Electric Circuits, 9th Edition P40 docx

366 Sinusoidal Steady-State Power Calculations
Example 10.1
Calculating Average and Reactive Power
a) Calculate the average power and the reactive
power at the terminals of the network shown in
Fig. 10.6 if
v = 100 cos (tat + 15") V,
i = 4sin(W - 15°) A.
b) State whether the network inside the box is
absorbing or delivering average power.
c) State whether the network inside the box is
absorbing or supplying magnetizing vars.
'-*•
+
V
Figure 10.6 A A pair of terminals used for calculating power.
Solution
a) Because i is expressed in terms of the sine func-
tion, the first step in the calculation for P and Q
is to rewrite i as a cosine function:
i = 4cos(o>f - 105°) A.
We now calculate P and Q directly from
Eqs.
10.10 and
10.11.
Thus
P = -(100)(4) cos [15 - (-105)] = -100 W,
Q = -100(4) sin [15 - (-105)] = 173.21 VAR.
b) Note from Fig. 10.6 the use of the passive sign
convention. Because of this, the negative value
of -100 W means that the network inside the

box is delivering average power to the terminals.
c) The passive sign convention means that, because
Q is positive, the network inside the box is
absorbing magnetizing vars at its terminals.
^ASSESSMENT PR0BLE
Objective 1—Understand ac power concepts, their relationships to one another, and how to calcuate them in a circuit
10.1 For each of the following sets of voltage and
current, calculate the real and reactive power
in the line between networks A and B in the
circuit shown. In each case, state whether the
power flow is from A to B or vice versa. Also
state whether magnetizing vars are being trans-
ferred from A to B or vice versa.
a) v = 100 cos (at - 45°) V;
i = 20cos(wr + 15°) A.
b) v = 100 cos
(cot
- 45°) V;
i = 20cos(&rf + 165°) A.
c) v =
100
cos (at - 45°) V;
i = 20 cos
(w*
- 105°) A.
d) V = 100 cos at V;
i =
20
cos (art + 120°) A.
A

i
—^-
+
V
B
Answer:
(a)P
Q
500 W
-866.03 VAR
(b) P = -866.03 W
Q = 500 VAR
(c) P = 500 W
Q = 866.03 VAR
(d) P = -500 W
Q = -866.03 VAR
A to B),
B to A);
B to A),
A to B);
A to B),
AtoB);
BtoA),
BtoA).
10.2 Compute the power factor and the reactive fac-
tor for the network inside the box in Fig. 10.6,
whose voltage and current are described in
Example 10.1.
Hint: Use -i to calculate the power and reac-
tive factors.

Answer: pf = 0.5 leading; rf = -0.866.
NOTE: Also try Chapter Problem 10.1,
10.2 Average and Reactive Power
367
Appliance Ratings
Average power is used
to
quantify the power needs
of
household appliances.
The average power rating
and
estimated annual kilowatt-hour consumption
of some common appliances
are
presented
in
Table
10.1.
The
energy con-
sumption values
are
obtained
by
estimating
the
number
of
hours annually

that
the
appliances
are in
use.
For
example,
a
coffeemaker
has an
estimated
annual consumption
of
140 kWh
and an
average power consumption during
operation
of 1.2
kW.
Therefore
a
coffeemaker
is
assumed
to be in
operation
140/1.2,
or
116.67,
hours

per
year,
or
approximately 19 minutes
per
day.
Example
10.2
uses Table
10.1 to
determine whether four common
appliances
can all be in
operation without exceeding
the
current-carrying
capacity
of the
household.
| Making Power
The branch circuit supplying
the
Calculations Involving Household Appliances
: outlets
in a
typical Solution
home kitchen
is
wired with
#12

conductor
and is
protected
by
either
a 20 A
fuse
or a 20 A
circuit
breaker. Assume that
the
following
120 V
appli-
ances
are in
operation
at the
same time:
a cof-
feemaker,
egg
cooker, frying
pan, and
toaster. Will
the circuit
be
interrupted
by the
protective device?

