the last leg the ups and downs of circuits chapter 31
TRANSCRIPT
The Last LegThe Last LegThe Last LegThe Last LegThe Ups and Downs of CircuitsThe Ups and Downs of Circuits
Chapter 31Chapter 31
The End is Near!
• Examination #3 – Friday (April 15th)• Taxes Due – Friday (April 15th)• Watch for new WebAssigns• 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 Resistors Capacitors Inductors
We looked at Steady State DC behaviors Transient DC behaviors.
We have not looked at sources that varied with time.
Example LR Circuit
i
0
equationcapacitor
theas form same
0
:0 drops voltageof sum
dt
dqR
C
qE
dt
diLiRE
Steady Source
Time Dependent Result:
R
L
eR
Ei LRt
constant time
)1( /
R
L
-1.5
-1
-0.5
0
0.5
1
1.5
0 1 2 3 4 5 6 7 8 9 10
Time
Vo
lts
Variable Emf Applied
emf
Sinusoidal
DC
Sinusoidal Stuff
)sin( tAemf
“Angle”
Phase Angle
Same Frequencywith
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
x x x x x x x x x x x x x x x x x x x x x x xx x x x x x x x x x x x x x x x x x x x x x xx x x x x x x x x x x x x x x x x x x x x x xx x x x x x x x x x x x x x x x x x x x x x xx x x x x x x x x x x x x x x x x x x x x x xx x x x x x x x x x x x x x x x x x x x x x xx x x x x x x x x x x x x x x x x x x x x x xx x x x x x x x x x x x x x x x x x x x x x xx x x x x x x x x x x x x x x x x x x x x x xx x x x x x x x x x x x x x x x x x x x x x xx x x x x x x x x x x x x x x x x x x x x x xx x x x x x x x x x x x x x x x x x x x x x x
The Real World
A
The Flux:
tAR
emfi
tBAemf
t
BA
bulb
sin
sin
cos
AB
OUTPUT
)sin(0 tVVemf
WHAT IS AVERAGE VALUE OF THE EMF ??
Average value of anything:
Area under the curve = area under in the average box
T
T
dttfT
h
dttfTh
0
0
)(1
)(
T
h
Average Value
T
dttVT
V0
)(1
0sin1
0
0 T
dttVT
V
For AC:
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
0)(1
0
T
dttVT
V
2VVrms
RMS
2
2)(
2
2)
2(
2
1
)2
(1
0
02
0
20
0
20
0
20
220
VV
VdSin
VV
tT
dtT
SinT
TVV
dttT
SinT
VtSinVV
rms
rms
T
rms
T
rms
Usually Written as:
2
2
rmspeak
peakrms
VV
VV
Example: What Is the RMS AVERAGE of the power delivered to the resistor in the circuit:
E
R
~
Power
tR
VRt
R
VRitP
tR
V
R
Vi
tVV
22
0
2
02
0
0
sin)sin()(
)sin(
)sin(
More Power - Details
R
VVV
RR
VP
R
VdSin
R
VP
tdtSinR
VP
dttSinTR
VP
tSinR
VtSin
R
VP
rms
T
T
200
20
20
2
0
22
0
0
22
0
0
22
0
22
022
0
22
1
2
1
2
1)(
2
1
)(1
2
)(1
Resistive Circuit
We apply an AC voltage to the circuit. Ohm’s Law Applies
Con
sid
er
this
cir
cuit
CURRENT ANDVOLTAGE IN PHASE
R
emfi
iRe
Alternating Current Circuits
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)] f
V = VP sin (t -v ) I = IP sin (t -I )
An “AC” circuit is one in which the driving voltage andhence the current are sinusoidal in time.
v
V(t)
t
Vp
-Vp
v
V(t)
t
Vp
-Vp
V = VP sin (wt - v )Phase Term
Vp and Ip are the peak current and voltage. We also use the
“root-mean-square” values: Vrms = Vp / and Irms=Ip /
v and I are called phase differences (these determine whenV and I are zero). Usually we’re free to set v=0 (but not I).
2 2
Alternating Current Circuits
V = VP sin (t -v ) I = IP sin (t -I )
v
V(t)
t
Vp
-Vp
Vrms
I/
I(t)
t
Ip
-Ip
Irms
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 = 170 V.2
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 = 170 V.This 60 Hz is the frequency f: so =2f=377 s -1.
