Complex Power

in
Steady-State AC Circuits

ECET 3500 – Survey of Electric Machines

Steady-State AC Voltage Sources

The voltage potential produced by an AC source may
be defined as: i(t)
+
v ( t )  V peak  sin(  t   ) v(t)

where: V peak is the peak value of the voltage,

 is the angular frequency (2πƒ) of the waveform, and
 is the phase angle of the voltage waveform.

v (t )
V peak

time


1
 Steady-State AC Voltage Sources
Similarly, the current produced by the AC source may
be defined as: i(t)
+
i ( t )  I peak  sin(  t   ) v(t)

where: I peak is the peak value of the current,

 is the angular frequency (2πƒ) of the waveform, and
 is the phase angle of the current waveform.

i (t )
I peak

time


Power in AC Circuits
Electric Power is defined as the rate at which
electric energy is either produced or consumed i(t)
by an element within the circuit. +
v(t)
The power produced or consumed by a circuit
element can be determined from the voltage and
current waveforms associated with that element:
i(t)
p( t )  v ( t )  i ( t ) (Watts) +

v(t)
where: p(t ) is the instantaneous rate that a circuit
element either produces or consumes -
energy at time t.

2
 Source vs. Load Convention
Note that the expression:
i(t)
p( t )  v ( t )  i ( t ) +
v(t)
defines the power “PRODUCED” by an element
when the current is defined in the same direction
as the voltage-rise across the element.
i(t)
But, if the current is defined in the opposite direction +

as the voltage-rise across an element, then p(t) v(t)

defines the power “CONSUMED” by that element.
-

Power from an AC Source

In the case of an AC source where:
v ( t )  V peak  sin(  t   )
i(t) i(t)
i ( t )  I peak  sin(  t   ) + +

v(t) v(t)
the general expression for power produced by the -
source is:
p( t )  V peak  I peak  sin(  t   )  sin(  t   )

To better understand the nature of this expression, it

may by useful to first consider the case where the
voltage source is supplying a resistive load.

3
 AC Sources and Resistive Loads
Given the voltage, vR, across a resistive load:
v R ( t )  V peak  sin(  t ) i(t) iR(t)
+ +
the resultant current, iR, that flows through the v(t) vR(t) R
resistor is defined by Ohm’s Law as: -

i R ( t )  I peak  sin(  t   )
v R (t )

R
V peak
  sin(  t )
R

AC Sources and Resistive Loads

Thus, for a resistive load:
V peak
v R ( t )  V peak  sin(  t ) iR (t )   sin(  t ) i(t) iR(t)
R + +

v(t) vR(t) R
the peak value of the current also adheres
to Ohm’s Law: -

V peak
I peak 
R
and the phase angle of the current equals the phase angle of the
applied voltage…
    0
There is no phase shift between the voltage and current waveforms relating to a purely resistive load.

4
 AC Power and Resistors
Given a resistor’s voltage and current waveforms:
V peak
v R ( t )  V peak  sin(  t ) iR (t )   sin(  t ) i(t) iR(t)
R + +

v(t) vR(t) R
the power consumed by the resistor is:
-
2
V peak
pR ( t )   sin 2 (  t )
R
pR(t) vR(t) iR(t)

AC Power and Resistors

Note that the resistor power varies periodically
but at a frequency that is 2x larger than that of
the resistor’s voltage and current waveforms. i(t) iR(t)
+ +

v(t) vR(t) R
Also note that the power consumed by the resistor
is always non-negative, which is expected since -

any negative power values would imply that the

resistor is instantaneously “producing” power.
pR(t) vR(t) iR(t)

5
 AC Power and Resistors
Although the power consumed by a resistor varies
periodically between zero and its peak value
under steady-state conditions: i(t) iR(t)
+ +
Ppeak  V peak  I peak v(t) vR(t) R

a resistor’s operation is often characterized in -

terms of the average power that it consumes.

pR(t) vR(t) iR(t)

Ppeak

Vpeak
Ipeak

AC Power and Resistors

To better understand the resistor power waveform,
it is useful to rewrite the power equation into the
following form: i(t) iR(t)
+ +

pR ( t )  V peak  I peak  sin 2 (  t ) v(t) vR(t) R

V I
-
 peak peak  [1  cos( 2    t )]
2
V I V I
 peak peak  peak peak  cos( 2    t )
2 2

