Copyright R. Janow – Spring 2015 Physics 121 - Electricity and Magnetism Lecture 14 - AC Circuits, Resonance Y&F Chapter 31, Sec. 3 - 8 The Series RLC

Copyright R. Janow – Spring 2015

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


Current & voltage phases in pure R, C, and L circuits

• Apply sinusoidal current i (t) = Imcos(Dt)• For pure R, L, or C loads, phase angles for voltage drops are 0, /2, -/2• “Reactance” means ratio of peak voltage to peak current (generalized resistances).

VR& Im in phase Resistance

RI/V mR C1I/V

DCmC LI/V DLmL

VL leads Im by /2Inductive Reactance

VC lags Im by /2Capacitive Reactance

Phases of voltages in series components are referenced to the current phasor

currrent

SamePhase


Same frequency dependance as E (t)• Same phase for the current in E, R, L, & C, but...... • Current leads or lags E (t) by a constant phase angle

Phasor Model for a Series LCR circuit, AC voltage

RL

C

E

vR

vC

vL

Apply EMF: )tcos()t( Dmax EE

Im

Em

Dt

VL

VC

VR

Show component voltage phasors rotating at D with phases relative to the current phasor Im and magnitudes below :

• VR has same phase as Im

• VC lags Im by /2• VL leads Im by /2

RIV mR

CmC XIV

LmL XIV

0byLlagsCCLR m 180 V V )VV( V

Ealong Im perpendicular to Im

ImEm

Dt+Dt

Phasors all rotate CCW at frequency D

• Lengths of phasors are the peak values (amplitudes) • The “x” components are instantaneous values measured

0)t(v)t(v)t(v)t( CLR EKirchoff Loop rule for series LRC:

)tcos(I)t(i Dmax The current i(t) is the same everywhere in the singlebranch in the circuit:


• R ~ 0 tiny losses, no power absorbed Im normal to Em ~ +/- /2

• XL=XC Im parallel to Em Z=R maximum current (resonance)

measures the power absorbed by the circuit: )cos(I I P mmmm E E

2CL

2R 2

m )V(V V EMagnitude of Em

in series circuit:

RR VRI CCC VI LLL VI

CD

C

1LDL Reactances:

CLRm IIII Same current amplitude in each component:

Z

Im

Em

Dt

VL-VC

VRXL-XC

R

For series LRC circuit, divide each voltage in |Em| by (same) peak current

ohms]Z[ I

Z

m

mEpeak applied voltage

peak current

])( R[ Z 1/22CL

2 Magnitude of Z: Applies to a single series

branch with L, C, R

Phase angle : R

V

VV )tan( CL

R

CL

See phasor diagram

Impedance is the ratio of peak EMF to peak current

Voltage addition rule for series LRC circuit


Summary: AC Series LCR Circuit

VL = ImXL+90º (/2)Lags VL by 90º

XL=dLLInductor

VC = ImXC-90º (-/2)Leads VC by 90º

XC=1/dCCCapacitor

VR = ImR0º (0 rad)In phase with VR

RRResistor

Amplitude Relation

Phase Angle

Phase of Current

Resistance or Reactance

SymbolCircuit Element

RL

C

E

vR

vC

vL

Z

im

Em

Dt

VL-VC

VR

XL-XC

Rsketch showsXL > XC

])X(X R[ Z 1/22CL

2

R

XX

V

VV )tan( CL

R

CL

)tcos()t( Dmax EE

)tcos(I)t(i Dm

Z

I m

m

E

)cos(IP P rms rmsrmsav E


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.



It is connected to an AC source with f = 60.0 Hz and εm=150 V.

Example 1: Analyzing a Series RLC circuit

137706022 s Hz ).( fD

513) 758 471() 425()XX(RZ 222CL

2

)H .)(s (LDL 471251377 1

)F .)(s /(C/ DC 7581050337711 61

R 425

(A) Determine the impedance of the circuit.

Angular frequency:

Resistance:

Inductive reactance:

Capacitive reactance:

A 292.0 513

V 150

ZI mm

(B) Find the peak current amplitude:

.rad 593.00.34) 425

758 471(tan)

R

XX(tan 1CL1

Current phasor Im leads the Voltage Em

Phase angle should be negative

XC > XL (Capacitive)

(C) Find the phase angle between the current and voltage.


