Example 10.7 Balancing Power Delivered with Power Absorbed in an ac Circuit a Calculate the total average and reactive power delivered to each impedance in the circuit shown in Fig.. Ma
Trang 1376 Sinusoidal Steady-State Power Calculations
The average power lost in the line results from
the current flowing through the line resistance:
* U = | I J2* = (89.44)2(0.05) = 400 W
Note that the power supplied totals 20,000 + 400
= 20,400 W, even though the loads require a
total of only 20,000 W
c) As we can see from the power triangle in
Fig 10.15(c), we can correct the power factor to 1
if we place a capacitor in parallel with the existing
loads such that the capacitor supplies 10 kVAR
of magnetizing reactive power The value of the
capacitor is calculated as follows First, find the
capacitive reactance from Eq 10.37:
X = ivy2
Q
(250)2 -10,000
the power factor is 1, the apparent power and the average power are the same, as seen from the power triangle in Fig 10.16(c) Therefore, the apparent power once the power factor has been corrected is
\S\ = P = 20 kVA
The magnitude of the current that supplies this apparent power is
20,000
250 = 80 A
The average power lost in the line is thus reduced to
Pimc = \h\ 2 R = (80)2(0.05) = 320 W
Now, the power supplied totals 20,000 + 320
= 20,320 W Note that the addition of the capaci-tor has reduced the line loss from 400 W to 320 W
= -6.25 a
Recall that the reactive impedance of a capacitor
is -l/o)C, and w = 2TT(60) = 376.99 rad/s, if
the source frequency is 60 Hz Thus,
o)X (376.99)(-6.25) 424.4 ^ F
The addition of the capacitor as the third load is
represented in geometric form as the sum of the
two power triangles shown in Fig 10.16 When
22.36 k V A ^ ~
-^ -^ t
20 kW (a)
lOkVAR +
20 kW (c)
- l O k V A R
(b)
Figure 10.16 A (a) The sum of the power triangles for loads 1 and 2 (b) The power triangle for a 424.4 ^tF capacitor at 60 Hz (c) The sum of the power triangles in (a) and (b)
Example 10.7 Balancing Power Delivered with Power Absorbed in an ac Circuit
a) Calculate the total average and reactive power
delivered to each impedance in the circuit shown
in Fig 10.17
b) Calculate the average and reactive powers
asso-ciated with each source in the circuit
c) Verify that the average power delivered equals
the average power absorbed, and that the
izing reactive power delivered equals the
magnet-izing reactive power absorbed
'VW orv-v->
i a / 2 0 +
-/16 n ;
+ v
•—wv
in
| ,
/3 0
39 I t
V s = 150/0°V
V, = (78 - /104) V Ij = ( - 2 6 - /52) A
V 2 = (72 + /104) V l v = ( - 2 + /6) A
V 3 = (150 - /130) V I 2 = ( - 2 4 - /58) A Figure 10.17 A The circuit, with solution, for Example 10.7
Trang 2Solution
a) The complex power delivered to the (1 + /2) ft
impedance is
Si = | v , I I = /», + /Qi
= ^(78 - /104)(-26 + /52)
= -(3380 + /6760)
= 1690 +/3380 V A
Thus this impedance is absorbing an average
power of 1690 W and a reactive power of
3380 VAR The complex power delivered to the
(12 - /16) ft impedance is
S 2 = 2V2^ = Pi + jQi
= -(72 + /104)(-2 - /6)
= 240 - /320 VA
Therefore the impedance in the vertical branch
is absorbing 240 W and delivering 320 VAR The
complex power delivered to the (I + /3) ft
impedance is
S 3 = ~V3IS = P 3 + jQ 3
= -(150 - /130)(-24 + /58)
= 1970 + /5910 V A
This impedance is absorbing 1970 W and
5910 VAR
b) The complex power associated with the inde-pendent voltage source is
ss = -|v,ii = ps + / a
