Transcript 35 AC Circuits

Ch – 35 AC Circuits
Reading Quiz – Ch. 35
1. The analysis of AC circuits uses a rotating
vector called a :
a. unit circle vector
b. phasor
c. emf vector
d. sinusoidal vector
2. In a capacitor, the peak current and peak
voltage are related by the
a. capacitive resistance.
b. capacitive reactance.
c. capacitive impedance.
d. capacitive inductance.
Learning Objectives – Ch 35
• To use a phasor analysis to analyze AC
circuits.
• To understand RC filter circuits.
• To understand the series RLC circuit and
resonance.
Alternating current - Circuits
powered by a sinusoidal emf are
• Power plants produce
oscillating emf and
currents.
• Steam or falling water
turning a turbine,
which in turn, causes
a coil of wire in a
magnetic field.
ε = ε0 cos ωt
ε0 is peak emf
ω is angular frequency
in rads/s or
ω = 2πf, where f is the
frequency in Hz
(cycles per second) or
s-1.
Phasor Diagram
• A phasor is a vector that
rotates counterclockwise
around the origin at
angular frequency ω.
• The length of the
phasor(radius) represents
the peak value of the
quantity.
• ωt is a phase angle. If
there are more than one
phasor in the diagram,
there can be multiple
phase angles.
Phasor Diagram
• The instantaneous value
is the actual value you
would measure at time t.
This value is never
greater than ε0.
• The instantaneous value
can be represented as
the projection of the
phasor vector onto the
horizontal axis of the
circle.
Here is a graphical representation
along the t axis:
Stop to Think
The magnitude of the
instantaneous value
of the emf
represented by this
phasor is:
a. Increasing
b. Decreasing
c. Constant
d. Need to know t
Stop to Think
The magnitude of the
instantaneous value
of the emf
represented by this
phasor is:
a. Increasing, since it is
traveling ccw
Resistor Circuits
• Use of lowercase for
instantaneous values,
e.g. vR, iR
• Uppercase for peak
values
vR = VR cos ωt
iR = IR cos ωt
• In an AC circuit,
resistor voltage and
current oscillate in
phase.
Resistor Circuits
• Representation of
resistor current and
voltage on a phasor
diagram.
• Vectors rotate at the
same frequency and
have the same phase
angle with the origin
• Cannot compare
magnitudes, since
units are different.
Capacitor Circuits
• For the capacitor
circuit shown at right:
vC = ε = VC cos ωt
i = dq/dt and q = CvC:
iC = - ωCVC sin ωt,
which we write as:
iC = ωCVC cos (ωt +
π/2)
Capacitor Circuits - ICE
vC = VC cos ωt
iC = ωCVC cos (ωt + π/2)
The consequence of this is
that the capacitor voltage
and current do not
oscillate in phase.
The current leads voltage by
π/2 rads, or by T/4.
ICE
Capacitor Circuits - ICE
• Current reaches peak
value IC the instant
the capacitor is fully
discharged and vC =0.
The current is zero
the instant the
capacitor is fully
charged.
Capacitive Reactance – Relationship
between peak current and voltage in a
capacitor
• IC = ωCVC, although
they don’t happen at
the same time.
• Define the capacitive
reactance XC = 1/ ωC,
then:
• IC = VC/XC
• This is analogous to
Ohm’s Law for DC.
Properties of Capacitive Reactance
• Only used for peak values
iC DOES NOT EQUAL vC/XC
• Dependent on emf frequency,
unlike resistance, which is a
property of the resistor
independent of circuit
frequency.
• decreases as frequency
increases.
• at very high frequencies, XC
approaches 0 and the
capacitor acts like a wire.
RC Filter Circuits
• resistor and capacitor are
in series: IC = IR
• at low frequencies, XC
will be large, limiting I.
• Since VR = IR, voltage
across resistor will small.
• At high frequencies, XC
will be small, so the
capacitor will act more
like an ideal wire, with
very little voltage drop.
Analyzing an RC circuit
• Draw the current phasor.
All circuit elements in
series have the same
current at any time. Angle
is arbitrary.
• Current is in phase with
VR, so draw that phasor
in phase with I. Current
leads Vc by 900, so draw
the capacitor voltage
phasor 900 behind (i.e.
clockwise.
Analyzing an RC circuit
• Kirchoff’s loop law says
vR + vC = ε, for the
instantaneous values.
The addition of peak
values is a vector
addition. Therefore ε0 is
drawn as the resultant
vector as shown.
• ε = ε0 cos ωt; the angle
between emf and x-axis
is ωt.
Analyzing an RC circuit
• ε02 = VR2 + VC2
• This relationship is for
peak values.
• Peak currents are
related to peak
voltages by:
• VR = IR
• VC = IXC
• ε02 = (R2 + 1/ω2C2)I2
Crossover Frequency
For low frequencies,
where XC>>R, the
circuit is primarily
capacitive. A load in
parallel with the
capacitor will get most
of the potential
difference and
therefore the power.
Crossover Frequency
For high frequencies,
where XC<<R, the
circuit is primarily
resistive. A load in
parallel with the
capacitor will be
blocked from getting
any power.
Crossover Frequency
Crossover frequency is
found where VR and
VC are equal:
ωc = 1/RC
fc = ωc /2π
Filters
• Low pass filter (top):
– connect the output in
parallel to capacitor
– at frequencies well
below ωc the signal is
transmitted with little
loss.
