Download t - Physics

The Ups and Downs of
Circuits
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
r
VG
For the next problem, recall that
i
E
 Rt / L
i  (1  e
)
R
time constant

L

R
R
L
For the circuit of Figure 30-19, assume that = 11.0 V, R = 6.00W ,
and L = 5.50 H. The battery is connected at time t = 0.
6W
5.5H
(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
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….
Energy Density in Capacitor
1
2
u  0E
2
Low
High
High
Low
Energy Flows from Capacitor to the Inductor’s Magnetic Field
Energy Flow
1
2
uC   0 E
2
Energy
uL 
1
20
B
2
LC Circuit
Loop Equation
Low
High
High
Low
q
di
L 0
C
dt
2
1
d i
iL 2 0
C
dt
2
d i
2
 i  0
2
dt
1
2
 
LC
2
d i
2
 i  0
2
dt
1
2
 
LC
i  A sin( t )  B cos(t )
di
 A cos(t )  B sin( t )
dt
When t=0, i=0
so
B=0
When t=0, voltage
across the inductor
= Q0/C
d 2i
2


i0
2
dt
1
2
 
LC
i  A sin( t )  B cos(t )
di
 A cos(t )  B sin( t )
dt
At t  0
di Q0
L   L[A]
dt C
Q0
Q0
Q0
A


LC LC / LC
LC
The Math Solution:
Q0
1
i(t ) 
Sin (
t)
LC
LC
Energy
Capacitor :
2
1
1Q
2
EC  CV 
2
2 C
Inductor :
1 2
E L  Li
2
Inductor
Q0
1
i (t ) 
Sin (
t)
LC
LC
2
1  Q0
1

EC  L 
Sin (
t )
2  LC
LC 
1
1 2 2
2 2
2
EC  LQ0  Sin t 
Q0 Sin t
2
2C
The Capacitor
Q0
1
i (t ) 
Sin (
t )  Q0 Sin (t )
LC
LC
For the Capacitor :
Q  Cv
From Diff Eq :
Q
di
 L  v  L 2Q0Cos(t )
C
dt
1 2 1 2 4 2
1
 1 
Cv  CL  Q0 Cos 2 (t )  CL2  2 2 Q02Cos 2 (t )
2
2
2
LC 
1 2

Q0 Cos 2 (t )
2C
Add ‘em Up …


1 2
2
2
Etotal 
Q0 Sin t  Cos t  Constant
2C
Add Resistance
Actual RLC:
2
d q
dq q
L 2 R
 0
dt
dt C
 Rt / 2 L
q  Q0 e
Cos( ' t   )
 '    ( R / 2 L)
2
2
Total Energy
Q02  Rt / L
U total 
e
Cos( ' t   )
2C 0
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
Alternating
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
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 
 2.65MW
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