Electric Circuits, 9th Edition P41 potx

376 Sinusoidal Steady-State Power Calculations
The average power lost in the line results from
the current flowing through the line resistance:
*U = |IJ
2
* = (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 =
ivy
2
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,
C =
-1 -1
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 kVA^~-
^^ t
20 kW
(a)

lOkVAR +
20 kW
(c)
-lOkVAR
(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 magnet-
izing reactive power delivered equals the magnet-
izing reactive power absorbed.
'VW orv-v->
i a /20
+
1 nn
-/16 n;
+ v
•—wv
in
|.,
/3 0
39 I
t

V
s
= 150/0°V
V, = (78 - /104) V Ij = (-26 - /52) A
V
2
= (72 + /104) V l
v
= (-2+ /6) A
V
3
= (150 - /130) V I
2
= (-24 - /58) A
Figure 10.17 A The circuit, with solution, for Example 10.7.
10.5
Power Calculations
377
Solution
a) The complex power delivered to the (1 + /2) ft
impedance is
Si = |v,II = /», + /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
= 2
V
2^ = 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
= ~V
3
IS = 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
s

s
= -|v,ii =
p
s
+ /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
= |(39I
r
)(I|) = P
x
+ jQ
x
= |(-78 + /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 = 5850 W.
The dependent voltage source is the only circuit
element delivering average

power.
Thus
delivered = 5850 W.
Magnetizing reactive power is being absorbed
by the two horizontal
branches.
Thus
^absorbed = Q\ + Qi = 9290 VAR.
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 = 9290 VAR.
378 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 -52 0,. Calculate:
a) the rms phasors V
L
and I
L
,
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
•AW-
/4 0
-TTTA.
O
250/0!
V(rms)
+
V,.
39 n
./26
a
Source
Line—
—***- Load
(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 V
v
in polar form.
Answer: (a) 252.20 /-4.54° V (rms),
5.38 /-38.23° A (rms);
(b) 1129.09 W, 752.73 VAR;
NOTE: Also try Chapter Problems 10.18,10.26, and 10.27.
/10
+
(_")
20070° V
(rms)
—n—

I
<2
v
s
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 Z
L
that results in the delivery of maximum average power to termi-
nals a and
b.
Any linear network may be viewed from the terminals of the
load in terms of a Thevenin equivalent circuit. Thus the task reduces to
finding the value of Z
L

that results in maximum average power delivered
to Z
L
in the circuit shown in Fig. 10.19.
For maximum average power transfer, Z
L
must equal the conjugate of
the Thevenin impedance; that is,
Condition for maximum average power
transfer •
Z
T
= Z
Th'
(10.38)
10.6 Maximum Power Transfer
379
We derive Eq. 10.38
by a
straightforward application
of
elementary calcu-
lus.
We begin
by
expressing
Z
Th
and
Z

L
in
rectangular form:
Z
Th
-
i?
Th
+ jX
Th
,
(10.39)
Z
L
= R
L
+ jX
L
,
(10.40)
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
P
=
HI
2
/*,,
(10.42)
Substituting Eq. 10.41 into Eq. 10.42 yields
P
=

|v
Th
|X
(R
Th
+ R
L
)
2
+ (X
Th
+ X
L
)
2
(10.43)
When working with
Eq.
10.43,
always remember that
Vr
h
,
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)
X,
= -X
riv
(10.46)
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
L
= Zj
h
.
380 Sinusoidal Steady-State Power Calculations
The Maximum Average Power Absorbed
The maximum average power that can be delivered to Z
L
when it is set

equal to the conjugate of Z
Th
is calculated directly from the circuit in
Fig. 10.19. When Z
L
= Zj
h
, the rms load current is V
Th
/2JR
L
, and the max-
imum average power delivered to the load is
P =
1
max
4Rr
1 ghj
4 R,
(10.48)
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 Z
L
only if Z

L
can be set
equal to the conjugate of
Z
Th
.
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 Z
L
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 Z
L
is
set equal to the magnitude of Z
Th
; 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.
Example 10.8 Determining Maximum Power Transfer without Load Restrictions
a) For the circuit shown in Fig.
10.20,
determine the

impedance Z
L
that results in maximum average
power transferred to Z
L
.
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
=
16 /0°
"Th
(-/6)
4 + /3 - /6
= 19.2 /-53.13° = 11.52 - /15.36 V.

