Transcript Chapter 18

Chapter 33
Alternating Current Circuits
AC Circuit
• An AC circuit consists of a combination of circuit
elements and an AC generator or source
• The output of an AC power source is sinusoidal and
varies with time according to the following equation
Δv = ΔVmax sin ωt
• Δv: instantaneous voltage
• ΔVmax is the maximum voltage (amplitude) of the
generator
2π
ω  2π ƒ 
T
• ω is the angular frequency of the AC voltage
Resistors in an AC Circuit
• Consider a circuit consisting of
an AC source and a resistor
ΔvR = ΔVmax sin ωt
• ΔvR is the instantaneous
voltage across the resistor
• The instantaneous current in the resistor is
vR Vmax
iR 

sin ωt  I max sin ωt
R
R
• The instantaneous voltage across the resistor is also
given as ΔvR = ImaxR sin ωt
Resistors in an AC Circuit
• The graph shows the current
through and the voltage across
the resistor
• The current and the voltage
reach their maximum values at
the same time
• The current and the voltage are
said to be in phase
• The direction of the current has no effect on the
behavior of the resistor
Resistors in an AC Circuit
• The rate at which electrical
energy is dissipated in the circuit
is given by
  i 2R
• i: instantaneous current
• The heating effect produced by an
AC current with a maximum value
of Imax is not the same as that of a
DC current of the same value
• The maximum current occurs for a small amount of
time
rms Current and Voltage
• The rms current is the direct current that would
dissipate the same amount of energy in a resistor as is
actually dissipated by the AC current
Irms 
Imax
2
 0.707 Imax
• Alternating voltages can also be discussed in terms of
rms values
V
Vrms 
max
2
 0.707 Vmax
• The average power dissipated in resistor in an AC
circuit carrying a current I is
2
rms
av  I
R
Ohm’s Law in an AC Circuit
• rms values will be used when discussing AC currents
and voltages
• AC ammeters and voltmeters are designed to read
rms values
• Many of the equations will be in the same form as in
DC circuits
• Ohm’s Law for a resistor, R, in an AC circuit
ΔVR,rms = Irms R
• The same formula applies to the maximum values of v
and i
Capacitors in an AC Circuit
• Consider a circuit containing a
capacitor and an AC source
• Kirchhoff’s loop rule gives:
v  vC  0
q
v   0
C
• ΔvC: instantaneous voltage
across the capacitor
q(t )  CVmax sin t
dq


 CVmax cos t  CVmax sin  t  
iC 
dt
2

Vmax
I max 
1 / C
Capacitors in an AC Circuit
• The voltage across the capacitor
lags behind the current by 90°
• The impeding effect of a capacitor
on the current in an AC circuit is
called the capacitive reactance
(measured in ohms):
1
XC 
C
Vmax  I max X C
q(t )  CVmax sin t
dq


 CVmax cos t  CVmax sin  t  
iC 
dt
2

Vmax
I max 
1 / C
Inductors in an AC Circuit
• Consider an AC circuit with a
source and an inductor
• Kirchhoff’s loop rule gives:
v  vL  0
di
v  L  0
dt
• ΔvL: instantaneous voltage
across the inductor
diL
 Vmax sin t
v  L
dt
Vmax
Vmax
iL 
sin tdt  
cos t

Vmax


L

L

sin  t  
Vmax
I max 
L
2

L
Inductors in an AC Circuit
• The voltage across the inductor
always leads the current by 90°
• The effective resistance of a coil
in an AC circuit is called its
inductive reactance (measured in
ohms): X  L
L
diL
 Vmax sin t
v  L
Vmax  I max X C
dt
Vmax
Vmax
iL 
sin tdt  
cos t

Vmax


L

L

sin  t  
Vmax
I max 
L
2

L
Chapter 33
Problem 11
Determine the maximum magnetic flux through an
inductor connected to a standard electrical outlet (ΔVrms=
120 V, f = 60.0 Hz).
The RLC Series Circuit
• The resistor, inductor, and capacitor
can be combined in a circuit
• The current in the circuit is the same
at any time and varies sinusoidally
with time
The RLC Series Circuit
• The instantaneous voltage across the
resistor is in phase with the current
• The instantaneous voltage across the
inductor leads the current by 90°
• The instantaneous voltage across the
capacitor lags the current by 90°
v R  Imax R sin ωt  VR sin ωt
π

