10
Capacitors
10.1
INTRODUCTION
Thus far, the only passive device appearing in the text has been the
resistor. We will now consider two additional passive devices called the
capacitor and the inductor (the inductor is discussed in detail in Chapter 12), which are quite different from the resistor in purpose, operation,
and construction.
Unlike the resistor, both elements display their total characteristics
only when a change in voltage or current is made in the circuit in which
they exist. In addition, if we consider the ideal situation, they do not
dissipate energy as does the resistor but store it in a form that can be
returned to the circuit whenever required by the circuit design.
Proper treatment of each requires that we devote this entire chapter
to the capacitor and, as mentioned above, Chapter 12 to the inductor.
Since electromagnetic effects are a major consideration in the design of
inductors, this topic will be covered in Chapter 11.
10.2
THE ELECTRIC FIELD
Recall from Chapter 2 that a force of attraction or repulsion exists
between two charged bodies. We shall now examine this phenomenon
in greater detail by considering the electric field that exists in the region
around any charged body. This electric field is represented by electric
flux lines, which are drawn to indicate the strength of the electric field
at any point around the charged body; that is, the denser the lines of
flux, the stronger the electric field. In Fig. 10.1, the electric field
strength is stronger at position a than at position b because the flux lines
are denser at a than at b. The symbol for electric flux is the Greek letter w (psi). The flux per unit area (flux density) is represented by the
capital letter D and is determined by
w
D A
(flux/unit area)
(10.1)
376

CAPACITORS
a
r1
b
r2
+
Flux lines radiate
outward for positive
charges and inward
for negative charges.
Positive charge
Electric
flux lines
FIG. 10.1
Flux distribution from an isolated positive charge.
The larger the charge Q in coulombs, the greater the number of flux
lines extending or terminating per unit area, independent of the surrounding medium. Twice the charge will produce twice the flux per unit
area. The two can therefore be equated:
wQ
(coulombs, C)
(10.2)
By definition, the electric field strength at a point is the force acting on a unit positive charge at that point; that is,
F
Q
(newtons/coulomb, N/C)
(10.3)
The force exerted on a unit positive charge (Q2 1 C), by a charge
Q1, r meters away, as determined by Coulomb’s law is
kQ Q2
kQ (1)
kQ1
F 1
1
2
2
r
r
r2
(k 9 109 N.m2/C2)
Substituting this force F into Eq. (10.3) yields
F
kQ1/r2
Q2
1
+
kQ1
r2
+
(a)
–
+
(b)
FIG. 10.2
Electric flux distribution: (a) like charges;
(b) opposite charges.
(N/C)
(10.4)
We can therefore conclude that the electric field strength at any point
distance r from a point charge of Q coulombs is directly proportional to
the magnitude of the charge and inversely proportional to the distance
squared from the charge. The squared term in the denominator will
result in a rapid decrease in the strength of the electric field with distance from the point charge. In Fig. 10.1, substituting distances r1 and
r2 into Eq. (10.4) will verify our previous conclusion that the electric
field strength is greater at a than at b.
Electric flux lines always extend from a positively charged body to a
negatively charged body, always extend or terminate perpendicular to
the charged surfaces, and never intersect.
For two charges of similar and opposite polarities, the flux distribution
would appear as shown in Fig. 10.2.

CAPACITANCE
The attraction and repulsion between charges can now be explained
in terms of the electric field and its flux lines. In Fig. 10.2(a), the flux
lines are not interlocked but tend to act as a buffer, preventing attraction
and causing repulsion. Since the electric field strength is stronger (flux
lines denser) for each charge the closer we are to the charge, the more
we try to bring the two charges together, the stronger will be the force
of repulsion between them. In Fig. 10.2(b), the flux lines extending
from the positive charge are terminated at the negative charge. A basic
law of physics states that electric flux lines always tend to be as short
as possible. The two charges will therefore be drawn to each other.
Again, the closer the two charges, the stronger the attraction between
the two charges due to the increased field strengths.
10.3
CAPACITANCE
Up to this point we have considered only isolated positive and negative
spherical charges, but the analysis can be extended to charged surfaces
of any shape and size. In Fig. 10.3, for example, two parallel plates of
a conducting material separated by an air gap have been connected
through a switch and a resistor to a battery. If the parallel plates are initially uncharged and the switch is left open, no net positive or negative
charge will exist on either plate. The instant the switch is closed, however, electrons are drawn from the upper plate through the resistor to
the positive terminal of the battery. There will be a surge of current at
first, limited in magnitude by the resistance present. The level of flow
will then decline, as will be demonstrated in the sections to follow. This
action creates a net positive charge on the top plate. Electrons are being
repelled by the negative terminal through the lower conductor to the
bottom plate at the same rate they are being drawn to the positive terminal. This transfer of electrons continues until the potential difference
across the parallel plates is exactly equal to the battery voltage. The
final result is a net positive charge on the top plate and a negative
charge on the bottom plate, very similar in many respects to the two
isolated charges of Fig. 10.2(b).
This element, constructed simply of two parallel conducting plates
separated by an insulating material (in this case, air), is called a capacitor. Capacitance is a measure of a capacitor’s ability to store charge
on its plates—in other words, its storage capacity.
A capacitor has a capacitance of 1 farad if 1 coulomb of charge is
deposited on the plates by a potential difference of 1 volt across the
plates.
The farad is named after Michael Faraday (Fig. 10.4), a nineteenthcentury English chemist and physicist. The farad, however, is generally too large a measure of capacitance for most practical applications,
so the microfarad (106) or picofarad (1012) is more commonly used.
Expressed as an equation, the capacitance is determined by
Q
C V
C farads (F)
Q coulombs (C)
V volts (V)
377
e
R
e
e
+
+ + +
– – –
E
e
e
e
V = E
–
e
FIG. 10.3
Fundamental charging network.
English (London)
(1791–1867)
Chemist and
Electrical
Experimenter
Honory Doctorate,
Oxford University,
1832
Courtesy of the
Smithsonian Institution
Photo No. 51,147
An experimenter with no formal education, he began
his research career at the Royal Institute in London
as a laboratory assistant. Intrigued by the interaction
between electrical and magnetic effects, he discovered electromagnetic induction, demonstrating that
electrical effects can be generated from a magnetic
field (the birth of the generator as we know it today).
He also discovered self-induced currents and introduced the concept of lines and fields of magnetic
force. Having received over one hundred academic
and scientific honors, he became a Fellow of the
Royal Society in 1824 at the young age of 32.
FIG. 10.4
Michael Faraday.
+ + + + + + + + +
– – – – – – – – –
Fringing
(a)
+++++++++
(10.5)
Different capacitors for the same voltage across their plates will
acquire greater or lesser amounts of charge on their plates. Hence the
capacitors have a greater or lesser capacitance, respectively.
A cross-sectional view of the parallel plates is shown with the distribution of electric flux lines in Fig. 10.5(a). The number of flux lines per
– – – – – – – – –
(b)
FIG. 10.5
Electric flux distribution between the plates of
a capacitor: (a) including fringing; (b) ideal.
378

