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Chapter 3
AC Power Analysis
1
AC Power Analysis
Chapter 3
3.1
3.2
3.3
3.4
3.5
3.6
3.7
3.8
Instantaneous and Average Power
Maximum Average Power Transfer
Effective or RMS Value
Apparent Power and Power Factor
Complex Power
Conservation of AC Power
Power Factor Correction
Power Measurement
2
AC Power Analysis
Chapter 3
Mostly all electric energy is supplied in the form of
sinusoidal voltages and currents
After discussion of sinusoidal circuits, now is to
consider sinusoidal steady state power calculations.
We begin with defining and deriving instantaneous
power and average power
3
3.1 Instantaneous and Average
Power
The power at any instant of time is p = vi
v  Vm cos(t  v )
i  I m cos(t  i )
where Vm and Im are the amplitudes and θv and θi are the
phase angles of the voltage and current, respectively
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3.1 Instantaneous and Average
Power
Use a zero time corresponding to the instant the
current is passing through a positive maximum
This reference system requires a shift of both the
voltage and current by θi
v  Vm cos(t  v  i )
i  I m cos(t )
p(t )  v(t ) i(t )  Vm I m cos ( t  v  i ) cos ( t )
1
cos A cos B  cos( A  B)  cos( A  B)
2
5
3.1 Instantaneous and Average
Power
p(t )  v(t ) i(t )  Vm I m cos ( t   v  i ) cos ( t )
1
1
 Vm I m cos ( v  i )  Vm I m cos (2 t   v  i )
2
2
p(t) > 0: power is absorbed by the circuit; p(t) < 0: power is absorbed by the source.
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3.1 Instantaneous and Average
Power
This shows the instantaneous power has two parts
p(t )  v(t ) i(t )  Vm I m cos ( t   v  i ) cos ( t )
1
1
 Vm I m cos ( v  i )  Vm I m cos (2 t   v  i )
2
2
Constant power
Sinusoidal power at 2t
The first part is constant or time independent. Its
value depends on the phase difference between the
voltage and the current
The second part is a sinusoidal function whose
frequency is 2, which is twice the frequency of the
voltage or current
Therefore, the instantaneous power goes through two
complete cycles for every cycle of either the voltage 7
or the current
3.1 Instantaneous and Average
Power
This shows the instantaneous power has two parts
p(t )  v(t ) i(t )  Vm I m cos ( t   v  i ) cos ( t )
1
1
 Vm I m cos ( v  i )  Vm I m cos (2 t   v  i )
2
2
Constant power
Sinusoidal power at 2t
Also note that the instantaneous power may be
negative for a portion of each cycle, even if the
network between the terminal is passive.
In a completely passive network, negative power
implies that energy stored in the inductors or
capacitors is now being extracted
8
3.1 Instantaneous and Average
Power
• Use the cos(   )  cos  cos   sin  sin  to expand the second
right term on the RHS eq.
1
1
p(t )  v(t ) i (t )  Vm I m cos ( v   i )  Vm I m cos (2 t   v   i )
2
2
which gives
1
1
1
p(t )  v(t ) i (t )  Vm I m cos ( v   i )  Vm I m cos ( v   i ) cos 2t  Vm I m sin  v   i sin 2t
2
2
2
p(t )  P  P cos 2t  Q sin 2t
1
where P = Vm I m cos ( v   i )
2
1
Q = Vm I m sin ( v   i )
2
9
3.1 Instantaneous and Average
Power
• P is called the average power, and Q is called the reactive
power
• Average power is sometimes called real power, because it
describes the power in a circuit that is transformed from
electric to non electric energy
• The average power associated with sinusoidal signals is the
average of the instantaneous power over one period, or, in
eq. form,
1 t0 T
P   t0 p(t ) dt
T
• where T is the period of the sinusoidal function
• The limits imply that we can initiate the integration process
at any convenient time to but must terminate the integration
10
exactly one period later
3.1 Instantaneous and Average
Power
• We could find the average power by substituting eq.
• p(t )  P  P cos 2t  Q sin 2t directly into
1 t0 T
P
T
 t0
p(t ) dt
• and then performing the integration.
• But note that the average value of p is given by the first
term on the RHS of eq. p(t )  P  P cos 2t  Q sin 2t
• because the integration of both cos2t and sin2t produces
a zero over a cycle
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3.1 Instantaneous and Average
Power
• The average power, P, is the average of the instantaneous
power over one period.
1 T
1
P   p(t ) dt  Vm I m cos ( v   i )
T 0
2
1. P is not time dependent.
2. When θv = θi , it is a purely
resistive load case.
3. When θv– θi = ±90o, it is a
purely reactive load case.
4. P = 0 means that the circuit
absorbs no average power.
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3.1 Instantaneous and Average
Power
Example 1
Calculate the instantaneous power and average
power absorbed by a passive linear network if:
v(t )  80 cos (10 t  20)
i (t )  15 sin (10 t  60)
Answer: 385.7  600cos(20t  10 )W, 385.7W
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3.1 Instantaneous and Average
Power
Example 2
A current I  10  30 flows through an
impedance Z  20  22Ω . Find the average
power delivered to the impedance.
Answer: 1854.36W
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3.2 Maximum Average Power
Transfer
• Maximum power is transferred to the load when the load
resistance equals the Thevenin resistance as seen from the
load (RL = RTh)
• We are now extend that result to ac circuit.
• Consider the circuit below, where an ac circuit is connected
to a load ZL and is represented by its Thevenin equivalent.
• In rectangular form, the Thevenin impedance
• ZTh and the load impedance ZL are
ZTH  R TH  j XTH
ZL  R L  j XL
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3.2 Maximum Average Power
Transfer
• The current through the load is
VTh
VTh
I