(by utilizing the trigonometric identity sin2x = ½·[1 – cos2x])

6
 AC Power and Resistors
Looking at the resultant resistor power waveform:
V peak  I peak V peak  I peak
pR ( t )    cos( 2    t ) i(t) iR(t)
2 2 + +

v(t) vR(t) R
It can be seen that the expression has two terms: -
• The first term is a constant that provides the value
of the average power consumed by the resistor.
pR(t)
Ppeak
V peak  I peak Ppeak
PAvg  
PAvg 2 2

AC Power and Resistors

Looking at the resultant resistor power waveform:
V peak  I peak V peak  I peak
pR ( t )    cos( 2    t ) i(t) iR(t)
2 2 + +

v(t) vR(t) R
It can be seen that the expression has two terms: -
• The second term is a purely “sinusoidal” term
that has a zero average value and varies with a
frequency that is 2x larger than that of the source.

7
 Real Power
In an AC system, the Real Power, P, produced or
i(t) iR(t)
consumed by a circuit element is defined in terms + +
of the average power produced or consumed by v(t) vR(t) R
that element:
-

P  Avg[ p( t )]  Avg[v ( t )  i ( t )] (Watts)

Note – since power is constant in a DC circuit, this + +

IR
definition for real power also applies to the
IDC

power produced or consumed by any VDC

+
VR R
element in a DC circuit.
- -

AC vs. DC Power in Resistors

Given a resistor supplied by an AC source, the
i(t) iR(t)
real power, PR(AC), consumed by the resistor is + +
the average value of the its power waveform, v(t) vR(t) R
which is only ½ that of its peak value:
-

Ppeak V peak  I peak

PR ( AC )   (Watts)
2 2
+ +
Yet, this result may cause confusion if the AC IDC IR

real power value is compared to the constant +

power supplied to a resistor by a DC source. VDC VR R

- -

8
 AC vs. DC Power in Resistors
If the peak value of the AC source is equal to the
i(t) iR(t)
magnitude of a separate DC source (Vpeak= VDC) + +
and both sources supply similar resistors, then: v(t) vR(t) R
the real power consumed by the AC-supplied -
resistor will only be ½ that of the power
consumed by the DC-supplied resistor .

V peak  I peak V peak

2
+ +
PR ( AC )   (Watts) IDC IR
2 2 R +
VDC VR R
2
VDC
PR ( DC )  VDC  I DC  (Watts)
R
- -

AC vs. DC Power in Resistors

In other words:
i(t) iR(t)
Given an AC source whose peak value is equal + +

to the magnitude of a DC source, v(t) vR(t) R

-
If both sources supply similar resistors,

Then the AC source will be ½ as effective as

the DC source in terms of power supplied to + +
IR
a resistor. IDC

+
R
P VDC VR

If Vpeak=VDC  PR ( AC )  R ( DC ) (Watts)
2
- -

9
 AC vs. DC Power in Resistors
For example:
i(t) iR(t)
The real power consumed by a 500 resistor + +

supplied by a 100Vpeak AC source is: v(t) vR(t) R

-
2
V peak 1002
PR ( AC )    10 Watts
2  R 2  500
And the power consumed by a 500 resistor + +
IR
supplied by a 100V DC source is: IDC

+
VDC VR R
V2 1002
PR ( DC )  DC   20 Watts
R 500
- -

Effective Voltage
But, given an AC source with peak voltage Vpeak,
i(t) iR(t)
+ +

If an effective voltage magnitude, Veff , is defined v(t) vR(t) R

for the AC source in terms of the magnitude of a -
DC source that would deliver the same average
power to a resistor, then:

“What is the effective voltage magnitude + +

IR
of an AC source with peak voltage Vpeak?” IDC

+
VDC VR R

- -

10
 Effective Voltage
Given:
2 i(t) iR(t)
V peak 2
VDC + +
PR ( AC )  PR ( DC ) 
2 R R v(t) vR(t) R
if: -
V peak
VDC 
2
then the AC and DC sources will both supply +
IDC
+
IR
equal average (real) powers to the resistor. +
VDC VR R

Thus, the effective voltage of the AC source is:

V peak - -
Veffective 
2

Effective Voltage
For example:
i(t) iR(t)
Given: Vpeak = 100V and R = 500, + +

v(t) vR(t) R
2
V peak 100 2
PR ( AC )    10 (Watts) -
2 R 2  500
The 100Vpeak
but: AC source is
V peak 100 just as effective
Veffective    70.7 V
+ +
as the 70.7V IDC IR
2 2 DC source +
VDC VR R
if: VDC = 70.7V and R = 500:
2
VDC 70.7 2 - -
PR ( DC )    10 (Watts)
R 500

11
 RMS Voltage Magnitude
It turns out that the effective voltage magnitude
i(t) iR(t)
of a sinusoidal AC source: + +

V peak v(t) vR(t) R

v ( t )  V peak  sin(  t   ) Veffective  -
2

is equal to the RMS (root-mean-squared) value

of its AC waveform, as defined by the function:
T
1
Veff  VRMS    v 2 ( t )  dt Note – this function can also be
T 0 used to determine the
effective magnitude of
any arbitrary periodic
(AC) waveform.

RMS Magnitudes
The voltage potential and current produced by
an AC source can also be expressed in terms of i(t)
their RMS magnitudes, V and I respectively: +
v(t)
v ( t )  V peak  sin(  t   )  2  V  sin(  t   )

i ( t )  I peak  sin(  t   )  2  I  sin(  t   )

V peak
where: V is the RMS magnitude of the AC voltage, and
2
I peak
I is the RMS magnitude of the AC current.
2

12
 RMS Magnitudes & Resistor Power
When expressed in terms of their RMS magnitudes:
v ( t )  2  V  sin(  t )
i(t) iR(t)
+ +
i ( t )  2  I  sin(  t ) v(t) vR(t) R
the power delivered to a resistor is: -

p R ( t )  V  I  V  I  cos( 2    t )

which has an (average) Real Power value of:

V2
PR ( AC )  V  I 
R

RMS Magnitudes & Resistor Power

The result:
V2
PR ( AC )  V  I  i(t) iR(t)
R + +

is similar to the DC formula for power: v(t) vR(t) R

2 -
VDC
PR ( DC )  VDC  I DC 
R
which provides an advantage for defining the AC
waveforms in terms of their RMS magnitudes
instead of their peak values.

13
 Real Power and Resistors
Real Power (P ) is the average power produced
or consumed by an element in an AC circuit.
i(t) iR(t)
Real Power is defined in units of Watts. + +

v(t) vR(t) R
In a purely resistive AC circuit, if the voltages -
and currents are expressed in terms of their
RMS magnitudes, then real power can be
calculated as:

P  V  I Watts

AC Sources and Reactive Loads

What if the AC source is supplying a load that is
purely reactive… iC(t)
+
I.e. – either Capacitive or Inductive? vC(t) C
-
Although a sinusoidal (AC) voltage source will cause
a sinusoidal (AC) current to flow through both
capacitors and inductors, their voltage and current
waveforms do not follow the linear Ohm’s Law iL(t)
+
relationship. Instead, their voltage and current
waveforms are governed by a differential vR(t) L
relationship. -

14
 AC Sources and Capacitors
For an ideal capacitor, the voltage-current relationship
is defined by the following equations:
i(t) iC(t)
dv ( t ) + +
iC ( t )  C  C v(t) vC(t) C
dt
t t -
1 1
vC ( t ) 
C iC (t )dt  C 0 iC (t )dt  Vo
We may obtain a solution for steady-state AC operation
from these relationships.

AC Sources and Capacitors

Given the voltage applied across a capacitor:

vC ( t )  2  V  sin(  t ) i(t) iC(t)

+ +
the resultant current will be: v(t) vC(t) C

iC ( t )  2  V    C  sin(  t  90 ) -

Note that:
• The capacitor current is phase-shifted by +90°
compared to the capacitor voltage, and
• The voltage and current magnitudes do not 1
V  I
follow a linear relationship w.r.t. capacitance.  C

15
 AC Power and Capacitors
Given a capacitor’s voltage and current waveforms:
vC ( t )  2  V  sin(  t )
i(t) iC(t)
+
iC ( t )  2  V    C  sin(  t  90 ) +
v(t) vC(t) C
the power consumed by the capacitor is: -

pC ( t )  V 2    C  sin( 2    t )

pC(t) vC(t) iC(t)

