Chapter 31: RLC Circuits - University of · PDF fileÎDamped oscillations in RLC circuits ... RMS quantities ÎForced oscillations Resistance, reactance, impedance Phase shift Resonant

PHY2049: Chapter 31 1

Chapter 31: RLC Circuits


TopicsLC Oscillations

Conservation of energy

Damped oscillations in RLC circuitsEnergy loss

AC currentRMS quantities

Forced oscillationsResistance, reactance, impedancePhase shiftResonant frequencyPower

TransformersImpedance matching


LC OscillationsWork out equation for LC circuit (loop rule)

Rewrite using i = dq/dt

ω (angular frequency) has dimensions of 1/t

Identical to equation of mass on spring

LC0q diLC dt

− − =

2 22

2 20 0d q q d qL qCdt dt

ω+ = ⇒ + =

2 22

2 20 0d x d xm kx xdt dt

ω+ = ⇒ + =

1LC

ω =

km

ω =


LC Oscillations (2)Solution is same as mass on spring ⇒ oscillations

qmax is the maximum charge on capacitorθ is an unknown phase (depends on initial conditions)

Calculate current: i = dq/dt

Thus both charge and current oscillateAngular frequency ω, frequency f = ω/2πPeriod: T = 2π/ω

( )max cosq q tω θ= +

( ) ( )max maxsin sini q t i tω ω θ ω θ= − + = − +

km

ω =


Plot Charge and Current vs t

( )q t

( )i t

1 2Tω π= =


Energy OscillationsTotal energy in circuit is conserved. Let’s see why

0di qLdt C+ =

0di q dqL idt C dt

+ = Multiply by i = dq/dt

Equation of LC circuit

( ) ( )2 21 02 2L d di q

dt C dt+ = Use

22dx dxx

dt dt=

221 1

2 2 0d qLidt C⎛ ⎞

+ =⎜ ⎟⎜ ⎟⎝ ⎠

UL + UC = const

221 1

2 2 constqLiC

+ =


Oscillation of EnergiesEnergies can be written as (using ω2 = 1/LC)

Conservation of energy:

Energy oscillates between capacitor and inductorEndless oscillation between electrical and magnetic energyJust like oscillation between potential energy and kinetic energy for mass on spring

( )22

2max cos2 2C

qqU tC C

ω θ= = +

( ) ( )2

2 2 2 2 2max1 1max2 2 sin sin

2LqU Li L q t t

Cω ω θ ω θ= = + = +

2max const2C L

qU UC

+ = =


Plot Energies vs t( )CU t ( )LU t Sum


LC Circuit ExampleParameters

C = 20μFL = 200 mHCapacitor initially charged to 40V, no current initially

Calculate ω, f and Tω = 500 rad/sf = ω/2π = 79.6 HzT = 1/f = 0.0126 sec

Calculate qmax and imaxqmax = CV = 800 μC = 8 × 10-4 Cimax = ωqmax = 500 × 8 × 10-4 = 0.4 A

Calculate maximum energiesUC = q2

max/2C = 0.016J UL = Li2max/2 = 0.016J

( )( )51/ 1/ 2 10 0.2 500LCω −= = × =


LC Circuit Example (2)Charge and current

Energies

Voltages

Note how voltages sum to zero, as they must!

( )0.0008cos 500q t= ( )0.4sin 500i t= −

( ) ( )2 20.016cos 500 0.016sin 500C LU t U t= =

( )/ 40cos 500CV q C t= =

( ) ( )max/ cos 500 40cos 500LV Ldi dt L i t tω= = − = −


RLC CircuitWork out equation using loop rule

Rewrite using i = dq/dt

Solution slightly more complicated than LC case

This is a damped oscillator (similar to mechanical case)Amplitude of oscillations falls exponentially

0di qL Ridt C+ + =

2

2 0d q R dq qL dt LCdt

+ + =

( ) ( )2/ 2max cos 1/ / 2tR Lq q e t LC R Lω θ ω− ′ ′= + = −


Charge and Current vs t in RLC Circuit

( )q t( )i t / 2tR Le−


RLC Circuit ExampleCircuit parameters

L = 12mL, C = 1.6μF, R = 1.5Ω

Calculate ω, ω’, f and Tω = 7220 rad/sω’ = 7220 rad/sf = ω/2π = 1150 HzT = 1/f = 0.00087 sec

Time for qmax to fall to ½ its initial valuet = (2L/R) * ln2 = 0.0111s = 11.1 ms# periods = 0.0111/.00087 ≈ 13

