Transcript L12_Alternating_Current

Alternating Current
Kirchoff’s rules for loops and junctions may be used to analyze
complicated circuits such as the one below, powered by an alternating
current (A.C.) source. But the analysis can quickly become much
more complicated. In this chapter we consider the behavior of basic,
but useful, circuits driven by A.C. sources.
Symbol for
A.C. source.
A.C.
 Demo
Simple alternating current (A.C.) generator
In a basic A.C. generator, a permanent magnet provides a reasonably uniform
magnetic field. As the generator loop turns in this field at frequency w (rad/s),
the flux through the loop changes sinusoidally with time. This causes the output
voltage (induced emf) to also change sinusoidally, 90 degrees out of phase with
the flux. We can see why this happens by applying Faraday’s Law of Induction
to the flux:
Let:
 B   0 cos(wt )
Then:
d
[ 0 cos(wt )]
dt
 w 0 sin( wt )
emf  
 VMAX sin( wt )
Motional emf
Resistor connected to A.C. source
VMAX sin( wt )
V  RI
VMAX sin( wt )  VR  RI R
VR VMAX

sin( wt )  I MAX sin( wt )
So the current in the resistor is: I R 
R
R
Use Kirchoff’s loop equation:
Note that the resistor voltage and current are in phase. (See plot.)
2
P  RI R2  RI MAX
sin 2 (wt )
Instantaneous power dissipation:
To find the average power (root mean square, “RMS”), integrate over one cycle:
PRMS  
2 / w
0
So that:
I
2
RMS

2
I MAX
RI
2
2
MAX

1 2
2
sin( wt )dt  RI MAX  RI RMS
2
I MAX
2
 .707  I MAX  I RMS
Inductor connected to A.C. source
VMAX sin( wt )
VL  L
dI L
dt
Use Kirchoff’s loop equation: L
dI L
 VMAX sin( wt )
dt
Integrate to solve for IL:
I L
VMAX
VMAX
VMAX

sin(
w
t
)
dt


cos(
w
t
)

sin(
w
t

)

L
wL
wL
2
From either the equations or the graph we see that, for A.C.
sources, the current in an inductor lags behind its voltage by 90o
Inductive Reactance, XL
Notice that whenever cos(wt) is +1 or -1, the
magnitude of the current in the inductor is at the
maximum. We can express this as follows:
I MAX
IL  
VMAX
cos(wt )
wL
VMAX

wL
This equation has the same structure as Ohm’s Law, and we can identify the
factor in the denominator as setting the ratio between V and I. We call this
factor the inductive reactance, and define it as follows:
X L  wL
And since RMS voltage and current are one-half their values at maximum, XL
can be used in both equations:
VMAX  X L I MAX
VRMS  X L I RMS
•It’s clear from these equations that XL must have units of ohms.
•Unlike resistance, inductive reactance changes with frequency!
Example
Capacitor connected to A.C. source
VMAX sin( wt )
VC 
Q
C
Q
 VMAX sin( wt )  Q  CVMAX sin( wt )
C
dQ

Differentiate to find IC: I C 
 wCVMAX cos(wt )  wCVMAX sin( wt  )
dt
2
Kirchoff’s loop equation:
From either the equations or the graph we see that, for A.C.
sources, the current in a capacitor leads its voltage by 90o
Capacitive Reactance, XC
Again, whenever cos(wt) is +1 or -1, the
magnitude of the current in the capacitor is at the
maximum. We can express this as follows:
I C  wCVMAX cos(wt )
I MAX  wCVMAX
This equation has the same structure as Ohm’s Law, and we can identify the
factor multiplying V as setting the ratio between V and I. We define the
capacitive reactance as the inverse of this factor:
1
XC 
wC
And since RMS voltage and current are one-half their values at maximum, XC
can be used in both equations:
VMAX  X C I MAX
VRMS  X C I RMS
•As with XL , XC must have units of ohms.
•And as before, this capacitive reactance changes with frequency!
Example
Frequency dependence of R, XL, and XC
X L  wL
V  RI
XC 
1
wC
The Driven RLC Circuit
We have already studied the damped oscillations of an RLC circuit which has
been energized with an initial charge or current. But now we consider the
response of this circuit when it is driven by an AC source, which feeds energy
into the circuit. We will see how this reaches an equilibrium, with the power
from the source being dissipated by the resistor.
Analyzing the Driven RLC Circuit
Since this circuit consists of one loop, the
same current, I, passes through every element
in the circuit. Once again, we can use
Kirchoff’s loop equation for voltages. Using V
for the source voltage:
V  VL  VC  VR
Now we insert all the equations from previous pages. With this more
complicated circuit the phase of the source will no longer be the same as that
for the resistor. So we let the source voltage be displaced by –f.
VMAX sin( wt   )  RI MAX sin( wt )  X L I MAX cos(wt )  X C I MAX cos(wt )
For t = 0:
For wt = /2:
 VMAX sin(  )  X L I MAX  X C I MAX  ( X L  X C ) I MAX
VMAX sin(  / 2   )  VMAX cos( )  RI MAX
Square and add:

2
2
VMAX
 I MAX
R 2  ( X L  X C )2
Define “impedance”, Z:

Z  R 2  ( X L  X C ) 2  VMAX  ZI MAX
We can think of “impedance”, Z, as “total resistance at a given frequency w.
Measuring voltages in an RLC Circuit.
Voltages as a function of time in a typical
RLC Circuit.
Frequency dependence of reactances
and impedance in RLC Circuit
At the frequency where XL= XC , the
impedance of the circuit is at a
minimum, and Z = R.
Since Z is minimum at the point where
XL= XC the average current is
maximum. The power dissipation is
also maximum at this point, meaning
that the power delivered to the circuit
by the AC source is maximum. At
what frequency does this occur?
X L  X C  wL 
1
1
 w2 
w 
wC
LC
1
LC
But this is the frequency of an LC circuit when there is no resistor present! At this
frequency, the circuit is in resonance, with the driving frequency, w, equal to
the natural frequency of the LC oscillator, wo.
Current as a function of w in a driven RLC circuit
w0 
1
LC
The plot at right shows current
resonance curves for several values
of R. The greater the resistance, the
lower the maximum current.
We can calculate the shapes of these resonance curves as follows:
I RMS 
VRMS

Z
VRMS
R 2  ( X L  X C )2
VRMS

L2
1 2
R  2 (w 
)
w
LC
2
2
VRMS

R 
2
L2
w
2
(w 2  wo2 ) 2
Resonance curves for a damped, driven harmonic
oscillator: mass + spring + damping.
Collapse of Tacoma Narrows bridge in 1940 
driven into resonance by wind. Too little damping!
Coils sharing the same magnetic flux, BA
These can be used
as “transformers”.
In air
With iron yoke
Raising and Lowering A.C. Voltage: Transformers
Faraday’s Law of Induction tells us the
relationship between the magnitudes of the
voltage at the terminals and the rate of change
of magnetic flux through each coil:
d 2
d 1
V2  N 2
dt
dt
But the flux through each coil is the same:
V1  N1
1   2
d1 d 2

dt
dt
V1 V2

N1 N 2
So the output (secondary) voltage can be raised or
lowered compared to the input (primary) voltage, by
the ratio of turns. A transformer that raises (lowers)
voltage is called a “step-up” (“step-down”) transformer.
Transformers lose very little power:
P  V1I1  V2 I 2
N2
V2 
V1
N1
V1
N1
I 2  I1 
I1
V2
N2
As the voltage is increased, the current is lowered by the same factor.
Transformers are not perfect,
but they dissipate little power.
Nikola Tesla (Serbian Cyrillic: Никола
Тесла; 10 July 1856 – 7 January 1943)
was a Serbian-American inventor,
mechanical engineer, and electrical
engineer. He was an important
contributor to the birth of commercial
electricity, and is best known for his
many revolutionary developments in the
field of electromagnetism in the late
19th and early 20th centuries building
on the theories of electromagnetic
technology discovered by Michael
Faraday and used in direct current (DC)
applications. Tesla's patents and
theoretical work formed the basis of
modern alternating current (AC) electric
power systems, including the polyphase
system of electrical distribution and the
AC motor.
Tesla Coil Schematic
Tesla Never Thought Small
Tesla’s A.C. Dynamo
(generator) used to generate
alternating currrent electricity,
which became the technology
of choice for electrification
across the U.S. (power grids)
and ultimately, the whole
planet. U.S. Patent 390721
Wardenclyffe Tower facility ~ 1915
Mark Twain in Tesla’s Lab.
A.C. power networks
Transmission lines:
155,000-765,000 V
Local substation:
7,200 V
House:
110/220 V
For a given length of power line,
P=RI2. If the power line voltage is
550,000 V, the current is reduced
by a factor of 5000 compared to
that at 110 V, and the power loss is
reduced by a factor of 25 million.