Chapter 10
Engr228
Circuit Analysis
Dr Curtis Nelson
Chapter 10 Objectives
• Understand the following ac power concepts, their
relationships to one another, and how to calculate them in
a circuit:
• Instantaneous power;
• Average (real) power;
• Root Mean Squared (RMS) value;
• Reactive power;
• Complex power;
• Power factor; and
• Power factor correction.
• Understand the condition for maximum real power
delivered to a load in an AC circuit and be able to
calculate the load impedance required to deliver maximum
real power to the load.
Engr228 - Chapter 10, Nilsson 9E
1
Instantaneous Power
• Instantaneous power is the power measured at any given
instant in time.
• In DC circuits, power is measured in watts as
P = vi = i 2 R =
v2
R
• In AC circuits, voltage and current are time-varying, so
instantaneous power (also measured in watts) is also timevarying. Power is still measured as
P = vi = i 2 R =
v2
R
Instantaneous AC Power
p( t ) = v( t ) × i( t )
• In an AC circuit, voltage and current are expressed in
general form as
v ( t ) = V m cos( ω t + θ v )
i ( t ) = I m cos( ω t + θ i )
• Instantaneous power is then
p (t ) = Vm I m cos(ωt + θ v ) cos(ωt + θ i )
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Instantaneous Power - continued
p (t ) = Vm I m cos(ωt + θ v ) cos(ωt + θ i )
• Using the trig identity
cos( x) cos( y ) = 0.5 cos( x − y ) + 0.5 cos( x + y )
it can be shown that instantaneous power is
p (t ) =
Vm I m
{cos( θ v − θ i ) + cos( 2ω t + θ v + θ i )}
2
• Note that the instantaneous power contains a constant
term as well as a component that varies with time at
twice the input frequency.
Instantaneous Power Plot
p (t ) =
Engr228 - Chapter 10, Nilsson 9E
Vm I m
{cos( θ v − θ i ) + cos( 2ω t + θ v + θ i )}
2
3
Instantaneous Power - continued
p (t ) =
Vm I m
{cos( θ v − θ i ) + cos( 2ω t + θ v + θ i )}
2
• The equation above can be further manipulated as follows:
cos(α + β ) = cosα cos β − sin α sin β
cos(2ωt +θv − θi ) = cos(θv − θi )cos(2ωt ) − sin(θv − θi )sin(2ωt )
p (t ) = V m I m cos(θv − θi ) + V m I m cos(θv − θi )cos(2ωt ) −V m I m sin(θv − θi )sin(2ωt )
2
2
2
• Note that the instantaneous power still contains a constant
term as well as components that vary with time at twice the
input frequency.
Instantaneous Power Plot
p (t ) = V m I m cos(θv − θi ) + V m I m cos(θv − θi )cos(2ωt ) −V m I m sin(θv − θi )sin(2ωt )
2
2
2
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Instantaneous vs. Average Power
• While instantaneous power is an interesting mathematical
development for AC circuits, it is not all that useful for
determining how much power a device, circuit, or system
generates or absorbs since it varies with time.
• Average Power is a better indicator of power in AC circuits.
Average power is sometimes called real power, because it
describes the power in a circuit that is transformed from
electric to nonelectric energy.
Average Power
The average power over an arbitrary interval from t1 to t2 is
where p(t) is the instantaneous power. When the power is
periodic with period T, the average power is calculated over any
one period:
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Average Power - continued
p (t ) = V m I m cos(θv − θi ) + V m I m cos(θv − θi )cos(2ωt ) −V m I m sin(θv − θi )sin(2ωt )
2
2
2
• To calculate average power, note that the last two terms in the
equation above will integrate to zero since the average value
of sin and cosine signals over one period is zero. Therefore
average power in AC circuits is calculated as
PAVG =
Vm I m
cos(θ v − θ i )
2
Average Power for Elements
PAVG =
Vm I m
cos(θ v − θ i )
2
• Since the voltage and current are in phase across a resistor, the
average power absorbed by the resistor R is
PR , AVG
1 Vm2
=
2 R
• The average power absorbed by a purely reactive element is
zero, since the current and voltage are 90 degrees out of phase:
PC , AVG = PL , AVG = 0
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Example: Power Calculations
• Calculate the instantaneous and average power if
v (t) = 80cos(10t + 20º)V
i(t) = 15cos(10t + 60º)A
Pinst (t) = 459.6 + 600cos(20t + 80º)W
Pavg = 459.6W
Textbook P. 429 (Hayt 8th)
Find the average power absorbed by each element.