( )( )61/ 0.012 1.6 10 7220ω −= × =

( )227220 1.5/ 0.024ω ω′ = −

/ 2 1/ 2tR Le− =


RLC Circuit (Energy)

0di qL Ridt C+ + = Basic RLC equation

Multiply by i = dq/dt2 0di q dqL i Ridt C dt

+ + =

22 21 1

2 2d qLi i Rdt C⎛ ⎞

+ = −⎜ ⎟⎜ ⎟⎝ ⎠

Collect terms(similar to LC circuit)

( ) 2L C

d U U i Rdt

+ = −Total energy in circuitdecreases at rate of i2R(dissipation of energy)

/tot

tR LU e−∼


Energy in RLC Circuit

( )CU t( )LU t

Sum

/tR Le−


QuizBelow are shown 3 LC circuits. Which one takes the least time to fully discharge the capacitors during the oscillations?

(1) A(2) B(3) C

A B C

C C CC

C

1/ LCω =C has smallest capacitance, therefore highestfrequency, therefore shortest period


AC CircuitsEnormous impact of AC circuits

Power deliveryRadio transmitters and receiversTunersFiltersTransformers

Basic componentsRLCDriving emf

Now we will study the basic principles


AC Circuits and Forced OscillationsRLC + “driving” EMF with angular frequency ωd

General solution for current is sum of two terms

sinm dtε ε ω=

sinm ddi qL Ri tdt C

ε ω+ + =

“Transient”: Fallsexponentially & disappears

“Steady state”:Constant amplitude

Ignore

/ 2 costR Li e tω− ′∼


Steady State SolutionAssume steady state solution of form

Im is current amplitudeφ is phase by which current “lags” the driving EMFMust determine Im and φ

Plug in solution: differentiate & integrate sin(ωt-φ)

( )sinm di I tω φ= −

( ) ( ) ( )cos sin cos sinmm d d m d d m d

d

II L I R t t tC

ω ω τ φ ω φ ω φ ε ωω

− + − − − =

sinmdi qL Ri tdt C

ε ω+ + =


( )cosd m ddi I tdt

ω ω φ= −

( )cosmd

d

Iq tω φω

= − −

Substitute


Steady State Solution for AC Current (2)

Expand sin & cos expressions

Collect sinωdt & cosωdt terms separately

These equations can be solved for Im and φ (next slide)

( )( )

1/ cos sin 0

1/ sin cosd d

m d d m m

L C R

I L C I R

ω ω φ φ

ω ω φ φ ε

− − =

− + =

( )( )

sin sin cos cos sin

cos cos cos sin sind d d

d d d

t t t

t t t

ω φ ω φ ω φ

ω φ ω φ ω φ

− = −

− = +High school trig!

cosωdt terms

sinωdt terms

( ) ( ) ( )cos sin cos sinmm d d m d d m d

d

II L I R t t tC

ω ω τ φ ω φ ω φ ε ωω

− + − − − =


Solve for φ and Im in terms of

R, XL, XC and Z have dimensions of resistance

Let’s try to understand this solution using “phasors”

Steady State Solution for AC Current (3)

1/tan d d L CL C X XR R

ω ωφ − −= ≡ m

mIZε

=

( )22L CZ R X X= + −

L dX Lω=

1/C dX Cω=

Inductive “reactance”

Capacitive “reactance”

Total “impedance”

( )( )

1/ cos sin 0

1/ sin cosd d

m d d m m

L C R

I L C I R

ω ω φ φ

ω ω φ φ ε

− − =

− + =Same equations


Understanding AC Circuits Using PhasorsPhasor

Voltage or current represented by “phasor”Phasor rotates counterclockwise with angular velocity = ωd

Length of phasor is amplitude of voltage (V) or current (I)y component is instantaneous value of voltage (v) or current (i)

εmIm

ωdt − φ


sinm dtε ε ω= i

ε

Current “lags” voltage by φ


AC Source and Resistor OnlyVoltage is

Relation of current and voltage

Current is in phase with voltage (φ = 0)

IR

i

ε R~sin /R d R Ri I t I V Rω= =

VR

sinR R dv iR V tω= =

ωdt


AC Source and Capacitor OnlyVoltage is

Differentiate to find current

Rewrite using phase


“Capacitive reactance”:Current “leads” voltage by 90°

sinC dq CV tω=

VC

IC

i

ε C~/ cosd C di dq dt CV tω ω= =

/ sinC C dv q C V tω= =

( )sin 90d C di CV tω ω= + °

( )sin 90 /C d C C Ci I t I V Xω= + ° =ωdt + 90ωdt

1/C dX Cω=


AC Source and Inductor OnlyVoltage is

Integrate di/dt to find current:

Rewrite using phase


“Inductive reactance”:Current “lags” voltage by 90°

( )/ / sinL ddi dt V L tω=

VL

IL

i

ε L~( )/ cosL d di V L tω ω= −

/ sinL L dv Ldi dt V tω= =

( ) ( )/ sin 90L d di V L tω ω= − °

ωdt − 90( )sin 90 /L d L L Li I t I V Xω= − ° = ωdt

L dX Lω=


What is Reactance?Think of it as a frequency-dependent resistance

ωd → 0, XC → ∞- Capacitor looks like a breakωd → ∞, XC → 0- Capacitor looks like a wire (“short”)

1C

d

XCω

=

L dX Lω=

( " " )RX R=

ωd → 0, XL → 0- Inductor looks like a wire (“short”)

ωd → ∞, XL → ∞- Inductor looks like a break

Independent of ωd


QuizThree identical EMF sources are hooked to a single circuit element, a resistor, a capacitor, or an inductor. The current amplitude is then measured as a function of frequency. Which one of the following curves corresponds to an inductive circuit?

(1) a(2) b(3) c(4) Can’t tell without more info

fd

Im

a

c

b

L dX Lω= For inductor, higher frequency gives higherreactance, therefore lower current


AC Source and RLC CircuitVoltage is


Current “lags” voltage by φImpedance: Due to R, XC and XL

Calculate Im and φ using geometrySee next slide

VR

VC

sinm dtε ε ω=

VL

εm

Im

ωdt − φ



AC Source and RLC Circuit (2)Right triangle with sides VR, VL-VC and εm

Solve for current: (Magnitude = Im, lags emf by phase φ)

VR

( ) ( )2 22 2

/1/tan

1/

m m

L C d d

L C d d

I ZX X L C

R R

Z R X X R L C

εω ωφ

ω ω

=− −

= =

= + − = + −

VL−VC

εm

φ

tan L C

R

V VV

φ −=

( )22 2m R L CV V Vε = + −

R m

L m L

C m C

V I RV I XV I X

=

=

=



AC Source and RLC Circuit (3)

Only XL − XC is relevant, reactances cancel each other

When XL = XC, then φ = 0Current in phase with emf, “Resonant circuit”:Z = R (minimum impedance, maximum current)

When XL < XC, then φ < 0Current leads emf, “Capacitive circuit”: ωd < ω0

When XL > XC, then φ > 0Current lags emf, “Inductive circuit”: ωd > ω0

tan L CX XR

φ −=

0 1/d LCω ω= =

( )22L CZ R X X= + −


RLC Example 1Below are shown the driving emf and current vs time of an RLC circuit. We can conclude the following

Current “leads” the driving emf (φ<0)Circuit is capacitive (XC > XL)ωd < ω0

εI

t


QuizWhich one of these phasor diagrams corresponds to an RLC circuit dominated by inductance?

(1) Circuit 1(2) Circuit 2(3) Circuit 3

εm εm εm1 32

Inductive: Current lags emf, φ>0


QuizWhich one of these phasor diagrams corresponds to an RLC circuit dominated by capacitance?

(1) Circuit 1(2) Circuit 2(3) Circuit 3

εm εm εm1 32

Capacitive: Current leads emf, φ<0


RLC Example 2R = 200Ω, C = 15μF, L = 230mH, εm = 36v, fd = 60 Hz

ωd = 120π = 377 rad/s

Natural frequency ( )( )60 1/ 0.230 15 10 538rad/sω −= × =

377 0.23 86.7LX = × = Ω

( )61/ 377 15 10 177CX −= × × = Ω

( )22200 86.7 177 219Z = + − = Ω

/ 36 / 219 0.164Am mI Zε= = =

XL < XCCapacitive circuit

1 86.7 177tan 24.3200

φ − −⎛ ⎞= = − °⎜ ⎟⎝ ⎠

Current leads emf(as expected)


RLC Example 3Circuit parameters: C = 2.5μF, L = 4mH, εm = 10v

ω0 = (1/LC)1/2 = 104 rad/sPlot Im vs ωd / ω0

R = 5Ω

R = 10Ω

R = 20ΩIm Resonance

0dω ω=


Power in AC CircuitsInstantaneous power emitted by circuit: P = i2R

More useful to calculate power averaged over a cycleUse <…> to indicate average over a cycle