= -^(150)(-26 + /52)
= 1950 - /3900 VA
Note that the independent voltage source is absorbing an average power of 1950 W and delivering 3900 VAR The complex power asso-ciated with the current-controlled voltage source is
S x = |(39Ir)(I|) = P x + jQ x
= | ( - 7 8 + /234)(-24 + /58)
= -5850 - /5070 VA
Both average power and magnetizing reactive power are being delivered by the dependent source
c) The total power absorbed by the passive imped-ances and the independent voltage source is
Absorbed = P\ + Pi + i>3 + A = 5 8 5 0 W
The dependent voltage source is the only circuit element delivering average power Thus
delivered = 5 8 5 0 W
Magnetizing reactive power is being absorbed
by the two horizontal branches Thus
^absorbed = Q\ + Qi = 9 2 9 0 V A R
Magnetizing reactive power is being delivered
by the independent voltage source, the capacitor
in the vertical impedance branch, and the dependent voltage source Therefore
Qdelivered = 9 2 9 0 V A R
Trang 3378 Sinusoidal Steady-State Power Calculations
I/ASSESSMENT PROBLEMS
Objective 1—Understand ac power concepts, their relationships to one another, and how to calculate them in a circuit
10.4 The load impedance in the circuit shown is
shunted by a capacitor having a capacitive
reac-tance of - 5 2 0, Calculate:
a) the rms phasors VL and IL,
b) the average power and magnetizing reactive
power absorbed by the (39 + /26) £1 load
impedance,
c) the average power and magnetizing reactive
power absorbed by the (1 + /4) O line
impedance,
d) the average power and magnetizing reactive
power delivered by the source, and
e) the magnetizing reactive power delivered by
the shunting capacitor
l a
• A W
-/4 0
- T T T A
O250/0!
V(rms)
+
V,
39 n
./26 a
(c) 23.52 W, 94.09 VAR;
(d) 1152.62 W, -376.36 VAR;
(e) 1223.18 VAR
10.5 The rms voltage at the terminals of a load is
250 V The load is absorbing an average power
of 40 kW and delivering a magnetizing reactive power of 30 kVAR Derive two equivalent impedance models of the load
Answer: 1 H in series with 0.75 Q, of capacitive
reactance; 1.5625 H in parallel with 2.083 H
of capacitive reactance
10.6 Find the phasor voltage Y s (rms) in the circuit
shown if loads L x and L 2 are absorbing 15 kVA
at 0.6 pf lagging and 6 kVA at 0.8 pf leading, respectively Express Vv in polar form
Answer: (a) 252.20 / - 4 5 4 ° V (rms),
5.38 / - 3 8 2 3 ° A (rms);
(b) 1129.09 W, 752.73 VAR;
NOTE: Also try Chapter Problems 10.18,10.26, and 10.27
/ 1 0 + ( _ " ) 20070° V (rms)
—n—
I <2
vs
Answer: 251.64 /15.91° V
10.6 Maximum Power Transfer
a»-Generalized linear
network operating
in the sinusoidal
steady state
b«
Figure 10.18 • A circuit describing maximum power
transfer
Recall from Chapter 4 that certain systems—for example, those that trans-mit information via electric signals—depend on being able to transfer a maximum amount of power from the source to the load We now reexam-ine maximum power transfer in the context of a sinusoidal steady-state network, beginning with Fig 10.18 We must determine the load imped-ance ZL that results in the delivery of maximum average power to termi-nals a and b Any linear network may be viewed from the termitermi-nals of the load in terms of a Thevenin equivalent circuit Thus the task reduces to finding the value of ZL that results in maximum average power delivered
to ZL in the circuit shown in Fig 10.19