Answer:
Engr228 - Chapter 10, Nilsson 9E
Pind=0 W
Pleft=-50 W
Pres=25 W
Pcap=0 W
Pright=25 W
7
Example: Average Power in a Resistor
Find the average power dissipated in a resistor R if the applied
voltage is v(t) = Asinθ.
The average power is found by integrating from 0 to 2π
p=
v 2 A2 sin 2 θ
=
R
R
P=
1
2π
2π
∫
2 2π
0
A2
sin 2 θ dθ
R
2π
=
A
A2
2
θ
θ
sin
d
=
2πR ∫0
2π
=
A2  1
sin 2θ 
A2
=
 θ−

2πR  2
4 θ =0
2R
1 − cos 2θ
dθ
2
0
∫
θ = 2π
Example: Average Power
• Calculate the average power absorbed by the resistor and
inductor. Find the average power supplied by the voltage source.
PR = 9.6W
PL = 0W
PSource = 9.6W
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Example: Average Power
Find the average power being delivered to an impedance ZL=
8 − j11 Ω by a current I= 5ej20° A.
Only the 8-Ω resistance enters the average-power calculation,
since the j11-Ω component will not absorb any average
power.
Thus,
P = (1/2)(52)8 = 100 W
Effective or RMS Value
• Sometimes using average power values can be confusing.
For instance, the average DC power absorbed by a resistor
is P = VM IM while the average AC power is P = VM IM /2
By introducing a new concept, called the effective value
of a periodic waveform, the formulas for the average
power absorbed by a resistor can be made the same for dc,
sinusoidal, or any general periodic waveform.
• The effective value of a periodic current or voltage is the
DC current or voltage that delivers the same average
power to a resistor as the periodic AC current or AC
voltage.
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Effective or RMS Value
• For any periodic function x(t) in general, the effective or
root mean squared (rms) value, is given by:
X rms =
1 T 2
x dt
T ∫0
Effective or RMS Value
• For example, the effective value of I = Imcosωt is
1 T 2
I m cos 2 ωtdt
∫
T 0
I rms =
I rms
I 2m
=
T
∫
T
0
1
I
(1 + cos 2ωt )dt = m
2
2
• Average power can then be written in two ways:
PAVG =
Vm I m
cos(θ v − θ i )
2
PAVG = Vrms I rms cos(θ v − θ i )
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Circuit Elements - Resistors
• In an ideal resistor, V and I are in phase so θv = θi.
2
PAVG
Veff
V I
V I
2
= m m cos 0 0 = m m = Veff I eff =
= (I eff ) R
2
2
R
• Average power is dissipated only in resistive elements.
Effective Values of Common Waveforms
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11
Example: RMS Power
Determine the effective (rms) value of the current waveform
in the figure below. If the current is passed through a 2Ω
resistor, find the average power absorbed by the resistor.
i(t)
10
0
2
4
6
8
10
t
-10
Irms = 8.165A
P = 133.3W
Reactive Power
• The average power in capacitors and inductors = 0 because
voltage and current have a 90º phase difference.
• However, there is still voltage across, and current through,
inductors and capacitors so they are part of the overall
power picture. We now investigate the part that inductors
and capacitors play in the power picture for general
circuits, leading us to define a new category called
Complex power (S).
• First, we must define the components of complex power.