From Table
10.1,
the total average power demanded
by
the
four appliances
is
P
= 1200 + 516 + 1196 4- 146
The total current
in the pro
4058
efr
"
120 '
= 4058
W.
ective device
is
33.82
A.
Yes,
the protective device will interrupt
the
circuit.
TABLE 10.1 Annual Energy Requirements
of
Electric Household Appliances
Appliance
Food preparation

Coffeemaker
Dishwasher
Egg cooker
Frying
pan
Mixer
Oven, microwave (only)
Range, with oven
Toaster
Laundry
Clothes dryer
Washing machine, automatic
Water heater
Quick recovery type
Comfort conditioning
Air conditioner (room)
Dehumidifier
Fan (circulating)
Heater (portable)
Average
Wattage
1,200
1,201
516
1,196
127
1,450
12,200
1,146
4,856

512
2,475
4,474
860
257
88
1,322
NOTE: Assess your understanding of this
Est. kWh
Consumed
Annually
3
140
165
14
100
2
190
596
39
993
103
4,219
4,811
860
b
377
43
176
Appliance

Health and beauty
Hair dryer
Shaver
Sunlamp
Home entertainment
Radio
Television, color, tube type
Solid-state type
Housewares
Clock
Vacuum cleaner
a) Based
on
normal
usage.
When using
such factors as the size
of
the specific
Average
Wattage
600
15
279
71
240
145
2
630
hese figure

appliance,
I
Est. kWh
Consumed
Annually
3
25
0.5
16
86
528
320
17
46
;
for
projections.
he geographical
area
of
use,
and individual usage should be taken into considera-
tion. Note that the wattages are not additive, since
all
units
are
normally not in operation
at
the same time.
b) Based

on
1000 hours
of
operation
pe
r
year.
This "igure will vary
widely depending on the area and the specific size
of
the unit. See
EEI-Pub
#76-2,
"Air Conditioning Usage Study,"
for
an estimate
for your location.
Source: Edison Electric Institute.
material
by
trying Chapter Problem
J0.2.
368 Sinusoidal Steady-State Power Calculations
K, „cos (tot
+ 6
r
) | R
Figure 10.7
•
A

sinusoidal voltage applied
to
the
terminals of a resistor.
10.3 The rms Value and Power
Calculations
In introducing the rms value of a sinusoidal voltage (or current) in
Section 9.1, we mentioned that it would play an important role in power
calculations. We can now discuss this role.
Assume that a sinusoidal voltage is applied to the terminals of a resis-
tor, as shown in Fig. 10.7, and that we want to determine the average
power delivered to the resistor. From Eq. 10.12,
P
= TL -R
dt
h+T
Vl
1
cos
2
(cot +
$
v
)dt
(10.18)
Comparing Eq. 10.18 with Eq. 9.5 reveals that the average power deliv-
ered to R is simply the rms value of the voltage squared divided by R, or
R
(10.19)
If the resistor is carrying a sinusoidal current, say,

I,,,
cos
(cot
+
<£,-),
the
average power delivered to the resistor is
P = IlnJt.
(10.20)
The rms value is also referred to as the effective value of the sinu-
soidal voltage (or current). The rms value has an interesting property:
Given an equivalent resistive load, R, and an equivalent time period, T,
the rms value of a sinusoidal source delivers the same energy to R as does
a dc source of the same value. For example, a dc source of 100 V delivers
the same energy in T seconds that a sinusoidal source of 100 V
nns
delivers,
assuming equivalent load resistances (see Problem 10.12). Figure 10.8
demonstrates this equivalence. Energywise, the effect of the two sources
is identical. This has led to the term effective value being used inter-
changeably with rms value.
The average power given by Eq. 10.10 and the reactive power given
by Eq. 10.11 can be written in terms of effective values:
V I
P =
cos
(0
V
- 0,-)
V