Example 1: Analyzing a series RLC circuit - continued

V 138) 471)(A 292.0(XIV Lmm,L

)2/t377cos()V 138()2/tcos(V)t(v Dm,LL VL leads VR by /2

V 222) 758)(A 292.0(XIV Cmm,C

)2/t377cos()V 222()2/tcos(V)t(v Dm,CC

VC lags VR by /2

mCLR V VVVV E 150483Note that: Why not?

Voltages add with proper phases: V VV V /

CLRm 1502122 E

V 124) 425)(A 292.0(RIV mm,R

)t377cos()V 124()tcos(V)t(v Dm,RR VR in phase with Im

VR leads Em by ||

(E) Find the peak and instantaneous voltages across each circuit element.


It is connected to an AC source with f = 60.0 Hz and εm=150 V.

)t377cos(292.0)tcos(I)t(i Dm

(D) Find the instantaneous current across the RLC circuit.


200 Hz Capacitive

Em lags Im

- 68.3º 8118 415 7957 3000

Frequency

f

Circuit Behavior

Phase Angle

Impedance Z

Reactance XL

Reactance XC

Resistance R

876 Hz Resistive

Max current

0º3000 Resonance

1817 1817 3000

2000 Hz Inductive

Em leads Im

+48.0º4498 4147 796 3000

RL

C

E

vR

vC

vL

ImEm

LC XX

ImEm

LC XX

ImEm

CL XX

C

XD

C

1LX DL Why should fD make

a difference?

])X(X R[ Z 1/22CL

2 )

R

XX(tan CL 1

Z

I mm

E

Example 2: Resonance in a series LCR Circuit:R = 3000 L = 0.33 H C = 0.10 mF Em = 100 V.

Find Z and for fD = 200 Hertz, fD = 876 Hz, & fD = 2000 Hz


Resonance in a series LCR circuit

R

LC

E

resistance R L C/ DLDC 1

R Z/ 2

CL2 21

Vary D: At resonance maximum current flows & impedance is minimized

0 ,R/I R,Z LC1/ X X MmresD henwLC E

inductance dominatescurrent lags voltage

capacitance dominatescurrent leads voltage

width of resonance (selectivity, “Q”) depends on R.Large R less selectivity, smaller current at peak

damped spring oscillatornear resonance

Z

I mm

E


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


Instantaneous and RMS (average) power)t(osc R IR )t(iP 22

m2

inst Instantaneous powerconsumption

• 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.• Pattern for power repeats every radians (T/2)

The RMS power, current, voltage are useful, DC-like quantities

rms )2τ( cycle wholea over inst of averageav PP P

Integrate:

R I dt)t(osc RI P 2m

T

0 2

12T

12mav

Pav is an “RMS” quantity:• “Root Mean Square”• Square a quantity (positive)• Average over a whole cycle• Compute square root.• In practice, divide peak value by sqrt(2) for Im or Emsince cos2(x) appears

RIPP rmsavrms2

2m

rmsI

I 2m

rmsE

E

2m

rmsV

V For any R, L, or C

Household power example: 120 volts RMS 170 volts peak

Z

I rmsrms

E

Integral = 1/2

cos2()


Power factor for AC CircuitsThe PHASE ANGLE determines the average RMS power actually absorbed due to the RMS current and applied voltage in the circuit.

Claim (proven below):Z

Irms

Erms

DtXL-XCR)cos(IIP P rms rmsrms rmsrmsav EE

ZR / )cos( factor" power" the is

Proof: start with instantaneous power (not very useful):

)tcos()tcos(I )t(I (t) )t(P DDmminst EE

Change variables:

{Integral}x I dt)tcos()tcos( I dt )t(P P mm0 DD1

mm0 inst1

av EE

Average it over one full period : 2 D

2 ,t x DD

dx)xcos()xcos( }Int{2

021

Use trig identity: )sin()x(ins )cos()x(osc )xcos(


Power factor for AC Circuits - continued

dx)xsin()xcos(2

)sin( dx)x(cos2

)cos( ntI2

0

2

0

121

2

)cos(I P mmav

E

)cos(IP P rms rmsrmsav E

Recall: RMS values = Peak values divided by sqrt(2)

