Download Lecture14: AC Circuits , Resonance

Physics 121 - Electricity and Magnetism
Lecture 14 - AC Circuits, Resonance
Y&F Chapter 31, Sec. 3 - 8
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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 – Fall 2015
Relative Current/Voltage Phases in pure R, C, and L elements
• Apply sinusoidal current i (t) = Imcos(wDt)
• For pure R, L, or C loads, phase angles for voltage drops are 0, p/2, -p/2
• “Reactance” means ratio of peak voltage to peak current (generalized resistances).
VR& Im in phase
Resistance
VR / Im  R
VC lags Im by p/2
Capacitive Reactance
VC / Im   C  1
wDC
VL leads Im by p/2
Inductive Reactance
VL / Im  L  wDL
Phases of voltages in series components are referenced to the current phasor
Same
Phase
currrent
Copyright R. Janow – Fall 2015
Phasors applied to a Series LCR circuit
E
vR
Applied EMF: E(t)  Emax cos(wDt  F)
R
Current: i(t)  Imax cos(wDt)
L
Same everywhere in the single essential branch
vL
• Same frequency dependance as E (t)
C
• Same phase for the current in E, R, L, & C, but......
• Current leads or lags E (t) by a constant phase angle F
vC
Phasors all rotate CCW at frequency wD
• Lengths of phasors are the peak values (amplitudes)
F
Em
Im
wDt+F
• The “x” components are instantaneous values measured
wDt
VL
Em
Im
F
wDt
VR
VC
Kirchoff Loop rule for series LRC:
E(t)  vR (t)  vL (t)  vC (t)  0
Component voltage phasors all rotate at wD with phases
relative to the current phasor Im and magnitudes below :
VR  ImR
• VR has same phase as Im
• VC lags Im by p/2
VC  ImXC
• VL leads Im by p/2
VL  ImXL




Em  VR  (VL  VC )
along Im


VC lags VL by 1800
perpendicular
to I
Copyright R. Janowm– Fall 2015
Voltage addition rule for series LRC circuit
2
Magnitude of Em Em
 VR2  (VL  VC )2
Em
in series circuit:
1


Reactances: L  wDL
C
VL-VC
wDC
IL L  VL
ICC  VC
IRR  VR
XL-XC
Same current amplitude in each component:
Im
F
Z
VR
R
Im  IR  IL  IC
Impedance magnitude is the
E
ratio of peak EMF to peak
|Z|  m
Im
current:
peak applied voltage
peak current
wDt
[ Z ]  ohms
For series LRC circuit, divide each voltage in |Em| by (same) peak current
2
2 1/2
Magnitude of Z: | Z |  [ R  (L  C ) ]
Phase angle F:
tan(F) 
VL  VC
  C
 L
VR
R
F measures the power absorbed by the circuit:
Applies to a single series
branch with L, C, R
See phasor diagram
 
P  Em  Im  EmIm cos(F)
• R ~ 0  tiny losses, no power absorbed  Im normal to Em  F ~ +/- p/2
Copyright(resonance)
R. Janow – Fall 2015
• XL=XC  Im parallel to Em  F  0  Z=R  maximum current
Summary: AC Series LCR Circuit
vR
E
R
L
vL
C
vC
Circuit
Element
Symbol
Resistance
or Reactance
Phase of
Current
Phase
Angle
Amplitude
Relation
Resistor
R
R
In phase
with VR
0º (0 rad)
VR = ImR
Capacitor
C
XC=1/wdC
Leads VC
by 90º
-90º (-p/2)
VC = ImXC
Inductor
L
XL=wdL
Lags VL
by 90º
+90º (p/2)
VL = ImXL
E(t)  Emax cos(wDt  F)
i(t)  Im cos(wDt)
Em
Im
F
Pav  Prms  ErmsI rms cos(F)
VL-VC
Z
VR
XL-XC
sketch shows
XL > XC
Em  Im | Z |
R
wDt
| Z |  [ R2  (XL  XC )2 ]1/2
tan(F ) 
VL  VC
X  XC
 L
VR
R
Copyright R. Janow – Fall 2015
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)
(B)
(C)
(D)
(E)
Determine the impedance of the circuit.
Find the amplitude of the current (peak value).
Find the phase angle between the current and voltage.
Find the instantaneous current across the RLC circuit.
Find the peak and instantaneous voltages across each circuit element.
Copyright R. Janow – Fall 2015
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.
Angular frequency:
wD  2pf  2p (60.0) Hz 
Resistance:
Inductive reactance:
Capacitive reactance:
377 s1
R  425 
L  wDL  (377 s1 )(1.25 H)  471 
C  1 / wDC  1 /( 377 s1 )(3.50  106 F)  758 
| Z | R 2  (XL  X C )2  (425 )2  (471   758 )2  513 
(B) Find the peak current amplitude:
Im 
m
150 V

