Download Chapter 21: Electric Charge and Electric Field

Chapter 28: Alternating Current
Phasors and Alternating Currents
 Alternating current (AC current)
• Current which varies sinusoidally in time is called alternating current (AC)
as opposed to direct current (DC). One example of AC current source is a
coil of wire rotating with constant angular velocity in a magnetic field.
The symbol
~
is used to denote an AC source. In general a source
means either a source of alternating current or voltage.
  V sin wt for alternatin g voltage, V  voltage amplitude
i  I sin wt for alternatin g current, I  current amplitude
• In the U.S. and Canada, commercial electric-power distribution system
uses a frequency of f = 60 Hz, corresponding to w = 377 rad/s. In much
of the rest of the world uses f = 50 Hz. In Japan, however, the country is
divided in two regions with f = 50 Hz and 60 Hz.
Note :  V sin wt, i  I sin wt  V  VP sin wt, I  I P sin wt in textboo k
Phasors and Alternating Currents
 Phasors
 Rectifier and rectified current
phasor
I=IP sin wt
• A convenient way to express a quantity
varying sinusoidally with time is by
a phasor in phasor diagram as shown.
IP
wt
O
+
+
-
w
Phasors and Alternating Currents
 Rectifier and rectified current (cont’d)
Phasors and Alternating Currents
 Root-mean-square current and voltage
• Root-mean-square current of a sinusoidal current
time averaged
1
I P2
2
I  I P sin wt  I  I sin wt  I (1  cos 2wt )  I 
2
2
2
I rms 
2
P
2
2
P
IP
2
• Root-mean-square voltage of a sinusoidal voltage
Vrms 
VP
2
For 120-volt AC, V=170 V.
Reluctance
 Resistance, inductance, capacitance and reactance
• Resistor in an AC circuit
Given: e  e m sin wt
em
VR  RI R  e m sin wt  I R 
x
R
s i n wt
R
Voltage across R in phase with current
r1
r1
x 0 , .. r1
0 , .. r1 through R
n
enm
em 1
R
VR
e
~
1
IR
I
IR V
R
f( x ) 0
0
0
f( x ) 0
 em 1
00

2
t
x
4
6
IR
em
1
0
R 0
wt
R
2
t
x
4
6
At time t
I=em/R
Reluctance
 Resistance, inductance, capacitance and reactance
• Inductor in an AC circuit
Given: e  e m sin wt
dI
VL  L L  e m sin wt 
dt

0,
r1
dI L 
em
L
L
IL
sin wtdt
e
~
e
e
I L   d I L   m co s w t  m sin wt   / 2
wL
wL
r1
Voltage across L leads currentx through
0 , .. r1 L by one-quarter cycle (90°).
.. r1
n
em 1
wL
n
e m1
IL
VL
e
0
f( x ) 0
0
f( x ) 0
 em

1
0
0
2
tx
4
6
em
1
wL 00
2
tx
4
6
I
IL
L
eV
m
wt
I=em/(wL)
At time t
0,
Reluctance
 Resistance, inductance, capacitance and reactance
• Capacitor in an AC circuit
Given: e  e m sin wt
C
Q
VC 
 e m sin wt  Q  Ce m sin wt
C
e
~
 I C  dQ  wCe m cos wt  wCe m sin( wt   )
dt
2
r1
IC
Voltage across C lags current through
C by one-quarter cycle (90°).
r1
x
.. r1
0,
.. r1
n
n
wCe m1
e m1
IC
VC
0
f( x ) 0
0
f( x ) 0
 em
wt
1
00
2
4
x
6
t
 wCe m1
I
IC
em
wt
00
2
t
x
4
6
e
VC
I=wCem
At time t
Reluctance
 LRC series circuit and reluctance
LRC circuit summary
Given: e  e m sin wt
Assume the solution for current: I (t )  I m sin( wt   )
VR  RI m sin( wt   )
1
VC  
I m cos(wt   )
wC
VL  wLI m cos(wt   )
(See derivation later)
amplitude
VR  I m R
1
VC  I m
wC
XC
VL  I mw L
XL
reactance
Reluctance
 LRC series circuit and reluctance (cont’d)
What is reactance?
fw/2
You can think of it as a frequency-dependent resistance.
1
XC 
wC
XL  wL
( " XR "  R )
For high ω, χC~0
- Capacitor looks like a wire (“short”)
For low ω, χC∞
- Capacitor looks like a break
For low ω, χL~0
- Inductor looks like a wire (“short”)
For high ω, χL∞
- Inductor looks like a break
(inductors resist change in current)
LRC Circuits
 LRC series circuit (cont’d)
• Given: e
• Assume:
R
 e m sin wt
I  I m sin( wt   )
amplitude
Q
Im
w
C
cos(wt   )
dI
 I mw cos(wt   )
dt
VR  RI  RI m sin( wt   )
 VC  Q   1 I m cos(wt   )
C
wC
dI
VL  L
 wLI m cos(wt   )
Note : sin( wt     / 2) 
cos(wt   )
The projections of these phasors along the
vertical axis are the actual values of the voltages
at the given time.
e
~

