# What Now

Document Sample

```					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
   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) Web-
Assignments
 Covers the entire
semester’s work
 May contain some
questions.
Induction                 3
Max Current Rate of
VR=iR                  increase = max emf
~current

Induction                         4
E
i   (1  e  Rt / L )
R
L
  (time constant)
R

Induction                       5
We also showed that

1
uinductor 
2
B
20
1
 0 E
2
ucapacitor
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
iR    L 0
C      dt
dQ Q        d 2Q
R      L 2 0
dt C        dt
Solution :
Rt

Q  Qm axe       2L
cos(d t )
where
2 1/ 2
 1  R               
d                     
 LC  2 L 
                      


Induction                      8
The Graph of that LR (no emf) circuit ..
Rt

2L
I                e

Induction               9
Induction   10
Mass on a Spring Result

   Energy will swap back and forth.
   Oscillation will slow down
   Not a perfect analogy

Induction   11
Induction   12
LC Circuit

High
Low

High                             Q/C
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

0.5

emf
DC
Volts

0
0   1   2   3    4           5   6    7     8     9   10

-0.5

-1

-1.5                                        Sinusoidal
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

xxxxxxxxxxxxxxxxxxxxxxx
xxxxxxxxxxxxxxxxxxxxxxx
xxxxxxxxxxxxxxxxxxxxxxx
xxxxxxxxxxxxxxxxxxxxxxx
xxxxxxxxxxxxxxxxxxxxxxx
xxxxxxxxxxxxxxxxxxxxxxx
xxxxxxxxxxxxxxxxxxxxxxx
xxxxxxxxxxxxxxxxxxxxxxx
xxxxxxxxxxxxxxxxxxxxxxx
xxxxxxxxxxxxxxxxxxxxxxx
xxxxxxxxxxxxxxxxxxxxxxx
xxxxxxxxxxxxxxxxxxxxxxx

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                                          T
1
h   f (t )dt
T 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
V   V0 sin t dt  0
1
T 0

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   2

Induction           38
RMS
2 2
T
1
Vrms       V02 Sin2t  V0           Sin ( T t )dt
T 0

1 T         2 2     2 
T
Vrms  V0             Sin ( t )d     t
T  2  0      T      T 
2
V0                     V0
          Sin ( )d  2           
2
Vrms
2      0

V0
Vrms   
2

Induction                       39
Usually Written as:

V peak
Vrms 
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
 V0            V02
P (t )  i R   sin(t )  R 
2
sin t
2

R               R

Induction                  42
More Power - Details
V02          V02
P      Sin2t      Sin2t
R            R
V02   1 T
T 0
P                Sin2 (t )dt
R
V02    T1
2 0 
P                    Sin2 (t )dt
R
V02    1 2                  V02 1
           0 Sin ( )d  R 2
2
P
R    2
V02   1 1  V0  V0  Vrms     2
P                          
R    2 R  2  2            R
Induction             43
Resistive Circuit

   We apply an AC voltage to the circuit.
   Ohm’s Law Applies

Induction         44
e  iR
emf
Consider this circuit

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 = VP sin (t - v )
        2    t          I = IP sin (t - I )
v
-Vp

 is the angular frequency (angular speed) [radians per second].
Sometimes instead of  we use the frequency f [cycles per second]
  2 f
Induction                           47
Frequency  f [cycles per second, or Hertz (Hz)]
V = Term
Phase VP 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
Vrms                                     Irms

      2   t                                     t
v                                     I/
-Vp                                    -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
V and I are zero). Usually we’re free to set v=0 (but not I).
Induction                           49
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      EMF (and also voltage across resistor):
E                  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
V and I
      2        t           “In-phase”

Induction                         54
Capacitors in AC Circuits
C              Start from:         q = C V [V=Vpsin(t)]
E                      Take derivative: dq/dt = C dV/dt
~                      So       I = C dV/dt = C VP  cos (t)
I = C  VP sin (t + /2)

V      This looks like IP=VP/R for a resistor
(except for the phase change).
I            So we call        Xc = 1/(C)
the Capacitive Reactance
          2 t
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” by 90º. I leads V by 90º.
Induction                              55
What the **(&@ does that mean??
2

V
1                       I

Phase=
-(/2)
I = C  VP sin (t + /2)

Current reaches it’s
maximum at an earlier time
Induction       than the voltage!      56
Capacitor Example
A 100 nF capacitor is                                   C
connected to an AC supply               E
of peak voltage 170V and
frequency 60 Hz.                          ~
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                           I=V/XC
XC      2.65M
C
Also, the current leads the voltage by 90o (phase difference).
