR (t) vL (t) vC (t) 0 instantaneous Voltage phasors for R, L, & C all rotate at wD : VL Em Im F wDt VR VC • VC lags Im by p/2 • VR has same phase as Im • VL leads Im by p/2 VR ImR VC ImXC VL ImXL Phasors add like vectors Em VR (VL VC ) VC lags VL by 1800 Copyright R. Janow along Im perpendicular to Im–Fall 2016 Voltage addition rule for series LRC circuit 2 Magnitude of Em Em VR2 (VL VC )2 Em in series circuit: 1 Reactances: L wDL C VL-VC wDC IL L VL ICC VC IRR VR XL-XC Same current amplitude in each component: Im F Z VR R Im IR IL IC Impedance magnitude is the E ratio of peak EMF to peak |Z| m Im current: peak applied voltage peak current wDt [ Z ] ohms For series LRC circuit, divide each voltage in |Em| by (same) peak current 2 2 1/2 Magnitude of Z: | Z | [ R (L C ) ] Phase angle F: tan(F) VL VC C L VR R F measures the power absorbed by the circuit: Applies to a single series branch with L, C, R See phasor diagram P Em Im EmIm cos(F) • R ~ 0 Im normal to Em F ~ +/- p/2 tiny losses, no power absorbed Copyright R. Janow –Fall 2016 • XL=XC Im parallel to Em F 0 Z=R maximum current (resonance) Summary: AC Series LCR Circuit vR E R L vL C vC Circuit Element Symbol Resistance or Reactance Phase of Current Phase Angle Amplitude Relation Resistor R R In phase with VR 0º (0 rad) VR = ImR Capacitor C XC=1/wdC Leads VC by 90º -90º (-p/2) VC = ImXC Inductor L XL=wdL Lags VL by 90º +90º (p/2) VL = ImXL E(t) Emax cos(wDt F) i(t) Im cos(wDt) Em Im F Pav Prms ErmsI rms cos(F) VL-VC Z VR XL-XC sketch shows XL > XC Em Im | Z | R wDt | Z | [ R2 (XL XC )2 ]1/2 tan(F ) VL VC X XC L VR R Copyright R. Janow –Fall 2016 Example 1: Analyzing a series RLC circuit A series RLC circuit has R = 425 Ω, L = 1.25 H, C = 3.50 μF. It is connected to an AC source with f = 60.0 Hz and εm= 150 V. (A) (B) (C) (D) (E) Determine the impedance of the circuit. Find the amplitude of the current (peak value). Find the phase angle between the current and voltage. Find the instantaneous current across the RLC circuit. Find the peak and instantaneous voltages across each circuit element. Copyright R. Janow –Fall 2016 Example 1: Analyzing a Series RLC circuit A series RLC circuit has R = 425 Ω, L = 1.25 H, C = 3.50 μF. It is connected to an AC source with f = 60.0 Hz and εm=150 V. (A) Determine the impedance of the circuit. Angular frequency: wD 2pf 2p (60.0) Hz Resistance: Inductive reactance: Capacitive reactance: 377 s1 R 425 L wDL (377 s1 )(1.25 H) 471 C 1 / wDC 1 /( 377 s1 )(3.50 106 F) 758 | Z | R 2 (XL X C )2 (425 )2 (471 758 )2 513 (B) Find the peak current amplitude: Im m 150 V 0.292 A | Z | 513 (C) Find the phase angle between the current and voltage. XC > XL (Capacitive) F tan1( Current phasor Im leads the Voltage Em Phase angle will be negative Copyright R. Janow –Fall 2016 XL X C 471 758 ) tan1( ) 34.0 0.593 rad. R 425 Example 1: Analyzing a series RLC circuit - continued A series RLC circuit has R = 425 Ω, L = 1.25 H, C = 3.50 μF. It is connected to an AC source with f = 60.0 Hz and εm=150 V. (D) Find the instantaneous current across the RLC circuit. i(t) Im cos(wDt) 0.292cos(377t) (E) Find the peak and instantaneous voltages across each circuit element. VR ,m ImR (0.292 A)( 425 ) 124 V VR in phase with Im VR leads Em by |F| vR (t) VR ,m cos(wD t) (124 V) cos(377t) VL,m ImXL (0.292 A)( 471 ) 138 V vL (t) VL,m cos(wD t p / 2) (138 V) cos(377t p / 2) VC,m ImX C (0.292 A)(758 ) 222 V VL leads VR by p/2 VC lags VR by p/2 vC (t) VC,m cos(wD t p / 2) (222 V) cos(377t p / 2) Add voltages above: VR VL VC 483V 150V Em What’s wrong? 