Transcript phys1444-lec23

PHYS 1444 – Section 04
Lecture #23
Wednesday November 30, 2011
Dr. Andrew Brandt
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AC Circuits
Maxwells Equations
Weds. November 30, 2011
Last HW Dec. 2
@10pm Ch 29,30
PHYS 1444-004, Dr. Andrew Brandt
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AC Circuit w/ Inductance only
• From Kirchhoff’s loop rule, we obtain
dI
V L 0
dt
• Thus
d  I 0 sin  t 
dI
  LI 0 cos t
V  L L
dt
dt

– Using the identity cos   sin   90
•

V  LI 0 sin  t  90
– where V0  LI 0
 V
0


sin  t  90

– Current and voltage are “out of phase by p/2 or 90o”.
– For an inductor we can write V0  I 0 X L
• Where XL is the inductive reactance of the inductor
XL  L
• The relationship V0  I 0 X L is not generally valid since V0 and
Vrms  I rms X L is valid
I0 do not occur at the same time, but
Weds. November 30, 2011
PHYS 1444-004, Dr. Andrew Brandt
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AC Circuit w/ Capacitance only
• From Kirchhoff’s loop rule, we obtain
Q
V 
C
• Current at any instant is
dQ
 I 0 sin  t
I
dt
• Thus, the charge Q on the plate at any instance is
Q

Q
Q 0
dQ 

t
t 0
I 0 sin  tdt  
• The voltage across the capacitor is
Q
1
V   I0
cos t
C
C

– Using the identity cos    sin   90


I0



cos t
1
V  I0
sin  t  90  V0 sin  t  90
C
– Where
I0
V0 
–
C
Weds. November 30, 2011
PHYS 1444-004, Dr. Andrew Brandt

3
AC Circuit w/ Capacitance only
So the voltage is V  V sin  t  90 
•
0
• What does this mean?
– Current and voltage are “out of phase by p/2 or 90o” but in this
case, the voltage reaches its peak ¼ cycle after the current
• What happens to the energy?
–
–
–
–
No energy is dissipated
The average power is 0 at all times
The energy is stored temporarily in the electric field
Then released back to the source
• Relationship between the peak voltage and the peak current
in the capacitor can be written as V0  I 0 X C
Infinite
1
XC 
C
– Where the capacitance reactance XC is defined as
– Again, this relationship is only valid for rms quantities
Weds. November 30, 2011
PHYS 1444-004, Dr. Andrew Brandt
Vrms  I rms X C
when
w=0.
4
Example 31 – 2
Capacitor reactance. What are the peak and rms current in
the circuit in the figure if C=1.0mF and Vrms=120V?
Calculate for f=60Hz.
The peak voltage is V0  2Vrms  120V  2  170V
The capacitance reactance is
1
1
1

XC 
 2.7 k 


1

6
 C 2p fC 2p  60s  1.0  10 F


Thus the peak current is
The rms current is
V0
170V

 63mA
I0 
X C 2.7k 
I rms
Weds. November 30, 2011
Vrms 120V

 44mA

X C 2.7k 
PHYS 1444-004, Dr. Andrew Brandt
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Drum Roll
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We’ve looked at L only circuits
We’ve looked at R only circuits
We’ve looked at C only circuits
We’ve looked at LR circuits
We’ve looked at RC circuits
…
Weds. November 30, 2011
PHYS 1444-004, Dr. Andrew Brandt
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AC Circuit w/ LRC
• The voltage across each element is
– VR is in phase with the current
– VL leads the current by 90o
– VC lags the current by 90o
• From Kirchhoff’s loop rule
• V=VR+VL+VC
– However since they do not reach the peak voltage at the
same time, the peak voltage of the source V0 will not equal
VR0+VL0+VC0
– The rms voltage also will not be the simple sum of the three
Weds. November 30, 2011
PHYS 1444-004, Dr. Andrew Brandt
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•
AC
Circuit
w/
LRC
The current at any instant is the same at all point in the circuit
– The currents in each elements are in phase
– Why?
• Since the elements are in series
– How about the voltage?
• They are not in phase.
• The current at any given time is
I  I 0 sin  t
• The analysis of LRC circuit can be done using the “phasor” diagram (but
not by us!)
Weds. November 30, 2011
PHYS 1444-004, Dr. Andrew Brandt
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Maxwell’s Equations
• The development of EM theory by Oersted, Ampere and others was not
done in terms of EM fields
– The idea of fields was introduced by Faraday
• Scottish physicist James C. Maxwell unified all the phenomena of
electricity and magnetism in one theory with only four equations
(Maxwell’s Equations) using the concept of fields
– This theory provided the prediction of EM waves
– As important as Newton’s law since it provides dynamics of electromagnetism
– This theory is also in agreement with Einstein’s special relativity
• The biggest achievement of 19th century electromagnetic theory is the
prediction and experimental verification that the electromagnetic waves
can travel through empty space
– This accomplishment
• Opened a new world of communication
• Yielded the prediction that the light is an EM wave
• Since all of Electromagnetism is contained in the four Maxwell’s
equations, this is considered as one of the greatest achievements of the
human intellect
Weds. November 30, 2011
PHYS 1444-004, Dr. Andrew Brandt
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Modifying Ampere’s Law
• A magnetic field is produced by an electric current

