copyright r. janow – spring 2015 physics 121 - electricity and magnetism lecture 14 - ac circuits,...
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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 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
Copyright R. Janow – Spring 2015
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
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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:
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• 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
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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
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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.
Copyright R. Janow – Spring 2015
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(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.
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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.
Copyright R. Janow – Spring 2015
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
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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
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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
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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()
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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(
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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
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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
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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
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TransformersDevices used to changeAC voltages. They have:• Primary• Secondary• Power ratings
power transformer
iron core
circuit symbol
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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.
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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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