Transcript Time Varying Circuits

Time Varying Circuits
2009
Induction
1
The Final Exam Approacheth
8-10 Problems similar
to Web-Assignments
 Covers the entire
semester’s work
 May contain some
short answer
(multiple choice)
questions.

Induction
2
Spring 2009 Final Exam Schedule
Tuesday, April 28 - Monday, May 4
No Exams on Sunday
EXAM
TIMES
Class Meeting Times
EXAM DAY 1
TUES 4/28
EXAM DAY 2
WED 4/29
EXAM DAY 3
THURS 4/30
EXAM DAY 4
FRI 5/1
EXAM DAY 5
SAT 5/2
7:00 a.m. 9:50 a.m.
7:30-10:20 F 9:007:30-10:20 W 8:3010:15 M/
7:30-10:20 T 9:007:30-8:45 TR 7:309:20 MWF 9:007:30-8:45 F 9:3010:15 TR (all a.m.)
10:20 R (all a.m.)
10:15 MW (all a.m.)
10:20 MWF (all
a.m.)
10:00 a.m. 12:50 p.m.
10:30-11:45 TR
10:30-1:20 T
1:00 p.m. 3:50 p.m.
1:30-4:20 W 2:301:30-2:45 TR 1:30- 3:20 MWF 3:001:30-4:20 R 3:004:20 T
4:15 WF 3:00-4:15 4:15 TR
MW
4:00 p.m. 6:50 p.m.
6:00-7:15 TR
6:00-7:15 MW
4:30-5:45 TR
4:30-7:15 F
FREE PERIOD
and Alternate Time and Alternate Time and Alternate Time and Alternate Time
7:00 p.m. 9:50 p.m.
6:00-8:50 T 7:3010:20 T (all p.m.)
10:30-1:20 W
11:30-12:20 MWF
12:00-1:15 MW
12:00-1:15 WF
6:00-8:50 W 7:007:50 MWF 7:308:45 MW 7:3010:20 W (all p.m.)
10:30-1:20 F 12:3010:30-1:20 R 12:00- 1:20 MWF 12:001:15 TR
1:15 M/
10:30-11:45 F
1:30-4:20 F 3:304:20 MWF 3:004:15 M/
1:30-2:45 F
EXAM DAY 6
MON 5/4
Finals For Saturday
7:30-8:20 MWF
Classes Are Held
7:30-8:45 MW 7:30During Regular
10:20 M (all a.m.)
Class Meeting Times
Finals For Saturday
10:30-11:20 MWF
Classes Are Held
10:30-11:45 MW
During Regular
10:30-1:20 M
Class Meeting Times
FREE PERIOD
6:00-8:50 R 7:306:00-8:50 F 8:008:45 TR 7:30-10:20
FREE PERIOD
Induction
8:50 MWF (all p.m.)
R (all p.m.)
1:30-2:20 MWF
1:30-2:45 MW 1:304:20 M
4:30-5:45 MW
and Alternate Time
6:00-6:50 MWF
6:00-8:50 M
7:303
10:20 M (all p.m.)
HowjaDo??
A.
B.
C.
D.
4
I done good
I done ok
I done not so ok
I screwed up major
Induction
The Test Itself Was
A.
B.
C.
D.
5
Fair
Not so fair.
Really Unfair.
The worst kind of unfair in the entire
universe.
Induction
Sort of like RC circuit issues.
Back to Circuits for a bit ….
Definition
Current in loop produces a magnetic field
in the coil and consequently a magnetic flux.
If we attempt to change the current, an emf
will be induced in the loops which will tend to
oppose the change in current.
This this acts like a “resistor” for changes in current!
Remember Faraday’s Law
d
emf  V   E  ds  
dt
Lentz
Look at the following circuit:



Switch is open
NO current flows in the circuit.
All is at peace!
Close the
circuit…



After the circuit has been close for a long time, the
current settles down.
Since the current is constant, the flux through the coil is
constant and there is no
.
Emf
Current is simply E/R (Ohm’s Law)
Close the
circuit…