I
= —•=—7= cos (0,,
- Bs)
V2 V2
K
'
= K-ff'eff cos (fl„
-0,.);
(10.21)
lOOV(rms)
R
V
s
= 100V(dc)
R
Figure 10.8
A
The effective value
of v,
(100
V
rms) delivers the
same power
to R
as the dc voltage V
s
(100
V
dc).
10.3 The rms Value and Power Calculations 369

and, by similar manipulation,
Q =
V
cff
I
cff
sm($
v
-ed.
(10.22)
The effective value of the sinusoidal signal in power calculations is so
widely used that voltage and current ratings of circuits and equipment
involved in power utilization are given in terms of rms values. For exam-
ple,
the voltage rating of residential electric wiring is often 240 V/120 V
service. These voltage levels are the rms values of the sinusoidal voltages
supplied by the utility company, which provides power at two voltage lev-
els to accommodate low-voltage appliances (such as televisions) and
higher voltage appliances (such as electric ranges). Appliances such as
electric lamps, irons, and toasters all carry rms ratings on their nameplates.
For example, a 120 V, 100 W lamp has a resistance of 120
2
/100, or 144 ft,
and draws an rms current of 120/144, or 0.833 A. The peak value of the
lamp current is 0.833
V2~,
or 1.18 A.
The phasor transform of a sinusoidal function may also be expressed
in terms of the rms
value.

The magnitude of the rms phasor is equal to the
rms value of the sinusoidal function. If a phasor is based on the rms value,
we indicate this by either an explicit statement, a parenthetical "rms" adja-
cent to the phasor quantity, or the subscript
"eff,"
as in Eq.
10.21.
In Example
10.3,
we illustrate the use of rms values for calculating power.
Example 10.3
Determining Average Power Delivered to a Resistor by Sinusoidal Voltage
a) A sinusoidal voltage having a maximum ampli-
tude of 625 V is applied to the terminals of a
50 fl resistor. Find the average power delivered
to the resistor.
b) Repeat (a) by first finding the current in the
resistor.
Solution
a) The rms value of the sinusoidal voltage is
625/V2,
or approximately 441.94 V. From
Eq. 10.19, the average power delivered to the
50
Cl
resistor is
P =
(441.94)2
50
= 3906.25 W.

b) The maximum amplitude of the current in the
resistor is 625/50, or 12.5 A. The rms value of
the current is 12.5/V2, or approximately
8.84 A. Hence the average power delivered to
the resistor is
P = (8.84)
2
50 = 3906.25 W.
i/ASSESSMENT PROBLEM
Objective 1—Understand ac power concepts, their relationships to one another, and how to calculate them in a drcuit
10.3 The periodic triangular current in Example 9.4,
repeated here, has a peak value of 180 mA.
Find the average power that this current deliv-
ers to a 5 kQ, resistor.
Answer: 54 W.
NOTE: Also try Chapter Problem 10.15.
370 Sinusoidal Steady-State Power Calculations
10.4 Complex Power
Complex power •
TABLE 10.2 Three Power Quantities and
Their Units
Quantity
Complex power
Average power
Reactive power
Units
volt-amps
watts
var
|.V|

- apparent power
reactive power
P - average power
Figure 10.9 •
A
power triangle.
Before proceeding to the various methods of calculating real and reactive
power in circuits operating in the sinusoidal steady state, we need to intro-
duce and define complex power. Complex power is the complex sum of
real power and reactive power, or
S = P + jQ.
(10.23)
As you will see, we can compute the complex power directly from the volt-
age and current phasors for a circuit. Equation 10.23 can then be used to
compute the average power and the reactive power, because P =
!R
{S}
and<2 = 3{S}.
Dimensionally, complex power is the same as average or reactive
power. However, to distinguish complex power from either average or
reactive power, we use the units volt-amps (VA).Thus we use volt-amps for
complex power, watts for average power, and vars for reactive power, as
summarized in Table 10.2.
Another advantage of using complex power is the geometric interpre-
tation it provides. When working with Eq.
10.23,
think of P, Q, and \S\ as
the sides of a right triangle, as shown in Fig. 10.9. It is easy to show that the
angle 6 in the power triangle is the power factor angle 0
I}

— 0,. For the
right triangle shown in Fig. 10.9,
Q
tan0 = |.
(10.24)
But from the definitions of P and Q (Eqs. [10.10] and [10.11 J, respectively),
Q _ (V
m
IJ2)sm(d
v
-dd
P (V
m
l
m
/2)cos(0
v
-ed
tan (0
V
- 0,-).
(10.25)
Therefore, 0 = 0
V
-
0,
The geometric relations for a right triangle mean
also that the four power triangle dimensions (the three sides and the
power factor angle) can be determined if any two of the four are known.
The magnitude of complex power is referred to as apparent power.

