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					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) Web-
       Assignments
      Covers the entire
       semester’s work
      May contain some
       short answer
       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.
   Add friction
       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

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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                                          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
          I Leads V???
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
       Steady State Solution for AC
                              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)
       Steady State Solution for AC
                              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
              they add as PHASORS.
                     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

				
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