ZTh  Z L ( RTh  jX Th )  RL  jX L 
• The average power delivered to the load is
2
VTh RL / 2
1 2
P  I RL 
2
2
2
RTh  RL    X Th  X L 
• Our objective is to adjust the load parameters RL and XL so
that P is maximum
• To do this set P / RL and P / X L equal to zero
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3.2 Maximum Average Power
Transfer
We obtain
VTh RL  X Th  X L 
P

X L
RTh  RL 2   X Th  X L 2
2

P

RL
• Setting
VTh
2
R
P / X L

2

Th  RL    X Th  X L   2 RL RTh  RL 
2

2
2 RTh  RL    X Th  X L 
2

2 2
to zero gives XL = -XTh
• and settingP / R to zero results in R 
L
L
2
RTH
  X Th  X L 
2
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3.2 Maximum Average Power
Transfer
Combining XL = -XTh and
RL 
2
RTH
  X Th  X L 
2
leads to
the conclusion that for maximum average power transfer,
XL = –XTH and RL = RTH

Z L  RL  jX L  RTh  jX Th  ZTh
For maximum average power transfer, the load impedance ZL
must be equal to the complex conjugate of the Thevenin
impedance ZTh
2
Pmax 
VTH
8 R TH
If the load is purely real, then R L 
2
2
R TH
 XTH
 ZTH
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3.2 Maximum Average Power
Transfer
Example 3
For the circuit shown below, find the load impedance ZL that
absorbs the maximum average power. Calculate that maximum
average power.
Answer: 3.415 – j0.7317W, 1.429W
19
3.3 Effective or RMS Value
• The idea of effective value arises from the need to measure
the effectiveness of a voltage or current source in delivering
power to a resistive load
The effective of a periodic current is the dc current that delivers the
same average power to a resistor as the periodic current.
• The circuit in (a) is ac while that of (b) is dc.
• To find Ieff that will transfer the same power to
resistor, R as the sinusoid i.
The average power absorbed by the resistor in
the ac circuit is
1
P
T

T
0
R T 2
2
i Rdt   i dt  I rms
R
0
T
2
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3.3 Effective or RMS Value
• While the power absorbed by the resistor in the dc circuit is
PI R
2
eff
• Equating the eq. and solving for Ieff ,we obtain
I eff 
1
T