For maximum average power transfer, ZL must equal the conjugate of the Thevenin impedance; that is,
Condition for maximum average power
Trang 4We derive Eq 10.38 by a straightforward application of elementary
calcu-lus We begin by expressing Z Th and ZL in rectangular form:
ZT h - i?Th + jX Th , (10.39)
In both Eqs 10.39 and 10.40, the reactance term carries its own algebraic
sign—positive for inductance and negative for capacitance Because we
are making an average-power calculation, we assume that the amplitude
of the Thevenin voltage is expressed in terms of its rms value We also use
the Thevenin voltage as the reference phasor Then, from Fig 10.19, the
rms value of the load current I is
I = *"rh
(R-i- h + R L ) + j(x Th + x L y (10.41)
Figure 10.19 • The circuit shown in Fig 10.18, with
the network replaced by its Thevenin equivalent
The average power delivered to the load is
Substituting Eq 10.41 into Eq 10.42 yields
(R Th + R L ) 2 + (X Th + X L ) 2
(10.43)
When working with Eq 10.43, always remember that Vrh, i?Th, and X Th
are fixed quantities, whereas R L and X L are independent variables
Therefore, to maximize P, we must find the values of /?L and X L where
dP/3R L and dP/BX h are both zero From Eq 10.43,
dP -\\ Th \ 2 2R L (X L + X Th )
dX L [(R L + R Th ) 2 + (X L + X Th ) 2 } 2 ' (10.44)
*P_ = IVlhPK^L + ^Th)2 + (XL + *Th)2 - 2R L (RL + *Th)]
dR L [(R L + R Th ) 2 + (X L + X Th ) 2 ] 2
From Eq 10.44, dP/dX L is zero when
(10.45)
From Eq 10.45,3P/dR L is zero when
R L = VWh + (*L + x Th y (10.47) Note that when we combine Eq 10.46 with Eq 10.47, both derivatives are
zero when Z = Z j
Trang 5380 Sinusoidal Steady-State Power Calculations
The Maximum Average Power Absorbed
The maximum average power that can be delivered to ZL when it is set equal to the conjugate of ZTh is calculated directly from the circuit in Fig 10.19 When ZL = Zjh, the rms load current is VTh/2JRL, and the max-imum average power delivered to the load is
P =
1 ghj
If the Thevenin voltage is expressed in terms of its maximum amplitude rather than its rms amplitude, Eq 10.48 becomes
1VJ,
8 R L
(10.49)
Maximum Power Transfer When Z is Restricted
Maximum average power can be delivered to ZL only if ZL can be set equal to the conjugate of ZTh There are situations in which this is not
pos-sible First, R L and X L may be restricted to a limited range of values In
this situation, the optimum condition for R L and X L is to adjust X L as
near to -X Th as possible and then adjust R L as close to
VRjh + (X L + X Th ) 2 as possible (see Example 10.9)
A second type of restriction occurs when the magnitude of ZL can be varied but its phase angle cannot Under this restriction, the greatest amount of power is transferred to the load when the magnitude of ZL is set equal to the magnitude of ZTh; that is, when
\Z L \ = |ZnJ (10.50)
The proof of Eq 10.50 is left to you as Problem 10.40
For purely resistive networks, maximum power transfer occurs when the load resistance equals the Thevenin resistance Note that we first derived this result in the introduction to maximum power transfer in Chapter 4
Examples 10.8-10.11 illustrate the problem of obtaining maximum power transfer in the situations just discussed
a) For the circuit shown in Fig 10.20, determine the
impedance ZL that results in maximum average
power transferred to ZL
b) What is the maximum average power transferred
to the load impedance determined in (a)?
