Transcript What Now??? - UCF Physics

Time Varying Circuits
2008
Induction
1
A look into the future

We have one more week after today (+ one
day)





Time Varying Circuits Including AC
Some additional topics leading to waves
A bit of review if there is time.
There will be one more Friday morning quiz.
I hope to be able to return the exams on
Monday at which time we will briefly review the
solutions.
Induction
2
The Final Exam
8-10 Problems similar
to (or exactly) WebAssignments
 Covers the entire
semester’s work
 May contain some
short answer
questions.

Induction
3
Max Current Rate of
increase = max emf
VR=iR
~current
Induction
4
E
(1  eRt / L )
R
L
  (time constant)
R
i
Induction
5
We also showed that
uinductor 
1
20
B
2
1
2
ucapacitor   0 E
2
Induction
6
At t=0, the charged capacitor is now
connected to the inductor. What would
you expect to happen??
Induction
7
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
d  
 LC
 R 
 
 2L 
Induction
2 1/ 2



8
The Graph of that LR (no emf) circuit ..
I
e
Induction
Rt

2L
9
Induction
10
Mass on a Spring Result


Energy will swap back and forth.
Add friction


Oscillation will slow down
Not a perfect analogy
Induction
11
Induction
12
LC Circuit
High
Low
Q/C
High
Low
Induction
13
The Math Solution (R=0):
  LC
Induction
14
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
15
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
16
Sinusoidal Stuff
emf  A sin( t   )
“Angle”
Phase Angle
Induction
17
Same Frequency
with
PHASE SHIFT

Induction
18
Different Frequencies
Induction
19
Note – Power is delivered to our homes
as an oscillating source (AC)
Induction
20
Producing AC Generator
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Induction
21
The Real World
Induction
22
A
Induction
23
Induction
24
The Flux:
  B  A  BA cos 
  t
emf  BA sin t
emf
i
A sin t
Rbulb
Induction
25
problems …
Induction
26
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
27
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
28
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
29
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
30
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
31
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
32
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
33
Source Voltage:
emf  V  V0 sin( t )
Induction
34
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
35
Average Value
T
1
V   V (t )dt
T0
For AC:
T
1
V   V0 sin t dt  0
T0
Induction
36
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
37
Average Value
T
1
V   V (t )dt  0
T 0
Vrms 
V
Induction
2
38
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

39
Usually Written as:
Vrms 
V peak
2
V peak  Vrms 2
Induction
40
Example: What Is the RMS AVERAGE
of the power delivered to the resistor in
the circuit:
R
E
~
Induction
41
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
42
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 (t )dt
2
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
43
Resistive Circuit


We apply an AC voltage to the circuit.
Ohm’s Law Applies
Induction
44
Consider this circuit
e  iR
emf
i
R
CURRENT AND
VOLTAGE
IN PHASE
Induction
45
Induction
46
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
Frequency  f [cycles per second, or Hertz (Hz)]
  2 f
47
Phase
Term
V= V
P sin (t - v )
V(t)
Vp

2
t
v
-Vp
Induction
48
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
49
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
50
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
51
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
52
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
53
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
V and I
“In-phase”
54
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).
V and I “out of phase”Induction
by 90º. I leads V by 90º.
55
I Leads V???
What the **(&@ does that mean??
2
V

I
1
Phase=
-(/2)
I = C  VP sin (t + /2)
Induction
Current reaches it’s
maximum at an earlier time
than the voltage!
56
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
57
Also, the current leadsInduction
the voltage by 90o (phase difference).
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.
V and I “out of phase”Induction
by 90º. I lags V by 90º.
58
Induction
59
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
60
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
Vp
61
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
Vp
Ip
t
Induction
Vp
Ip
62
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 equations can
be
solved
for
I
and

m
Induction
63
(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 equations can
be
solved
for
I
and

m
Induction
64
(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
65
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
Vp
66
Reactance - Phasor Diagrams
Resistor
Capacitor
Inductor
Vp
Ip
Vp
Ip
t
Ip
t
t
Induction
Vp
67
“Impedance” of an AC Circuit
R
L
~
C
The impedance, Z, of a circuit relates peak
current to peak voltage:
Ip 
Vp
Z
Induction
(Units: OHMS)
68
“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 AC equivalent
of Ohm’s law.)
Induction
69
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
70
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
71
Phasors for a Series RLC Circuit
Ip
VLp
VRp

(VCp- VLp)
Induction
VP
VCp
72
Phasors for a Series RLC Circuit
Ip
VLp
VRp

(VCp- VLp)
VP
VCp
By Pythagoras’ theorem:
(VP )2 = [ (VRp )2 + (VCp - VLp)2 ]
Induction
73
Phasors for a Series RLC Circuit
Ip
VLp
VRp

(VCp- VLp)
VP
VCp
By Pythagoras’ theorem:
(VP )2 = [ (VRp )2 + (VCp - VLp)2 ]
= Ip2 R2 +Induction
(Ip XC - Ip XL) 2
74
Impedance of an RLC Circuit
R
Solve for the current:
Ip 
~
L
C
Vp
Vp

Z
R2  (X c  X L )2
Induction
75
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
76
Impedance of an RLC Circuit
Vp
Ip 
Z
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:
2
1

2
R 
 L
C
Z
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
1 0
4
Induction 5
1 0

The current dies away
at both low and high
frequencies.
77
Phase in an RLC Circuit
Ip
VLp
We can also find the phase:
VRp
(VCp- VLp)

VP
tan  = (VCp - VLp)/ VRp
or;
or
VCp
Induction
tan  = (XC-XL)/R.
tan  = (1/C - L) / R
78
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
voltage are in phase).
Induction
79
Power in an AC Circuit
V
= 0

I
2
t
V(t) = VP sin (t)
I(t) = IP sin (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
80
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
(T=1/f )
81
Power in an AC Circuit
Now: sin(t   )  sin(t )cos   cos(t )sin 
Induction
82
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
83
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
84
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 write as InductionP
 IrmsVrms cos 
85
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
86
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
87
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
88
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
89
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
90