the ups and downs of circuits the end is near! quiz – nov 18 th – material since last quiz....
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The Ups and Downs of The Ups and Downs of CircuitsCircuits
The End is Near!
• Quiz – Nov 18th – Material since last quiz. (Induction)
• Exam #3 – Nov 23rd – WEDNESDAY• LAST CLASS – December 2nd • FINAL EXAM – 12/5 10:00-12:50 Room
MAP 359• Grades by end of week. Hopefully
Maybe.
A circular region in the xy plane is penetrated by a uniform magnetic field in the positive direction of the z axis. The field's magnitude B (in teslas) increases with time t (in seconds) according to B = at, where a is a constant. The magnitude E of the electric field set up by that increase in the magnetic field is given in the Figure as a function of the distance r from the center of the region. Find a.[0.030] T/s
VG
r
For the next problem, recall that
R
L
eR
Ei LRt
constant time
)1( /i
R
L
For the circuit of Figure 30-19, assume that = 11.0 V, R = 6.00 , and L = 5.50 H. The battery is connected at time t = 0.
(a) How much energy is delivered by the battery during the first 2.00 s?[23.9] J (b) How much of this energy is stored in the magnetic field of the inductor? [7.27] J(c) How much of this energy is dissipated in the resistor?[16.7] J
6
5.5H
Let’s put an inductor and a capacitor in the SAME circuit.
At t=0, the charged capacitor is connected to the inductor. What would you expect to happen??
Current would begin to flow….
High
Low
Low
High
202
1Eu Energy Density in Capacitor
Energy Flows from Capacitor to the Inductor’s Magnetic Field
Energy Flow
202
1EuC 2
02
1BuL
Energy
LC Circuit
High
Low
Low
High
LC
idt
id
dt
idLi
C
dt
diL
C
q
EquationLoop
1
0
01
0
2
22
2
2
2
)sin()cos(
)cos()sin(
1
0
2
22
2
tBtAdt
di
tBtAiLC
idt
id
When t=0,
i=0so
B=0When t=0, voltageacross the inductor= Q0/C
LC
Q
LCLC
Q
LC
QA
ALC
Q
dt
diL
000
0
/
][
0At t
)sin()cos(
)cos()sin(
1
0
2
22
2
tBtAdt
di
tBtAiLC
idt
id
The Math Solution:
)1
()( 0 tLC
SinLC
Qti
Energy
2
22
2
1
:2
1
2
1
:
LiE
InductorC
QCVE
Capacitor
L
C
Inductor
tSinQC
tSinLQE
tLC
SinLC
QLE
tLC
SinLC
Qti
C
C
220
2220
2
0
0
2
1
2
1
)1
(2
1
)1
()(
The Capacitor
)(2
1
)(1
2
1)(
2
1
2
1
)(
:Eq Diff
:Capacitor For the
)()1
()(
220
22022
2220
422
02
00
tCosQC
tCosQCL
CLtCosQCLCv
tCosQLvdt
diL
C
Q
From
CvQ
tSinQtLC
SinLC
Qti
Add ‘em Up …
Constant2
1 2220 tCostSinQ
CEtotal
Add Resistance
Actual RLC:
)'(2
Energy Total
)2/('
)'(
0
/
0
20
22
2/0
2
2
tCoseC
QU
LR
tCoseQq
C
q
dt
dqR
dt
qdL
LRttotal
LRt
New Feature of Circuits with L and C These circuits can produce oscillations in the currents and
voltages Without a resistance, the oscillations would continue in an
un-driven circuit. With resistance, the current will eventually die out.
The frequency of the oscillator is shifted slightly from its “natural frequency”
The total energy sloshing around the circuit decreases exponentially
There is ALWAYS resistance in a real circuit!
Types of Current
Direct Current Create New forms of life
Alternating Current Let there be light
-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
Alternating
emf
Sinusoidal
DC
Sinusoidal Stuff
)sin( tAemf
“Angle”
Phase Angle
Same Frequencywith
PHASE SHIFT
Different Frequencies
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 d
dt(1
2LI2 1
2
q2
C)0
LIdI
dtq
C
dq
dt0 L(
dq
dt)d2q
dt2 q
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 d
dt(1
2LI2 1
2
q2
C)0
LIdI
dtq
C
dq
dt0 L(
dq
dt)d2q
dt2 q
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 d
dt(1
2LI2 1
2
q2
C)0
LIdI
dtq
C
dq
dt0 L(
dq
dt)d2q
dt2 q
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 d
dt(1
2LI2 1
2
q2
C)0
LIdI
dtq
C
dq
dt0 L(
dq
dt)d2q
dt2 q
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