Solution
a) We begin by determining the Thevenin
equiva-lent with respect to the load terminals a, b After
two source transformations involving the 20 V
source, the 5 fl resistor, and the 20 il resistor, we
simplify the circuit shown in Fig 10.20 to the one shown in Fig 10.21 Then,
V"Th Th = 16 / 0 ° (-/6)
4 + /3 - /6
= 19.2 /-53.13° = 11.52 - /15.36 V
Figure 10.20 • The circuit for Example 10.8
Trang 6Figure 10.21 • A simplification of Fig 10.20 by source
transformations
which we replaced the original network with its Tlievenin equivalent From Fig 10.22, the rms magni-tude of the load current I is
19.2/V2
'•« = im = U785 A '
The average power delivered to the load is
We find the Thevenin impedance by
deactivat-ing the independent source and calculatdeactivat-ing the
impedance seen looking into the terminals a
and b Thus,
zTh = (7y 6 ) (!+ J? =5-7 6 - /i-68 a
m ~ n -/1.6811 5.76 ft , / a
19.2/-53.13Y +
V
It
5.76 0 4/1.68 0
-For maximum average power transfer, the load
impedance must be the conjugate of ZTh, so
ZL = 5.76 + /1.68 ft
b) We calculate the maximum average power
deliv-ered to ZL from the circuit shown in Fig 10.22, in
Figure 10.22 A The circuit shown in Fig 10.20, with the
original network replaced by its Thevenin equivalent
P = I 2cU (5.76) = 8 W
Example 10.9 Determining Maximum Power Transfer with Load Impedance Restriction
a) For the circuit shown in Fig 10.23, what value of
ZL results in maximum average power transfer to
ZL? What is the maximum power in milliwatts?
b) Assume that the load resistance can be varied
between 0 and 4000 ft and that the capacitive
reactance of the load can be varied between
0 and -2000 ft What settings of R L and X L
transfer the most average power to the load?
What is the maximum average power that can be
transferred under these restrictions?
Solution
a) If there are no restrictions on R L and X L , the
load impedance is set equal to the conjugate of
the output or the Thevenin impedance Therefore
we set
R L = 3000 ft and X L = -4000 ft,
or
3000 a /4000 a
-vw
10
V (rms
Figure 10.23 • The circuit for Examples 10.9 and 10.10
Because the source voltage is given in terms of its rms value, the average power delivered to ZL is
1 102 25
P = 4 3000 = Tm W = 8 3 3 m W
-b) Because R L and X L are restricted, we first set X L
as close to -4000 ft as possible; thus
Xj = -2000 ft Next, we set R L as close to
vRjh + (X L + X Th ) 2 as possible.Thus
ZL = 3000 - /4000 ft R = V3000 + (-2000 + 4000) = 3605.55 ft
Trang 7382 Sinusoidal Steady-State Power Calculations
Now, because R L can be varied from 0 to 4000 ft,
we can set R L to 3605.55 ft Therefore, the load
impedance is adjusted to a value of
ZL = 3605.55 - /2000 ft
With ZL set at this value, the value of the load
current is
10 / 0 °
leff
6605.55 + /2000 1.4489 / - 1 6 8 5 ° mA
The average power delivered to the load is
P = (1.4489 X Kr3)2(3605.55) = 7.57 mW
This quantity is the maximum power that we can
deliver to a load, given the restrictions on R L
and X L Note that this is less than the power that
can be delivered if there are no restrictions; in (a) we found that we can deliver 8.33 mW
Example 10.10 Finding Maximum Power Transfer with Impedance Angle Restrictions
A load impedance having a constant phase angle of
-36.87° is connected across the load terminals a and
b in the circuit shown in Fig 10.23 The magnitude of
ZL is varied until the average power delivered is the
most possible under the given restriction
a) Specify ZL in rectangular form
b) Calculate the average power delivered to ZL
Solution
a) From Eq 10.50, we know that the magnitude of
ZL must equal the magnitude of ZT h Therefore
\ZL\ = |ZThl = 13000 + /4000| = 5000 ft
Now, as we know that the phase angle of ZL is -36.87°, we have
ZL = 5000 / - 3 6 8 7 ° = 4000 - /3000 ft b) With ZL set equal to 4000 - /3000 ft, the load current is
10
!eff = ^ ^ ^7^ = 1-4142 / - 8 1 3 ° mA,
and the average power delivered to the load is
P = (1.4142 X 10"3)2(4000) = 8mW This quantity is the maximum power that can be delivered by this circuit to a load impedance whose angle is constant at —36.87° Again, this quantity is less than the maximum power that can
be delivered if there are no restrictions on ZL
^ A S S E S S M E N T P R O B L E M
Objective 2—Understand the condition for maximum real power delivered to a load in an ac circuit
10.7 The source current in the circuit shown is 3.6 mH
3cos5000r A
a) What impedance should be connected
across terminals a,b for maximum average
power transfer?
b) What is the average power transferred to
the impedance in (a)?
c) Assume that the load is restricted to pure
resistance What size resistor connected
across a,b will result in the maximum
aver-age power transferred?
d) What is the average power transferred to
the resistor in (c)?