Specifically,
Apparent power •
\S\ = 2 P
2
+ Q
2
(10.26)
Apparent power, like complex power, is measured in volt-amps. The
apparent power, or volt-amp, requirement of a device designed to convert
electric energy to a nonelectric form is more important than the average
power requirement. Although the average power represents the useful
output of the energy-converting device, the apparent power represents the
volt-amp capacity required to supply the average power. As you can see
from the power triangle in Fig. 10.9, unless the power factor angle is 0°
(that is, the device is purely resistive, pf = 1, and Q = 0), the volt-amp
capacity required by the device is larger than the average power used by
the device. As we will see in Example 10.6, it makes sense to operate
devices at a power factor close to 1.
Many useful appliances (such as refrigerators, fans, air conditioners,
fluorescent lighting fixtures, and washing machines) and most industrial
loads operate at a lagging power factor. The power factor of these loads
sometimes is corrected either by adding a capacitor to the device itself or
10.5 Power Calculations 371
by connecting capacitors across the line feeding the load; the latter
method is often used for large industrial loads. Many of the Chapter
Problems give you a chance to make some calculations that correct a lag-
ging power factor load and improve the operation of a circuit.
Example 10.4 uses a power triangle to calculate several quantities
associated with an electrical load.
Example 10.4

Calculating Complex Power
An electrical load operates at 240 V rms. The load
absorbs an average power of 8 kW at a lagging
power factor of 0.8.
a) Calculate the complex power of the load.
b) Calculate the impedance of the load.
Solution
a) The power factor is described as lagging, so we
know that the load is inductive and that the
algebraic sign of the reactive power is positive.
From the power triangle shown in Fig. 10.10,
P =
\S\
cos 0,
Q =
\S\
sin 0.
cos
0
- 0.8, sin
0
= 0.6.
Now, because
Therefore
Q =
SkW
cos0
10 sine
0.8
6 kVAR,

= lOkVA,
and
5 = 8 + /6 kVA.
b) From the computation of the complex power of
the load, we see that P = 8 kW. Using Eq.
10.21,
= K>f
f
4ffcos(0
t;
= (240)/
eff
(0.8)
= 8000 W.
0i)
Solving for /
efr
,
/
cff
= 41.67 A.
We already know the angle of the load imped-
ance,
because it is the power factor angle:
0 = cos
-1
(0-8) = 36.87°.
We also know that 0 is positive because the
power factor is lagging, indicating an inductive
load. We compute the magnitude of the load

impedance from its definition as the ratio of the
magnitude of the voltage to the magnitude of
the current:
\Z\ =
lK-a-1
l/effl
240
41.67
5.76.
Hence,
Z = 5.76 /36.87° D, = 4.608 + y'3.456 O.
Figure 10.10 • A power triangle.
10.5 Power Calculations
We are now ready to develop additional equations that can be used to cal-
culate real, reactive, and complex power. We begin by combining Eqs. 10.10,
10.11,
and 10.23 to get
VI VI
S =
~Y~cos
(0
V
- 6,) +
j—^—sm(e
v
- $i)
V I
r
in
1

m
[cos (0,,- 0i) +; sin
(0,,
- 0,)]
iggefflrtt o \v
m
I
m
/{6
n
- 9d.
(10.27)
372 Sinusoidal Steady-State Power Calculations
If we use the effective values of the sinusoidal voltage and current,
Eq. 10.27 becomes
S = KffW(0, ~ 0/). (10.28)
Equations 10.27 and 10.28 are important relationships in power calcula-
tions because they show that if the phasor current and voltage are known at
a pair of terminals, the complex power associated with that pair of terminals
is either one half the product of the voltage and the conjugate of the cur-
rent, or the product of the rms phasor voltage and the conjugate of the rms
phasor current. We can show this for the rms phasor voltage and current in
Fig. 10.11 as follows:
=
V
ea
ei
e
>-I
cii