1 / 2 2 Janow –Fall 2016 R. V 150 V Voltages add with proper phases: Em VR2 VCopyright L C Example 2: Resonance in a series LCR Circuit: Em = 100 V. R = 3000 L = 0.33 H C = 0.10 mF Find |Z| and F for fD = 200 Hertz, fD = 876 Hz, & fD = 2000 Hz vR E R L vL Why should fD make X w L D L a difference? XC C | Z | [ R2 (XL XC )2 ]1/2 vC Frequency f Resistance R Reactance XC 200 Hz 3000 7957 876 Hz 3000 2000 Hz 3000 X C XL Im F < 0 Reactance XL Im wD C Em |Z| XL X C ) R Impedance |Z| Phase Angle F Circuit Behavior 415 8118 - 68.3º Capacitive Em lags Im 1817 1817 3000 Resonance 0º Resistive Max current 796 4147 4498 +48.0º Inductive Em leads Im X C XL Em Em F tan1 ( 1 Im F0 XL X C Em F 0 Im Copyright R. Janow –Fall 2016 Resonance in a series LCR circuit E R L C 1 / wDC C Vary wD: XC XL | Z | R L C 2 2 1/ 2 L wDL Im Em |Z| R resistance At resonance maximum current, minimum impedance when wD 1/ LC wres | Z | R, Im EM / R, F 0 width of resonance (selectivity, “Q”) depends on R. Large R less selectivity, smaller current at peak capacitance dominates current leads voltage inductance dominates current lags voltage damped spring oscillator near resonance Copyright R. Janow –Fall 2016 Power in AC Circuits • Resistors always dissipate power, but the instantaneous rate varies as i2(t)R • No power is lost in pure capacitors and pure inductors in an AC circuit – Capacitor stores energy during two 1/4 cycle segments. During two other segments energy is returned to the circuit – Inductor stores energy when it produces opposition to current growth during two ¼ cycle segments (the source does work). When the current in the circuit begins to decrease, the energy is returned to the circuit Copyright R. Janow –Fall 2016 AC Power Dissipation in a Resistor 2 2 2 Instantaneous power P i ( t ) R I R c os (wt) inst m • Power is dissipated in R, not in L or C • cos2(x) is always positive, so Pinst is always positive. But, it is not constant. • Power pattern repeats every p radians (t/2) The RMS power is an AC equivalent to DC power Pav Prms averageof Pinst overa wholecycle (wτ 2p) Integrate Pinst in resistor over t: 1 t 1 2 2 Prms Im R cos2 (wt)dt Im R t 0 2 RMS means “Root Mean Square” • Square a quantity (positive) • Average over a whole cycle • Compute square root. COMPUTING RMS QUANTITIES: • For any RMS quantity divide peak value such as Im or Em by sqrt(2) Irms Im 2 Erms Vrms Integral = 1/2 2 Prms Pav Irms R Em 2 Vm 2 Irms Erms |Z| For any R, L, or C R. Janow –Fall 2016 Household power example: 120 volts RMS Copyright 170 volts peak Power factor for an AC LCR Circuit The PHASE ANGLE F determines the average RMS power actually absorbed due to the RMS current and applied voltage in the circuit. Erms F Claim (proven below): Pav Prms Erms Irms ErmsI rms cos(F) cos(F) R / | Z | is the " power factor" Irms |Z| XL-XC R wDt Proof: Start with instantaneous power (not very useful): Pinst (t) E(t) I(t) Em Im cos(wDt F) cos(wDt) Average it over one full period t: Pav t P t 1 0 inst (t) dt Em Im Change variables: wDt 2p t cos(w t F) cos(w t)dt t 1 0 D D Em Im x {Integral} x wDt, wDt 2p, dt dx/wD 1 2p {Integral} cos(x F) cos(x)dx 2p 0 Copyright R. Janow –Fall 2016 Use trig identity: cos(x F) cos(x) cos(F) sin(x) sin(F) Power factor for AC Circuits - continued {Integral} cos(F) Even integrand 1 2p 2p 0 cos2 (x)dx sin(F) 2p 2p 0 cos(x) sin(x)dx Odd integrand 2p 2 sin(F) sin (x) 0 2p 2 0 2p 2p cos(F) x sin(2x) cos(F) 2p 2 0 