Im
wC
w
Imw L
dt
This picture corresponds to a snapshot at t=0.
L

em

Im
Im R
LRC Circuits
 LRC series circuit (cont’d)
Problem: Given Vdrive = εm sin ωt,
find VR, VL, VC, IR, IL, IC
R
C
e
Strategy:
1. Draw Vdrive phasor at t=0
2. Guess iR phasor
iR  im sin( wt   )
 im sin(  ) at t  0
3. Since VR = iR R, this is also the
direction for the VR phasor.
~

-φ

(No L or C → f = 0)
4. Realize that due to Kirchhoff’s current law, iL = iC = iR (i.e., the same
current flows through each).
L
LRC Circuits
 LRC series circuit (cont’d)
5. The inductor current IL always lags VL  draw VL 90˚ further
counterclockwise.
6. The capacitor voltage VC always lags IC  draw VC 90˚ further
clockwise.
VL= I XL
VC =
I XC
-φ

VR = I R
The lengths of the phasors depend on R, L, C, and ω. The relative orientation
of the VR, VL, and VC phasors is always the way we have drawn it.
 is determined such that VR + VL + VC = ε (Kirchhoff’s voltage rule)
These are added like vectors.
LRC Circuits
 Phasor diagrams for LRC circuits: Example
y
Vout
ε
~
x
y
IR
 IR   ( IX C )2  e 2
2
ε
VC
x
I 2 ( R 2  X C2 )  e 2
I
e
R 2  X C2
amplitude of current
LRC Circuits
 Filters : Example
Vout
Vout  IR  e
~
Vout
e

R2  X C
R
R 
2
Ex.: C = 1 μF, R = 1Ω
1
wC

2
2
1

1
 
w0 2
w
1
w0 
RC
High-pass filter
1
"transmission"
R
0.8
0.6
High-pass filter
0.4
0.2
0
0.E+00
1.E+06
2.E+06
3.E+06
4.E+06
(Angular) frequency, om ega
5.E+06
6.E+06
Note: this is ω, f 
w
2
LRC Circuits
 Filters
Vout
~
~
ω=0
No current
Vout ≈ 0
ω=∞
Capacitor ~ wire
Vout ≈ ε
Vout
Vout
ω = ∞ No current
Vout ≈ 0
Highe pass filter
w0
Vout
e
Lowpass filter
ω = 0 Inductor ~ wire
Vout ≈ ε
ω = 0 No current because of capacitor
~
ω = ∞ No current because of inductor
(Conceptual sketch only)
w
w0
Vout
e
w0
Band-pass filter
w
LRC Circuits
 Phasor diagrams for LRC circuits: Example 2
Im XL


em

Im XC

em

X L  XC
tan  
R
Im R
Reluctance for inductor
X L  wL

e  I R  X L  X C 
Reluctance for capacitor
2
m
1
XC 
wC
amplitude
Impedance Z
Z  R  X L  X C 
2
Im R
Im (X L -X C)
2
Im 
2
m
2
2

