Download Time Varying Circuits

Time Varying Circuits
2006
Induction - Spring 2006
1
What is going on?

There are only 7 more classes and the
final is 3 weeks away.


Exam Issues




Scotty, beam me somewhere else!
Look Ashamed!
Inductor Circuits
Quiz Friday
AC Next week & Following Monday
Induction - Spring 2006
2
Other
Wire
Second Problem
B
q
h
q
a
Both Currents are going into
the page.
B total
a
0
 2 B Sin (q )  2 I
Sin (q )
2r
Sin (q ) 1 h
h
B~
~
~
r
r r a2  h2
d
h( 2h)  a 2  h 2
B~
 0(max)
dh
stuff
Induction - Spring 2006 h  a
3
First Problem
DV
1 2
qDV  mv
2
mv2
qvB 
R
2qDV
v
m
qBR qBR m
m

v
2qDV
q 2 B 2 R 2 qB 2 R 2
m

 eR 2 (const )
2qDV
2DV
M  2e(2 R) 2  8eR 2
M /m8
Induction - Spring 2006
4
The Last two Problems were
similar to WebAssigns that were
also reviewed in class.
Circular Arc – Easy Biot-Savart
Moving Rod
Induction - Spring 2006
5
Question


What about these
problems was “unfair”?
Why so many blank or
completely wrong
pages?
Induction - Spring 2006
6
And Now …..
From the past
Induction - Spring 2006
7
Max Current Rate of
increase = max emf
VR=iR
~current
Induction - Spring 2006
8
E
(1  e Rt / L )
R
L
  (time constant)
R
i
Induction - Spring 2006
9
We also showed that
1
2
uinductor 
B
20
1
2
ucapacitor   0 E
2
Induction - Spring 2006
10
LR Circuit
i
Steady Source
sum of voltage drops  0 :
di
 E  iR  L  0
dt
same form as the
capacitor equation
q
dq
E R
0
C
dt
Let E  0, then
Induction - Spring 2006
11
Time Dependent Result:
E
 Rt / L
i  (1  e
)
R
time constant

L

R
Induction - Spring 2006
12
R
L
Induction - Spring 2006
13
At t=0, the charged capacitor is now
connected to the inductor. What would
you expect to happen??
Induction - Spring 2006
14
The math …
For an RLC circuit with no driving potential (AC or DC source):
Q
di
L 0
C
dt
dQ Q
d 2Q
R
 L 2 0
dt C
dt
Solution :
iR 
Q  Qmax e

Rt
2L
cos(d t )
where
 1  R 
d  
 
 LC  2 L 
Induction - Spring 2006
2 1/ 2



15
The Graph of that LR (no emf) circuit ..
e
Rt

2L
Induction - Spring 2006
16
Induction - Spring 2006
17
Mass on a Spring Result


Energy will swap back and forth.
Add friction


Oscillation will slow down
Not a perfect analogy
Induction - Spring 2006
18
Induction - Spring 2006
19
LC Circuit
Low
High
Q/C
High
Low
Induction - Spring 2006
20
The Math Solution (R=0):
  LC
Induction - Spring 2006
21
New Feature of Circuits with L and C



These circuits produce oscillations in the
currents and voltages
Without a resistance, the oscillations would
continue in an un-driven circuit.
With resistance, the current would eventually
die out.
Induction - Spring 2006
22
Variable Emf Applied
1.5
1
Volts
emf
0.5
DC
0
0
1
2
3
4
5
6
7
8
9
10
-0.5
-1
Sinusoidal
-1.5
Tim e
Induction - Spring 2006
23
Sinusoidal Stuff
emf  A sin( t   )
“Angle”
Phase Angle
Induction - Spring 2006
24
Same Frequency
with
PHASE SHIFT