– at frequencies well
above ωc , the signal
is attenuated
Filters
• High pass filter:
– connect the output in
parallel to resistor, as
in the lower diagram
– at frequencies well
below ωc the signal is
attenuated
– at frequencies well
above ωc , the signal
is transmitted with little
loss.
Inductor Circuits
• An inductor is a
device that produces
a uniform magnetic
field when a current
passes through it. A
solenoid is an
inductor.
|∆VL| = L dI/dt
Inductor circuits (ELI)
vL = ε = VL cos ωt
Where VL = ε0, and
vL = L diL/dt
For iL, instantaneous
current:
diL = (vL/L) dt
di = VL cos ωt dt
L
Integrating, we get:
IL = VL /ωL sin ωt, which
we write as:
iL = VL /ωL cos (ωt - π/2)
iL = IL cos (ωt - π/2)
Inductor circuit - ELI
• The math is similar to
that of a capacitor:
iL = IL cos (ωt - π/2) and
iC = IC cos (ωt + π/2)
The difference is that
the current in the
circuit lags the
inductor voltage by
900 or T/4:
ELI
Inductive reactance
• We can define the
inductive reactance to be
XL = ωL, then:
IL = VL/XL
(valid for peak values of I, V
only)
• Compare to:
XC = 1/ ωC
• both reactances are
frequency dependent.
• inductive reactance
increases with frequency.
• capacitive reactance
decreases with
frequency.
The Series RLC circuit
• This circuit acts as
both a low pass and a
high pass filter at the
same time. It only
allows signal to pass
from a narrow range
of frequencies.
The Series RLC circuit
• Two important
observations:
– The instantaneous
current through all 3
elements is the same
at a given time:
i = iR + iC + iL
– The sum of the
instantaneous
voltages add up to the
emf at a given time:
ε = vR + vC + vL
Analyzing an RLC circuit
• Draw the current phasor.
All circuit elements in
series have the same
current at any time. Angle
is arbitrary.
• Current is in phase with
VR, so draw that phasor
in phase with I. Current
leads Vc by 900 (ICE), so
draw the capacitor
voltage phasor 900
behind (i.e. clockwise).
Current lags VLby the
same amount (ELI) so
draw it ahead.
RLC circuit analysis
• Kirchoff’s loop law says
vR + vC + vL = ε, for the
instantaneous values.
The addition of the peak
values is a vector.
addition. Therefore ε0 is
drawn as the resultant
vector as shown.
• VC and VL are in opposite
directions and so can be
represented as the vector
(VL – VC or vice versa).
RLC circuits
• The length of the emf
phasor is the
hypotenuse of a right
triangle:
ε02 = VR2 + (VL - VC)2
• This relationship is for
peak values.
Phase angle between I and ε0
• If VL > VC then the emf
leads current:
i = I cos(ωt –φ)
where ωt is the angle
between the emf and the
horizontal axis.
• If VC > VL then phasor
diagram would be below
horizontal axis and emf
lags current:
i = I cos[ωt – (-φ)] or
i = I cos(ωt +φ)
RLC Circuits
ε02 = VR2 + (VL - VC)2
• Peak currents are
related to peak
voltages by:
 VR = IR
 VC = IXC
ε02 = [R2 + (XL- XC )2]I2
Taking the square root
of both sides…
Impedance
The impedance of an
RLC circuit is defined
as:
Z = R 2  (X L - XC ) 2
and has units of ohms.
Ohm’s Law for ac
circuits can be written
as:
I = ε0/Z
This is for peak
values only.
Impedance
The impedance of an
RLC circuit is defined
as:
Z = R 2  (X L - XC ) 2
and has units of ohms.
Ohm’s Law for ac
circuits can be written
as:
I = ε0/Z
This is for peak
values only.
Phase angle, revisited
From the diagram
on the right we see
that:
tan φ = (VL - VC)/VR
tan φ = (XL - XC)I/IR
φ = tan-1 (XL - XC)/R
φ
Phase angle, revisited
The phasor diagram at the
right shows a case where
the current lags emf by:
φ = tan-1 (XL - XC)/R
We can express the peak
resistor voltage as:
VR = ε0 cos φ
Resistor voltage oscillates
in phase with emf only if
φ=0 rads, i.e there are no
capacitors or inductors in
the circuit.
Resonance Frequency
Z=
R  (X L - XC )
2
2
• I = ε0/Z
• In an RLC circuit, current will be limited at low frequency
by XC = 1/ωC being large and at high frequency by XL =
ωL being large.
• Current will be maximum when impedance, Z is
minimized.
• Any ideas what we can do to minimize that term?
Resonance Frequency, ω0
• ω0 = 1/ C
• this frequency will produce the maximum
current in the RLC circuit:
• Imax = ε0/R, as Z = 0 at this frequency.
• at ω0, energy is transferred back and forth
between inductor and capacitor.
Peak current as ω is varied.
• ω0 can be considered the
natural frequency of the circuit,
the frequency at which it would
“like” to oscillate.
• When the emf (acting as the
driving force) oscillates at the
same frequency, there will be a
large response in terms of
output.
• Note that decreasing the
resistance decreases the
damping, and narrows the
frequency range of large
response.
Graphs of emf and current for
frequencies below, at, above ω0
• When ω < ω0, current leads emf, φ<0
• When ω > ω0, current lags emf, φ>0
• When ω = ω0, circuit is purely resistive φ=0.
Numerical Problem
What is the phase angle
when the emf
frequency is
a. 14 kHz
b. 18 kHz
c. What is ω0 for this
circuit