Induction                            57
Inductors in AC Circuits
V = VP sin (t)

~           L    Loop law: V +VL= 0 where VL = -L dI/dt
Hence: dI/dt = (VP/L) sin(t).
Integrate: I = - (VP / L cos (t)
or    I = [VP /(L)] sin (t - /2)
V                   Again this looks like IP=VP/R for a
resistor (except for the phase change).
I
            t So we call         XL =  L
2
the       Inductive Reactance
Here the current lags the voltage by 90o.

V and I “out of phase” by 90º. I lags V by 90º.
Induction                           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
Ip
t

Induction        Vp                    62
Current (2)cos d t      m sin d t
I m d L cos  d     I m R sin  d t   
Im
d C

• Expand sin & cos expressions
sin d t     sin d t cos   cos d t sin 
High school trig!
cos d t     cos d t cos   sin d t sin 

• Collect sindt & cosdt terms separately
d L  1/ d C  cos   R sin   0           cosdt terms
I m d L  1/ d C  sin   I m R cos    m sindt terms


• These equations can be solved for Im and 63
Induction
(next slide)
Current (2)cos d t      m sin d t
I m d L cos  d     I m R sin  d t   
Im
d C

• Expand sin & cos expressions
sin d t     sin d t cos   cos d t sin 
High school trig!
cos d t     cos d t cos   sin d t sin 

• Collect sindt & cosdt terms separately
d L  1/ d C  cos   R sin   0           cosdt terms
I m d L  1/ d C  sin   I m R cos    m sindt terms


• These equations can be solved for Im and 64
Induction
(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
d L  1/ d C    X  XC                        m
tan                       L                    Im 
R              R                           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
Ip
t                                                  t
t
Induction        Vp                        66
Reactance - Phasor Diagrams

Resistor         Capacitor                       Inductor
Vp
Ip        Vp          Ip
Ip
t                                               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:
Vp
Ip             (Units: OHMS)
Z
Induction                   68
“Impedance” of an AC Circuit
R

L
~          C

The impedance, Z, of a circuit relates peak
current to peak voltage:
Vp
Ip           (Units: OHMS)
Z
(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
Induction                   71
Phasors for a Series RLC Circuit
Ip
VRp
VLp
             VP

(VCp- VLp)
VCp

Induction              72
Phasors for a Series RLC Circuit
Ip
VRp
VLp
             VP

(VCp- VLp)
VCp

By Pythagoras’ theorem:
(VP )2 = [ (VRp )2 + (VCp - VLp)2 ]

Induction              73
Phasors for a Series RLC Circuit
Ip
VRp
VLp
         VP

(VCp- VLp)
VCp

By Pythagoras’ theorem:
(VP )2 = [ (VRp )2 + (VCp - VLp)2 ]

= Ip2 R2 +InductionXC - Ip XL) 2
(Ip                    74
Impedance of an RLC Circuit
R

Solve for the current:                            L
~   C
Vp             Vp
Ip                        
R2  (X c  X L )2   Z

Induction           75
Impedance of an RLC Circuit
R

Solve for the current:                                 L
~    C
Vp             Vp
Ip                        
R2  (X c  X L )2   Z

 1
2
Impedance:          Z       R 
2
 L 

C
Induction                76
Impedance of an RLC Circuit
Vp
Ip                                         The current’s magnitude depends on
Z                                   the driving frequency. When Z is a
 1
2
 L
minimum, the current is a maximum.
Z           R 
2
 C                      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               L=1mH                    R =10

C=10F                                      The current dies away
R = 1 0 0 
at both low and high
r                           frequencies.
0
1 0
2
1 0
3
1 0
4    Induction 5
1 0                        77
Phase in an RLC Circuit
Ip      We can also find the phase:
VRp
VLp                                 tan  = (VCp - VLp)/ VRp
     VP
or;
tan  = (XC-XL)/R.
or
(VCp- VLp)       VCp                tan  = (1/C - L) / R

Induction                             78
Phase in an RLC Circuit
Ip      We can also find the phase:
VRp
VLp                                 tan  = (VCp - VLp)/ VRp
      VP
or;
tan  = (XC-XL)/R.
or
(VCp- VLp)          VCp             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                                V(t) = VP sin (t)
I
I(t) = IP sin (t)
          2          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   P( t )dt               (T=1/f )
T 0
Induction                     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 
1
Use:    sin ( t ) 
2
2
and:    sin( t ) cos( t )  0
1
So        P        I PV P cos 
2

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 
1
Use:    sin ( t ) 
2
2
and:    sin( t ) cos( t )  0
1
So        P         I PV P cos 
2

P
which we usually write as Induction     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

```
DOCUMENT INFO
Shared By:
Categories:
Tags:
Stats:
 views: 5 posted: 2/27/2012 language: English pages: 90