Induction - Spring 2006
25
Different Frequencies
Induction - Spring 2006
26
Note – Power is delivered to our homes
as an oscillating source (AC)
Induction - Spring 2006
27
Producing AC Generator
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Induction - Spring 2006
28
The Real World
Induction - Spring 2006
29
A
Induction - Spring 2006
30
Induction - Spring 2006
31
The Flux:
  B  A  BA cos 
  t
emf  BA sin t
emf
i
A sin t
Rbulb
Induction - Spring 2006
32
April 12, 2006
Induction - Spring 2006
33
Schedule

Today


Friday



Quiz on this weeks material
Some problems and then AC circuits
Monday



Finish Inductors
Last FULL week of classes
Following Monday is last day of class
FINAL IS LOOMING!
Induction - Spring 2006
34
Some Problems
Induction - Spring 2006
35
14.
Calculate the resistance in an RL circuit in
which L = 2.50 H and the current increases to
90.0% of its final value in 3.00 s.
Induction - Spring 2006
36
16. Show that I = I0 e – t/τ is a solution of
the differential equation where τ = L/R and
I0 is the current at t = 0.
Induction - Spring 2006
37
17.
Consider the circuit in Figure P32.17, taking ε =
6.00 V, L = 8.00 mH, and R = 4.00 Ω. (a) What is the
inductive time constant of the circuit? (b) Calculate the
current in the circuit 250 μs after the switch is closed.
(c) What is the value of the final steady-state current?
(d) How long does it take the current to reach 80.0% of
its maximum value?
Induction - Spring 2006
38
18.
In the circuit shown in Figure P32.17,
let L = 7.00 H, R = 9.00 Ω, and ε = 120 V.
What is the self-induced emf 0.200 s after the
switch is closed?
Induction - Spring 2006
39
27.
A 140-mH inductor
and a 4.90-Ω resistor are
connected with a switch to a
6.00-V battery as shown in
Figure P32.27. (a) If the
switch is thrown to the left
(connecting the battery), how
much time elapses before the
current reaches 220 mA? (b)
What is the current in the
inductor 10.0 s after the
switch is closed? (c) Now the
switch is quickly thrown from
a to b. How much time
elapses before the current
falls to 160 mA?
Induction - Spring 2006
40
32.
At t = 0, an emf of 500 V is applied to a
coil that has an inductance of 0.800 H and a
resistance of 30.0 Ω. (a) Find the energy stored
in the magnetic field when the current reaches
half its maximum value. (b) After the emf is
connected, how long does it take the current to
reach this value?
Induction - Spring 2006
41
52.
The switch in Figure P32.52 is connected to point a for a
long time. After the switch is thrown to point b, what are (a) the
frequency of oscillation of the LC circuit, (b) the maximum
charge that appears on the capacitor, (c) the maximum current in
the inductor, and (d) the total energy the circuit possesses at t =
3.00 s?
Induction - Spring 2006
42
Back to Variable
Sources
Induction - Spring 2006
43
Source Voltage:
emf  V  V0 sin( t )
Induction - Spring 2006
44
Average value of anything:
T
h T   f (t )dt
0
h
1
h 
T
T
 f (t )dt
0
T
Area under the curve = area under in the average box
Induction - Spring 2006
45
Average Value
T
1
V   V (t )dt
T0
For AC:
T
1
V   V0 sin t dt  0
T0
Induction - Spring 2006
46
So …



Average value of current will be zero.
Power is proportional to i2R and is ONLY
dissipated in the resistor,
The average value of i2 is NOT zero because
it is always POSITIVE
Induction - Spring 2006
47
Average Value
T
1
V   V (t )dt  0
T0
Vrms 
V
Induction - Spring 2006
2
48
RMS
Vrms 
V02 Sin 2t  V0
1
2 2
Sin ( t )dt

