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SIN(ωT)
V(T) = V
Chapter 10
Sinusoidal SteadyState Analysis
M
the amplitude of the wave is Vm
the argument is ωt
the radian or angular frequency is ω
note that sin() is periodic
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V(T) = V
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M
SIN(ωT)
2
A more general form of a sine wave includes a phase θ
V(T) = V SIN(ωT +θ)
M
the period of the wave is T
the frequency f is 1/T: units Hertz (Hz)
1 ω
F =
=
ω =2πF
T 2π
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The new wave (in red) is said to lead the original (in green)
by θ. The original sin(ωt) is said to lag the new wave by θ.
θ can be in degrees or radians, but the argument of sin() is always radians.
3
1.
When the source is sinusoidal, we often ignore
the transient/natural response and consider
only the forced or “steady-state” response.
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Apply KVL:
L
DI
+ RI = V
COS(ωT)
M
DT
2.
Make a good guess:
3.
Solve for the constants:
The source is assumed to exist forever: −∞<t<∞
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Apply superposition and use
E jθ = COS(θ) + J SIN(θ)
1.
Apply KVL, assume vs=Vmejωt.
DI
L +RI=V
S
DT
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2.
Find the complex response
i(t) = Imejωt+θ
3.
Find Im and θ, (discard the imaginary part)
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The term ejωt is common to all
voltages and currents and can
be ignored in all intermediate
steps, leading to the phasor:
Find the voltage on the capacitor.
I=I E
M
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In the frequency domain, Ohm’s Law
takes the same form:
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jφ
= I ∠φ
M
The phasor representation
of a current (or voltage) is
in the frequency domain
Answer: vc(t)=298.5 cos(5t − 84.3◦) mV
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Differentiation in time becomes multiplication in
phasor form: (calculus becomes algebra!)
11
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Differentiation in time becomes multiplication in
phasor form: (calculus becomes algebra!)
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Time Domain
Frequency Domain
Calculus (hard but real)
Algebra (easy but complex)
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Define impedance as Z=V/I, i.e. V=IZ
Applying KVL in time implies KVL for phasors:
V1+V2 +L +VN =0
ZR=R
Applying KCL in time implies KCL for phasors:
ZL=jωL
ZC=1/jωC
Impedance is the equivalent of
resistance in the frequency domain.
I1+I2 +L +IN =0
Impedance is a complex number (unit ohm).
Impedances in series or parallel can be
combined using “resistor rules.”
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the admittance is Y=1/Z
YR=1/R
YL=1/jωL
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Find the impedance of the network at 5 rad/s.
YC=jωC
if Z=R+jX; R is the resistance, X is
the reactance (unit ohm Ω)
if Y=G+jB; G is the conductance, B is
the susceptance: (unit siemen S)
Answer: 4.255 + j4.929 Ω
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Find the phasor voltages V1 and V2.
Find the currents i1(t) and i2(t).
Answer:
Answer: V1=1-j2 V and V2=-2+j4 V
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I1(T) = 1.24 COS(10 T + 29.7◦) A
3
I2(T) = 2.77 COS(10 T + 56.3◦) A
19
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Thévenin’s theorem also applies to
phasors; we can use it to find V1.
The setup is shown below:
The superposition principle applies to
phasors; use it to find V1.
Answer: V1=V1L +V1R =(2-j2)+(-1) = 1-j2 V
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If we assume I=1 ⁄ 0° A
The arrow for the phasor V on the
phasor diagram is a photograph,
taken at ωt = 0, of a rotating arrow
whose projection on the real axis is
the instantaneous voltage v(t).
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Chapter 11 AC
Circuit Power
Analysis
Assume V = 1 /0◦ V
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The instantaneous power
is p(t)=v(t)i(t).
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If in the same RL circuit, the source is Vmcos(ωt), then
and so the power will be
At all times t,
Constant
Term
power supplied =
power absorbed
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The average power over an arbitrary interval from t1 to t2 is
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Double
Frequency
Term
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If v(t)=Vmcos(ωt+θ) and i(t)=Imcos(ωt+ϕ), then
1
P = 2 VMIM COS(θ −φ)
When the power is periodic with period T, the average
power is calculated over any one period:
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The average power absorbed by a resistor R is
1V2
PR
Find the average power absorbed by
each element.
M
=
2R
The average power absorbed by a purely
reactive element(s) is zero, since the current
and voltage are 90 degrees out of phase:
Answer:
PX =0
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PL=0 W
Pleft=-50 W
PC=0 W,
PR=25 W
Pright=25 W
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First, solve for the load power:
An independent voltage source in series with
an impedance ZTH delivers a maximum
average power to that load impedance ZL
which is the conjugate of ZTH:
=
ZL = ZTH*
1
2
| Vth |2 RL
(Rth + RL )2 + ( Xth + XL )2
Clearly, P is largest when XL+Xth=0
Solving dP/dRL=0 will show that RL=Rth
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The effective value is often referred to
as the root-mean-square or RMS value.