Define RMS quantities to avoid ½ factors in AC circuits

House currentVrms = 110V ⇒ Vpeak = 156V

( )2 2sinm dP I tω φ= −

( )2 2 212sinm d mP I R t I Rω φ= − =

rms 2mII = rms 2

mεε = 2ave rmsP I R=

Instantaneous power oscillates


Power in AC Circuits (2)Recall power formula

Rewrite

Based on

cosφ is the “power factor”To maximize power delivered to circuit ⇒ make φ close to zeroMost power delivered to load happens near resonanceE.g., too much inductive reactance (XL) can be cancelled by increasing XC (decreasing C)

2ave rmsP I R=

rmsave rms rms rms cos

ZP I R Iε ε φ= =

cos mR

m m

I RV RI Z Z

φε

= = =

ave rms rms cosP Iε φ= cos RZ

φ =


Power Example 1R = 200Ω, XC = 150Ω, XL = 80Ω, εrms = 120v, fd = 60 Hz

ωd = 120π = 377 rad/s

( )22200 80 120 211.9Z = + − = Ω

1 80 150tan 19.3200

φ − −⎛ ⎞= = − °⎜ ⎟⎝ ⎠

cos 0.944φ =

ave rms rms cos 120 0.566 0.944 64.1WP Iε φ= = × × =

rms rms / 120 / 211.9 0.566AI Zε= = =

2 2ave rms 0.566 200 64.1WP I R= = × =

Current leads emfCapacitive circuit

Same


Power Example 1 (cont)R = 200Ω, XC = 150Ω, XL = 80Ω, εrms = 120v, fd = 60 Hz

How much capacitance must be added to maximize the power in the circuit (and thus bring it into resonance)?

Want XC = XL to minimize Z, so must decrease XC

So we must add 15.5μF capacitance to maximize power

150 1/ 17.7μFC dX C Cω= Ω = =

new new80 33.2μFC LX X C= = Ω =


Q FactorWe can quantify low damping situations using Q factor

Define using values at resonance

Can also be expressed (using resonance values) as

Energy in circuit2Energy lost per cycle

Q π≡

( )

2102

210 2

22 /

m

m

LI LQRI R

ωπ

π ω= =

×

CL XXQR R

= =


Power Example 2Circuit parameters: C = 2.5μF, L = 4mH, εm = 10v

ω0 = (1/LC)1/2 = 10000 rad/sPlot Power vs ωd / ω0 for different R values, using Q

R = 5Ω Q = 8

R = 10Ω Q = 4

R = 20Ω Q = 20dω ω=

Pave

R = 2Ω Q = 20

Resonance


Q Factor (cont)

For Q > 3 or so, can easily show that

( )

22 rms

ave rms 2 21/

RP I RL C R

ε

ω ω= =

− +

FWHM = Full Width at Half Maximum

0FWHM

Q ω≈


Radio TunerSet RLC tuner to 103.7 (ugh!)

Circuit response Q = 500. Maximized for f = 103.7Other radio stations.

RLC response is less


QuizA generator produces current at a frequency of 60 Hz with peak voltage and current amplitudes of 100V and 10A, respectively. What is the average power produced?

(1) 1000 W(2) 707 W(3) 1414 W(4) 500 W(5) 250 W

1ave peak peak rms rms2P I Iε ε= =


QuizThe figure shows the current and emf of a series RLC circuit. To increase the rate at which power is delivered to the resistive load, which option should be taken?

(1) Increase R(2) Decrease L(3) Increase L(4) Increase C

Current lags applied emf (peak occurs later, φ > 0), thus circuit is inductive. Max power is at φ = 0, so need to either (1) reduce XL by decreasing L or (2) cancel XL by increasing XC (decrease C).


If you wanted to increase the power delivered to this RLC circuit, which modification(s) would work?

(a) increase R

(b) increase C

(c) increase L

RLC Circuit Example

(d) decrease R

(e) decrease C

(f) decrease L

Again, current lags emf.See previous page for details.


Example (Prob. 31-48)Variable frequency EMF source with εm=6V connected to a resistor and inductor. R=80Ω and L=40mH.

At what frequency fd does VR = VL?

At that frequency, what is phase angle φ?

What is current amplitude?

VRVL

εm

Im2000rad/s

/ 2 318Hzd d

d d

L Rfω ω

ω π= ⇒ =

= =

tan 1 45φ φ= ⇒ = °

2 2/ 80 80 6 /113 0.053Am mI ε= + = =

Documents

Chapter 31: RLC Circuits - University of · PDF fileÎDamped oscillations in RLC circuits ... RMS quantities ÎForced oscillations Resistance, reactance, impedance Phase shift Resonant