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Lecture 14 - AC Circuits, Resonance Y&F Chapter 31, Sec

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Lecture 14 - AC Circuits, Resonance Y&F Chapter 31, Sec Physics 121 - Electricity and Magnetism Lecture 14 - AC Circuits, Resonance Y&F Chapter 31, Sec. 3 - 8 • The Series RLC Circuit. Amplitude and Phase Relations • Phasor Diagrams for Voltage and Current • Impedance and Phasors for Impedance • Resonance • Power in AC Circuits, Power Factor • Examples • Transformers • Summaries Copyright R. Janow – Fall 2013 Current & voltage phases in pure R, C, and L circuits Current is the same everywhere in a single branch (including phase) Phases of voltages in elements are referenced to the current phasor • Apply sinusoidal voltage E (t) = EmCos(wDt) • For pure R, L, or C loads, phase angles are 0, +p/2, -p/2 • Reactance” means ratio of peak voltage to peak current (generalized resistances). VR& iR in phase VC lags iC by p/2 VL leads iL by p/2 Resistance Capacitive Reactance Inductive Reactance 1 V /i R Vmax /iC C Vmax /iL L wDL max R wDC Copyright R. Janow – Fall 2013 The impedance is the ratio of peak EMF to peak current peak applied voltage Em Z [Z] ohms peak current that flows im 2 2 2 Magnitude of Em: Em VR (VL VC) 1 Reactances: L wDL C wDC L VL /iL C VC /iC R VR /iR im iR,max iL,max iC,max For series LRC circuit, divide Em by peak current 2 2 1/2 Applies to a single Magnitude of Z: Z [ R (L C) ] branch with L, C, R VL VC L C Phase angle F: tan(F) see diagram VR R F measures the power absorbed by the circuit: P Em im Em im cos(F) • R ~ 0 tiny losses, no power absorbed im normal to Em F ~ +/- p/2 • XL=XC im parallel to Em F 0 Z=R maximum currentCopyright (resonance) R. Janow – Fall 2013 Copyright R. Janow – Fall 2013 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. (B) Find the amplitude of the current (peak value). (C) Find the phase angle between the current and voltage. (D) Find the instantaneous current across the RLC circuit. (E) Find the peak and instantaneous voltages across each circuit element. Copyright R. Janow – Fall 2013 Series LCR circuit driven by an external AC voltage Apply EMF: VR E (t) = EmCos(wDt) wD is the driving frequency E R The current i(t) is the same everywhere in the circuit L VL i (t) = EmCos(wDt+F) C • Same frequency dependance as E (t) but... • Current leads or lags E (t) by a constant phase angle F VC • Same phase for the current in E, R, L, & C Phasors all rotate CCW at frequency wD F • Lengths of phasors are the peak values (amplitudes) E im m • The “y” components are the measured values. t wD wDt-F Plot voltages in components with phases relative to current phasor im : • VR has same phase as im VR imR • VC lags im by p/2 VC imXC V L • VL leads im by p/2 Em VL imXL im F Kirchoff Loop rule for potentials (measured along y) E (t) V (t) V (t) V (t) 0 wDt-F R L C V R 0 Em VR (VL VC) VC lags VL by 180 V C along im perpendicularCopyright R. toJanow im – Fall 2013 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. 1 Angular frequency: wD 2pf 2p (60.0) Hz 377 s Resistance: R 425 1 Inductive reactance: L wDL (377 s )(1.25 H) 471 Capacitive reactance: 1/ w C 1/(377 s1)(3.50106 F) 758 C D 2 2 2 2 Z R (XL XC ) (425 ) (471 758 ) 513 (B) Find the peak current amplitude: 150 V I m 0.293 A m Z 513 (C) Find the phase angle between the current and voltage. E X > X (Capacitive) Current vector Im leads the Voltage m C L Phase angle should be negative X X 471 758 tan1( L C ) tan1( ) 34.0 0.593 rad. R 425 Copyright R. Janow – Fall 2013 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. (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| V