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Apparent Power
• The apparent power (S) is the magnitude of the product of
the rms voltage and current. It is measured in volt-amperes
(VA) to distinguish it from the other types of power.
PAVG = Vrms I rms cos(θ v − θ i )
PAVG = S cos(θ v − θ i )
S = Vrms I rms
• S is known as the apparent power.
• cos(θV - θi) is know as the power factor.
Power Factor
• Power factor is the cosine of the phase difference between
voltage and current. It is also the cosine of the angle of the
load impedance (Ohm’s Law).
Power Factor = cos(θ v − θ i )
where
θ v − θ i is the Power Factor Angle
• For a resistive load, PF=1.
• For a purely reactive load, PF=0.
• Generally, 0 ≤ PF ≤ 1
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Leading and Lagging Power Factor
• pf is lagging if the current lags the voltage (θV - θI is positive
like in an inductive load).
• pf is leading if the current leads the voltage (θV - θI is
negative like in a capacitive load).
Textbook P. 440 (Hayt 8th)
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.
Pavg2 = 288 W
Pavg1 = 144 W
Papp = S = 720 VA
PF = 0.6 (lagging)
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Apparent Power & Power Factor
• Power factor can also be defined as:
PF =
average power
apparent power
=
Pavg
Veff I eff
which leads us to a graphical depiction of Complex Power.
Complex Power
Complex power is the product of the rms voltage phasor and the
complex conjugate of the rms current phasor. As a complex
quantity, its real part is real power P (watts) and its imaginary
part is reactive power Q (vars).
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Complex Power
• 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).
1
S = P + jQ = VI * = Vrms I rms∠θv − θi
2
*
= VrmsI rms
2
S = I rms
Z=
2
Vrms
Z*
Complex Power Triangle
S
Q
θ
IZI
θ
P
R
Power triangle
Impedance triangle
pf = cos θ =
Engr228 - Chapter 10, Nilsson 9E
X
P R
=
S Z
16
Complex Power - Summary
Real (Average) power:
P = Re(S) = S cos(θ v − θi ) Watts
Reactive power:
Q = Im(S) = S sin(θ v − θi )
VAR
Apparent power:
S = S = Vrms I rms = P 2 + Q 2 VA
Complex power:
S = P + jQ = VRMS I * RMS
VA
• Q = 0 for resistive loads (unity power factor).
• Q < 0 for capacitive loads (leading power factor).
• Q > 0 for inductive loads (lagging power factor).
Example: Power Factor
Calculate the power factor of the circuit below as seen by the
source. What is the average power supplied by the source?
pf = 0.936 (lagging)
PAVG = 118W
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Example: Apparant Power and Power Factor
• A series connected RC load draws a current of
i (t ) = 4 cos(100πt + 10° ) A
when the applied voltage is
v(t ) = 120 cos(100πt − 20° )V
• Find the apparent power and the power factor of the load.
Determine the element values that form the load.
S = 240 V
pf = 0.866 (leading)
C = 212.2µF
R = 26.0Ω
Complex Power
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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Complex Power
Complex powers to loads add:
S = VI* = V(I1 + I2 )* = V(I1* + I*2 ) = S1 + S2
Power Factor Correction
• The process of increasing the power factor without
altering the voltage or current to the original load is known
as power factor correction.
• Most loads are inductive. The power factor of the load can
be improved (to make it closer to unity) by installing a
capacitor in parallel with an inductive load.
a) Original inductive load
Engr228 - Chapter 10, Nilsson 9E
b) Inductive load with
improved power factor
19
Power Factor Correction Capacitors
Power Factor Correction Capacitors
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Power Factor Correction by Power Triangle
• The triangle below shows that the reactive power (Q1) in
a large inductive load can be reduced by an amount (QC)
by adding a capacitor in parallel to the load.