T 0
T
T
1 T 
 2 
2 2

  Sin ( t )d 
t
T  2  0
T
 T 
T
Vrms  V0
Vrms
V0

2
Vrms
V0

2
2
V0
0 Sin ( )d  2
2
Induction - Spring 2006

49
Usually Written as:
Vrms 
V peak
2
V peak  Vrms 2
Induction - Spring 2006
50
Example: What Is the RMS AVERAGE
of the power delivered to the resistor in
the circuit:
R
E
~
Induction - Spring 2006
51
Power
V  V0 sin( t )
V V0
i   sin( t )
R R
2
2
V
V
 0

2
2
0
P(t )  i R   sin( t )  R 
sin t
R
R

Induction - Spring 2006
52
More Power - Details
2
V02
V
P 
Sin 2t  0 Sin 2t
R
R
P
P
P
P
V02

R
2
V0

R
V02

R
V02

R
1
T


2
T
Sin 2 (t )dt
0
T
1
0


Sin 2 (t )dt
2
V
1 2
2
0 1
Sin ( )d 

2 0
R 2
2
1 1  V0  V0  Vrms
 


2 R  2  2 
R
Induction - Spring 2006
53
AC Circuits
April 17, 2006
Induction - Spring 2006
54
Last Days …

If you need to, file your taxes TODAY!



Due at midnight.
Note: File on Web has been updated.
This week



Monday & Wednesday – AC Circuits followed
by problem based review
Friday – Review problems Next Week
Monday – Complete Problem review.
Induction - Spring 2006
55
Final Examination




Will contain 8-10 problems. One will probably
be a collection of multiple choice questions.
Problems will be similar to WebAssign
problems. Class problems may also be a source.
You have 3 hours for the examination.
SCHEDULE: MONDAY, MAY 1 @ 10:00 AM

http://pegasus.cc.ucf.edu/%7Eenrsvc/registrar/calend
ar/exam/
Induction - Spring 2006
56
Back to AC
Induction - Spring 2006
57
Resistive Circuit


We apply an AC voltage to the circuit.
Ohm’s Law Applies
Induction - Spring 2006
58
Consider this circuit
e  iR
emf
i
R
CURRENT AND
VOLTAGE
IN PHASE
Induction - Spring 2006
59
Induction - Spring 2006
60
Alternating Current Circuits
An “AC” circuit is one in which the driving voltage and
hence the current are sinusoidal in time.
V(t)
Vp
v

2
t
V = VP sin (t - v )
I = IP sin (t - I )
-Vp
 is the angular frequency (angular speed) [radians per second].
Sometimes instead of  we use the frequency f [cycles per second]
Induction - Spring 2006
Frequency  f [cycles per second, or Hertz (Hz)]
  2 f
61
Phase
Term
V = VP
sin (wt - v )
V(t)
Vp

2
t
v
-Vp
Induction - Spring 2006
62
Alternating Current Circuits
V = VP sin (t - v )
I = IP sin (t - I )
I(t)
V(t)
Ip
Vp
Irms
Vrms
v
-Vp

2
t
I/
t
-Ip
Vp and Ip are the peak current and voltage. We also use the
“root-mean-square” values: Vrms = Vp / 2 and Irms=Ip / 2
v and I are called phase differences (these determine when
Induction - Spring 2006
63
V and I are zero). Usually we’re free to set v=0 (but not I).
Example: household voltage
In the U.S., standard wiring supplies 120 V at 60 Hz.
Write this in sinusoidal form, assuming V(t)=0 at t=0.
Induction - Spring 2006
64
Example: household voltage
In the U.S., standard wiring supplies 120 V at 60 Hz.
Write this in sinusoidal form, assuming V(t)=0 at t=0.
This 120 V is the RMS amplitude: so Vp=Vrms 2 = 170 V.
Induction - Spring 2006
65
Example: household voltage
In the U.S., standard wiring supplies 120 V at 60 Hz.
Write this in sinusoidal form, assuming V(t)=0 at t=0.
This 120 V is the RMS amplitude: so Vp=Vrms 2 = 170 V.
This 60 Hz is the frequency f: so =2 f=377 s -1.
Induction - Spring 2006
66
Example: household voltage
In the U.S., standard wiring supplies 120 V at 60 Hz.
Write this in sinusoidal form, assuming V(t)=0 at t=0.
This 120 V is the RMS amplitude: so Vp=Vrms 2 = 170 V.
This 60 Hz is the frequency f: so =2 f=377 s -1.
So V(t) = 170 sin(377t + v).
Choose v=0 so that V(t)=0 at t=0: V(t) = 170 sin(377t).
Induction - Spring 2006
67
Review: Resistors in AC Circuits
R
E
~
EMF (and also voltage across resistor):
V = VP sin (t)
Hence by Ohm’s law, I=V/R:
I = (VP /R) sin(t) = IP sin(t)
(with IP=VP/R)
V
I