AC Power and Capacitors

Looking at the resultant capacitor power waveform:
pC ( t )  V 2    C  sin( 2    t ) i(t) iC(t)
+ +

it can be seen that it varies sinusoidally at twice v(t) vC(t) C

the frequency of the capacitor’s voltage and current -
waveforms and that it has a zero-average value.
PC  0 (zero real power)
pC(t) vC(t) iC(t)

16
 AC Power and Capacitors
Despite the fact that the (average) real power
consumed by a capacitor is zero:
i(t) iC(t)
PC  0 + +

v(t) vC(t) C
there is energy flowing into and out of the
capacitor as it temporarily stores and releases -

a charge in a periodic manner.

pC(t)

storing energy storing energy

releasing energy releasing energy

AC Sources and Inductors

For an ideal inductor, the voltage-current relationship
is defined by the following equations:
i(t) iL(t)
di ( t ) + +
v L (t )  L  L v(t) vL(t) L
dt
t t -
1 1
i L ( t )   v L ( t )dt   v L ( t )dt  I o
L  L0

We may obtain a solution for steady-state AC operation

from these relationships.

17
 AC Sources and Inductors
Given the voltage applied across a inductor :

v L ( t )  2  V  sin(  t ) i(t) iL(t)

+ +
the resultant current will be: v(t) vL(t) L
V
iL (t )  2   sin(  t  90 ) -
L
Note that:
• The inductor current is phase-shifted by –90°
compared to the inductor voltage, and
• The voltage and current magnitudes do not V  I   L
follow a linear relationship w.r.t. inductance.

AC Power and Inductors

Given an inductor’s voltage and current waveforms:
v L ( t )  2  V  sin(  t )
i(t) iL(t)
V + +
iL (t )  2   sin(  t  90 )
L v(t) vL(t) L
the power consumed by the inductor is: -
2
V
pL ( t )    sin( 2    t )
L
pL(t) vL(t) iL(t)

18
 AC Power and Inductors
Looking at the resultant inductor power waveform:
V2
pL ( t )    sin( 2    t ) i(t) iL(t)
L + +

it can be seen that it varies sinusoidally at twice v(t) vL(t) L

the frequency of the inductor’s voltage and current -
waveforms and that it has a zero-average value.
PL  0 (zero real power)
pL(t) vL(t) iL(t)

AC Power and Inductors

Despite the fact that the (average) real power
consumed by an inductor is zero:
i(t) iL(t)
PL  0 + +

v(t) vL(t) L
the inductor also temporarily stores and
releases energy in a periodic manner. -

pL(t)

storing energy storing energy

releasing energy releasing energy

19
 AC Power – General Case
Given a source’s voltage and current waveforms
expressed in terms of their RMS magnitudes:
i(t) i(t)
v ( t )  2  V  sin(  t   ) + +

v(t) v(t)
i ( t )  2  I  sin(  t   ) -

the general expression for the instantaneous power

produced by the AC source is:
Note that this
p( t )  v ( t )  i ( t ) expression also
defines the
 2  V  I  sin(  t   )  sin(  t   ) power consumed
by the load

AC Power – General Case

The instantaneous power expression:

p( t )  2  V  I  sin(  t   )  sin(  t   ) i(t) i(t)

+ +

can be modified using several trigonometric v(t) v(t)

identities into the following form: -

p( t )  V  I  cos( )
 V  I  cos( )  cos( 2    t )
 V  I  sin( )  sin( 2    t )
where:
    (angle of the voltage minus angle of the current)

20
 AC Power – General Case
The expression for AC power has three terms:
p( t )  V  I  cos( )
i(t) i(t)
 V  I  cos( )  cos( 2    t ) + +

v(t) v(t)
 V  I  sin( )  sin( 2    t )
-

The first term is a constant that provides the average

or real power produced by the source:
P  V  I  cos( )

AC Power – General Case

The expression for AC power has three terms:
p( t )  V  I  cos( )
i(t) i(t)
 V  I  cos( )  cos( 2    t ) + +

v(t) v(t)
 V  I  sin( )  sin( 2    t )
-

The remaining terms are both purely sinusoidal and

vary at a frequency that is 2x greater than that of the
voltage or current waveforms.