Alternate form:

R I )cos(ZIP 2rms

2rmsrms

If R=0 (pure LC circuit) +/- /2 and Pav = Prms = 0

Also note: )cos( ZR ZI andrms rms E

2

)cos(

4

)x2sin(

2

x

2

)cos(

2

0

2

0

0 2

)x(sin

2

)sin(

2

0

2


fD = 200 Hz

Example 2 continued with RMS quantities:

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

RL

C

E

VR

VC

VL

Find Erms: V.71 /mrms 2EE

Find Irms at 200 Hz: before as 8118 Z

mA. 8.75 8118 V / 71 Z / I rmsrms E

Find the power factor:

0.369 8118

3000 Z/R)cos(

Find the phase angle :

as before 68 R

tan 0CL1

Find the average power:

Watts0.23 0.369 10 8.75 71 )cos(IP -3rms rmsav E

or Watts0.23 3000 10 8.75 RIP 2 3-2

rmsav

Recall: do not use arc-cos to find


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 F capacitor.

Find the capacitive reactance of the circuit:

Find the inductive reactance of the circuit:

The impedance of the circuit is:

The phase angle for the circuit is:

The RMS current in the circuit is:

The average power consumed in this circuit is:

If the inductance could be changed to maximize the current through the circuit, what would the new inductance L’ be?

How much RMS current would flow in that case?

Example 3 – Series LCR circuit analysis using RMS values

133)(377x2x10 / 1 C/X -5DC 1

188.5377x.5 LX DL

s/rad x .fD 377602862

74.7 ])X(X R[ Z 1/22CL

2

is positive since XL>XC (inductive)0.669)cos( ,48.0

R

XX )tan( 0CL

ZR / )cos( )cos(I P W.516 50 x(3.2) R I P wherermsrmsrmsor22

rmsrms E

A.3.2 74.7

240

ZI rmsrms

E

RESONANCE.at maximum a isCurrent

H. 0.352 10x2x377

1 C L'377

52 2

D

1

C'L

1D

A.4.8 50

V240

RI R Z rmsrmsresonancet A

E


TransformersDevices used to changeAC voltages. They have:• Primary• Secondary• Power ratings

power transformer

iron core

circuit symbol


Transformers

Faradays Law for primary and secondary:

dt

dNV

dt

dNV B

ssB

pp

Ideal Transformer

iron core• zero resistance in coils• no hysteresis losses in iron core• all field lines are inside core

Assume zero internal resistances,EMFs Ep, Es = terminal voltages Vp, Vs

Assume: The same amount of flux B cuts each turn in both primary and secondary windings in ideal transformer (counting self- and mutual-induction)

s

s

p

pB

N

V

N

V

dt

d

turn per

voltage induced

Assuming no losses: energy and power are conserved

p

s

s

p

N

N

I

I

pppconservedsss IVPIVP V N

NV p

p

ss

Turns ratio fixesthe step up or step down voltage ratio

Vp, Vs are instantaneous (time varying)but can also be regarded as RMS averages, as can be the power and current.


Example: A dimmer for lightsusing a variable inductance

f =60 Hz = 377 rad/sec

Light bulbR=50

Erms=30 VL

Without Inductor:0 Watts,18 R/P 2

rmsrms0, E

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

b) What would be the change in the RMS current?Without inductor: A 6.0

50

V30

RI rms

rms,0

E

P0,rms = 18 W.

With inductor: A 316.0527.0 A 6.0 )cos(R

I rmsrms

EPrms = 5 W.

Recall: )(cosP P 2rms0,rms R I P 2

rmsrms )cos(R

Z

I rmsrmsrms

EE

)(cosx Watts18 Watts5 2

0)(X ]R /X [ tan 58.2 CL-1o

L f 2 80.6 )tan(58.2 50 )tan(R X oL

mH 214 377 / 80.6 L

)58.2 cos( 0.527 )cos( o


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Copyright R. Janow – Spring 2015 Physics 121 - Electricity and Magnetism Lecture 14 - AC Circuits, Resonance Y&F Chapter 31, Sec. 3 - 8 The Series RLC