Part II - 3_AC Signals and Power
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Transcript of Part II - 3_AC Signals and Power
What you need to know
Identify the frequency, angular frequency, peak
value, root-mean-square (rms) value, and phase
of a sinusoidal signal
Determine rms value of periodic current or
voltage
Solve steady-state AC circuits using phasors
and complex impedances
Compute power for steady-state AC circuits
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Contents
1. Sinusoidal Signals
2. Phasors
3. Complex Impedances
4. Circuit Analysis with Phasors and
Complex Impedances
5. Power in AC Circuits
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1. Sinusoidal Signals
A sinusoidal voltage is given by
v(t) =Vm cos(ωt +θ )
where Vm is the maximum value of the voltage, ω is the
angular frequency in rad/s, and θ is the phase angle, in
radian.
Sinusoidal signals are periodic, repeating after each
period T (s): ωT = 2π.
The frequency of the signal is f=1/T Hz
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A sinusoidal voltage waveform given by v(t) = Vm cos (ωt + θ). Note: Assuming that θ is in
degrees, we have tmax = –θ/360 × T. For the waveform shown, θ is –45°.
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Phasors as Rotating Vectors
12
A sinusoid can be represented as
the real part of a vector rotating
counterclockwise in the complex
plane.
Phase Relationships
13
Since the vectors rotate
counterclockwise, V1 leads V2 by
60°(or, equivalently, V2 lags V1 by 60°)
v1(t) leads v2(t) by 60°= The peaks of v1(t) occur 60°before the peaks of v2(t).
3. Complex Impedances
Impedance: the ratio of the phasor of the sinusoidal
voltage to the phasor of the sinusoidal current
Complex numbers for inductance and capacitance
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4. Circuit Analysis with Phasors
and Complex Impedances
In a circuit with
sinusoidal sources,
inductors, capacitors
and resistors, the
sources can be
replaced by their
phasors, and the
inductance and
capacitance can be
replaced by their
impedance, follow by
the application of
Kirchoff’s laws
(KVL,KCL).
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Example: Steady-State AC Analysis of a
Series Circuit
Find the steady-state current for the following circuit. Also, find the phasor voltage
across each element and construct a phasor diagram.
21
Example: Steady-State AC Analysis of a
Series Circuit (continue)below.
23
100 + j100
= 141.4 [cos(45o)
+ j sin(45o)]
Three-Phase Power
Three-phase power is an arrangement
in which three sinusoidal voltages are
generated out of phase with each other.
The most common case is the balanced
voltage case with 3 voltages of equal
amplitude and frequency but with phase
120 degrees apart.
Most of the AC power is generated and
distributed as three-phase power
because
Efficient – less wiring needed
Total power delivery is constant
Three-phase motor has non-zero starting
torque.
Voltage-vs-time plot
28
Three Phase Source – Wye connected
(Y-connected)
Phase sequence can be important because
the direction of rotation of a three-phase
motor is opposite for the two phase
sequences. Hence, direction of the
motor can be reversed by interchanging the b
and c connection.30
Wye-Wye Connection
A three-phase source
is connected to a
balanced three-phase
load (i.e. the 3 loads
are equal).
Each source is referred
to as phase. The term
phase also refers to
the load.
Examples, phase A of
source is van(t); phase
A of load is the
impedance between A
and N.
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Six wires are needed to connect three single-phase sources to three
loads. In a three-phase system, the same power transfer can be
accomplished with three wires. (Four if the neutral wire is used, for the
three-phase connection to achieve the same power transfer.)
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Advantage 1: wiring for connecting the sources to the loads is less
expensive for three-phase AC compared to that of single-phase AC.
Power in Three-Phase Circuit
34
Advantage 2: the total power is constant (as a
function of time) rather than pulsating for
three-phase AC compared to that of single-
phase AC.
The reactive power is given by:
The phasor diagram is shown below. Note
that a different scale has been chosen for
the currents than for the voltages.
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Reactive power is the
power that flow back and
forth between the sources
and energy-storage
elements contained in a
three-phase or single-
phase load.
Example: Analysis of a Wye-Wye system
with 0.2 H inductance and 50 ohms
resistance
40
A balance positive-sequence wye-connected 60 Hz three-phase source has
line-to-line voltage of VL = 1000 V. This source is connected to a balanced
wye-connected load. Each phase of the load consists of a 0.2-H inductance in
series with a 100-ohms resistance. Find the line-to-neural voltages, the line
currents and the power delivered to the load. Assume that the phase of Van is
zero.