Copyright R. Janow – Fall 2015 Physics 121 - Electricity and Magnetism Lecture 14-15_E - AC Circuits, Resonance Y&F Chapter 31, Sec. 3 – 8, No Y&F reference

Copyright R. Janow – Fall 2015

Physics 121 - Electricity and Magnetism Lecture 14-15_E - AC Circuits, Resonance Y&F Chapter 31, Sec. 3 – 8, No Y&F reference for slides on Complex analysis

• The Series RLC Circuit. Amplitude and Phase Relations– Time domain, Real functions

• Phasor Diagrams for Voltage and Current• Impedance Magnitude and Phasors for Impedance• Resonance• Power in AC Circuits, Power Factor• Examples• Transformers• Series LCR Circuit Summary• Circuit Analysis using Complex Exponentials

– Imaginary and Complex Numbers– Complex Exponential Phasors and Rotations– Phasors as Solutions of Steady State Oscillator Equations– Phasor representation applied to current versus voltage in R. L. C.– Series LCR Circuit using Complex Phasors– Parallel LCR Circuit using Complex Phasors– Transient Solution of Damped Oscillator using Complex Phasors


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 circuitA 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.


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.

Example 1: Analyzing a Series RLC circuit

137706022 s Hz ).( fD

513) 758 471() 425()XX(R|Z| 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.

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.

)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 m

m

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|2/1 2

CL2

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

|Z|R / )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(|Z|IP 2rms

2rmsrms

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

Also note: )cos( |Z|R |Z|I 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

|Z|I 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| rmsrmsresonance tA

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


2bac4a2

1Im{z}

,a2

b}zRe{

Circuit Analysis using Complex Exponentials

1 roots has 1 z2

ac4ba2

1

a2

bz 2

• Roots of a quadratic equation az2 – bz +c = 0 are complex if b2<4ac:

i roots has 1 z2

1 i is “pure imaginary”

• EE’s use j instead of i (i is for current) • Complex conjugate: )zIm( i)zRe( z*

• Re{z} and Im{z} are both real numbers

)zIm( i )zRe( z • Complex number in rectangular form:

• Addition: )}zIm()zIm( { i )zRe()zRe( z z 112121

)zIm( 2 *zz )zRe( 2 *z z


Representation using the Complex Plane

)zIm( i)zRe( z

Im{z} y

)zRe( x

iy x z

Imaginary axis (y)

Real axis (x)x

y|z|

222 y x y)i(x iy)(x *zz r • Magnitude2:

• Polar form:

)sin( r y

)cos( r x

)sin( r i )cos( r z

|z| r

• Argument: ) y/x ( tan -1

“Argand diagram”

• Picture also displays complex-valued functions


Complex Exponentials: Euler’s Formula

• Taylor’s Series definition of exponential function:

complex) (including u all for 6!

u

5!

u

4!

u

3!

u

2!

u u 1 ue

65432

• Evaluate for u = i:

all for

6!

i

5!

i

4!

i

3!

i

2!

i i 1 ie

6655443322

...} 5!3!

{ i ...}6!4!2!

{1 ie53642

• Recognize series definitions of sine and cosine:

...6!4!2!

1 )cos(642

(radians) ... 5!3!

)sin(53

• Hence: )sin( i )cos( ie

Above is a complex number of magnitude 1, argument :

use i2n+1 = i(-1)2n

1 ie

ni2e 10e i2e

i 2/ie

*)i(e 1 ie

*)2/ie( i 2/ie

• Special Properties:


Complex Quantities in Polar & Rectangular Form

Multiply Euler’s Formula by magnitude r:

Product:

)( ie r r

ie r

ie r z z 21

212

21

121

•Magnitude of a product = product of individual magnitudes•Argument of a product = sum of individual arguments•Two complex entities are equal if and only if their amplitudes (magnitudes) are equal and their arguments (phases) are equal

DeMoivre’s Theorem: )}sin(ni){cos(nnrn ienrn} ier { nz

ie ien1n]i2e[ ie)n2( ie Periodicity 2:

n/)k2( ie1/nr 1/nz for integer n & k = 0,1….n-1

Sum:

)yi(y )x(x z z 212121 2/12

212

2121 })y(y )x(x { |z z|

)sin( r i )cos( r ie r z

)sin( r i )cos( r ie r *z

Polar form of arbitrary complex z

Complex conjugate of z


Rotation Operators, Evolution Operators, Phasors:

)( ier iez z'