Answer: (a) 20 - /10 ft;
(b)18W;
(c) 22.36 ft;
(d) 17.00 W
4H
NOTE: Also try Chapter Problems 10.44,10.47, and 10.48
Trang 8Example 10.11 Finding Maximum Power Transfer in a Circuit with an Ideal Transformer
The variable resistor in the circuit in Fig 10.24 is
adjusted until maximum average power is delivered
to R L
a) What is the value of R L in ohms?
b) What is the maximum average power (in watts)
delivered to R { 1
60O ideal a
^ ^ T l 4:1
O840/Q!
20 n
Figure 10.24 • The circuit for Example 10.11
Solution
a) We first find the Thevenin equivalent with
respect to the terminals of R L The circuit for
determining the open circuit voltage in shown in
Fig 10.25 The variables V], V2, l\, and I2 have
been added to expedite the discussion
60 n
- V A —
< • a
I, +
«~Ii
840/0!^+
V (rms)
4:1
Ideal
- • a
V 2
rh
20 n
»b
Figure 10.25 A The circuit used to find the Thevenin voltage
First we note the ideal transformer imposes the
following constraints on the variables Vj, V2, Ii, and I2:
The open circuit value of I? is zero, hence I] is zero It follows that
Vi = 840 /QP_ v , V2 = 210 / 0 ° V
From Fig 10.25 we note that \ r h is the negative
of V2, hence
VTh = - 2 1 0 / 0 ° V
The circuit shown in Fig 10.26 is used to
deter-mine the short circuit current Viewing l { and I2
as mesh currents, the two mesh equations are
840 / 0 ° = goij - 20I2 + \ h
0 = 20I2 - 201! + V2
60(1
I, + k
V ' I
840Z0°/ + \
r(rms)W
4
- I '
1
c v2
Ideal ] • +
(
:20O
i II
Figure 10.26 • The circuit used to calculate the short circuit
current
When these two mesh current equations are com-bined with the constraint equations we get
840 / ( F = - 4 0 I2 + Vj,
Vi
0 = 25I2 + - i
4 Solving for the short circuit value of I2 yields
I2 = - 6 A
Therefore the Thevenin resistance is
-210
R Th
6 = 35 n
Trang 9384 Sinusoidal Steady-State Power Calculations
Maximum power will be delivered to R L when
R L equals 35 ft
b) The maximum power delivered to R L is most
easily determined using theThevenin equivalent
From the circuit shown in Fig 10.27 we have
P =
1 max
- 2 1 0
70 (35) = 315 W Figure 10.27 • The Thevenin equivalent loaded for maximum
power transfer
^ A S S E S S M E N T P R O B L E M S
Objective 3—Be able to calculate all forms of ac power in ac circuits with linear transformers and ideal
transformers
10.8 Find the average power delivered to the
100 Q resistor in the circuit shown if
v g = 660 cos 5000r V
10 mH
100 ft
Answer: 612.5 W
10.9 a) Find the average power delivered to the
400 fl resistor in the circuit shown if
v t = 248 cos 10,000f V
b) Find the average power delivered to the
375 H resistor
c) Find the power developed by the ideal
volt-age source Check your result by showing the
power absorbed equals the power developed
NOTE: Also try Chapter Problems 10.64 and 10.65
40 mH 50 mH lOOmH
— l ' Y W ^ —
Answer: (a) 50 W;
(b)49.2W;
(c) 99.2 W, 50 + 49.2 = 99.2 W
10.10 Solve Example 10.11 if the polarity dot on the
coil connected to terminal a is at the top
Answer: (a) 15 Q,;
(b) 735 W
10.11 Solve Example 10.11 if the voltage source is
reduced to 146 / 0 ° V rms and the turns ratio is reversed to 1:4
Answer: (a) 1460 H;
(b)58.4W
Practical Perspective
Heating Appliances
A handheld hair dryer contains a heating element, which is just a resistor
heated by the sinusoidal current passing through it, and a fan that blows
the warm air surrounding the resistor out the front of the unit This is