2
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 = 170 V.This 60 Hz is the frequency f: so =2f=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).
2
Resistors in AC Circuits
ER
~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 and I“In-phase”
V
t
I
This looks like IP=VP/R for a resistor (except for the phase change). So we call Xc = 1/(C) the Capacitive Reactance
Capacitors in AC Circuits
E
~C Start from: q = C V [V=Vpsin(t)]
Take derivative: dq/dt = C dV/dtSo I = C dV/dt = C VP cos (t)
I = C VP sin (t + /2)
The reactance is sort of like resistance in that IP=VP/Xc. Also, the current leads the voltage by 90o (phase difference).
V
t
I
V and I “out of phase” by 90º. I leads V by 90º.
I Leads V???What the **(&@ does that mean??
I
V
Current reaches it’s maximum at
an earlier time than the voltage!
1
2
I = C VP sin (t +/2)
Capacitor Example
E
~
CA 100 nF capacitor isconnected to an AC supply of peak voltage 170V and frequency 60 Hz.
What is the peak current?What is the phase of the current?
MX
f
C 65.2C
1
1077.3C
rad/sec 77.360227
Also, the current leads the voltage by 90o (phase difference).
Again this looks like IP=VP/R for aresistor (except for the phase change).
So we call XL = L the Inductive Reactance
Inductors in AC Circuits
LV = VP sin (t)Loop law: V +VL= 0 where VL = -L dI/dtHence: dI/dt = (VP/L) sin(t).Integrate: I = - (VP / L cos (t)
or I = [VP /(L)] sin (t - /2)
~
Here the current lags the voltage by 90o.
V
t
I
V and I “out of phase” by 90º. I lags V by 90º.
Phasor Diagrams
Vp
Ipt
Resistor
A phasor is an arrow whose length represents the amplitude ofan 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.
A phasor is an arrow whose length represents the amplitude ofan 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.
Phasor Diagrams
Vp
Ipt
Vp
Ip
t
Resistor Capacitor
A phasor is an arrow whose length represents the amplitude ofan 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.
A phasor is an arrow whose length represents the amplitude ofan 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.
Phasor Diagrams
Vp
Ipt
Vp
Ip
t
Vp Ip
t
Resistor Capacitor Inductor
A phasor is an arrow whose length represents the amplitude ofan 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.
A phasor is an arrow whose length represents the amplitude ofan 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.
+ +i
+++
+
i
i
i
i
i
LC Circuit
time
U UB UE 12LI2 1
2q2
C
dU
dt ddt
(1
2LI2 1
2
q2
C) 0
LIdI
dtq
C
dq
dt0 L(
dq
dt)d2q
dt2q
C
dq
dt
Ld2q
dt 2
1
Cq 0
Analyzing the L-C Circuit
Total energy in the circuit:
Differentiate : N o change in energy
U UB UE 12LI2 1
2q2
C
dU
dt ddt
(1
2LI2 1
2
q2
C) 0
LIdI
dtq
C
dq
dt0 L(
dq
dt)d2q
dt2q
C
dq
dt
Ld2q
dt 2
1
Cq 0
Analyzing the L-C Circuit
Total energy in the circuit:
Differentiate : N o change in energy
U UB UE 12LI2 1
2q2
C
dU
dt ddt
(1
2LI2 1
2
q2
C) 0
LIdI
dtq
C
dq
dt0 L(
dq
dt)d2q
dt2q
C
dq
dt
Ld2q
dt 2
1
Cq 0
Analyzing the L-C Circuit
Total energy in the circuit:
Differentiate : N o change in energy
U UB UE 12LI2 1
2q2
C
dU
dt ddt
(1
2LI2 1
2
q2
C) 0
LIdI
dtq
C
dq
dt0 L(
dq
dt)d2q
dt2q
C
dq
dt
Ld2q
dt 2
1
Cq 0
Analyzing the L-C Circuit
Total energy in the circuit:
Differentiate : N o change in energy
The charge sloshes back andforth with frequency = (LC)-1/2
The charge sloshes back andforth with frequency = (LC)-1/2
tqq
qdt
qd
p
cos
022
2