T
i 2 dt  I rms
0
• The effective value of the voltage is found in the same way
as current, that is,
Veff 
1 T 2
 v dt  Vrms
T 0
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3.3 Effective or RMS Value
• This indicates that the effective value is the square root of
the mean (or average) of the square of the periodic signal.
• Thus the effective value is often known as the root-meansquare value, or rms value
• For any periodic function x(t) in general, the rms value is
given by
X rms 
1 T 2
 x dt
T 0
The effective value of a periodic signal is its root mean square (rms)
value
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3.3 Effective or RMS Value
• For the sinusoid i(t)=Imcost, the effective or rms value is
I rms 

1 T 2 2
 I m cos t dt
T 0
I m2 T 1
Im


1

cos
2

t
dt


T 02
2
The average power can be written in terms of the rms values:
1
P  Vm I m cos (θv  θi )  Vrms I rms cos (θv  θi )
2
Similarly, the average power absorbed by a resistor, R
P I
2
rms
2
Vrms
R
R
Note: If you express amplitude of a phasor source(s) in rms, then all the
answer as a result of this phasor source(s) must also be in rms value.
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3.3 Effective or RMS Value
• Power engineers conventionally say a voltage and current in
terms of its RMS value
• So if power engineers say the voltage is 240 V, it is a RMS
value of 240 V and it means that it peak value is 339.4 V
• This is only valid for sinusoidal signal
• For non-sinusoidal waveform, the expression must be
modified to include the non-sinusoidal waveform such as by
using Fourier series to represent the non-sinusoids.
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3.4 Apparent Power and Power
Factor
• Apparent Power, S, is the product of the r.m.s. values of
voltage and current.
• It is measured in volt-amperes or VA to distinguish it from
the average or real power which is measured in watts.
P  Vrms I rms cos (θ v  θi )  S cos (θ v  θi )
Apparent Power, S
Power Factor, pf
• Power factor is the cosine of the phase difference between
the voltage and current. It is also the cosine of the angle
of the load impedance.
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3.4 Apparent Power and Power
Factor
Purely resistive
load (R)
θv– θi = 0,
Pf = 1
Purely reactive
load (L or C)
θv– θi =
±90o,
pf = 0
θv– θi > 0
θv– θi < 0
Resistive and
reactive load
(R and L/C)
P/S = 1, all power are
consumed
Power cannot be extracted(all
electric energy is dissipated in
the form of thermal energy)
P (average power)= 0, no real
power consumption
• Lagging - inductive
load(demand magnetizing
vars)
• Leading - capacitive
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load(deliver magnetizing vars)
3.5 Complex Power
Complex power S is the product of the voltage and the
complex conjugate of the current:
The complex sum of real and reactive power
V  Vmθ v
I  I mθi
1

V I  Vrms I rms  θ v  θ i
2
27
3.5 Complex Power
1
S  V I  Vrms I rms  θ v  θ i
2
 S  Vrms I rms cos (θ v  θi )  j Vrms I rms sin (θ v  θi )
S =
P
+ j
Q
P: is the average power in watts delivered to a load and it is
the only useful power.
Q: is the reactive power exchange between the source and
the reactive part of the load. It is measured in VAR.
• Q = 0 for resistive loads (unity pf).
• Q < 0 for capacitive loads (leading pf).
• Q > 0 for inductive loads (lagging pf).
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3.5 Complex Power
 S  Vrms I rms cos (θ v  θi )  j Vrms I rms sin (θ v  θi )
S =
P
+ j
Q
Apparent Power, S = |S| = Vrms*Irms = P 2  Q 2
Real power,
P = Re(S) = S cos(θv – θi)
Reactive Power, Q = Im(S) = S sin(θv – θi)
Power factor,
pf = P/S = cos(θv – θi)
•Complex power contains all the information pertaining to the
power absorbed by a load
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3.5 Complex Power
 S  Vrms I rms cos (θ v  θi )  j Vrms I rms sin (θ v  θi )
S =
Power Triangle Impedance Triangle
P
+ j
Power Factor
Q
30
There are 3 types of electric power:
– Real power (watts, kilowatts, or kW)
– Reactive power (kilovolt-amps reactive or kVAR)
– Apparent power (kilovolt-amps or kVA)
Real/Active power is said to be the flow of real energy
which produces a measureable output such as heat, light
or mechanical energy
– Household meters measure real power – the power we use
at home
– Real power does the work – it lights the lights and runs the
motors
– Example: A light bulb is 100 watts
Reactive power provides the magnetic field to make
motors operate
• The magnetic field is needed for real power (MW) to flow
through transformers
• The magnetic field is the invisible force of magnetism
– Example: The head on a “teh-tarik”
Reactive Power MVAr
Total Power
MVA
Real Power
MW
Apparent power is the actual/total power
the generator must supply to the system
– It includes both real power (kilowatts) and
reactive power (kVARS)
• The bulk power system needs both real
power and reactive power to be reliable
• Power Factor (p.f.) is the ratio of real
power (kW) to apparent power (kVA) in a
circuit
3.6 Conservation of AC Power
The principle applies to both ac and dc circuits
The complex real, and reactive powers of the sources
equal the respective sums of the complex, real, and
reactive powers of the individual loads.
The complex power supplied by the source equals the
complex powers delivered to loads Z1 and Z2