Exponential factor ei rotates complex number z by in complex planeLet = t-t0)

• A(t)is often called the “time domain phasor” corresponding to A0cos(t+)• Re {A} = A0cos(t+) is the measurable instantaneous value of A(t)• Time dependent exponentials like eit are sometimes called evolution operators• The corresponding “frequency domain phasor” A (often simply called a “phasor” by EE’s) is the complex quantity:

)tt(i

e 0 evolves z from t0 to t for tf > t0

Im

Re

iez

iez

z

Let a(t) be a sinusoidally varying real quantity to be pictured as a vector in the complex plane rotating (counterclockwise) at angular frequency

})t( ie {A Re {A(t)} Re )tcos( A a(t) 00

ie A A 0


Complex exponentials (time domain phasors) are solutions of oscillator differential equations

Derivatives:

ie i d/)ied( ))/d(df()(fe d/))(f ed(

can be complex

chain rule

Complex exponentials are alternative solutions to sines or cosines in circuit differential equations describing steady state oscillations. Example: Simple Harmonic Oscillator

k/m x dt

xd 0

202

2

Second derivatives: set = t+amplitude A0

Ai )t)/dt(d( Ai dt/])t( ieA[d dA/dt 0

A dt/]Ai[d A/dtd 222

Solutions:

solution a also x*)/2(x cos(x) :note

ex (t)*x or ex x(t) )ti(0

)ti(0

)'tsin(x x(t) or )tcos(x x(t) 00


Phasor Definitions, continued

A phasor represents a sinusoidal, steady state signal whose amplitude A0, phase and frequency are time invariant.

}tie ) ie (A { Re a(t) 0

peak value of a(t), real

frequency domain phasor A, complex

time domain phasor, complex

sinusoidal value, time domain, real

Phasor Transform P:forward - time to frequency domain

inverse - frequency to time domain

ie A A :e.g. } a(t) { A 0 P

}tieARe{ a(t) :e.g. } A { a(t) P 1-

Advantage of phasor transform: For signals as above (sinusoidal, no transients), replace differentiation and integration in time domain differential equations by simple algebraic operations on phasors in frequency domain ( complex coefficients, common factors of eit cancel).

Process: Replace real variables in time dependent analysis problem with variables written using complex exponentials. Cancel common factors of eit

and solve remaining algebraic problem. Then return to real solution in time domain.


Complex Exponential Representation applied to Passive Circuit Elements: Revisit AC Voltage vs. Current in L, C, and

R

Note: now using j = sqrt(-1)

)t(jm

De)t( EE tjm

DeI)t(i The applied voltage & currentas time domain phasors:

0(t )vR - )t(ε

vR( t )ε

i(t))tDj(

m

tDjω

mR e eR )t(i)t(v R I E

equate magnitudes:

equate arguments:

R Imm E0 jtjtj DD

Resistor:

0(t )vL - )t(ε

vL(t)Lε

i(t)Inductor:

)t(jm

)tjDm

)tjmL

DDD eeLjI ]e[dt

dI L

dt

i dL (t ) v E

equate magnitudes:

equate arguments:

LmmDLmDm I L LI EE

2/ je e je jtjjtj DD

ε

i(t)

vC ( t )C

0(t )vC - )t(ε Capacitor:

)t(jm

tj

D

mtjmC

DDD eeCj

I dte

C

I dt)t(i

C

1

C

q(t) (t ) v

E

equate magnitudes:

equate arguments:

I C/1 C/I CmmDCmDm EE

2/ je e je jtjjtj DD


Revisit AC Voltage vs. Current in L, C, and Rusing Complex Exponential Representation - Continued

)t(jm

De)t( EE tjm

DeI)t(i & current:Time domain applied voltage:

For resistor: For inductor: For capacitor:

VR& IRm in phase Resistance

VL leads ILm by /2Inductive Reactance

RI/V RmR LI/V DLLmL C1I/V

DCCmC

VC lags ICm by /2Capacitive Reactance

Individual passive circuit elements:

e0 e+j/2 e-j/2

Relative phasor rotations of voltage drops relative to currents:

tDjω

RmR e)t(v R I 2/jtDj

LLmL eeI (t ) v 2/jtDj

CCmC eeI (t ) v

Voltage drop phasors (time domain) relative to currents:

Define (complex) impedance by z = v(t)/i(t) and |z| = Vmax/Imax

Impedances of simple circuit elements:

R Rz LL j z CC j z

R |z| R LL |z| CC |z|

Magnitudes of Impedances |z| = [zz*]1/2


Revisiting the series LCR circuit using phasors

RL

C

E

vR

vC

vL

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

AC voltage: )t(jm

De)t( EE tjm

DeI)t(i Current:

Use preceding formulas for vR, vL, vC. Currents are the same everywhere in an essential branch (series LCR):

CmLmRmm I I I I

(1) eeI eeI ee 2/jtj

Cm2/jtj

Lm

tDjω

m)t(j

mDDD R I E

2/jC

2/jL

j e e e|z| z

)t(i

(t) R E

j e j e :apply 2/j2/j

(1) ) j( e|z| z CLj R

Divide equation (1) above by Im and cancel common factor of ejt:

Im

Em

Dt

VL

VC

VR

time domain phasors

General definition: The complex impedance (or simply impedance) is the ratio of the voltage phasor to the current phasor (time or frequency domain).

]) ( [ *zz|z| 1/22CL00 2R

Magnitude of Z: Sketch shows > 0 for VL>VC XL>XC

At t = 0 sketch would showPhasors in frequency domain


Revisiting the series LCR circuit, continued #1

Phase angle for the circuit:

Re{z}

Im{z}

R

V

VV )tan( CL

R

CL

Impedance z above (Equation 1) is also the sum of the impedances of the 3 series elements

(1) ) j( |z|z CLj R

Impedance for Series LCR circuit section:

Z

XL-XC

R Re

Im

Sketch shows >0 for XL > XC

VL > VC

Z is complex but not a phasor, as it does not represent rotation; impedance is time independent.Previous “phasor diagrams” may have shown z rotating with Em.

Generalize: the equivalent impedance for basic circuit elements in series (arbitrarily many) is the sum of the individual (complex) impedances.

|z|

R

V )(osc

m

R E

Resonance: As before, z is real for XL= XC (2 = 1/LC). |z| is minimized.Current amplitude Im is maximized at resonance


Parallel LCR circuit using complex phasors

Kirchhoff Rules Application (time domain):

AC voltage: )t(jm

De)t( EEtj

mDeI)t(i Instantaneous Current:

All steady state voltages and currents oscillate at driving frequency D vCvL

R L C

E vR

iRiL iC

ia

b

Instantaneous voltages across parallel branches

have the same magnitude and phase

• 2 essential nodes “a” & “b” • 4 essential branches,• not all independent

(1) )t(i )t(i )t(i )t(i CLR Current Rule:

0)t(v)t( L E

0)t(v)t( R E0)t(v)t( C E

Loop Rule: • E-R-E• E-L-E• E-C-E )t(v )t(v )t(v )t( CLR E

Current Amplitudes within the Branches:

RR

V I Rm

Rm

E

LL

LmLm

V I

E

CC

CmCm

V I

EC1 DC

L DL


Parallel LCR circuit, continued #1Instantaneous currents in branches, phases relative to voltages:

2/j)t(jLmL eeI)t(i D

)t(jRmR

DeI)t(i

2/j)t(jCmC eeI)t(i D

• in phase with E(t)

• lags E(t) by /2

• leads E(t) by /2

IRm

Im

Em, VR, VL, VC

Dt

ICm-ILm

Re

Im

Sketch shows >0 for ILm > ICm XC>XL

Substitute currents into junction rule equation:

(2) eeI eeI eI eI )t(i 2/j)t(jCm

2/j)t(jLm

)t(jRm

tjm

DDDD

In current equation (2), cancel ejt factor and multiply by e-j2/j

C

m2/j

L

mm2/jCm

2/jLmRm

jm e e

R eI eI I eI

EEE

])I I ( I [ I 1/22CmLmRm

2m

Current amplitude addition rule (Pythagorean)

|Z|

I m

m

EAs before, the magnitude of the complex impedance

z is the ratio of peak EMF to peak currentand the impedance

I

|Z|

m

mE

e z phasor current

phasor voltage z

j


Parallel LCR circuit, continued #2

Substitute, then divide above by Em

z

1

e|z|

1

|z|

e

j

j

Note that: j e je 2/j2/j

Find 1/|z|: multiply 1/z by complex conjugate (1/z)* and take square root

1

1

j[ R

1

1

1 j[

R

1

*zz

1

|z|

1

LCLC002

(4)

