Download Lecture 12 - AC Circuits

The Last Leg
The Ups and Downs of Circuits
Chapter 31
The End is Near!
• Examination #3 – Friday (April
15th)
• Taxes Due – Friday (April 15th)
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• Final Exam is Wednesday, April
27th.
• Grades will be submitted as
quickly as possible.
That’s Two Weeks. That’s Two Weeks. That’s Two Weeks.
So far we have considered

DC Circuits




We looked at



Resistors
Capacitors
Inductors
Steady State DC behaviors
Transient DC behaviors.
We have not looked at sources that varied
with time.
Example LR Circuit
i
Steady Source
sum of voltage drops  0 :
di
 E  iR  L  0
dt
same form as the
capacitor equation
q
dq
E R
0
C
dt
Time Dependent Result:
E
 Rt / L
i  (1  e
)
R
time constant

L

R
R
L
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
9
10
Sinusoidal Stuff
emf  A sin( t   )
“Angle”
Phase Angle
Same Frequency
with
PHASE SHIFT

Different Frequencies
At t=0, the charged capacitor is connected
to the inductor. What would you expect
to happen??
The Math Solution:
  LC
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.
The Graph
Note – Power is delivered to our homes
as an oscillating source (AC)
Producing AC Generator
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The Real World
A
The Flux:
  B  A  BA cos 
  t
emf  BA sin t
emf
i
A sin t
Rbulb
OUTPUT
emf  V  V0 sin( t )
WHAT IS AVERAGE VALUE OF THE EMF ??
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
Average Value
T
1
V   V (t )dt
T0
For AC:
T
1
V   V0 sin t dt  0
T0
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
Average Value
T
1
V   V (t )dt  0
T0
Vrms 
V
2
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

Usually Written as:
Vrms 
V peak
2
V peak  Vrms 2
Example: What Is the RMS AVERAGE
of the power delivered to the resistor in
the circuit:
R
E
~
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

More Power - Details
2
V02
V
P 
Sin 2t  0 Sin 2t
R
R
P
P
P
P
V02

R
2
V0

R
V02

R
V02

R
1
T


2
T
Sin 2 (t )dt
0
T
1
0


Sin 2 (t )dt
2
V
1 2
2
0 1
Sin ( )d 

2 0
R 2
2
1 1  V0  V0  Vrms
 


2 R  2  2 
R
Resistive Circuit


We apply an AC voltage to the circuit.
Ohm’s Law Applies
Consider this circuit
e  iR
emf
i
R
CURRENT AND
VOLTAGE
IN PHASE
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]
Frequency  f [cycles per second, or Hertz (Hz)]
  2 f
Phase Term
V = VP sin (wt - v )
V(t)
Vp

v
-Vp
2
t
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
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.
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.
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.
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).
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
V and I
“In-phase”
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” by 90º. I leads V by 90º.
I Leads V???
What the **(&@ does that mean??
2
V

1
I
I = C  VP sin (t + /2)
Current reaches it’s maximum at
an earlier time than the voltage!
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
Also, the current leads 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” by 90º. I lags V by 90º.
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
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
Vp
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
Ip
t
t
Vp
i
i
+
+
+
time
i
i
LC Circuit
i
i
+
+
+
Analyzing the L-C Circuit
Total energy in the circuit:
1 2 1 q2
U  UB  UE  LI 
2
2 C
2
Differentiate :
dU d 1 2 1 q
 ( LI 
)  0 N o change
in energy
dt dt 2
2 C
dI q dq
dq d 2 q q dq
 LI 
 0  L( ) 2 
dt C dt
dt dt
C dt
d 2q 1
L 2  q  0
dt
C
Analyzing the L-C Circuit
Total energy in the circuit:
1 2 1 q2
U  UB  UE  LI 
2
2 C
2
Differentiate :
dU d 1 2 1 q
 ( LI 
)  0 N o change
in energy
dt dt 2
2 C
dI q dq
dq d 2 q q dq
 LI 
 0  L( ) 2 
dt C dt
dt dt
C dt
d 2q 1
L 2  q  0
dt
C
Analyzing the L-C Circuit
Total energy in the circuit:
1 2 1 q2
U  UB  UE  LI 
2
2 C
2
Differentiate :
dU d 1 2 1 q
 ( LI 
)  0 N o change
in energy
dt dt 2
2 C
dI q dq
dq d 2 q q dq
 LI 
 0  L( ) 2 
dt C dt
dt dt
C dt
d 2q 1
L 2  q  0
dt
C
Analyzing the L-C Circuit
Total energy in the circuit:
1 2 1 q2
U  UB  UE  LI 
2
2 C
2
Differentiate :
d 2q
2


q0
2
dt
q  q p cos t
dU d 1 2 1 q
 ( LI 
)  0 N o change
in energy
dt dt 2
2 C
dI q dq
dq d 2 q q dq
 LI 
 0  L( ) 2 
dt C dt
dt dt
C dt
d 2q 1
L 2  q  0
dt
C
The charge sloshes back and
forth with frequency  = (LC)-1/2