P = S1cosθ1
Q1 = Ptanθ1
Q2 = Ptanθ2
QC = Q1 – Q2
= P(tan θ1 – tan θ2)
Power Factor Correction by Power Triangle
• But QC is also the amount of reactive power the
capacitor must provide
QC =
2
Vrms
2
= ωCVrms
XC
• The value of required shunt
capacitance is then:
C=
QC
P(tan θ1 − tan θ2 )
=
2
2
ωVrms
ωVrms
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Textbook Problem 11.42 Hayt 7E
1. Find the power factor at which the source is operating.
2. Find PAVG being supplied by the source.
3. What size capacitor should be placed in parallel with the
source to cause its power factor to be unity?
pf = 0.8969 (lagging)
PAVG = 992W
C = 90.09µF
Example: PF Correction #1
When connected to a 120 VRMS, 60Hz power line, a load
absorbs 4kW at a lagging power factor of 0.8. Find the value
of capacitance necessary to raise the pf to 0.95.
C = 310.4µF
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Example: PF Correction #2
Find the value of parallel capacitance needed to correct a
load of 140kVAR at 0.85 lagging pf to unity pf. Assume
that the load is supplied by a 110VRMS, 60Hz line.
C = 30,691µF
Textbook Problem 11.46 Hayt 7E
The load below draws 10 kVA at a PF = 0.8 lagging. If the
magnitude of the load current IL = 40ARMS, what must be the
value of C to cause the source to operate at a PF = 0.9 lagging?
C = 79.48µF
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Textbook Problem 11.52 Hayt 7E
A source of 230 VRMS is supplying three loads in parallel:
1.2kVA at a lagging PF of 0.8, 1.6kVA at a lagging PF of 0.9,
and 900W at unity PF. Find
(a) The amplitude of the source current.
(b) The PF at which the source is operating.
(c) The complex power being furnished by the source.
(a) IS = 15.62 A rms
(b) PFS = 0.9188
(c) S = 3300 + j1417 VA
Maximum Power Transfer
An independent voltage source in series with an impedance Zth
delivers a maximum average power to that load impedance ZL
when it is the conjugate of Zth.
ZL = Zth*
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Maximum Power Transfer Derivation
First, solve for the load power:
| Vth |2 RL
=
(Rth + RL )2 + (X th + X L )2
1
2
Clearly, P is largest when XL+Xth=0.
Solving dP/dRL=0 will show that RL=Rth.
Maximum Average Power Transfer
• In rectangular form, the Thevenin impedance and Load
impedance are:
Z Th = R Th + jX Th
Z L = R L + jX L
• For maximum average power transfer, the load impedance
ZL must be equal to the complex conjugate of the Thevenin
impedance ZTh.
Z L = R L + jX L = R Th − jX Th = Z*Th
Z L = Z*Th
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Maximum Average Power Transfer
• When ZL = Z*TH, the maximum power delivered to the load
can be shown to be
Pmax
V 2TH , RMS
=
4 RTh
• If the load is purely resistive, then the value of RL that
maximizes PL is the magnitude of the Thevenin impedance.
• Note that the 4 factor in the denominator above becomes 8 if
the voltage is expressed as an average or peak value.
Example: Maximum Average Power #1
Determine the load impedance ZL that maximizes the average
power drawn from the circuit below. Calculate the maximum
average power.
Z L = Z *Th = 2.9333 − j 4.4667Ω
Pmax = 2.3675W
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Example: Maximum Average Power #2
Find the load impedance ZL that absorbs the maximum
average power for the circuit shown below. Calculate the
maximum average power.
Z L = Z *Th = 3.415 − j 0.7317Ω
Pmax = 1.429W
Chapter 10 Summary
• Understand the following ac power concepts, their
relationships to one another, and how to calculate them in
a circuit:
• Instantaneous power;
• Average (real) power;
• Root Mean Squared (RMS) value;
• Reactive power;
• Complex power;
• Power factor; and
• Power factor correction.
• Understand the condition for maximum real power
delivered to a load in an AC circuit and be able to
calculate the load impedance required to deliver maximum
real power to the load.
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