2
t
Induction - Spring 2006
V and I
“In-phase”
68
Capacitors in AC Circuits
C
Start from:
q = C V [V=Vpsin(t)]
Take derivative: dq/dt = C dV/dt
So
I = C dV/dt = C VP  cos (t)
E
~
I = C  VP sin (t + /2)
V
I

2 t
This looks like IP=VP/R for a resistor
(except for the phase change).
So we call
Xc = 1/(C)
the Capacitive Reactance
The reactance is sort of like resistance in
that IP=VP/Xc. Also, the current leads
the voltage by 90o (phase difference).
Induction - Spring 2006
V and I “out of phase”
by 90º. I leads V by 90º.
69
I Leads V???
What the **(&@ does that mean??
2
V

1
I
Phase=
-(/2)
I = C  VP sin (t + /2)
Induction - Spring 2006
Current reaches it’s
maximum at an earlier time
than the voltage!
70
Capacitor Example
A 100 nF capacitor is
connected to an AC supply
of peak voltage 170V and
frequency 60 Hz.
C
E
~
What is the peak current?
What is the phase of the current?
  2f  2  60  3.77 rad/sec
C  3.77 107
1
XC 
 2.65M
C
I=V/XC
- Spring
2006 by 90o (phase difference).
71
Also, the current Induction
leads the
voltage
Inductors in AC Circuits
~
L
V = VP sin (t)
Loop law: V +VL= 0 where VL = -L dI/dt
Hence:
dI/dt = (VP/L) sin(t).
Integrate: I = - (VP / L cos (t)
or
V
Again this looks like IP=VP/R for a
resistor (except for the phase change).
I