21
 AC Power and Resistors
For a purely resistive load (  0), the general
power waveform:
i(t) iR(t)
p( t )  V  I  cos( ) + +

v(t) vR(t)
 V  I  cos( )  cos( 2    t ) R
-
 V  I  sin( )  sin( 2    t )
simplifies to:

p( t )  V  I  V  I  cos( 2    t )

which is equivalent to the previously determined

result for a purely resistive load.

AC Power and Capacitors or Inductors

For a purely capacitive load (  90 ) or a purely
inductive load (  90 ), the waveform: i(t)
+
iC(t)
+
p( t )  V  I  cos( ) v(t) vC(t) C

 V  I  cos( )  cos( 2    t ) -

 V  I  sin( )  sin( 2    t )
simplifies to: i(t) iL(t)
+ +
p( t )   V  I  sin( 2    t ) v(t) vL(t) L

which is similar to the previously determined -

results for capacitive and inductive loads.

22
 Reactive Power
The term Reactive Power is used to characterize
the amount of energy that is temporarily stored i(t)
+
iC(t)
and released by reactive loads. (I.e. – capacitive +
v(t) vC(t) C
or inductive loads).
-

Reactive Power, Q, is defined as the peak value of

the power that flows in and out of a reactive load.
i(t) iL(t)
Since, for a purely reactive load: + +

v(t) vL(t) L
p( t )   V  I  sin( 2    t ) -
then:
QC  V  I or Q L  V  I (VARs) VARs ≡ VoltAmpsReactive

Real Power and Reactive Power

Although a voltage source can produce (or consume) +
i(t)
+
iR(t)

both real power and reactive power, in terms of v(t) vR(t) R

passive loads (resistors, capacitors and inductors): -

Real power is consumed only by resistive loads. i(t) iC(t)

+ +
(Ideal capacitors and inductors consume zero-average power)
v(t) vC(t) C
PC  0 PL  0 -

Reactive power only relates to reactive loads. i(t) iL(t)

+
(Resistors do not temporarily store and release energy) +
v(t) vL(t) L
QR  0 -

23
 AC Power in Mixed (R-L-C) Circuits
If a source is connected to a circuit that contains a
i(t) i(t)
combination of resistive, capacitive, or inductive + +
loads, then the angle difference,  , between the v(t) v(t)
phase angles of the voltage and current will be:
-

 90    90
In a circuit with multiple load-types, the angle difference:
θ ≠ 0°, –90°, or +90°
thus all three terms will exist in the power waveform:
p( t )  V  I  cos( )
 V  I  cos( )  cos( 2    t )
 V  I  sin( )  sin( 2    t )

Real Power in R-L-C Circuits

Given a source that is connected to a circuit that
i(t) i(t)
contains multiple load-types and the general + +
power waveform: v(t) v(t)
p( t )  V  I  cos( ) -

 V  I  cos( )  cos( 2    t )
v ( t )  2  V  sin(  t   )
 V  I  sin( )  sin( 2    t )
i ( t )  2  I  sin(  t   )
Real power is defined as the average value of the    
power waveform:  90    90
P  V  I  cos( ) Watts

24
 Reactive Power in R-L-C Circuits
Given a source that is connected to a circuit that
i(t) i(t)
contains multiple load-types and the general + +
power waveform: v(t) v(t)
p( t )  V  I  cos( ) -

 V  I  cos( )  cos( 2    t )
v ( t )  2  V  sin(  t   )
 V  I  sin( )  sin( 2    t )
i ( t )  2  I  sin(  t   )
Reactive power is magnitude of the third term    
which relates to the power that flows in and out of  90    90
a reactive elements in the circuit:
Q  V  I  sin( ) VARs

Reactive Power – Capacitors vs. Inductors

Although Reactive Power, Q, is defined as the
i(t) i(t)
magnitude of the power that flows in and out of + +
the reactive loads: v(t) v(t)

Q  V  I  sin( ) -

The reactive power equation will return a Despite being defined as a

negative value for a circuit that is primarily magnitude, the sign is often
included to characterize the
capacitive (–90° ≤ θ ≤ 0°), and type of the load (capacitive or
inductive) to which the
reactive power relates.
The reactive power equation will return a Due to the resultant signs, capacitors are
positive value for a circuit that is primarily often characterized as “producing”
inductive (–90° ≤ θ ≤ 0°). reactive power and inductors are often
characterized as “consuming” reactive
power despite neither actually consuming
or producing a net amount of energy