 0.292 A
| Z | 513 
(C) Find the phase angle between the current and voltage.
XC > XL (Capacitive)
F  tan1(
Current phasor Im leads the Voltage Em
Phase angle should be negative
XL  X C
471   758 
)  tan1(
)  34.0  0.593 rad.
Copyright R. Janow – Fall 2015
R
425 
Example 1:
Analyzing a series RLC circuit - continued
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.
(D)
Find the instantaneous current across the RLC circuit.
i(t)  Im cos(wDt)  0.292cos(377t)
(E) Find the peak and instantaneous voltages across each circuit element.
VR ,m  ImR  (0.292 A)( 425 )  124 V
VR in phase with Im
VR leads Em by |F|
vR (t)  VR ,m cos(wD t)  (124 V) cos(377t)
VL,m  ImXL  (0.292 A)( 471 )  138 V
vL (t)  VL,m cos(wD t  p / 2)  (138 V) cos(377t  p / 2)
VC,m  ImX C  (0.292 A)(758 )  222 V
VL leads VR by p/2
VC lags VR by p/2
vC (t)  VC,m cos(wD t  p / 2)  (222 V) cos(377t  p / 2)
Note that:
VR  VL  VC  483V  150V  Em
Voltages add with proper phases:

Why not?


2 1/ 2
2
Em  VR  VLCopyright
 VC R. Janow
 150
V
– Fall 2015
Example 2: Resonance in a series LCR Circuit:
Em = 100 V.
R = 3000  L = 0.33 H
C = 0.10 mF
Find Z and F for fD = 200 Hertz, fD = 876 Hz, & fD = 2000 Hz
vR
E
R
L
vL
Why should fD make X  w L
D
L
a difference?
XC 
C
| Z |  [ R2  (XL  XC )2 ]1/2
vC
Frequency
f
Resistance
R
Reactance
XC
200 Hz
3000 
7957 
876 Hz
3000 
2000 Hz
3000 
X C  XL
Im F < 0
Reactance
XL
Im 
wD C
Em
|Z|
XL  X C
)
R
Impedance
|Z|
Phase
Angle F
Circuit
Behavior
415 
8118 
- 68.3º
Capacitive
Em lags Im
1817 
1817 
3000 
Resonance
0º
Resistive
Max current
796 
4147 
4498 
+48.0º
Inductive
Em leads Im
X C  XL
Em
Em
F  tan1 (
1
Im
F0
XL  X C
Em F  0
Im
Copyright R. Janow – Fall 2015
Resonance in a series LCR circuit
E
R
L

| Z |  R  L   C 
2
C  1 / wDC
C

2 1/ 2
L  wDL
Im 
Em
|Z|
R  resistance
Vary wD: At resonance maximum current flows & impedance is minimized
XC  XL
when
wD  1/ LC  wres
 | Z |  R, Im  EM / R, F  0
width of resonance (selectivity, “Q”) depends on R.
Large R  less selectivity, smaller current at peak
capacitance dominates
current leads voltage
inductance dominates
current lags voltage
damped spring oscillator
near resonance
Copyright R. Janow – Fall 2015
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
Copyright R. Janow – Fall 2015
Power Dissipation in a Resistor
2
2
2
Instantaneous power  P

i
(
t
)
R

I
R
c
os
(wt)
inst
m
dissipation
2
cos (q)
• 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.
• Power pattern repeats every p radians (T/2)
The RMS power, current, voltage are useful, DC-like quantities
Pav  Prms  rms averageof Pinst overa wholecycle(wτ  2p)
Integrate Pinst:
1
2
Prms  ImR
T

T
1 2
cos (wt)dt  Im R Integral = 1/2
2
0
Pav is an “RMS” quantity:
• “Root Mean Square”
• Square a quantity (positive)
• Average over a whole cycle
• Compute square root.
• For other rms quantities,
divide peak value
such as Im or Em by sqrt(2)
2
2
Prms  Pav  Irms
R
Erms 
Em
2
Irms 
Im
2
Irms 
Erms
|Z|
Vm
For any R, L, or C
2
Copyright
R. Janow
– Fall 2015
Household power example: 120 volts RMS  170
volts
peak
Vrms 
Power factor for AC Circuits
The PHASE ANGLE F determines the average RMS
power actually absorbed due to the RMS current
and applied voltage in the circuit.
Erms
F
Claim (proven below):

Pav  Prms  Erms  Irms  ErmsI rms cos(F)
cos(F)  R / | Z | is the " power factor"

Irms
|Z|
XL-XC
R
wDt
Proof: start with instantaneous power (not very useful):
Pinst (t)  E(t) I(t)  Em Im cos(wDt  F) cos(wDt)
Average it over one full period :
Pav 

P


1
0
inst
(t) dt  Em Im
Change variables:
wD  2p

cos(w t  F) cos(w t)dt


1
0
D
D
 Em Im x {Integral}
x  wDt, wD  2p
1 2p
{Integral} 
cos(x  F) cos(x)dx
2p 0

Use trig identity: cos(x  F)  cos(x) cos(F)  sin(xCopyright
) sin(F)R. Janow – Fall 2015
Power factor for AC Circuits continued
1 2p
1 2p
2
Integral  cos(F)
2p 0
cos (x)dx  sin(F)
cos(x) sin(x)dx
2p
2

sin(F) sin (x) 