em
R
2
2


 X L  XC

em

Z
LRC Circuits
 Phasor diagrams for LRC circuits: Tips
• This phasor diagram was drawn as a
snapshot of time t=0 with the voltages being
given as the projections along the y-axis.
y
f
• Sometimes, in working problems, it is
easier to draw the diagram at a time when
the current is along the x-axis (when I=0).
f
I
ImXL
em
f
ImR
ImXC
“Full Phasor Diagram”
I X
m L
em
f
X
m C
x
I R
m
From this diagram, we can also create a
triangle which allows us to calculate the
impedance Z:
ImZ
| f |
Im X L  X C
ImR
“Impedance Triangle”
Resonance in Alternating Current Circuits
 Resonance
For fixed R, C, L the current Im will be a maximum at the resonant
frequency w0 which makes the impedance Z purely resistive.
i.e.:
em
em
Im  
2
Z
R2  X L  X C 
reaches a maximum when:
R
C
X = XC
L
This condition is obtained when:
wo L 
1
wo

w oC
e
~
L
resonance frequency
1
1
; f0 
LC
2 LC
• Note that this resonant frequency is identical to the natural frequency
of the LC circuit by itself!
• At this frequency, the current and the driving voltage are in phase!
tan  
X L  XC
0
R
Resonance in Alternating Current Circuits
 Resonance (cont’d)
XL
Im 
em
Z
Im
Z
R
Z
cos 

em
cos 
R
X L  XC
tan  
R
Plot the current versus w, the frequency
of the voltage source: →
XL - XC

R
x
0.0 ,
r1
n
em
1
R0
.. r1
XC
R=Ro
f( x )
Im0.5
g( x )
R=2Ro
00
00
1
wx
2w2o
Resonance in Alternating Current Circuits
 Resonance (cont’d)
R
On Resonance: 0 and Z=R
VR  IR  e
VL  IX L 
e XL
R
 eQ
I
C
e
R
VC  IX C 
e XC
R
e
~
L
 eQ
On resonance, the voltage across the reactive elements is
amplified by Q!
Necessary to pick up weak radio signals, cell phone
transmissions, etc.
Power in Alternating Current Circuits
 Power
• The instantaneous power (for some frequency, w) delivered at time t is
given by:
P(t )  e (t ) I (t )  e m sin wt I m sin( wt   )
• The most useful quantity to consider here is not the instantaneous power
but rather the average power delivered in a cycle.
 P(t )  e m I m sin wt sin( wt   )
• To evaluate the average on the right, we first expand the sin(wt-) term.
Power in Alternating Current Circuits
 Power
• Expanding,
sin wt sin( wt   )  sin wt (sin wt cos   cos wt sin  )
• Taking the averages,
sin wt cos wt   0
1
1.01+1
sinwtcoswt
(Product of even and odd function = 0)
h( x ) 0
0
• Generally:
2
1
1
2
sin 2 x 
sin
xdx

2 0
2
1.01 1
-1
00
2
0
x
0.0 ,
4
2
6
6.28
.. r1
n
• Putting it all back together again,
1/2
0
2
 P(t )  e m I m cos  sin wt   sin  sin wt cos wt 
r1
wx t
1
+1
sin2wt
h( x ) 0
0
 P(t ) 
1
e m I m cos 
2
-11
00
2
wx t
4
2
6
Power in Alternating Current Circuits
 Power
This result is often rewritten in terms of rms values:
e rms 
1
em
2
I rms 
1
Im
2
 P(t )  e rms I rms cos 
Power delivered depends on the phase, f, the “power factor”
Phase depends on the values of L, C, R, and w
Therefore...
 P(t )  e rms I rms cos 
Power in Alternating Current Circuits
 Power
Power, as well as current, peaks at w = w0. The sharpness of the resonance
depends on the values of the components.
Recall:
Im 
em
R
cos 
e 2 rms 2
2
 P(t ) 
cos   I rms
R
R
We can write this in the following manner (which we won’t try to prove):
e 2 rms
x2
 P(t ) 
R x 2  Q 2 ( x 2  1) 2
…introducing the curious factors Q and x...
Resonance in Alternating Current Circuits
 Power and resonance
A parameter “Q” is often defined to describe the sharpness of
resonance peaks in both mechanical and electrical oscillating systems.
“Q” is defined as
U
Q  2 max
DU
where Umax is max energy stored in the system and DU is the energy
dissipated in one cycle
1 2
LI max
2
For RLC circuit, Umax is
U max 
Losses only come from R:
1 2
1 2  2 

