Transcript Lect18

0,
r1
.. r1
n
VC
1.01 1
UE
0
f( x ) 0
C
1.01 1
0
1.01 1
0
2
VL
4
x
t
6
6.28
LC
Circuits
0
f( x ) 0
1.01 1
L
0
0
2
tx
4
6
6.28
UB
t
Today...
• Oscillating voltage and current
• Transformers
• Qualitative descriptions:
• LC circuits (ideal inductor)
• LC circuits (L with finite R)
• Quantitative descriptions:
• LC circuits (ideal inductor)
• Frequency of oscillations
• Energy conservation
Oscillating Current and Voltage
Q. What does eosinwt
~
mean??
A. It is an A.C. voltage source. Output voltage
appears at the terminals and is sinusoidal in
time with an angular frequency w.
Oscillating circuits have both
AC voltage and current.
I(t)
Simple for resistors, but...
eosinwt
~
R

εo
I(t)  sin ωt 
R
Transformers
• AC voltages can be stepped up or
stepped down by the use of
transformers.
• The AC current in the primary
circuit creates a time-varying
magnetic field in the iron
e
• This induces an emf on the
secondary windings due to the
mutual inductance of the two sets
of coils.
iron
~
V1
V2
N1
(primary)
N2
(secondary)
• The iron is used to maximize the mutual inductance. We
assume that the entire flux produced by each turn of the
primary is trapped in the iron. (Recall from magnetism lab
how the ferromagnet “sucks in” the B-field.)
Ideal
No resistance losses
Transformers (no load)
All flux contained in iron
Nothing connected on secondary
• The primary circuit is just an AC voltage
source in series with an inductor. The change
in flux produced in each turn is given by:
iron
e ~
dturn V1

dt
N1
• The change in flux per turn in the secondary coil is the
same as the change in flux per turn in the primary coil
(ideal case). The induced voltage appearing across the
secondary coil is given by:
V2  N 2
V1
V2
N1
N2
(primary) (secondary)
dturn N 2

V1
dt
N1
• Therefore,
• N2 > N1  secondary V2 is larger than primary V1 (step-up)
• N1 > N2  secondary V2 is smaller than primary V1 (step-down)
• Note: “no load” means no current in secondary. The primary current,
termed “the magnetizing current” is small!
Ideal Transformer (with a load)
• Changing flux produced by primary coil
induces an emf in secondary. When we
connect a resistive load to secondary coil, e ~
emf in secondary  current I2 in secondary
Power is dissipated only in the load
resistor R.
2
Pdissipated
iron
V1
V2
N1
V2
2
 I2 R 
 V2 I 2
R
(primary)
N2
(secondary)
Where did this power come from?
It could come only from the voltage source in the primary:
Pgenerated  V1I1
I1 V2

I 2 V1
=
N2
N1
V1
V1
N2

N1

I1 
V1I1  V2 I 2
N2
I2
N1
1
R
Lecture 18, ACT 1
• The primary coil of an ideal transformer is
connected to an AC voltage source as shown.
e ~
There are 50 turns in the primary and 200
turns in the secondary.
1A – If V1 = 120 V, what is the potential drop
across the resistor R ?
iron
V1
V2
N1
(primary)
(a) 30 V
1B
(b) 120 V
N2
(secondary)
(c) 480 V
– If 960 W are dissipated in the resistor R,
what is the current in the primary ?
(a) 8 A
(b) 16 A
(c) 32 A
R
Lecture 18, ACT 1
• The primary coil of an ideal transformer is
connected to an AC voltage source as shown.
e ~
There are 50 turns in the primary and 200
turns in the secondary.
1A – If V1 = 120 V, what is the potential drop
across the resistor R ?
iron
V1
V2
N1
(primary)
(a) 30 V
(b) 120 V
N2
(secondary)
(c) 480 V
The ratio of turns, (N2/N1) = (200/50) = 4
The ratio of secondary voltage to primary voltage is equal to the
ratio of turns, (V2/V1) = (N2/N1)
Therefore, V2= 480 V
R
iron
Lecture 18, ACT 1
• The primary coil of an ideal transformer is
connected to an AC voltage source as shown.e ~
There are 50 turns in the primary and 200
turns in the secondary.
– If V1 = 120 V, what is the potential drop
across the resistor R ?
1A
(a) 30 V
(b) 120 V
V1
V2
N1
(primary)
N2
(secondary)
(c) 480 V
The ratio of turns, (N2/N1) = (200/50) = 4
The ratio (V2/V1) = (N2/N1). Therefore, V2= 480 V
1B
– If 960 W are dissipated in the resistor R,
what is the current in the primary ?
(a) 8 A
(b) 16 A
(c) 32 A
Energy is conserved—960 W should be produced in the primary
P1 = V1 I1 implies that 960W / 120V = 8 A
R
Preflight 18:
An ideal transformer has N1 = 4, N2 = 6.
Side 1 is connected to a generator with
e = Vmax sin ( w t )
2) What is the maximum EMF on side 2?
a) V2max = 2e1/3
b) V2max = e1
c) V2max = 3e1/2
1
2
An ideal transformer steps down the
voltage in the secondary circuit. The
number of loops on each side is
unknown.
4) Compare the currents in the primary and secondary circuits
a) I1 < I2
b) I1 = I2
c)
I1 > I2
Where are we going?
• Oscillating circuits
• radio, TV, cell phone, ultrasound,
clocks, computers, GPS
What’s Next?
• Why and how do oscillations occur
in circuits containing capacitors and
inductors?
• naturally occurring, not driven for now
• stored energy
• capacitive <-> inductive
Energy in the Electric
and Magnetic Fields
Energy stored in a capacitor ...
U
1
CV 2
2
… energy density ...
Energy stored in an inductor ….
+++ +++
--- ---
1
uelectric  e 0 E 2
2
B
1 2
U  LI
2
… energy density ...
E
umagnetic
1 B2