I X (0.292 A)(471 ) 138 V L,m m L o VL leads VR by 90 o VC,m ImXC (0.293 A)(758 ) 222 V VC lags VR by 90 Note that: VR VL VC 483V 150V Em Why not? 1/ 2 V2 V V 2 150 V Voltages add with proper phases: Em R LCopyrightC R. Janow – Fall 2013 Example 2: Resonance in a series LCR Circuit: R = 3000 L = 0.33 H C = 0.10 mF Em = 100 V. Find Z and F for fD = 200 Hertz, fD = 876 Hz, & fD = 2000 Hz VR Why should f make 1 E R D X w L X Em L D C Im a difference? w C L VL D Z C 2 2 1/2 1 XL XC Z [ R (XL XC) ] F tan ( ) R VC Frequency Resistance Reactance Reactance Impedance Phase Circuit f R XC XL Z Angle F Behavior 200 Hz 3000 7957 415 8118 - 68.3º Capacitive Em lags Im 876 Hz 3000 1817 1817 3000 0º Resistive Resonance Max current 2000 Hz 3000 796 4147 4498 +48.0º Inductive Em leads Im XC XL XC XL XL XC Em I I Im F < 0 m Em F 0 m Em F0 Copyright R. Janow – Fall 2013 Resonance 1/ 2 E R 2 2 Em Z R L C Im L Z C C 1/ wDC L wDL R resistance Vary wD: At resonance maximum current flows & impedance is minimized XC XL 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 damped spring oscillator capacitance dominates inductance dominates near resonance current leads voltage current lags voltage Copyright R. Janow – Fall 2013 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 – In a capacitor, during one-half of a cycle energy is stored and during the other half the energy is returned to the circuit – In an inductor, the source does work against the back emf of the inductor and energy is stored in the inductor. When the current in the circuit begins to decrease, the energy is returned to the circuit Copyright R. Janow – Fall 2013 Instantaneous and RMS (average) power instantaneous power P i2(t) R I2 R sin2(wt F) consumed by circuit inst m • Power is dissipated in R, not in L or C 2 • Sin (x) is always positive, so Pinst is always positive. But, it is not constant. • Pattern for power repeats every p radians (T/2) The RMS power, current, voltage are useful, DC-like quantities Pav average of Pinst over a whole cycle (wτ 2p) Integrate T 2 1 2 1 2 P I R sin (wt F)dt I R (the integral 1/2) av m T 0 2 m P is an “RMS” quantity: 2 av Prms Pav IrmsR • “Root Mean Square” • Square a quantity (positive) E I E • Average over a whole cycle m m rms Erms Irms Irms • Compute square root. 2 2 Z • In practice, divide peak value Vm by sqrt(2) for Im or Emsince Vrms any component 2 sin (x) appears 2 Household power example: 120 volts RMS 170Copyright volts R. Janowpeak – Fall 2013 Power factor for AC Circuits The PHASE ANGLE F determines the average RMS power actually absorbed by the RMS current and Erms applied voltage in the circuit. Irms F Result (proven below): Z XL-XC wDt-F Pav Prms Erms I rms ErmsI rms cos(F) R cos(F) R / Z is the "power factor" Proof: start with instantaneous power (not very useful, shift cos(x) by pi/2): Pinst(t) E(t) I(t) Em Im sin(wDt)sin(wDt F) Average it over one full cycle: P 1 P (t) dt E I 1 sin(w t)sin(w t F)dt av 0 inst m m 0 D D Note trig identities: sin(wDt F) sin(wDt)cos(F) cos(wDt)sin(F) 1 sin(wDt)cos(wDt) sin(2wDt) 2 Copyright R. Janow – Fall 2013 Power factor for AC Circuits - continued Substitute trig identities: P E I cos(F) 1 sin2 (w t)dt E I sin(F) 1 sin(2w t) av m m 0 D m m 2 0 D Over a full period: 1 2 1 1 wD 2p sin (w t)dt sin(2wDt) 0 0 D 2 2 0 cos(F) Pav Em Im 0 2 Recall: RMS values = Peak values divided by sqrt(2) Pav Prms ErmsI rms cos(F) Also note: Erms I rmsZ and R Z cos(F) 2 2 Alternate form: Prms IrmsZcos(F) IrmsR If R=0 (pure LC circuit) F +/- p/2 and Pav = Prms = 0 Copyright R.
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