When switch is first closed, current begins to flow rapidly.
The flux through the inductor changes rapidly.
An emf is created in the coil that opposes the increase in
current.
The net potential difference across the resistor is the
battery emf opposed by the emf of the coil.
Close the
circuit…
d
emf  
dt
Ebattery  V (notation)
d
 V  iR 
0
dt
Moving right along …
Ebattery  V (notation)
d
 V  iR 
0
dt
The flux is proportion al to the current
as well as to the number of turns, N.
For a solonoid,
  i  Li  N B
d
di
L
dt
dt
di
 V  iR  L  0
dt
Definition of Inductance L
N B
L
i
UNIT of Inductance = 1 henry = 1 T- m2/A
B is the flux near the center of one of the coils
making the inductor
Consider a Solenoid
l
 B  ds   i
0 enclosed
 Bl   0 nli
n turns per unit length
or
B   0 ni
So….
N B nlBA nl 0 niA
L


i
i
i
or
L   0 n 2 Al
or
inductance
2
L/l 
 n A
unit length
Depends only on geometry just like C and
is independent of current.
Inductive Circuit

i




Switch to “a”.
Inductor seems like a
short so current rises
quickly.
Field increases in L and
reverse emf is generated.
Eventually, i maxes out and
back emf ceases.
Steady State Current
after this.
THE BIG INDUCTION




As we begin to increase the current in the coil
The current in the first coil produces a magnetic
field in the second coil
Which tries to create a current which will reduce
the field it is experiences
And so resists the increase in current.
Back to the real world…
Switch to “a”
i
sum of voltage drops  0 :
di
 E  iR  L  0
dt
same form as the
capacitor equation
q
dq
E R
0
C
dt
Solution (See textbook)
E
 Rt / L
i  (1  e
)
R
time constant
L

R
Switch position “b”
E0
di
L  iR  0
dt
E t / 
i e
R
VR=iR
~current
Max Current Rate of
increase = max emf
E
(1  eRt / L )
R
L
  (time constant)
R
i
IMPORTANT QUESTION





Switch closes.
No emf
Current flows for a while
It flows through R
Energy is conserved (i2R)
WHERE DOES THE ENERGY COME FROM??
For an answer
Return to the Big C

E=e0A/d


+dq
+q
-q

We move a charge dq
from the (-) plate to the
(+) one.
The (-) plate becomes
more (-)
The (+) plate becomes
more (+).
dW=Fd=dq x E x d
The calc

q
dW  (dq ) Ed  (dq ) d  (dq )
d
e0
e0 A
d
d q2
W
qdq 

e0 A
e0 A 2
or
2
2


1

Ad
1

1
2
2


W
(A) 
 e 0  2  Ad  e 0 E Ad
2e 0 A
2 e0
2  e0 
2
d
energy
1
2
u
 e0 E
unit volum e 2
The energy is in
the FIELD !!!
What about POWER??
di
E  L  iR
dt
i :
di 2
iE  Li  i R
dt
power
to
circuit
Must be dWL/dt
power
dissipated
by resistor
So
dWL
di
 Li
dt
dt
1 2
WL  L  idi  Li
2
1
WC  CV 2
2
Energy
stored
in the
Capacitor
WHERE is the energy??
l
 B  ds   i
0 enclosed
0l  Bl   0 nil
B   0 ni
or
B
 0 Ni
l
  BA 
 0 Ni
l
A
Remember the Inductor??
N
L
i
N  Number of turns in inductor
i  current.
Φ  Magnetic flux throu gh one turn.
So …
N
L
i
N
i
L
1 2 1 2 N 1
W  Li  i
 N i
2
2
i
2
 0 NiA

l
1   0 NiA 
1
2
2 2 A
W  Ni
0 N i

2  l  2 0
l
1
A
W
 N i
2 0
l
2
0
2 2
From before :
 0 Ni
B
l
1 2 2A
1 2
W
Bl

B V (volume)
2 0
l 2 0
or
W
1 2
u

B
V 2 0
ENERGY IN THE
FIELD TOO!
IMPORTANT CONCLUSION


A region of space that contains either a magnetic or
an electric field contains electromagnetic energy.
The energy density of either is proportional to the
square of the field strength.
Solution (From Before)
E
 Rt / L
i  (1  e
)
R
time constant
L

R
At t=0, the charged capacitor is now connected to the
inductor. What would you expect to happen??
35
Induction
The math …
For an RLC circuit with no driving potential (AC or DC source):
Q
di
iR 
L
0
C
dt
dQ Q
d 2Q
R