I = [VP /(L)] sin (t - /2)
2
t So we call
the
XL =  L
Inductive Reactance
Here the current lags the voltage by 90o.
Induction by
- Spring
2006 I lags V by 90º.
V and I “out of phase”
90º.
72
Induction - Spring 2006
73
Phasor Diagrams
A phasor is an arrow whose length represents the amplitude of
an AC voltage or current.
The phasor rotates counterclockwise about the origin with the
angular frequency of the AC quantity.
Phasor diagrams are useful in solving complex AC circuits.
The “y component” is the actual voltage or current.
Resistor
Vp
Ip
t
Induction - Spring 2006
74
Phasor Diagrams
A phasor is an arrow whose length represents the amplitude of
an AC voltage or current.
The phasor rotates counterclockwise about the origin with the
angular frequency of the AC quantity.
Phasor diagrams are useful in solving complex AC circuits.
The “y component” is the actual voltage or current.
Resistor
Capacitor
Vp
Ip
Ip
t
t
Induction - Spring 2006
Vp
75
Phasor Diagrams
A phasor is an arrow whose length represents the amplitude of
an AC voltage or current.
The phasor rotates counterclockwise about the origin with the
angular frequency of the AC quantity.
Phasor diagrams are useful in solving complex AC circuits.
The “y component” is the actual voltage or current.
Resistor
Capacitor
Inductor
Vp
Ip
t
Vp
Ip
Induction - Spring 2006
Vp
Ip
76
Steady State Solution for AC
Im
Current
(2)
I m d L cos  d     I m R sin  d t    
cos  d t      m sin  d t
d C
• Expand sin & cos expressions
sin d t     sin d t cos   cos  d t sin 
cos d t     cos d t cos   sin  d t sin 
High school trig!
• Collect sindt & cosdt terms separately
cosdt terms
d L  1/ d C  cos   R sin   0
sindt terms
I m d L  1/ d C  sin   I m R cos    m
• These equationsInduction
can- Spring
be2006solved for Im and 77
(next slide)
Steady State Solution for AC
Im
Current
(2)
I m d L cos  d     I m R sin  d t    
cos  d t      m sin  d t
d C
• Expand sin & cos expressions
sin d t     sin d t cos   cos  d t sin 
cos d t     cos d t cos   sin  d t sin 
High school trig!
• Collect sindt & cosdt terms separately
cosdt terms
d L  1/ d C  cos   R sin   0
sindt terms
I m d L  1/ d C  sin   I m R cos    m
• These equationsInduction
can- Spring
be2006solved for Im and 78
(next slide)
Steady State Solution for AC Current (3)
d L  1/ d C  cos   R sin   0
I m d L  1/ d C  sin   I m R cos    m
• Solve for  and Im in terms of
tan  
d L  1/ d C
R
X  XC
 L
R
Im 
m
Z
• R, XL, XC and Z have dimensions of resistance
X L  d L
Inductive “reactance”
X C  1/ d C
Capacitive “reactance”
Z  R2   X L  X C 
2
Total “impedance”
• Let’s try to understand this solution using
“phasors”
Induction - Spring 2006
79
REMEMBER Phasor Diagrams?
A phasor is an arrow whose length represents the amplitude of
an AC voltage or current.
The phasor rotates counterclockwise about the origin with the
angular frequency of the AC quantity.
Phasor diagrams are useful in solving complex AC circuits.
Resistor
Capacitor
Inductor
Vp
Ip
Vp
Ip
t
Ip
t
t
Induction - Spring 2006
Vp
80
Reactance - Phasor Diagrams
Resistor
Capacitor
Inductor
Vp
Ip
Vp
Ip
t
Ip
t
t
Induction - Spring 2006
Vp
81
“Impedance” of an AC Circuit
R
L
~
C
The impedance, Z, of a circuit relates peak
current to peak voltage:
Ip 
Vp
Z
Induction - Spring 2006
(Units: OHMS)
82
“Impedance” of an AC Circuit
R
L
~
C
The impedance, Z, of a circuit relates peak
current to peak voltage:
Ip 
Vp
Z
(Units: OHMS)
(This is the ACInduction
equivalent
of Ohm’s law.)
- Spring 2006
83
Impedance of an RLC Circuit
R
E
~
L
C
As in DC circuits, we can use the loop method:
E - V R - VC - VL = 0
I is same through all components.
Induction - Spring 2006
84
Impedance of an RLC Circuit
R
E
~
L
C
As in DC circuits, we can use the loop method:
E - V R - VC - VL = 0
I is same through all components.
BUT: Voltages have different PHASES
 they add as PHASORS.
Induction - Spring 2006
85
Phasors for a Series RLC Circuit
Ip
VLp
VRp

(VCp- VLp)
VP
VCp
Induction - Spring 2006
86
Phasors for a Series RLC Circuit
Ip
VLp
VRp

(VCp- VLp)
VP
VCp
By Pythagoras’ theorem:
(VP )2 = [ (VRp )2 + (VCp - VLp)2 ]
Induction - Spring 2006
87
Phasors for a Series RLC Circuit
Ip
VLp
VRp

(VCp- VLp)
VP
VCp
By Pythagoras’ theorem:
(VP )2 = [ (VRp )2 + (VCp - VLp)2 ]
2 + (I
2
- Spring
2006- I X )
= Ip2 RInduction
X
p C
p L
88
Impedance of an RLC Circuit
R
Solve for the current:
Ip 
~
L
C
Vp
Vp