2 0
LC Circuits
• Consider the RC and LC
series circuits shown:
++++
----
C R
++++
----
C
L
• Suppose that the circuits are
formed at t=0 with the
capacitor charged to value Q.
There is a qualitative difference in the time development of the
currents produced in these two cases. Why??
• Consider from point of view of energy!
• In the RC circuit, any current developed will cause energy
to be dissipated in the resistor.
• In the LC circuit, there is NO mechanism for energy
dissipation; energy can be stored both in the capacitor and
the inductor!
RC/LC Circuits
I
Q+++
---
I
Q+++
---
C R
I
I
0
0
t
L
LC:
current oscillates
RC:
current decays exponentially
0
C
0
t
LC Oscillations
(qualitative)
I 0
+ +
- -
C
I  I0
L

C
Q0
Q  Q0


I 0
I  I0
C
Q0
L
L

-
-
+ +
C
L
Q  Q0
Alternate way to draw:
C
VC = Q/C
V=0
L
VL = L dI/dt
VC+VL = 0
 VC = -VL
r1
0,
.. r1
LC Oscillations
n
1.01 1
r1
x 0 , .. r1
(qualitative)
n
Q
0
VC
f( x ) 0
1.01 1
1.01 1
1.01 1 0
0
f( x ) 0
0
2
4
x
I
6
6.28
1.01 1
0
1.01 1
0
0
f( x ) 0
VL
1.01 1
1.01 1 0
0
dI
dt
2
4
x
0
f( x ) 0
These voltages are
f( x ) 0
0
opposite, since the
6.28
cap and ind are
traversed in “opposite”
1.01 1
directions
0
0
0
2
tx
4
4
x
6
6.28
6
0
1.01 1
2
6
6.28
2
tx
4
6
6.28
2
Lecture 18, Act
2
t=0
• At t=0, the capacitor in the LC
circuit shown has a total charge
+ +
Q0. At t = t1, the capacitor is
= Q0
Q
uncharged.
- C
– What is the value of Vab=Vb-Va,
2A the voltage across the
inductor at time t1?
(a) Vab < 0
(b) Vab = 0
t=t 1
a
L
Q =0
C
L
b
(c) Vab > 0
2B – What is the relation between UL1, the energy stored in the
inductor at t=t1, and UC1, the energy stored in the capacitor at
t=t1?
(a) UL1 < UC1
(b) UL1 = UC1
(c) UL1 > UC1
Lecture 18, Act
2
t=0
• At t=0, the capacitor in the LC
circuit shown has a total charge
+ +
Q0. At t = t1, the capacitor is
= Q0
Q
uncharged.
- C
– What is the value of Vab=Vb-Va,
2A the voltage across the
inductor at time t1?
(a) Vab < 0
(b) Vab = 0
t=t 1
a
L
Q =0
C
L
b
(c) Vab > 0
• Vab is the voltage across the inductor, but it is also
(minus) the voltage across the capacitor!
• Since the charge on the capacitor is zero, the voltage
across the capacitor is zero!
Lecture 18, Actt=02
• At t=0, the capacitor in the LC
circuit shown has a total charge
+ +
Q0. At t = t1, the capacitor is
= Q0
Q
uncharged.
- C
– What is the relation between UL1,
2B
the energy stored in the inductor
at t=t1, and UC1, the energy stored
in the capacitor at t=t1?
(a) UL1 < UC1
(b) UL1 = UC1
• At t=t1, the charge on
the capacitor is zero.
Q12
U C1 
0
2C
t=t 1
a
L
Q =0
C
L
b
(c) UL1 > UC1
• At t=t1, the current is a
maximum.
1 2 Q02
U L1  LI1 
0
2
2C
Preflight 18:
At time t = 0 the capacitor is fully charged
with Qmax, and the current through the circuit
is 0.
2) What is the potential difference across the inductor at t = 0?
a) VL = 0
b) VL = Qmax/C
c) VL = Qmax/2C
3) What is the potential difference across the inductor when the current
is maximum?
a) VL = 0
b) VL = Qmax/C
c) VL = Qmax/2C
LC Oscillations
(L with finite R)
• If L has finite R, then
0,
will
be damped.
r1– energy will be dissipated in R r1the oscillations
r1 10
n 100
.. r1
x 0 , .. r1
n
n
R=0
R0
1
1
Q
Q
f( x ) 00
f( x ) 00
1
0
t
5
x
1
10
0
t5
10
x
• The number of oscillations is described by the “Q” of the oscillator
(we will return to this in Lect. 20) [NOTE: Q here is not charge!]
Q  2
U max
DU
Umax is max energy stored in the system
DU is the energy dissipated in one cycle
Review of Voltage Drops
