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.

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 s1)(3.50106 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    tan1( L C )  tan1( )  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 1 E Why should fD make Em R XL  wDL XC  I  a difference? w C m 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 F0 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

• 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

R = 3000  L = 0.33 H C = 0.10 mF Em = 100 V. VR

fD = 200 Hz E R

L VL Find Erms: Erms  Em / 2  71 V. C

Find Irms at 200 Hz: Z  8118  as before VC Irms  Erms / Z  71 V / 8118   8.75 mA. Find the power factor: 3000 Recall: do not use cos(F)  R / Z   0.369 arc-cos to find F 8118 Find the phase angle F:     F  tan1 L C   680 as before  R  Find the average power: -3 Pav  ErmsI rms cos(F)  71  8.75  10  0.369  0.23 Watts

or 2 -3 2 Pav  IrmsR  8.75  10   3000  0.23 Watts

Copyright R. Janow – Fall 2013 Example 3 – LCR circuit analysis using RMS values A 240 V (RMS), 60 Hz voltage source is applied to a series LCR circuit consisting of a 50- ohm resistor, a 0.5 H inductor and a 20 mF capacitor. wD  2pf  6.28 x 60  377 rad/ s

-5 Find the capacitive reactance of the circuit: XC  1/ wDC  1 / (377x2x10 )  133 

Find the inductive reactance of the circuit: XL  wDL  377x.5  188.5 

The impedance of the circuit is: 2 2 1/2 Z  [ R  (XL  XC) ]  74.7 

X  X The phase angle for the circuit is: tan(F)  L C  F  48.00, cos(F)  0.669 R F is positive since XL>XC (inductive)

Erms 240 The RMS current in the circuit is: Irms    3.2 A. Z 74.7 The average power consumed in this circuit is: 2 2 Prms  IrmsR  (3.2) 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 would the new inductance L’ be? Current is a maximum at RESONANCE. 1 1 1 wD  377   L'   0.352 H. LC w2C 3772 x2x105 D How much RMS current would flow in that case? Erms 240 V At resonance Z  R  Irms    4.8 A. R 50  Copyright R. Janow – Fall 2013 Transformers Devices used to change power transformer AC voltages. They have: • Primary • Secondary • Power ratings

iron core

circuit symbol Copyright R. Janow – Fall 2013 Transformers Ideal Transformer Assume zero internal resistances,

EMFs Ep, Es = terminal voltages Vp, Vs Faradays Law for primary and secondary: dF dF V  N B V  N B p p dt s s dt

The same flux FB cuts each turn in both primary and secondary windings of an iron core ideal transformer (counting self- and • zero resistance in coils mutual-induction) • no hysteresis losses in iron core • all field lines are inside core induced voltage dFB Vp Vs    turn dt Np Ns Assume no losses: energy and power are conserved Ns Turns ratio fixes  Vs  Vp the step up or step Ps  VsIs  conserved Pp  VpIp Np down voltage ratio Ip Ns Vp, Vs are instantaneous (time varying)   but can also be regarded as RMS Is Np averages, as can be the power and Copyright R. Janow – Fall 2013 current. Example: A dimmer for lights Light using a variable inductance R=50  f =60 Hz w = 377 rad/sec Erms=30 V L Without Inductor: 2 Prms  Erms /R  18 Watts, F  0

a) What value of the inductance would dim the lights to 5 Watts?

2 E E 2 P  I R rms rms Erms 2 Recall: rms rms Irms   cos(F)  P  cos (F) Z R rms R

5 Watts  18 Watts x cos2(F) cos(F)  0.527  cos( 58.2o ) F  58.2o  tan-1 [ X / R ] L o XL  R tan(F)  50 tan(58.2 )  80.6   2p f L  L  80.6 / 377  214 mH

b) What is the change in the RMS current? E 30 V Without inductor: I  rms   0.6 A P = 18 W. rms Z 50  rms ' Erms With inductor: Irms  cos(F)  0.6 A  0.527  0.316 A Prms = 5 W. R Copyright R. Janow – Fall 2013 SUMMARY – AC CIRCUIT RELATIONS