L
0
2
dt
C
dt
Solution :
Q  Qmax e

Rt
2L
cos( d t )
where
 1
 R 
d  



 LC  2 L 
36
2




1/ 2
Induction
The Graph of that LR (no emf) circuit
I
37
e
Rt

2L
Induction
38
Induction
Mass on a Spring Result


Energy will swap back and forth.
Add friction


39
Oscillation will slow down
Not a perfect analogy
Induction
40
Induction
LC Circuit
Low
High
Q/C
High
Low
41
Induction
The Math Solution (R=0):
  LC
Induction
42
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.
43
Induction
Variable Emf Applied
1.5
1
Volts
emf
0.5
DC
0
0
1
2
3
4
5
6
7
8
-0.5
-1
Sinusoidal
-1.5
Tim e
44
Induction
9
10
Sinusoidal Stuff
emf  A sin( t   )
“Angle”
Phase Angle
45
Induction
Same Frequency
with
PHASE SHIFT

46
Induction
Different Frequencies
47
Induction
Note – Power is delivered to our
homes as an oscillating source (AC)
48
Induction
Producing AC Generator
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49
Induction
The Real World
50
Induction
A
51
Induction
52
Induction
The Flux:
  B  A  BA cos 
  t
emf  BA sin t
emf
i
A sin t
Rbulb
53
Induction
problems …
54
Induction
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
55
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
56
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
57
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
58
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
59
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
60
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
61
Source Voltage:
emf  V  V0 sin( t )
62
Induction
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
63
Induction
Average Value
T
1
V   V (t )dt
T0
For AC:
64
T
1
V   V0 sin t dt  0
T0
Induction
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
65
Induction
Average Value
T
1
V   V (t )dt  0
T 0
Vrms 
66
V
2
Induction
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
67
Vrms
V0

2
Vrms
V0

2
2
V0
0 Sin ( )d  2
2
Induction

Usually Written as:
Vrms 
V peak
2
V peak  Vrms 2
68
Induction
in the circuit:
R
E
~
69
Induction
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

70
Induction
More Power - Details
2
V02
V
P 
Sin 2t  0 Sin 2t
R
R
P
P
P
P
71
V02

R
V02

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
Resistive Circuit


We apply an AC voltage to the circuit.
Ohm’s Law Applies
Induction
72
Consider this circuit
73
e  iR
emf
i
R
CURRENT AND
VOLTAGE
IN PHASE
Induction
74
Induction
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
75
Phase
Term
V= V
P sin (t - v )
V(t)
Vp

2
t
v
-Vp
Induction
76
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
77
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
78
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
79
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
80
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
81
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”
82
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.
83
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!
84
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
85
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.
86
Induction
87
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
88
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
89
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
90
Steady State Solution for AC
Im
Current
(2)
I m d L cos  d     I m R sin  d t    
cos  d t     e 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   e m
• These equations can
be
solved
for
I
and

m
Induction
91
(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     e 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   e m
• These equations can
be
solved
for
I
and

m
Induction
92
(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   e m
• Solve for  and Im in terms of
tan  
d L  1/ d C
R
X  XC
 L
R
Im 
em
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
93
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
94
Reactance - Phasor Diagrams
Resistor
Capacitor
Inductor
Vp
Ip
Vp
Ip
t
Ip
t
t
Induction
Vp
95
“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)
96
“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
97
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
98
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
99
Phasors for a Series RLC Circuit
Ip
VLp
VRp

(VCp- VLp)
Induction
VP
VCp
100
Phasors for a Series RLC Circuit
Ip
VLp
VRp

(VCp- VLp)
VP
VCp
By Pythagoras’ theorem:
(VP )2 = [ (VRp )2 + (VCp - VLp)2 ]
Induction
101
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
102
Impedance of an RLC Circuit
R
Solve for the current:
Ip 
~
L
C
Vp
Vp

Z
R2  (X c  X L )2
Induction
103
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
104
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.
105
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
106
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
107
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
108
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 )
109
Power in an AC Circuit
Now: sin(t   )  sin(t )cos   cos(t )sin 
Induction
110
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
111
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
112
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 
113
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
114
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
115
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
116
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
117
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
118