  0
2p  2 0 


2p
2p
cos(F)  x
sin(2x) 
cos(F)


 
2p  2 0
4
2
0 

 Pav
2p 0
cos(F)
 Em Im
2
Recall: RMS values = Peak values divided by sqrt(2)
 Pav  Prms  ErmsI rms cos(F)
Also note: Erms  I rms | Z |
Alternate form:
and
R  | Z | cos(F)
2
2
Prms  Irms
| Z | cos(F)  Irms
R
If R=0 (pure LC circuit) F  +/- p/2 and Pav = Prms = 0
Copyright R. Janow – Fall 2015
Example 2 continued with RMS quantities:
R = 3000 
L = 0.33 H
C = 0.10 mF
Em = 100 V.
VR
fD = 200 Hz
Find Erms:
E
L
Erms  Em / 2  71 V.
Find Irms at 200 Hz:
R
VL
C
| Z |  8118 
as before
VC
Irms  Erms / | Z |  71 V / 8118   8.75 mA.
Find the power factor:
3000
cos(F)  R / | Z | 
 0.369
8118
Find the phase angle F:
   C 
0
F  tan 1  L


68
as before

 R

Recall: do not use
arc-cos to find F
Find the average power:
Pav  ErmsI rms cos(F)  71 8.75  10-3  0.369  0.23 Watts
or

2
Pav  Irms
R  8.75  10- 3

2
 3000  0.23 Watts
Copyright R. Janow – Fall 2015
Example 3 – Series LCR circuit analysis using RMS values
A 240 V (RMS), 60 Hz voltage source is applied to a series LCR circuit consisting of a 50ohm resistor, a 0.5 H inductor and a 20 mF capacitor. w  2pf  6.28 x 60  377 rad / s
D
Find the capacitive reactance of the circuit:
XC  1 / wDC  1 / (377x2x10-5 )  133 
Find the inductive reactance of the circuit:
XL  wDL  377x.5  188.5 
The impedance of the circuit is:
| Z |  [ R2  (XL  XC )2 ]1/2  74.7 
The phase angle for the circuit is:
tan(F) 
F is positive since XL>XC (inductive)
Irms
The RMS current in the circuit is:
XL  X C
 F  48.0 0 , cos(F )  0.669
R
E
240
 rms 
 3.2 A.
| Z | 74.7
The average power consumed in this circuit is:
2
Prms  Irms
R  (3.2) 2 x 50  516 W.
or
Prms  Erms Irms cos(F)
where
cos(F)  R / Z
If the inductance could be changed to maximize the current through the circuit, what
would the new inductance L’ be? Current is a maximum at RESONANCE.
wD  377 
1

L 'C
L' 
1
2
D
wC

1
 0.352 H.
2
5
377 x2x10
How much RMS current would flow in that case?
At resonance
| Z |  R  Irms 
Erms
R

240 V
 4.8 A.
50 
Copyright R. Janow – Fall 2015
Transformers
Devices used to change
AC voltages. They have:
• Primary
• Secondary
• Power ratings
power transformer
iron core
circuit
symbol
Copyright R. Janow – Fall 2015
Transformers
Ideal Transformer
Assume zero internal resistances,
EMFs Ep, Es = terminal voltages Vp, Vs
Faradays Law for primary and secondary:
Vp  Np
dFB
dt
Vs  Ns
dFB
dt
Assume: The same amount of flux FB cuts
each turn in both primary and secondary
windings in ideal transformer (counting
self- and mutual-induction)
induced voltage
per turn
N
 Vs  s Vp
Np

dFB Vp Vs


dt
Np Ns
Turns ratio fixes
the 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.
iron core
• zero resistance in coils
• no hysteresis losses in iron core
• all field lines are inside core
Assuming no losses: energy and
power are conserved
Ps  VsIs  conserved  Pp  VpIp

Ip
Is

Ns
Np
Copyright R. Janow – Fall 2015
Example: A dimmer for lights
using a variable inductance
w = 377 rad/sec
f =60 Hz
Without Inductor:
2
P0,rms  Erms
/ R  18 Watts,
Light bulb
R=50 
Erms=30 V
L
F0
a) What value of the inductance would dim the lights to 5 Watts?
2
R
Recall: Prms  Irms
E
E
Irms  rms  rms cos(F)
Z
R
 Prms  P0,rmscos2 (F)
5 Watts  18 Watts x cos2 (F)
cos(F)  0.527  cos( 58.2o )
F  58.2o  tan-1 [ XL / R ]
(X C  0)
o
XL  R tan(F)  50 tan(58.2 )  80.6   2p f L
 L  80.6 / 377  214 mH
b) What would be the change in the RMS current?
E
30 V
I0,rms  rms 
 0.6 A
P0,rms = 18 W.
Without inductor:
R
With inductor:
Irms 
Erms
R
50 
cos(F)  0.6 A  0.527  0.316 A
Prms = 5 W.
Copyright R. Janow – Fall 2015
Copyright R. Janow – Fall 2015