DU  I max
RT  I max
R
2
2
 w res 
This gives
Q
w res L
period
R
And for completeness, note
x 
w
w res
Resonance in Alternating Current Circuits
 Power and resonance
wres
For Q > few, Q 
FWHM
e 2 rms
Q=3
R0
FWHM
Full Width at Half Maximum
FWHM
<P>
0
Q
Quality of the peak
Higher Q = sharper peak = better quality
R=Ro
R=2Ro
0
w
2wo
Transformers
 Transformers
• AC voltages can be stepped up or
stepped down by the use of
transformers.
The AC current in the primary circuit
creates a time-varying magnetic field
in the iron
e
iron
~
This induces an emf on the secondary
windings due to the mutual inductance of
the two sets of coils.
V1
V2
N
1
(primary)
N
2
(secondary)
• The iron is used to maximize the mutual inductance. We assume that the
entire flux produced by each turn of the primary is trapped in the iron.
Transformers
 Ideal transformer without a load
No resistance losses
All flux contained in iron
Nothing connected on secondary
The primary circuit is just an AC voltage
source in series with an inductor. The
change in flux produced in each turn is given by:
dturn V1

dt
N1
iron
e ~
V1
V2
N1
N2
(primary) (secondary)
• The change in flux per turn in the secondary
coil is the same as the change in flux per turn in the primary coil (ideal case).
The induced voltage appearing across the secondary coil is given by:
dturn N 2
V2  N 2

V1
dt
N1
• Therefore,
• N2 > N1  secondary V2 is larger than primary V1 (step-up)
• N1 > N2  secondary V2 is smaller than primary V1 (step-down)
• Note: “no load” means no current in secondary. The primary current,
termed “the magnetizing current” is small!
Transformers
 Ideal transformer with a load
What happens when we connect a resistive load
to the secondary coil?
Changing flux produced by primary coil induces
an emf in secondary which produces current I2
I2 
iron
e ~
V2
R
V1
V2
N1
(primary)
R
N2
(secondary)
This current produces a flux in the secondary coil
µ N2I2, which opposes the change in the original
flux -- Lenz’s law
This induced changing flux appears in the primary
circuit as well; the sense of it is to reduce the emf in
the primary, to “fight” the voltage source. However,
V1 is assumed to be a voltage source. Therefore, there
must be an increased current I1 (supplied by the voltage
source) in the primary which produces a flux µ N1I1
which exactly cancels the flux produced by I2.
I1 
N2
I2
N1
Transformers
 Ideal transformer with a load (cont’d) e
iron
~
Power is dissipated only in the load resistor R.
V22
2
Pdissipated  I 2 R 
 V2 I 2
R
Where did this power come from?
It could come only from the voltage source in the primary:

Pgenerated  V1I1
I1 V2

I 2 V1
I1  I 2
=
N2
N1
N2
N1
V1
V1
=

V1
V2
N1
(primary)
N2
(secondary)
V1I1  V2 I 2
N2
N1
V2 N 2 V1  N 2 
 

R N1 R  N1 
2
The primary circuit has
to drive the resistance R
of the secondary.
R
Exercises
 Exercise 1
Suppose em = 100 volts, f=1000 Hz, R=10 Ohms, L=4.22 mH,
Find XL, Z, I, VR, and Vl.
XL  wL  6.28 1000  0.00422H  26.5
Z  R 2  (wL) 2
Z  10 2  (26.5) 2  28.3
em
100
I

 3.53 A.
Z 28.3
VR  RI  10  3.53  35.3 V.
VL  X L I  26.5  3.53  93.5 V.
Exercises
 Exercise 2: Calculate power lost in R in Exercise 1
2
Pavg  Irms
R
I
3.53A
Irms 

 2.50A
2 1.414
Pavg  (2.50A)210  62.5Watts
To calculate power produced by the generator you need to take
account of the phase difference between the voltage and the
current. In general you can write:
Pavg  ermsIrms cos 
For an inductor P = 0 because the phase difference between current through
the inductor and voltage across the inductor is 90 degrees