Figure 10.20 • The circuit for Example 10.8.
10.6 Maximum Power Transfer 381
Figure 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 calculating the
impedance seen looking into the terminals a
and
b.
Thus,
z
Th =
(
7
y6)(
!
+
J
? =

5
-
76
- /i-
68 a
-
Th
4 + /3 - /6
J
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 Z
T
h, so
Z
L
= 5.76 + /1.68 ft.
b) We calculate the maximum average power deliv-
ered to Z
L
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
2
cU
(5.76) = 8W.
Example 10.9
Determining Maximum Power Transfer with Load Impedance Restriction
a) For the circuit shown in Fig.
10.23,
what value of
Z
L
results in maximum average power transfer to
Z
L
? 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 Z
L
is
1 10
2
25
P
= 4
3000
=

T
mW = 833mW
-
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
Z
L
= 3000 - /4000 ft.
R
L
= V3000
2
+ (-2000 + 4000)
2

= 3605.55 ft.
382
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
Z
L
= 3605.55 - /2000 ft.
With Z
L
set at this value, the value of the load
current is
10 /0°
leff
6605.55 + /2000
1.4489
/-16.85° mA.
The average power delivered to the load is
P = (1.4489 X Kr
3
)
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
Z
L
is varied until the average power delivered is the
most possible under the given restriction.
a) Specify Z
L
in rectangular form.
b) Calculate the average power delivered to Z
L
.
Solution
a) From Eq. 10.50, we know that the magnitude of
Z
L
must equal the magnitude of Z
Th

. Therefore
\ZL\
= |Z
T
hl = 13000 + /4000| = 5000 ft.
Now, as we know that the phase angle of Z
L
is
-36.87°,
we have
Z
L
= 5000 /-36.87° = 4000 - /3000 ft.
b) With Z
L
set equal to 4000 - /3000 ft, the load
current is
10
!eff = ^^ ^7^ =
1-4142
/-8.13° mA,
611
7000 + /1000
L
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 Z
L
.
^ASSESSMENT PROBLEM
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.
10.6 Maximum Power Transfer
383

Example 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
^^Tl 4:1
O
840/Q!
V (rms)
r*L
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],
V
2
, l\, and I
2
have
been added
to
expedite

the
discussion.
60
n
-VA—
<•
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,
V
2
, Ii, and I
2
:
V
2
= Jv,.
I,
=
-\h.
The open circuit value
of I? is
zero, hence
I] is
zero.
It
follows that
Vi
= 840
/QP_
v, V
2
= 210 /0°V.
From Fig. 10.25
we
note that
\
rh

is the
negative
of V
2
, hence
V
Th
= -210 /0° V.
The circuit shown
in
Fig. 10.26
is
used
to
deter-
mine
the
short circuit current. Viewing
l
{
and I
2
as mesh currents,
the
two mesh equations
are
840
/0° =
goij
-

20I
2
+ \
h
0
=
20I
2
-
201!
+ V
2
.
60(1
I,
+ k
V
'
I
840Z0°/
+ \
r
(rms)W
4
-I'
1
c
v
2
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 =
-40I
2
+ Vj,
Vi
0
=
25I

2
+ -i.
4
Solving
for the
short circuit value
of I
2
yields
I
2
= -6 A.
Therefore the Thevenin resistance
is
-210
R
Th
-
-6
= 35
n.
384 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
210/0°/-
V(rms")U
35
O
P
=
1
max
-210
70
(35)
= 315 W.
Figure 10.27 •
The
Thevenin equivalent loaded for maximum
power transfer.
^ASSESSMENT PROBLEMS
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'YW^—
:37512
400 O
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 it 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
Hot
air
4
Controls
Fan
and motor
Power
cord
Figure 10.28 • Schematic representation of
a
handheld hair dryer.
Practical

Perspective
385
•
1
Thermal fuse
; xfc
R^
<
R->
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, it 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
I * * * IJ_
ft,
R
2

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.