Across Circuit Elements
I(t)
C
Q  Id t
V 
C
C
Voltage determined by
integral of current and
capacitance
I(t)
2
L
dI
d Q
V L
L 2
dt
dt
Voltage determined by
derivative of current and
inductance
LC Oscillations
(quantitative, but only for R=0)
I
• What is the oscillation frequency ω0?
• Begin with the loop rule:
Q
d 2Q Q
L 2  0
dt
C
+ +
- -
C
L
• Guess solution: (just harmonic oscillator!)
Q  Q0 cos(w0t   )
where
remember:
, Q0 determined from initial conditions
d2x
m 2  kx  0
dt
• Procedure: differentiate above form for Q and substitute into
loop equation to find w0.
1
• Note: Dimensional analysis  w0 
LC
LC Oscillations
(quantitative)
• General solution:
Q  Q0 cos(w0t   )
+ +
- -
C
L
• Differentiate:
dQ
 w 0Q0 sin( w 0t   )
dt
d 2Q Q
L 2  0
dt
C
d 2Q
2


w
0 Q0 cos(w 0t   )
2
dt
• Substitute into loop eqn:


1
1
2


w
L

0
L  w Q0 cos(w 0t   )  Q0 cos(w 0t   )   0
0
C
C
2
0
Therefore,
1
w0 
LC
which we could have determined
from the mass on a spring result:
k
1/ C
1
w0 


m
L
LC
3
Lecture 18, Act 3
t=0
• At t=0 the capacitor has charge Q0; the resulting
oscillations have frequency w0. The maximum
current in the circuit during these oscillations has
value I0.
3A – What is the relation between w0 and w2, the
frequency of oscillations when the initial charge
= 2Q0?
(a) w2 = 1/2 w0
(b) w2 = w0
+ +
Q  Q0
- -
C
L
(c) w2 = 2w0
3B – What is the relation between I0 and I2, the maximum current in
the circuit when the initial charge = 2Q0?
(a) I2 = I0
(b) I2 = 2I0
(c) I2 = 4I0
Lecture 18, Act 3
t=0
• At t=0 the capacitor has charge Q0; the resulting
oscillations have frequency w0. The maximum
current in the circuit during these oscillations has
value I0.
3A – What is the relation between w0 and w2, the
frequency of oscillations when the initial charge
= 2Q0?
(a) w2 = 1/2 w0
(b) w2 = w0
+ +
Q  Q0
- -
C
L
(c) w2 = 2w0
• Q0 determines the amplitude of the oscillations (initial
condition)
• The frequency of the oscillations is determined by the circuit
parameters (L, C), just as the frequency of oscillations of a mass
on a spring was determined by the physical parameters (k, m)!
Lecture 18, Act 3
• At t=0 the capacitor has charge Q0; the resulting
oscillations have frequency w0. The maximum
current in the circuit during these oscillations has
value I0.
3B – What is the relation between I0 and I2, the
maximum current in the circuit when the initial
charge = 2Q0?
(a) I2 = I0
(b) I2 = 2I0
t=0
+ +
Q  Q0
- -
C
L
(c) I2 = 4I0
• The initial charge determines the total energy in the circuit:
U0 = Q02/2C
• The maximum current occurs when Q=0!
• At this time, all the energy is in the inductor: U = 1/2 LIo2
• Therefore, doubling the initial charge quadruples the total
energy.
• To quadruple the total energy, the max current must double!
Preflight 18:
The current in a LC circuit is a
sinusoidal oscillation, with
frequency ω.
5) If the inductance of the circuit is increased, what will happen
to the frequency ω?
a) increase
b) decrease
c) doesn’t change
6) If the capacitance of the circuit is increased, what will happen
to the frequency?
a) increase
b) decrease
c) doesn’t change
LC Oscillations
Energy Check
• Oscillation frequency w 0  1 has been found from the
LC
loop equation.
• The other unknowns ( Q0,  ) are found from the initial
conditions. E.g., in our original example we assumed initial
values for the charge (Qi) and current (0). For these values:
Q0 = Qi,  = 0.
• Question: Does this solution conserve energy?
1 Q 2 (t ) 1 2
U E (t ) 