shown schematically in Fig 10.28 The heater tube in this figure is a
resis-tor made of coiled nichrome wire Nichrome is an alloy of iron, chromium,
and nickel Two properties make i t ideal for use in heaters First, it is more
resistive than most other metals, so less material is required to achieve the
needed resistance This allows the heater to be very compact Second,
unlike many other metals, nichrome does not oxidize when heated red hot in
air Thus the heater element lasts a long time
A circuit diagram for the hair dryer controls is shown in Fig 10.29 This
is the only part of the hair dryer circuit that gives you control over the heat
Heater tube
Controls
Fan and motor
Power cord
Figure 10.28 • Schematic representation of a
handheld hair dryer
Trang 10• 1 Thermal fuse
f t T t
OFF L M H
Figure 10.29 A A circuit diagram for the hair dryer
controls
\ Thermal fuse
" * is? " 1 5 ft 2
t t t t
OFF L M H
(a)
R\ > R 2
Figure 10.30 • (a) The circuit in Fig 10.29 redrawn
for the LOW switch setting, (b) A simplified
equiva-lent circuit for (a)
setting The rest of the circuit provides power to the fan motor and is not of interest here The coiled wire that comprises the heater tube has a connec-tion partway along the coil, dividing the coil into two pieces We model this
in Fig 10.29 with two series resistors, R^ and R 2 The controls to turn the
dryer on and select the heat setting use a four-position switch in which two pairs of terminals in the circuit will be shorted together by a pair of sliding metal bars The position of the switch determines which pairs of terminals are shorted together The metal bars are connected by an insulator, so there
is no conduction path between the pairs of shorted terminals
The circuit in Fig 10.29 contains a thermal fuse This is a protective device that normally acts like a short circuit But if the temperature near the heater becomes dangerously high, the thermal fuse becomes an open circuit, discontinuing the flow of current and reducing the risk of fire or injury The thermal fuse provides protection in case the motor fails or the airflow becomes blocked While the design of the protection system is not part of this example, i t is important to point out that safety analysis is an essential part of an electrical engineer's work
Now that we have modeled the controls for the hair dryer, let's deter-mine the circuit components that are present for the three switch settings
To begin, the circuit in Fig 10.29 is redrawn in Fig 10.30(a) for the LOW switch setting The open-circuited wires have been removed for clarity A simplified equivalent circuit is shown in Fig 10.30(b) A similar pair of fig-ures is shown for the MEDIUM setting (Fig 10.31) and the HIGH setting (Fig 10.32) Note from these figures that at the LOW setting, the voltage
source sees the resistors R± and R 2 in series; at the MEDIUM setting, the
volt-age source sees only the R 2 resistor; and at the HIGH setting, the voltage source sees the resistors in parallel
Thermal fuse
• • • • •
/¾
t t t t
OFF L M H
(a)
ft,
(b) Figure 10.31 • (a) The circuit in Fig 10.29 redrawn
for the MEDIUM switch setting, (b) A simplified
equiv-alent circuit for (a)
Thermal fuse
v • • • 11
t t t t
OFFL M H
(a)
ft, ft?
(b)
Figure 10.32 A (a) The circuit in Fig 10.29 redrawn
for the HIGH switch setting, (b) A simplified equiva-lent circuit for (a)
NOTE: Assess your understanding of this Practical Perspective by trying Chapter Problems 10.66-10.68