ac circuits with rlc componentsphy2049: chapter 31 18 ac circuits with rlc components Îenormous...
TRANSCRIPT
PHY2049: Chapter 31 18
AC Circuits with RLC ComponentsEnormous impact of AC circuits
Power deliveryRadio transmitters and receiversTunersFiltersTransformers
Basic componentsRLCDriving emf
Now we will study the basic principles
PHY2049: Chapter 31 19
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 20
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 21
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 22
Solve for φ and Im
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”
PHY2049: Chapter 31 23
Graphical Representation: R, XL, XC, Z
tan L CX XR
φ−
=
L CX X−φ
( )22L CZ R X X= + −
PHY2049: Chapter 31 24
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 25
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 26
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=
VCIC
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 90C di I tω= + ° ωdt + 90
ωdt1/C dX Cω=
PHY2049: Chapter 31 27
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 90L di I tω= − °ωdt
L dX Lω=
PHY2049: Chapter 31 28
What is Reactance?Think of it as a frequency-dependent resistance
Shrinks with increasing ωd
1C
d
XCω
=
L dX Lω=
( " " )RX R=
Grows with increasing ωd
Independent of ωd
PHY2049: Chapter 31 29
QuizThree identical EMF sources are hooked to a single circuit element, a resistor, a capacitor, or an inductor. The current amplitude is then measured vs frequency. Which curve 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 30
AC Source and RLC CircuitVoltage is
R, L, C all present
Relation of current and voltage
Current “lags” voltage by φImpedance: Due to R, XC and XL
Calculate Im and φSee next slide
VR
VC
sinm dtε ε ω=
VL
εm
Im
ωdt − φ
( )sinm di I tω φ= −
I
PHY2049: Chapter 31 31
AC Source and RLC Circuit (2)Solution using trig (or geometry)
Magnitude = Im, lags emf by phase φ)
( ) ( )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
=− −
= =
= + − = + −
( )sinm di I tω φ= −
PHY2049: Chapter 31 32
AC Source and RLC Circuit (3)
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 33
Graphical Representation: VR, VL, VC, εm
tan L C L C
R
X X V VR V
φ− −
= =
L CV V−φ
mε
( )22m R L CV V Vε = + −
PHY2049: Chapter 31 34
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 35
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 36
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 37
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 38
RLC Example 2 (cont)
0.164 86.7 14.2VL m LV I X= = × =
0.164 177 29.0VC m CV I X= = × =
0.164 200 32.8VR mV I R= = × =
( )2232.8 14.2 29.0 36.0 mε+ − = =
L CV V−φ
mε
PHY2049: Chapter 31 39
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=
0/dω ω
PHY2049: Chapter 31 40
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 R 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 41
Power in AC Circuits (2)Recall power formula
Rewrite
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 rms rms cos
ZRP I R I IZ
εε ε φ⎛ ⎞= = =⎜ ⎟
⎝ ⎠
ave rms rms cosP Iε φ=
R
L CX X−Z
φcos R
Zφ =