e-!
9
>
Complex powers =
\
cfi
l*^
(10.29)
Note that left = h&
eJ$i
follows from Euler's identity and the trigonomet-
ric identities cos(—0) = cos(fl) and sin(-0) =
—
sin (#):
I^e*
- /
cff
cos (Si) + //
eff
sin
(-0,-)
= /
eff
cos (0/) -
jl
ei{
sin (6i)
=
Ifo
The same derivation technique could be applied to Eq. 10.27 to yield

S = -VI*. (10.30)
Both Eqs. 10.29 and 10.30 are based on the passive sign convention. If the
current reference is in the direction of the voltage rise across the termi-
nals,
we insert a minus sign on the right-hand side of each equation.
To illustrate the use of Eq. 10.30 in a power calculation, let's use the
same circuit that we used in Example
10.1.
Expressed in terms of the pha-
sor representation of the terminal voltage and current,
V = 100 /15° V,
I = 4/-105° A.
Therefore
S = -(100 /15°)(4 / + 105°) = 200 /120°
•ell
•*-
4-
V
t
.ff
.
Circuit
Figure 10.11 • The phasor voltage and current associ-
ated with a pair of terminals.
= -100 + /173.21 VA.
10.5 Power Calculations 373
Once we calculate the complex power, we can read off both the real and
reactive powers, because S = P + jQ. Thus
P = -100 W,
Q = 173.21 VAR.

The interpretations of the algebraic signs on P and Q are identical to those
given in the solution of Example 10.1.
let
+
v
cff
—
z
1
Figure 10.12 A
The
general circuit of
Fig.
10.11
replaced with an equivalent impedance.
V
cf
f =
Zltf.
(10.31)
Substituting Eq. 10.31 into Eq. 10.29 yields
S - ZI
efr
Ie[f
= Heff|
2
Z
= |I
eff
|

2
(fl + IX)
= HeffPi? + /|I
cff
|
2
X = P + jQ, (10.32)
from which
P = lleffl
2
^ = \l\fr (10.33)
Q =
Ihd
2
*
= \l\X- (10.34)
In Eq. 10.34, X is the reactance of either the equivalent inductance or
equivalent capacitance of the circuit. Recall from our earlier discussion of
reactance that it is positive for inductive circuits and negative for capaci-
tive circuits.
A second useful variation of Eq. 10.29 comes from replacing the cur-
rent with the voltage divided by the impedance:
S = Veff(^y = ^ = P + jQ- (10.35)
Alternate Forms for Complex Power
Equations 10.29 and 10.30 have several useful variations. Here, we use the
rms value form of the equations, because rms values are the most common
type of representation for voltages and currents in power computations.
The first variation of Eq. 10.29 is to replace the voltage with the prod-
uct of the current times the impedance. In other words, we can always rep-
resent the circuit inside the box of Fig. 10.11 by an equivalent impedance,

as shown in Fig.
10.12.
Then,
374 Sinusoidal Steady-State Power Calculations
Note that if Z is a pure resistive element
P =
and if Z is a pure reactive element,
Q =
R '
|V
eff
|
2
(10.36)
X
(10.37)
In Eq. 10.37, X is positive for an inductor and negative for a capacitor.
The following examples demonstrate various power calculations in
circuits operating in the sinusoidal steady state.
Example 10.5
Calculating Average and Reactive Power
In the circuit shown in Fig.
10.13,
a load having an
impedance of 39 + /26 O is fed from a voltage
source through a line having an impedance of
1 + /4 O. The effective, or rms, value of the source
voltage is 250 V.
a) Calculate the load current I
L

and voltage V
L
.
b) Calculate the average and reactive power deliv-
ered to the load.
c) Calculate the average and reactive power deliv-
ered to the line.
d) Calculate the average and reactive power sup-
plied by the source.
Thus the load is absorbing an average power of
975 W and a reactive power of 650 VAR.
in
/4 a
JTYY\ »_
6
D
250/0°
V(rms)
Line Source
Figure 10.13 • The circuit for Example 10.5.
3912
II
/26 ft;
Load
Solution
a) The line and load impedances are in series across
the voltage source, so the load current equals the
voltage divided by the total impedance, or
I,
250 /0°

