Transcript Phys132 Lecture 5 - University of Connecticut

Physics 1502: Lecture 23
Today’s Agenda
• Announcements:
– RL - RV - RLC circuits
• Homework 06: due next Wednesday …
• Maxwell’s equations / AC current
Induction
Self-Inductance, RL Circuits
a
I
I
R
XXX
XXXX
XX
b
L

long
solenoid
Energy and
energy density
Inductors in Series and Parallel
• In series (like resistors)
a
a
L1
Leq
L2
And:
b
b
• In parallel (like resistors)
a
L1
And finally:
a
L2
b
Leq
b
Summary of E&M
• J. C. Maxwell (~1860) summarized all of the work on
electric and magnetic fields into four equations, all
of which you now know.
• However, he realized that the equations of electricity
& magnetism as then known (and now known by
you) have an inconsistency related to the
conservation of charge!
Gauss’ Law
Faraday’s Law
Gauss’ Law
For Magnetism
Ampere’s Law
I don’t expect you to see that these equations are inconsistent
with conservation of charge, but you should see a lack of
symmetry here!
Ampere’s Law is the Culprit!
• Gauss’ Law:
• Symmetry: both E and B obey the same kind of equation
(the difference is that magnetic charge does not exist!)
• Ampere’s Law and Faraday’s Law:
!
• If Ampere’s Law were correct, the right hand side of Faraday’s
Law should be equal to zero -- since no magnetic current.
• Therefore(?), maybe there is a problem with Ampere’s Law.
• In fact, Maxwell proposes a modification of Ampere’s Law by
adding another term (the “displacement” current) to the right
hand side of the equation! ie
Displacement current
Remember:
Iin
FE
Iout
changing electric
flux
Maxwell’s Displacement Current
• Can we understand why this “displacement current” has the
form it does?
• Consider applying Ampere’s
Law to the current shown in
the diagram.
• If the surface is chosen as
1, 2 or 4, the enclosed
current = I
• If the surface is chosen as
3, the enclosed current = 0!
(ie there is no current
between the plates of the
capacitor)
circuit
Big Idea: The Electric field between the plates changes in time.
“displacement current” ID = 0 (dfE/dt) = the real current I in the
wire.
Maxwell’s Equations
• These equations describe all of Electricity and
Magnetism.
• They are consistent with modern ideas such as
relativity.
• They even describe light
R
C

~
L
w
i m wL
f
im
wC
f
m
f
im R
Power Production
An Application of Faraday’s Law
• You all know that we produce power from many sources. For
example, hydroelectric power is somehow connected to the
release of water from a dam. How does that work?
• Let’s start by applying Faraday’s Law to the following
configuration:
S
S
S
N
N
N
Side View
End View
Power Production
An Application of Faraday’s Law
• Apply Faraday’s Law
Power Production
An Application of Faraday’s Law
• A design schematic
Water
RLC - Circuits
R
• Add resistance to circuit:
C
x
0,
r1
r1
.. r1
L

10
n
100
n
1
• Qualitatively: Oscillations
created by L and C are
damped (energy dissipation!)
• Compare to damped
oscillations in classical
mechanics:
f( x ) 0
1
0
5
x
10
AC Circuits - Intro
• Last time we discovered that a LC circuit
was a natural oscillator.
• However, any real attempt to construct a
LC circuit must account for the
resistance of the inductor. This
resistance will cause oscillations to
damp out.
R
+ +
- x
0,
r1
C
r1
.. r1
L
10
n
100
n
1
f( x ) 0
1
• Question: Is there any way to modify the circuit above to
0
5
10
x
sustain the oscillations without damping?
• Answer: Yes, if we can supply energy at the rate the resistor
dissipates it! How? A sinusoidally varying emf (AC generator)
will sustain sinusoidal current oscillations!
AC Circuits
Series LCR
R
• Statement of problem:
Given  = msinwt , find i(t).
Everything else will follow.
C

~
L
• Procedure: start with loop equation?
• We could solve this equation in the same manner we did
for the LCR damped circuit. Rather than slog through the
algebra, we will take a different approach which uses
phasors.
R Circuit
• Before introducing phasors, per se, begin by considering
simple circuits with one element (R,C, or L) in addition to the
driving emf.
• Begin with R: Loop eqn gives:
R