S  VI *  V I1*  I 2*  VI1*  VI 2*  S1  S2
I  I1  I 2
34
3.6 Conservation of AC Power
The same results can be obtained for a series
connection.

V  V1  V2

S  VI *  V1  V2 I *  V1I *  V2 I *  S1  S2
Whether
the
loads
are
connected
in
series/parallel/general, the total power supplied by the
source equals the total power delivered to the load.
This means that the total complex power in a network is
the sum of the complex powers of the individual
elements, BUT not true of apparent power.
.
35
3.7 Power Factor Correction
Most all practical loads are inductive (motors, fluorescent
lights, etc.) and operate at a lagging power factor.
Power factor correction is necessary for economic reason.
36
3.7 Power Factor Correction
The design of transmission system is very sensitive to the
magnitude of the current as determined by the applied load.
Increased currents result in increased power losses due to
the resistance of the lines
Heavier currents also require larger conductors, increasing
the amount of copper and obviously the require increased
generating capacities
Since the voltage is fixed, the apparent power is directly
related to the current level
37
3.7 Power Factor Correction
The smaller the apparent power, the smaller is the current
drawn from the supply
Minimum current is therefore drawn from a supply when
S=P and Q=0.
PF angle approaches zero degrees and PF approaches 1,
revealing that the network is appearing more and more
resistive at the input terminals.
38
3.7 Power Factor Correction
Power factor correction is the process of increasing the
power factor without altering the voltage or current to
the original load.
Because most loads are inductive, power factor can be
corrected by installing a capacitor in parallel with the
load.
Inductive load with improved pf
39
3.7 Power Factor Correction
Qc = Q1 – Q2
= P (tan θ1 - tan θ2)
= ωCV2rms
Q1 = S1 sin θ1
= P tan θ1
P = S1 cos θ1
Q2 = P tan θ2
C 
Phasor diagram of
currents relative to V
Qc
P (tan θ1  tan θ 2 )

2
2
ωVrms
ω Vrms
40
3.7 Power Factor Correction
When connected to a 120-Vrms, 60-Hz power line, a load
absorbs 4 kW at a lagging power factor of 0.8. Find the
value of capacitance necessary to raise the pf to 0.95.
41
3.8 Power Measurement
The wattmeter is the instrument for measuring the average
power.
Consists of two coils: the current coil and the voltage coil
A current coil(very low impedance) is connected in series
with the load and responds to the load current
The voltage coil(very high impedance) is connected in
parallel with the load and respond s to the load voltage
The basic structure
42
Equivalent Circuit with load
3.8 Power Measurement
When the two coils are energized, the mechanical inertia of
the moving system produces a deflection angle that is
proportional to the average value of the product v(t)i(t)
If
v(t )  Vm cos(t   v ) and i(t )  I m cos(t   i )
P  Vrms Irms cos (θ v  θi ) 
The basic structure
1
2
Vm Im cos (θ v  θi )
43
Equivalent Circuit with load