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PHY2049: Chapter 31 1
Chapter 31: RLC Circuits
PHY2049: Chapter 31 2
TopicsLC Oscillations
Conservation of energy
Damped oscillations in RLC circuitsEnergy loss
AC currentRMS quantities
Forced oscillationsResistance, reactance, impedancePhase shiftResonant frequencyPower
TransformersImpedance matching
PHY2049: Chapter 31 3
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 =
PHY2049: Chapter 31 4
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 =
PHY2049: Chapter 31 5
Plot Charge and Current vs t
( )q t
( )i t
1 2Tω π= =
PHY2049: Chapter 31 6
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
+ =
PHY2049: Chapter 31 7
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
+ = =
PHY2049: Chapter 31 8
Plot Energies vs t( )CU t ( )LU t Sum
PHY2049: Chapter 31 9
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 −= = × =
PHY2049: Chapter 31 10
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= = − = −
PHY2049: Chapter 31 11
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− ′ ′= + = −
PHY2049: Chapter 31 12
Charge and Current vs t in RLC Circuit
( )q t( )i t / 2tR Le−
PHY2049: Chapter 31 13
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− =
PHY2049: Chapter 31 14
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−∼
PHY2049: Chapter 31 15
Energy in RLC Circuit
( )CU t( )LU t
Sum
/tR Le−
PHY2049: Chapter 31 16
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
PHY2049: Chapter 31 17
AC CircuitsEnormous impact of AC circuits
Power deliveryRadio transmitters and receiversTunersFiltersTransformers
Basic componentsRLCDriving emf
Now we will study the basic principles
PHY2049: Chapter 31 18
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− ′∼
PHY2049: Chapter 31 19
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+ + =
( )sinm di I tω φ= −
( )cosd m ddi I tdt
ω ω φ= −
( )cosmd
d
Iq tω φω
= − −
Substitute
PHY2049: Chapter 31 20
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
− + − − − =
PHY2049: Chapter 31 21
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
PHY2049: Chapter 31 22
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 di I tω φ= −
sinm dtε ε ω= i
ε
Current “lags” voltage by φ
PHY2049: Chapter 31 23
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
PHY2049: Chapter 31 24
AC Source and Capacitor OnlyVoltage is
Differentiate to find current
Rewrite using phase
Relation of current and voltage
“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=
PHY2049: Chapter 31 25
AC Source and Inductor OnlyVoltage is
Integrate di/dt to find current:
Rewrite using phase
Relation of current and voltage
“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=
PHY2049: Chapter 31 26
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
PHY2049: Chapter 31 27
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
PHY2049: Chapter 31 28
AC Source and RLC CircuitVoltage is
Relation of current and voltage
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
( )sinm di I tω φ= −
PHY2049: Chapter 31 29
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
=
=
=
( )sinm di I tω φ= −
PHY2049: Chapter 31 30
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= + −
PHY2049: Chapter 31 31
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
PHY2049: Chapter 31 32
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
PHY2049: Chapter 31 33
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
PHY2049: Chapter 31 34
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)
PHY2049: Chapter 31 35
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=
PHY2049: Chapter 31 36
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
PHY2049: Chapter 31 37
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 =
PHY2049: Chapter 31 38
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
PHY2049: Chapter 31 39
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 =
PHY2049: Chapter 31 40
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
= =
PHY2049: Chapter 31 41
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
PHY2049: Chapter 31 42
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≈
PHY2049: Chapter 31 43
Radio TunerSet RLC tuner to 103.7 (ugh!)
Circuit response Q = 500. Maximized for f = 103.7Other radio stations.
RLC response is less
PHY2049: Chapter 31 44
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= =
PHY2049: Chapter 31 45
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).
PHY2049: Chapter 31 46
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.
PHY2049: Chapter 31 47
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= + = =