Download Lecture14: AC Circuits , Resonance

Physics 121 - Electricity and Magnetism
Lecture 14 - AC Circuits, Resonance
Y&F Chapter 31, Sec. 3 - 8
•
•
•
•
•
•
•
Phasor Diagrams for Voltage and Current
The Series RLC Circuit. Amplitude and Phase Relations
Impedance and Phasors for Impedance
Resonance
Power in AC Circuits, Power Factor
Examples
Transformers
Copyright R. Janow –Fall 2016
Current/Voltage Phases in pure R, C, and L circuit elements
• Sinusoidal current i (t) = Imcos(wDt). Peak is Im
• Peak voltage drops in R, L, or C loads lead/lag current by 0, p/2, -p/2 radians
• Reactances (generalized resistances) are ratios of peak voltages to peak currents
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 2016
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 current everywhere in the single 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
• F in interval +/- p/2
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
wDt
Apply Kirchhoff Loop rule to series LRC:
E(t)  vR (t)  vL (t)  vC (t)  0 instantaneous
Voltage phasors for R, L, & C all rotate at wD :
VL
Em
Im
F
wDt
VR
VC
• VC lags Im by p/2
• VR has same phase as Im
• VL leads Im by p/2
VR  ImR
VC  ImXC
VL  ImXL
Phasors add like vectors






Em  VR  (VL  VC )
VC lags VL by 1800
Copyright R. Janow
along Im
perpendicular
to Im–Fall 2016
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  Im normal to Em  F ~ +/- p/2  tiny losses, no power absorbed
Copyright R. Janow –Fall 2016
• XL=XC  Im parallel to Em  F  0  Z=R  maximum current (resonance)
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 2016
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 2016
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 will be negative
Copyright R. Janow –Fall 2016
XL  X C
471   758 
)  tan1(
)  34.0  0.593 rad.
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)
Add voltages above: VR  VL  VC  483V  150V  Em


What’s wrong?
 1 / 2
2 Janow –Fall 2016
R.

V
 150 V
Voltages add with proper phases: Em  VR2  VCopyright
L
C
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 2016
Resonance in a series LCR circuit
E
R
L
C  1 / wDC
C
Vary wD:
XC  XL

| Z |  R  L   C 
2

2 1/ 2
L  wDL
Im 
Em
|Z|
R  resistance
At resonance maximum current, minimum impedance
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 2016
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 2016
AC Power Dissipation in a Resistor
2
2
2
Instantaneous power  P

i
(
t
)
R

I
R
c
os
(wt)
inst
m
• 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 is an AC equivalent to DC power
Pav  Prms  averageof Pinst overa wholecycle (wτ  2p)
Integrate Pinst in resistor over t:
1 t
1 2
2
Prms  Im
R
cos2 (wt)dt  Im
R
t 0
2

RMS means “Root Mean Square”
• Square a quantity (positive)
• Average over a whole cycle
• Compute square root.
COMPUTING RMS QUANTITIES:
• For any RMS quantity divide peak
value such as Im or Em by sqrt(2)
Irms 
Im
2
Erms 
Vrms 
Integral = 1/2
2
Prms  Pav  Irms
R
Em
2
Vm
2
Irms 
Erms
|Z|
For any R, L, or C
R. Janow –Fall 2016
Household power example: 120 volts RMS  Copyright
170 volts
peak
Power factor for an AC LCR Circuit
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 t:
Pav 
t
P

t
1
0
inst
(t) dt  Em Im
Change variables:
wDt  2p
t
cos(w t  F) cos(w t)dt

t
1
0
D
D
 Em Im x {Integral}
x  wDt, wDt  2p, dt  dx/wD
1 2p
{Integral} 
cos(x  F) cos(x)dx
2p 0

Copyright R. Janow –Fall 2016
Use trig identity: cos(x  F)  cos(x) cos(F)  sin(x) sin(F)
Power factor for AC Circuits - continued
{Integral}  cos(F)
Even integrand
1
2p
2p 0
cos2 (x)dx  sin(F)
2p
2p 0
cos(x) sin(x)dx
Odd integrand
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
1
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 2016
Example 2 continued with RMS quantities:
R = 3000 
L = 0.33 H
C = 0.10 mF
Em = 100 V.
fD = 200 Hz
Find Erms:
E
R
L
Erms  Em / 2  71 V.
Find Irms at 200 Hz:
VR
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 directly:
   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 2016
VL
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
Current is a maximum at RESONANCE.
would the new inductance L’ be?
wD  377 
1
1
1
 L' 

 0.352 H.
2
2

5
L' C
w C
377 x 2x10
D
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 2016
Transformers
Devices used to change
AC voltages. They have:
• Primary
• Secondary
• Power ratings
power transformer
iron core
circuit
symbol
Copyright R. Janow –Fall 2016
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)
or 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 2016
Example: A dimmer for lights
using a variable inductance
f =60 Hz
w = 377 rad/sec
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?
Erms Erms
2
Irms 

cos(F)
cos(F)  R/ | z |
Recall: Prms  IrmsR
 Prms
|Z|
R
 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)
XL  R tan(F)  50 tan(58.2 o )  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 
Prms = 5 W.
Copyright R. Janow
–Fall 2016
cos(F)  0.6 A  0.527  0.316 A
Copyright R. Janow –Fall 2016