40 + /30
= 4 - /3 = 5 /-36.87° A (rms).
Because the voltage is given in terms of its
rms value, the current also is rms. The load volt-
age is the product of the load current and load
impedance:
V
L
= (39 + /26)1
L
= 234 -/13
= 234.36 /-3.18° V (rms).
b) The average and reactive power delivered to the
load can be computed using Eq.
10.29.
Therefore
S = \Jl = (234 -/13)(4 +/3)
= 975 + /650 VA.
c) The average and reactive power delivered to the
line are most easily calculated from Eqs. 10.33
and 10.34 because the line current is known. Thus
P = (5)
2
(1) = 25 W,
Q = (5)2(4) = 100 VAR.
Note that the reactive power associated with the
line is positive because the line reactance is
inductive.
d) One way to calculate the average and reactive
power delivered by the source is to add the com-

plex power delivered to the line to that delivered
to the load, or
S = 25 + /100 + 975 + /650
= 1000 +/750VA.
The complex power at the source can also be cal-
culated from Eq. 10.29:
S
s
= -250IL-
10.5 Power Calculations 375
The minus sign is inserted in Eq. 10.29 whenever
the current reference is in the direction of a volt-
age
rise.
Thus
5,
= -250(4 + /3) = -(1000 4- /750) VA.
The minus sign implies that both average power
and magnetizing reactive power are being deliv-
ered by the source. Note that this result agrees
with the previous calculation of 5, as it must,
because the source must furnish all the average
and reactive power absorbed by the line and load.
Example 10.6
Calculating Power in Parallel Loads
The two loads in the circuit shown in Fig. 10.14 can
be described as follows: Load 1 absorbs an average
power of
8
kW at a leading power factor of

0.8.
Load
2 absorbs 20 kVA at a lagging power factor of 0.6.
0.05 n
• VW-
+
/0.50 a
v,
250/0°
V (rms)
I., L
Figure 10.14 • The circuit for Example 10.6.
a) Determine the power factor of the two loads in
parallel.
b) Determine the apparent power required to sup-
ply the loads, the magnitude of the current, I
v
,
and the average power loss in the transmission
line.
c) Given that the frequency of the source is 60 Hz,
compute the value of the capacitor that would
correct the power factor to 1 if placed in parallel
with the two loads. Recompute the values in (b)
for the load with the corrected power factor.
Solution
a) All voltage and current phasors in this problem
are assumed to represent effective values. Note
from the circuit diagram in Fig. 10.14 that
I

v
= l
{
+ I
2
. The total complex power absorbed
by the two loads is
S = (250)i;
=
(250)(1,
+ I
2
)
8
=
(250)1^
+ (250)I
2
=
s
x
+ s
2
.
We can sum the complex powers geometrically,
using the power triangles for each load, as shown
in Fig.
10.15.
By hypothesis,
8000(.6)

= 8000 - /6000 VA,
S
2
= 20,000(.6) + /20,000(.8)
= 12,000 +/16,000 VA.
-36.87°
16kVAR
10
kVAR
Figure 10.15 • (a)
The power
triangle for load 1. (b) The
power triangle for load 2. (c) The sum of the power triangles.
It follows that
S = 20,000 + /10,000 VA,
and
, 20,000 + /10,000
Therefore
I
v
= 80 - /40 = 89.44 /-26.57° A.
Thus the power factor of the combined load is
pf = cos(0 + 26.57°) = 0.8944 lagging.
The power factor of the two loads in parallel is
lagging because the net reactive power is positive.
b) The apparent power which must be supplied to
these loads is
\S\ = |20 + /10| = 22.36 kVA.
The magnitude of the current that supplies this
apparent power is

II.sl = 180 - j40| = 89.44 A.