1/221

1

2

R

1

|z|

1

LC

Phase angle :As before, is positive when applied voltage Em leads the current Im

R/1

X/11/X

I

II )tan( Lc

R

LmCmsketches in adjacentsketches in opposite

(5) ) X/11/X R( )tan( LC > 0 for X C > XL

(3) ]1

1

j[ R

1

z

1

LC

Sketch shows 0 for 1/XC > 1/XL

IC > IL XL > XC

1/z

1/XC-1/XL

1/RE

Im


Parallel LCR circuit, continued #3

Resonance: Maximize or minimize as function of frequency

(4)

1/221

1

2

R

1

|z|

1

LC

|Z|

I mm

E

Minimum of Im when XL = XC, i.e. when 2 = 1/LC Same frequency as series LCR, but current is

minimizedR |z| before as

R

1 as

|z|

1 Lim res

0R z also, R

1 as

z

1 Lim res

At resonance, the current amplitudes ILm and ICm in the L & C branches are equal, but 180o apart in phase. These cancel at all times at nodes a and b.

Generalize: the equivalent impedance for circuit elements or branches in parallel (arbitrarily many) is the sum of the reciprocals of the individual (complex) branch impedances.


Example: Use Complex Exponentials to solve a Differential Equation

a

L CE

+b

RRevisit Damped Oscillator: After “step response” to switch at ‘a’ saturates, turn switch to ‘b’. Decaying natural oscillations start for weak damping

R)t(idt

dUtot 2

(1) 1/LC 0)t(Qdt

dQ

L

R

dt

)t(Qd 0

202

2

• dissipation of potential energy• not a steady state system• solutions not simple sinusoids

second order equation for Q(t)

Try solution… (2) eQ)t(Q )t(j0

… but if s real solution oscillates forever

)t(QjejQdt

)t(dQ )t(j0 )t(Qe)j(Q

dt

)t(Qd 2)t(j202

2

Derivatives:

Substitute into Equation (1): 0)t(QQ L

Rj Q 2

02

(3) 0 L

Rj 2

02 Cancel common factors of Q: “characteristic

equation”

Phasor-like trial solution (2) turned differential equation into polynomial equation

yx j x and y assumed real Assume complex frequency


Use Complex Exponentials to solve Differential Equations #2

Equation (3) becomes 2 separate equations for real and imaginary terms

j2 2yyx

2x

2 Expand

(4.2) 0 L

R 2

0y2y

2x Re{Eq 3}

Substitute wy:

(4.1) 0 L

R2 xyx Im{Eq 3}

2L

R y

0 L2

R 2

0

22x

(5) 2

2LR

20x

Shifted natural frequency x

is

real for underdamping

imaginary for overdamping

2

2LR

20

2

2LR

20

For real x , underdamping: tjω

t)jj0

jt)j(j0

xy2

yx ee eQ e eQ)t(Q

(6) ee Q )t(Q )tj(ω

tL2

R

0x

damping oscillation

(7) )tcos(e Q )t(Q Re x t

L2

R

0

ee Q )t(Q t

L2

Rj

0

(8) e )cos( Q )t(Q Re t

L2

R

0

For critical damping, x = 0: 2

2LR

20 No oscillations, decay

only:



Use Complex Exponentials to solve Differential Equations #3

For overdamping: 2

2LR

20 x above imaginary. Equations 4.1 & 4.2 invalid

Return to Eq. 3.0, assume frequency ispure imaginary ( no oscillation)

2y

2yy )'( real ' 'j

Eq. 3.0 becomes: 0 'L

R )'( 2

0y2

y quadratic, real coefficients

Solution: (9) j 2L

Rj 'j 2

02

2LR

y two rootsboth pure imaginary

Both roots lead to decay w/o oscillation + root: + implies damping faster than e-rt/2L

- root: - implies damping slower than e-rt/2L but not growth

Most general solution: linear combination of Q+ and Q-, each of form of Eq. 2.0

t 2LR

t 2LR

21L2/Rtj

0 21

20

220

2

e e eeQ )t(Q )t(Q )t(Q

Recall definition, hyperbolic cosine:

cosh(x) e e 2

1 xx

(10) ) t 20

2 ( cosh e )cos(Q )t(QRe 2L

RRt/2L-0

Correctly reduces to critical damped Eq. 8.0

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Copyright R. Janow – Fall 2015 Physics 121 - Electricity and Magnetism Lecture 14-15_E - AC Circuits, Resonance Y&F Chapter 31, Sec. 3 – 8, No Y&F reference