B  dl  m 0 I encl
I
• This equation represents the general form of Ampere’s
law:
 B  dl  m I
0 encl
d E
 m0 0
dt
Extra term
from
Maxwell
• This means that a magnetic field can be caused not only by an ordinary
electric current but also by a changing electric flux
Weds. November 30, 2011
PHYS 1444-004, Dr. Andrew Brandt
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Displacement Current
• Maxwell interpreted the second term in the generalized
Ampere’s law equivalent of an electric current
– He called this term the displacement current, ID
– While the other term is called as the conduction current, I
• Ampere’s law then can be written as

– Where
B  dl  m0  I  I D 
d E
I D  0
dt
– While it is in effect equivalent to an electric current, a flow of electric
charge, this actually does not have anything to do with the flow itself
Weds. November 30, 2011
PHYS 1444-004, Dr. Andrew Brandt
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•
Gauss’ Law for Magnetism
•
If there is symmetry between electricity and magnetism, there must be an equivalent
law in magnetism as Gauss’ Law in electricity
The magnetic flux through a closed surface which completely encloses a volume is
•
B  dA

What was Gauss’ law in the electric case?
B 
– The electric flux through a closed surface is equal to the total net charge Q enclosed by the
surface divided by 0.
Q

•
•
E  dA 
encl
0
Gauss’ Law
for electricity
Similarly, we can write Gauss’ law for magnetism as
Why is result of the integral zero?

B  dA  0
Gauss’ Law for
magnetism
– There are no isolated magnetic poles, the magnetic equivalent of single electric charges
Weds. November 30, 2011
PHYS 1444-004, Dr. Andrew Brandt
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Gauss’ Law for Magnetism
• What does Gauss’ law in magnetism mean
physically?

B  dA  0
– There are as many magnetic flux lines that enter the
enclosed volume as leave it
– If magnetic monopoles do not exist, there is no starting
or stopping point of the flux lines
• Electricity has sources and sinks
– Magnetic field lines must be continuous
– Even for bar magnets, the field lines exist both insides
and outside of the magnet
Weds. November 30, 2011
PHYS 1444-004, Dr. Andrew Brandt
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Maxwell’s Equations
• In the absence of dielectric or magnetic materials,
the four equations developed by Maxwell are:
Gauss’ Law for electricity
Qencl
E  dA 
A generalized form of Coulomb’s law relating

0
electric field to its sources, the electric charge
Gauss’ Law for magnetism
 B  dA  0


A magnetic equivalent of Coulomb’s law, relating magnetic field
to its sources. This says there are no magnetic monopoles.
d B
E  dl  
dt
B  dl  m0 I encl
Weds. November 30, 2011
Faraday’s Law
An electric field is produced by a changing magnetic field
d E
 m0 0
dt
PHYS 1444-004, Dr. Andrew Brandt
Ampére’s Law
A magnetic field is produced by an
electric current or by a changing
electric field
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Example 32 – 1
Charging capacitor. A 30-pF air-gap capacitor has circular plates of area A=100cm2. It
is charged by a 70-V battery through a 2.0- resistor. At the instant the battery is
connected, the electric field between the plates is changing most rapidly. At this instance,
calculate (a) the current into the plates, and (b) the rate of change of electric field
between the plates. (c) Determine the magnetic field induced between the plates.
Assume E is uniform between the plates at any instant and is zero at all points beyond
the edges of the plates.
Since this is an RC circuit, the charge on the plates is: Q  CV0 1  et RC 
For the initial current (t=0), we differentiate the charge with respect to time.
CV0 t
dQ
e

I0 
RC
dt t 0
RC
t 0
V0
70V

 35 A

R 2.0 
The electric field is E    Q A
0
0
Change of the dE dQ dt


electric field is dt A 0 8.85  1012 C 2
Weds. November 30, 2011
35 A

N  m 2  1.0  102 m 2
PHYS 1444-004, Dr. Andrew Brandt

 4.0  1014 V m  s
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Example 32 – 1
(c) Determine the magnetic field induced between the plates. Assume E is uniform
between the plates at any instant and is zero at all points beyond the edges of the plates.
The magnetic field lines generated by changing electric field is
perpendicular to E and is circular due to symmetry
d E
Whose law can we use to determine B?
B  dl  m0  0
Extended Ampere’s Law w/ Iencl=0!
dt
We choose a circular path of radius r, centered at the center of the plane, following the B.
2
For r<rplate, the electric flux is  E  EA  Ep r since E is uniform throughout the plate
d Ep r 2
dE
 m0  0 p r 2
So from Ampere’s law, we obtain B   2p r   m0  0
dt
dt
r dE
B  m0  0
For r<rplate
Solving for B
2 dt
2
Since we assume E=0 for r>rplate, the electric flux beyond  E  EA  Ep rplate
the plate is fully contained inside the surface.
2
d Ep rplate
dE
2
So from Ampere’s law, we obtain B   2p r   m0  0
 m0  0p rplate
dt
dt
2
m0  0 rplate dE
Solving for B
For r>rplate
B
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2PHYS
r 1444-004,
dt Dr. Andrew Brandt
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