Z
R2  (X c  X L )2
Induction - Spring 2006
89
Impedance of an RLC Circuit
R
Solve for the current:
Ip 
~
L
C
Vp

Z
R2  (X c  X L )2
Impedance:
Vp
Z
 1

R 
 L
C
2
2
Induction - Spring 2006
90
Impedance of an RLC Circuit
Vp
Ip 
Z
 1
R 
 L
C
Z
2
2
The current’s magnitude depends on
the driving frequency. When Z is a
minimum, the current is a maximum.
This happens at a resonance frequency:
The circuit hits resonance when 1/C-L=0:  r=1/ LC
When this happens the capacitor and inductor cancel each other
and the circuit behaves purely resistively: IP=VP/R.
IP
R =10
L=1mH
C=10F
R = 1 0 0 
0
1 0
r
2
1 0
3

Induction
- Spring5 2006
4
1 0
1 0
The current dies away
at both low and high
frequencies.
91
Phase in an RLC Circuit
Ip
VLp
We can also find the phase:
VRp
(VCp- VLp)

VP
tan  = (VCp - VLp)/ VRp
or;
or
VCp
tan  = (XC-XL)/R.
tan  = (1/C - L) / R
Induction - Spring 2006
92
Phase in an RLC Circuit
Ip
VLp
We can also find the phase:
VRp
(VCp- VLp)

VP
tan  = (VCp - VLp)/ VRp
or;
or
VCp
tan  = (XC-XL)/R.
tan  = (1/C - L) / R
More generally, in terms of impedance:
cos   R/Z
At resonance the phase goes to zero (when the circuit becomes
purely resistive, the current
and- Spring
voltage
Induction
2006 are in phase).
93
Power in an AC Circuit
V
= 0

V(t) = VP sin (t)
I
2
I(t) = IP sin (t)
t
(This is for a purely
resistive circuit.)
P
P(t) = IV = IP VP sin 2(t)
Note this oscillates
twice as fast.

2
t
Induction - Spring 2006
94
Power in an AC Circuit
The power is P=IV. Since both I and V vary in time, so
does the power: P is a function of time.
Use, V = VP sin (t) and I = IP sin ( t+ ) :
P(t) = IpVpsin(t) sin ( t+ )
This wiggles in time, usually very fast. What we usually
care about is the time average of this:
1 T
P  0 P( t )dt
T
Induction - Spring 2006
(T=1/f )
95
Power in an AC Circuit
Now: sin( t   )  sin( t )cos   cos(t )sin 
Induction - Spring 2006
96
Power in an AC Circuit
Now: sin( t   )  sin( t )cos   cos(t )sin 
P( t )  I PVP sin(  t )sin(  t   )
 I PVP sin 2(  t )cos   sin(  t )cos(  t )sin 
Induction - Spring 2006
97
Power in an AC Circuit
Now: sin( t   )  sin( t )cos   cos(t )sin 
P( t )  I PVP sin(  t )sin(  t   )
 I PVP sin 2(  t )cos   sin(  t )cos(  t )sin 
Use:
and:
So
sin ( t ) 
2
1
2
sin( t ) cos( t )  0
P 
1
2
I PV P cos 
Induction - Spring 2006
98
Power in an AC Circuit
Now: sin( t   )  sin( t )cos   cos(t )sin 
P( t )  I PVP sin(  t )sin(  t   )
 I PVP sin 2(  t )cos   sin(  t )cos(  t )sin 
Use:
and:
So
sin ( t ) 
2
1
2
sin( t ) cos( t )  0
P 
1
2
I PV P cos 
which we usually writeInduction
as - Spring
P2006 IrmsVrms cos 99
Power in an AC Circuit
P  IrmsVrms cos 
 goes from -900 to 900, so the average power is positive)
cos( is called the power factor.
For a purely resistive circuit the power factor is 1.
When R=0, cos()=0 (energy is traded but not dissipated).
Usually the power factor depends on frequency.
Induction - Spring 2006
100