v L  Imax X L sin  ωt    VL cos ωt
2

π

v C  Imax X C sin  ωt    VC cos ωt
2

Phasor Diagrams
• Because of the different phase
relationships with the current, the voltages
cannot be added directly
• To simplify the analysis of AC circuits, a graphical
constructor called a phasor diagram can be used
• A phasor is a vector rotating CCW; its length is
proportional to the maximum value of the variable it
represents
• The vector rotates at an angular speed equal to the
angular frequency associated with the variable, and the
projection of the phasor onto the vertical axis
represents the instantaneous value of the quantity
Phasor Diagrams
• The voltage across the resistor is in phase with the
current
• The voltage across the inductor leads the current by 90°
• The voltage across the capacitor lags behind the current
by 90°
Phasor Diagrams
• The phasors are added as vectors
to account for the phase
differences in the voltages
• ΔVL and ΔVC are on the same line
and so the net y component is
ΔVL - ΔVC
Phasor Diagrams
• The voltages are not in phase, so
they cannot simply be added to
get the voltage across the
combination of the elements or
the voltage source
Vmax 
VR2  (VL  VC )2
VL  VC
tan  
VR
•  is the phase angle between the
current and the maximum voltage
• The equations also apply to rms
values
Phasor Diagrams
ΔVR = Imax R
ΔVL = Imax XL
ΔVC = Imax XC
Vmax 
VR2  (VL  VC )2
VL  VC
tan  
VR
Vmax  I max R  ( X L  X C )
2
2
Impedance of a Circuit
• The impedance, Z, can also be represented in a phasor
diagram
Z 
R2  ( X L  X C )2
X L  XC
tan  
R
• φ: phase angle
• Ohm’s Law can be applied to the impedance
ΔVmax = Imax Z
• This can be regarded as a generalized form of Ohm’s
Law applied to a series AC circuit
Summary of Circuit Elements,
Impedance and Phase Angles
Problem Solving for AC Circuits
• Calculate as many unknown quantities as possible
(e.g., find XL and XC)
• Be careful with units – use F, H, Ω
• Apply Ohm’s Law to the portion of the circuit that is
of interest
• Determine all the unknowns asked for in the problem
Chapter 33
Problem 24
An AC source with Vmax = 150 V and f = 50.0 Hz is connected between
points a and d in the figure. Calculate the maximum voltages between
(a) points a and b, (b) points b and c, (c) points c and d, and (d) points
b and d.
Power in an AC Circuit
• No power losses are associated with pure capacitors
and pure inductors in an AC circuit
• In a capacitor, during 1/2 of a cycle energy is stored and
during the other half the energy is returned to the circuit
• In an inductor, the source does work against the back
emf of the inductor and energy is stored in the inductor,
but when the current begins to decrease in the circuit,
the energy is returned to the circuit
Power in an AC Circuit
• The average power delivered by the generator is
converted to internal energy in the resistor
Pav = Irms ΔVR,rms
ΔVR, rms = ΔVrms cos 
Pav = Irms ΔVrms cos 
•
cos  is called the power factor of the circuit
• Phase shifts can be used to maximize power outputs
Chapter 33
Problem 28
A series RLC circuit has a resistance of 45.0 Ω and an impedance of
75.0 Ω. What average power is delivered to this circuit when ΔVrms =
210 V?
Resonance in an AC Circuit
• Resonance occurs at the frequency,
ω0, where the current has its maximum
value
I rms 
Vrms
R2  ( X L  X C )2
• To achieve maximum current, the
impedance must have a minimum
value
• This occurs when XL = XC and
1
0 L 
0C
0 
1
LC
Resonance in an AC Circuit
• Theoretically, if R = 0 the current
would be infinite at resonance
• Real circuits always have some
resistance
• Tuning a radio: a varying capacitor
changes the resonance frequency of
the tuning circuit in your radio to
match the station to be received
Transformers
• An AC transformer consists of two coils of wire wound
around a core of soft iron
• The side connected to the input AC voltage source is
called the primary and has N1 turns
• The other side, called the secondary, is connected to a
resistor and has N2 turns
• The core is used to increase the
magnetic flux and to provide a
medium for the flux to pass
from one coil to the other
 B Transformers
 B
V1   N1
V2   N 2
t
t
• The rate of change of the flux is the same for both coils,
so the voltages are related by
N2
V2 
N1
V1
• When N2 > N1, the transformer is referred to as a step
up transformer and when N2 < N1, the transformer is
referred to as a step down transformer
• The power input into the primary
equals the power output at the
secondary
I1V1  I 2 V2
Chapter 33
Problem 43
A transmission line that has a resistance per unit length of 4.50 × 10-4
Ω/m is to be used to transmit 5.00 MW across 400 miles (6.44 × 105 m).
The output voltage of the generator is 4.50 kV. (a) What is the line loss
if a transformer is used to step up the voltage to 500 kV? (b) What
fraction of the input power is lost to the line under these
circumstances?
Answers to Even Numbered Problems
Chapter 33:
Problem 4
(a) 25.3 rad/s
(b) 0.114 s
Answers to Even Numbered Problems
Chapter 33:
Problem 10
3.80 J