EE111 Electrical Circuit Analysis
Lecture 26
Dr. Oliver Faust1
November 29, 2014
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School of Science & Engineering
Habib University
Motivation
Introducing complex power. This analysis method is necessary to evaluate
AC power networks.
Contents
1 Complex Power
1.1 The Power Triangle . . . . . . . . . . . . . . . . . . . . . . . . . .
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2 Example
2.1 RC circuit with an AC source . . . . . . . . . . . . . . . . . . . .
2.2 RL circuit with an AC source . . . . . . . . . . . . . . . . . . . .
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1
Complex Power
In a previous discussion we concluded that “complex” numbers simplify the
analysis. By allowing us to carry two pieces of information together through
a series of calculations via the “real” and “imaginary” components, they often
greatly simplify what might otherwise be tedious calculations. This is particularly true with power, since we have resistive as well as inductive and capacitive
elements in a general load. In this section, we define complex power to allow
us to calculate the various contributions to the total power in a clean, efficient
fashion. The magnitude of the complex power is simply the apparent power.
The real part is the average power and – as we are about to see – the imaginary
part is a new quantity, termed the reactive power, which describes the rate of
energy transfer into and out of reactive load components (e.g., inductors and
capacitors).
We define complex power with reference to a general sinusoidal voltage
Veff = Veff θ across a pair of terminals and a general sinusoidal current Ieff =
Ieff φ flowing into one of the terminals in such a way as to satisfy the passive
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sign convention. The average power P absorbed by the two-terminal network is
thus
P = Veff Ieff cos(θ − φ)
(1)
Complex nomenclature is next introduced by making use of Euler’s formula in
the same way as we did in introducing phasors. We express P as
o
n
(2)
P = Veff Ieff < ej(θ−φ)
or
o
n
P = < Veff ejθ Ieff e−jφ
(3)
The phasor voltage may now be recognized as the first two factors within the
brackets in the preceding equation, but the second two factors do not quite
correspond to the phasor current, because the angle includes a minus sign,
which is not present in the expression for the phasor current. That is, the
phasor current is
Ieff = Ieff ejφ
(4)
and we therefore must make use of conjugate notation:
I∗eff = Ieff e−jφ
(5)
P = < {Veff I∗eff }
(6)
Hence
and we may now let power become complex by defining the complex power S as
S = Veff I∗eff
(7)
If we first inspect the polar or exponential form of the complex power,
S = Veff Ieff ej(θ−φ)
(8)
we see that the magnitude of S, Veff Ieff , is the apparent power. The angle of S,
(θ − φ), is the PF angle (i.e., the angle by which the voltage leads the current).
In rectangular form, we have
S = P + jQ
(9)
where P is the average power, as before. The imaginary part of the complex
power is symbolized as Q and is termed the reactive power. The dimensions
of Q are the same as those of the real power P , the complex power S, and the
apparent power |S|. In order to avoid confusion with these other quantities, the
unit of Q is defined as the volt-ampere-reactive (abbreviated VAR). From Eqs.
7 and 9, it is seen that
Q = Veff Ieff sin(θ − φ)
(10)
The physical interpretation of reactive power is the time rate of energy flow
back and forth between the source (i.e., the utility company) and the reactive
components of the load (i.e., inductances and capacitances). These components
alternately charge and discharge, which leads to current flow from and to the
source, respectively.
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1.1
The Power Triangle
Power in the Cartesian coordinate system
=
S
Q
θ−φ
<
P
A commonly employed graphical representation of complex power is known
as the power triangle, and is illustrated in the figure above. The diagram shows
that only two of the three power quantities are required, as the third may be
obtained by trigonometric relationships. If the power triangle lies in the first
quadrant (θ − φ > 0); the power factor is lagging (corresponding to an inductive
load); and if the power triangle lies in the fourth quadrant (θ − φ < 0), the
power factor is leading (corresponding to a capacitive load). A great deal of
qualitative information concerning our load is therefore available at a glance.
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Example
2.1
RC circuit with an AC source
We use PSpice to discuss an RC circuit with an AC source. All components are
in series and the particular values follow as:
• R = 10 Ω,
• C = 200 µF,
• AC amplitude A = 10 V.
2.2
RL circuit with an AC source
We use PSpice to discuss an RL circuit with an AC source. All components are
in series and the particular values follow as:
• R = 10 Ω,
• L = 200 µH,
• AC amplitude A = 10 V.
Outlook
Lecture 27 will cover:
• Summary,
• Review.
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