Q0 cos 2 (w 0t   )
2 C
2C
1
1
U B (t )  Li 2 (t )  Lw 02Q02 sin 2 (w 0t   )
2
2
r1
Energyx Check
0 , .. r1
n
Energy in Capacitor
U E (t ) 
1 2
Q0 cos 2 (w 0t   )
2C
UE
1
f( x ) 0.5
Energy in Inductor
LC
U B (t ) 

1
U B (t )  Lw 02Q02 sin 2 (w 0t   ) x 04, r1 ..00r1
0
2
n
1
1
w0 
1 2
Q0 sin 2 (w 0t   )
2C
Therefore,
Q02
U E (t )  U B (t ) 
2C
2
t
4
6
4
6
x
UB
f( x ) 0.5
00
0
2
t
Lecture 18, Act 4
I
• At t=0 the current flowing through the circuit is 1/2
of its maximum value.
– Which of the following plots best represents
UB, the energy stored in the inductor as a
4
function of time?
(a)
x
0,
r1
x
(b)
.. r1
n
0,
r1
n
.. r1
1
1
1
UBf( x) 0.5
Uf(Bx ) 0.5
UBf( x) 0.5
00 0
0
2
4
x
time
6
00 0
0
2
4
x
time
6
Q
00
00
+ +
- -
C
L
(c)
2
4
x
time
6
I
Lecture 18, Act 4
• At t=0 the current flowing through the circuit is 1/2
+ +
of its maximum value.
Q
- – Which of the following plots best represents UB,
the energy stored in the inductor as a function of
4
time?
r1
(a)
x
0,
r1
x
(b)
.. r1
n
0,
n
.. r1
1
1
1
UBf( x) 0.5
Uf(Bx ) 0.5
UBf( x) 0.5
00 0
0
2
4
x
time
6
00 0
0
2
4
x
time
6
00
00
C
L
(c)
2
4
6
x
time
• The key here is to realize that the energy stored in the
inductor is proportional to the CURRENT SQUARED.
• Therefore, if the current at t=0 is 1/2 its maximum value, the
energy stored in the inductor will be 1/4 of its maximum value!!
Summary
• Transformers used to step up/down voltage
V2 
N2
V1
N1
I1V1  I 2V2
I1 
N2
I2
N1
• Oscillating voltage and current
x
0,
•.. r1 Qualitative description
VC
V
r1
n
1.01 1
1.01 1
L
0
0
f( x ) 0
f( x ) 0
1.01 1
•
0
0
2
4
x
6
6.28
1.01 1
0
2
4
Quantitative description
0
x
6
6.28
– Frequency of oscillations
– Energy conservation
w0 
1
LC
Appendix: LCR Damping
r1
r1 10
n 100
For your interest, we do not derive
but
only illustrate
the
x 0 , here,
.. r1
n
following behavior
1
R
+ +
- -
C
L
0
f( x ) 0
Q  Q0e  t cos(w 'o t   )
 
R  R0
Q
R
2L
t
1
0
r1circuit,
r1depends
10 5 nalso
100on10R !
In anx LRC
w
0 , .. r1
x
n
1
Q
 1 R 
w 'o    2 
 LC 4 L 
2
R
R0
4
f( x ) 0
0
1
0
t5
10