4 2 0 Pav 1 cos(F) Em Im 2 Recall: RMS values = Peak values divided by sqrt(2) Pav Prms ErmsI rms cos(F) Also note: Erms I rms | Z | Alternate form: and R | Z | cos(F) 2 2 Prms Irms | Z | cos(F) Irms R If R=0 (pure LC circuit) F +/- p/2 and Pav = Prms = 0 Copyright R. Janow –Fall 2016 Example 2 continued with RMS quantities: R = 3000 L = 0.33 H C = 0.10 mF Em = 100 V. fD = 200 Hz Find Erms: E R L Erms Em / 2 71 V. Find Irms at 200 Hz: VR C | Z | 8118 as before VC Irms Erms / | Z | 71 V / 8118 8.75 mA. Find the power factor: 3000 cos(F) R / | Z | 0.369 8118 Find the phase angle F directly: C 0 F tan 1 L 68 as before R Recall: do not use arc-cos to find F Find the average power: Pav ErmsI rms cos(F) 71 8.75 10-3 0.369 0.23 Watts or 2 Pav Irms R 8.75 10- 3 2 3000 0.23 Watts Copyright R. Janow –Fall 2016 VL Example 3 – Series LCR circuit analysis using RMS values A 240 V (RMS), 60 Hz voltage source is applied to a series LCR circuit consisting of a 50ohm resistor, a 0.5 H inductor and a 20 mF capacitor. w 2pf 6.28 x 60 377 rad / s D Find the capacitive reactance of the circuit: XC 1 / wDC 1 / (377x2x10-5 ) 133 Find the inductive reactance of the circuit: XL wDL 377x.5 188.5 The impedance of the circuit is: | Z | [ R2 (XL XC )2 ]1/2 74.7 The phase angle for the circuit is: tan(F) F is positive since XL>XC (inductive) Irms The RMS current in the circuit is: XL X C F 48.0 0 , cos(F ) 0.669 R E 240 rms 3.2 A. | Z | 74.7 The average power consumed in this circuit is: 2 Prms Irms R (3.2) 2 x 50 516 W. or Prms Erms Irms cos(F) where cos(F) R / Z If the inductance could be changed to maximize the current through the circuit, what Current is a maximum at RESONANCE. would the new inductance L’ be? wD 377 1 1 1 L' 0.352 H. 2 2 5 L' C w C 377 x 2x10 D How much RMS current would flow in that case? At resonance | Z | R Irms Erms R 240 V 4.8 A. 50 Copyright R. Janow –Fall 2016 Transformers Devices used to change AC voltages. They have: • Primary • Secondary • Power ratings power transformer iron core circuit symbol Copyright R. Janow –Fall 2016 Transformers Ideal Transformer Assume zero internal resistances, EMFs Ep, Es = terminal voltages Vp, Vs Faradays Law for primary and secondary: Vp Np dFB dt Vs Ns dFB dt Assume: The same amount of flux FB cuts each turn in both primary and secondary windings in ideal transformer (counting self- and mutual-induction) induced voltage per turn N Vs s Vp Np dFB Vp Vs dt Np Ns Turns ratio fixes the step up or step down voltage ratio Vp, Vs are instantaneous (time varying) or RMS averages, as can be the power and current. iron core • zero resistance in coils • no hysteresis losses in iron core • all field lines are inside core Assuming no losses: energy and power are conserved Ps VsIs conserved Pp VpIp Ip Is Ns Np Copyright R. Janow –Fall 2016 Example: A dimmer for lights using a variable inductance f =60 Hz w = 377 rad/sec Without Inductor: 2 P0,rms Erms / R 18 Watts, Light bulb R=50 Erms=30 V L F0 a) What value of the inductance would dim the lights to 5 Watts? Erms Erms 2 Irms cos(F) cos(F) R/ | z | Recall: Prms IrmsR Prms |Z| R P0,rmscos2 (F) 5 Watts 18 Watts x cos2 (F) cos(F) 0.527 cos( 58.2o ) F 58.2o tan-1 [ XL / R ] (X C 0) XL R tan(F) 50 tan(58.2 o ) 80.6 2p f L L 80.6 / 377 214 mH b) What would be the change in the RMS current? E 30 V I0,rms rms 0.6 A P0,rms = 18 W. Without inductor: R With inductor: Irms Erms R 50 Prms = 5 W. Copyright R. Janow –Fall 2016 cos(F) 0.6 A 0.527 0.316 A Copyright R. Janow –Fall 2016