~
Voltage across R in phase with
current through R
m
m / R
0
0
m
m / R
0
t
x
Note:
this is
always,
always,
true…
always.
iR
VR
0
iR
t
Lecture 23, ACT 1a
• Consider a simple AC circuit with a
purely resistive load. For this circuit the
voltage source is  = 10V sin (2p50(Hz)t)
and R = 5 W. What is the average
current in the circuit?
A) 10 A
B) 5 A
C) 2 A
D) √2 A
R

~
E) 0 A
Chapter 28, ACT 1b
• Consider a simple AC circuit with a
purely resistive load. For this circuit the
voltage source is  = 10V sin (2p50(Hz)t)
and R = 5 W. What is the average power
in the circuit?
A) 0 W
B) 20 W
C) 10 W
D) 10 √2 W
R

~
RMS Values
• Average values for I,V are not that helpful (they are zero).
• Thus we introduce the idea of the Root of the Mean
Squared.
• In general,
So Average Power is,
C Circuit
• Now consider C: Loop eqn gives:
C



~
iC
Voltage across C lags current through C by
one-quarter cycle (90).
wCm
m
iC
VR
0
0
m
wCm
0
t
x
Is this
always
true?
YES
0
t
Lecture 23 , ACT 2
• A circuit consisting of capacitor C and voltage
source  is constructed as shown. The graph
shows the voltage presented to the capacitor as a
function of time.
– Which of the following graphs best represents
the time dependence of the current i in the
circuit?
i
i
(a)
t
i
(b)
t

t
C
(c)
t

L Circuit
• Now consider L: Loop eqn gives:
iL



~
L
Voltage across L leads current through L by onequarter cycle (90).
m
m wL
iL
VR
0
0
m
0
t
x
m wL
0
tx
Yes, yes,
but how to
remember?
Phasors
• R: V in phase with i

• C: V lags i by 90

• L: V leads i

by 90
• A phasor is a vector whose magnitude is the maximum value
of a quantity (eg V or I) and which rotates counterclockwise in
a 2-d plane with angular velocity w. Recall uniform circular
motion:
The projections of r (on
the vertical y axis)
execute sinusoidal
oscillation.
y
y
w
x
Suppose:
xx
00,,
r1
r1
.... r1
r1
nn
Phasors for L,C,R
11

V
i
R
V
f(f(xx))000
xx
00
22
44
66
xx
V
r1
0 , .. r1
n 1
w
i
f( x ) 00
0
1.01 1
wt
C
V
2
4
V
i
f( x ) 0
0
V
C
66
x
1.01
1
R
t
i
x
w
wt
r1
00,, r1 .... r1
r1
n 11
11
i
L
w
i
wt
L
2
0
2
4
6
Lecture 23, ACT 3
• A series LCR circuit driven by emf  = 0sinwt
produces a current i=imsin(wt-f). The phasor
diagram for the current at t=0 is shown to the right.
– At which of the following times is VC, the
magnitude of the voltage across the capacitor, a
maximum?
t=0
f
i
R
i
C
t=0
t=tc
t=tb
i
(a)
i
(b)
(c)

~
L
Series LCR
AC Circuit
R
• Consider the circuit shown here:
the loop equation gives:
C

~
L
• Assume a solution of the form:
• Here all unknowns, (im,f) , must be found from the loop eqn;
the initial conditions have been taken care of by taking the emf
to be: =m sinwt.
• To solve this problem graphically, first write down expressions
for the voltages across R,C, and L and then plot the
appropriate phasor diagram.
Phasors: LCR
• Given:
• Assume:
R

C
L

~

• From these equations, we can draw
the phasor diagram to the right.
• This picture corresponds to a
snapshot at t=0. The projections of
these phasors along the vertical axis
are the actual values of the voltages
at the given time.
w
i m wL
f
im
wC
f
m
f
im R
Phasors: LCR
w
i m XL
R
C

~
m
f
L
f
i m XC
f
im R
• The phasor diagram has been relabeled in terms of the
reactances defined from:
The unknowns (im,f) can now be solved for
graphically since the vector sum of the voltages
VL + VC + VR must sum to the driving emf .
Phasors:LCR
i m XL
m
f
f
i m XC
f

im R
i m (XL -X C)
m
im R

f
Phasors:Tips
y
• This phasor diagram was drawn as a
snapshot of time t=0 with the voltages
being given as the projections along the
y-axis.
• 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).
m
f
i mR
imXC
m
f
f
i m XC
i mX L
i m XL
f
x
im R
From this diagram, we can also create a
triangle which allows us to calculate the
impedance Z:
Z
| XL-XC |
| f
R
“Full Phasor Diagram”
“ Impedance Triangle”