The same power is delivered to the
resistor in the circuits shown.
For sine waves:
Power is now P
periodic, period T
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V
=
EFF
=I
2
1
2 VM ≅ 0.707VM
R
EFF
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If v(t)=Vmcos(ωt+θ) and
i(t)=Imcos(ωt+ϕ), then
1
P = 2 VM IM COS(θ −φ) = VEFF IEFF
Find the average power being delivered to an
impedance ZL= 8 − j11 Ω by a current I= 5ej20° A.
COS(θ
Only the 8-Ω resistance enters the averagepower calculation, since the j11-Ω component
will not absorb any average power.
−φ)
the apparent power is defined as VeffIeff
and is given the units volt-ampere V A
Thus,
P = (1/2)(52)8 = 100 W
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Power factor is defined as
PF =
=
V
the information as to whether current
leads or lags voltage is lost, so we add
the adjective to the power factor term.
I
EFF
EFF
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An inductive load has a lagging PF.
A capacitive load has a leading PF.
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S = Veff I*eff = Veff Ieff E
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j (θ−φ )
= P + JQ
the real part of S is P, the average power
the imaginary part of S is Q, the reactive power,
which represents the flow of energy back and forth
from the source (utility company) to the inductors
and capacitors of the load (customer)
Answer: 288 W, 144 W, 720 VA, PF=0.6 (lagging)
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Define the complex power S as
Find the average power delivered to each of the two
loads, the apparent power supplied by the source,
and the power factor of the combined loads.
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PF = COS(θ −φ)
P
for a resistive load, PF=1
for a purely reactive load, PF=0
generally, 0 ≤ PF ≤ 1
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Since the power factor for sine waves is
average power
apparent power
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Complex powers to loads add:
S = VI* = V(I1 + I2)* = V(I1* + I*2) = S1 + S2
Splitting the current phasor IEFF into in-phase
and out-of-phase components is another
way of visualizing the complex power.
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Chapter 12
Polyphase Circuits
An industrial consumer is operating a 50 kW induction motor at a lagging PF
of 0.8. The source voltage is 230 V rms. In order to obtain lower electrical
rates, the customer wishes to raise the PF to 0.95 lagging.
Specify a suitable solution.
Answer: deploy a capacitor in parallel with the motor, as shown above.
At 60 Hz, C=1.056 mF
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Polyphase (in practice,
three-phase) voltage
sources are important
for transmission over
the grid, and for large
industrial loads.
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The notation Vab indicates
the voltage between point a
(+) and point b (−).
Indicating the + and −
terminals is redundant
if double-subscript
voltages are used.
These three sources are 120°
out of phase with each other.
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KVL equations can be written correctly
without reference to a circuit diagram:
Vad = Vab +Vbd
Vab = Van +Vnb
Vad = Vab +Vbc +Vcd
Vab = −Vba
This is an example three-phase
source with a neutral.
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= Van −Vbn
=173.2 30
O
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The typical North American household
is provided single-phase 3-wire
power, where V1≅110 V and VAB=2V1.
The double subscript notation Iab means
the current from a to b by the direct path
When the direct path
is not obvious
(e.g. between c and d)
we need to use a
different notation.
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When the A and B loads
are balanced, the
neutral wire from n to
N carries no current:
IaA =
Van
Zp
= IBb =
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Determine the power delivered to each of the
three loads as well as the power lost in the
neutral wire and each of the two lines.
Answer:
P50 = 206 W
P100 =117 W
P20+j10 = 1763 W
PaA = 126 W
PbB = 108 W
PnN =3 W
Vnb
Zp
I NN =IBB+IAA =IBB−IAA =0
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A balanced three phase source has
The abc (positive) and cba (negative) sequences:
Van = Vbn = Vcn
and
Van +Vbn +Vcn = 0
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The line voltages have amplitude
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VL =
55
3VP
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For a balanced source and balanced loads, the
neutral wire current INN=0 and so the neutral
wire could have any impedance, including ∞.
57
Find the total power delivered to the loads.
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A balanced three-phase system with a line
voltage of 300 V is supplying a balanced
Y-connected load with 1200 W at a
leading PF of 0.8.
Find the line current and the perphase load impedance.
Answer: IL = 2.89 A, Z P =
Answer: P=600 W
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60e 36.9◦ Ω
J
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A delta-connected load is also commonly used
(note the absence of the neutral wire).
An example phasor
diagram for an
inductive phase
impedance.
Note that
I L = 3IP
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The sum of the
powers measured
by the wattmeters
is the total power
delivered.
The wattmeter is a four-terminal device
that measures power delivered to the
network if connected as shown:
The common node x
is arbitrary.
the wattmeter
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One wattmeter
can be
eliminated if
the point x is
moved to a line
(as shown, B).
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