25
 Phasor Analysis of an AC Circuit
When performing a phasor analysis on an AC
i(t) i(t)
circuit, the sinusoidal voltages and currents: + +

v ( t )  2  V  sin(  t   ) v(t) v(t)

-
i ( t )  2  I  sin(  t   )

are defined by their phasor values:

When expressed in polar form, the angles
The phasor values are also ~ j may be defined either in degrees or radians.
shown in complex exponential V  Ve  V But, when expressed in complex exponential
form because some calculators form, most calculators require the angles to
do not allow complex numbers ~ j be defined in radians.
to be expressed in polar form. I  Ie  I
such that they are expressed as complex numbers in polar form
with the RMS magnitude and phase angle of their respective
sinusoidal waveform.

Phasor Analysis of an AC Circuit

Additionally, when performing a phasor analysis on
i(t) i(t)
an AC circuit, the individual load elements: + +

R, L, and C v(t) v(t)

-
are defined in terms of their impedance values, Z,
such that for:
Circuit Elements Impedance Values

Resistors  ZR  R
Note that the impedance of a
resistor is purely real, while
Inductors  Z L  j   L the impedance of either an
inductor or a capacitor is
purely imaginary.
 1 
Capacitors  Z C   j 
  C 

26
Phasor Analysis with Complex Impedances
When performing a phasor analysis, multiple load
i(t) i(t)
elements are often combined into single equivalent + +
impedances that have both resistive and reactive v(t) v(t)
components, such that:
-
Complex number expressed
Z  R  jX in rectangular form

where: R is the resistive component of the load, and

X is the reactive component of the load.

Note that, when expressed in polar form, the angle of the impedance
is the difference angle  .
~
V V V V
Z ~   (   )   Z 
I I  I I

Complex Power
The term Complex Power is used to characterize
i(t) i(t)
both the real power and the reactive power + +
produced or consumed by a single element in an v(t) v(t)
AC circuit.
-

Complex Power, S, is typically expressed as a ~

complex number in rectangular form: V  V
~
I  I
S  P jQ    
Z  R  jX
where: P is real power, and
Q is reactive power.

27
 Phasor Analysis and Complex Power
The Complex Power produced or consumed by a
i(t) i(t)
single element in an AC circuit can be defined in + +
terms of that element’s phasor voltage and current: v(t) v(t)
~ ~
S  P  jQ  V  I  -

 (V )  ( I   ) ~
V  V
 V  I  (   )  V  I   ~
I  I
 V  I  cos   j V  I  sin     
~ ~ Z  R  jX
where I  is the complex conjugate of the current I :
~ S  P jQ
I   ( I  )   ( I   )

Apparent Power and Power Factor

Two other quantities related to complex power are
i(t) i(t)
often utilized when characterizing AC systems: + +

v(t) v(t)
Apparent Power, |S|, is defined as the magnitude -
of complex power:
Apparent Power is
often used when rating ~
S  V  I  P 2  Q2 an AC device: V  V
|S|rated = Vrated  Irated ~
I  I
Power Factor, pf, is defined as the ratio of an    
elements real power over its apparent power:
Z  R  jX
P V  I  cos  S  P jQ
pf    cos 
S V I ~ ~
S V  I 

28
 Leading or Lagging Power Factor
Power Factor is often characterized by a qualifier,
i(t) i(t)
either leading or lagging. + +

v(t) v(t)
A leading power factor exists for a capacitive load -
where the current waveform is “leading” the
voltage, resulting in a negative difference angle θ: ~
V  V
 90    0 ~
I  I
A lagging power factor exists for an inductive load    
where the current waveform is “lagging” the For a purely resistive load,
voltage, resulting in a positive difference angle θ: the difference angle
θ = 0°
0    90 resulting in a “unity”
power factor
pf = cosθ = cos0° = 1

Summary of Complex Power Equations

~ ~ i(t) i(t)
Complex Power (S): S  P  jQ  V  I  + +

v(t) v(t)
Real Power (P): P  V  I  cos -

Reactive Power (Q): Q  V  I  sin  ~

V  V
~
Apparent Power (|S|): S  V  I  P 2  Q2 I  I
   
Power Factor (pf): pf  cos Z  R  jX

29