CAPACITORS
unit area (D) between the two plates is quite uniform. At the edges, the
flux lines extend outside the common surface area of the plates, producing an effect known as fringing. This effect, which reduces the
capacitance somewhat, can be neglected for most practical applications.
For the analysis to follow, we will assume that all the flux lines leaving
the positive plate will pass directly to the negative plate within the common surface area of the plates [Fig. 10.5(b)].
If a potential difference of V volts is applied across the two plates
separated by a distance of d, the electric field strength between the
plates is determined by
+
+ +++ + ++
+ +++ + ++
–
–
–
–
–
–
–
+ +++ + ++
+ + + + + + +
V (volts)
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
+
V
d
–
–
–
–
–
–
–
–
Dielectric
d
+
(a)
–
V (volts)
–
– – (Dielectric)
–
–
– – (Resultant)
+ + + + + + +
+ + + + + + +
(Dielectric = air)
–
–
–
–
–
–
–
d
(b)
FIG. 10.6
Effect of a dielectric on the field distribution
between the plates of a capacitor:
(a) alignment of dipoles in the dielectric;
(b) electric field components between the
plates of a capacitor with a dielectric present.
(volts/meter, V/m)
(10.6)
The uniformity of the flux distribution in Fig. 10.5(b) also indicates
that the electric field strength is the same at any point between the two
plates.
Many values of capacitance can be obtained for the same set of parallel plates by the addition of certain insulating materials between the
plates. In Fig. 10.6(a), an insulating material has been placed between a
set of parallel plates having a potential difference of V volts across
them.
Since the material is an insulator, the electrons within the insulator
are unable to leave the parent atom and travel to the positive plate. The
positive components (protons) and negative components (electrons) of
each atom do shift, however [as shown in Fig. 10.6(a)], to form dipoles.
When the dipoles align themselves as shown in Fig. 10.6(a), the material is polarized. A close examination within this polarized material will
indicate that the positive and negative components of adjoining dipoles
are neutralizing the effects of each other [note the dashed area in Fig.
10.6(a)]. The layer of positive charge on one surface and the negative
charge on the other are not neutralized, however, resulting in the establishment of an electric field within the insulator [dielectric; Fig. 10.6(b)].
The net electric field between the plates (resultant air dielectric)
would therefore be reduced due to the insertion of the dielectric.
The purpose of the dielectric, therefore, is to create an electric field
to oppose the electric field set up by free charges on the parallel plates.
For this reason, the insulating material is referred to as a dielectric, di
for “opposing” and electric for “electric field.”
In either case—with or without the dielectric—if the potential across
the plates is kept constant and the distance between the plates is fixed,
the net electric field within the plates must remain the same, as determined by the equation V/d. We just ascertained, however, that the
net electric field between the plates would decrease with insertion of the
dielectric for a fixed amount of free charge on the plates. To compensate
and keep the net electric field equal to the value determined by V and d,
more charge must be deposited on the plates. [Look ahead to Eq.
(10.11).] This additional charge for the same potential across the plates
increases the capacitance, as determined by the following equation:
Q
C D D
V
For different dielectric materials between the same two parallel
plates, different amounts of charge will be deposited on the plates. But
CAPACITANCE
w Q, so the dielectric is also determining the number of flux lines
between the two plates and consequently the flux density (D w/A)
since A is fixed.
The ratio of the flux density to the electric field intensity in the
dielectric is called the permittivity of the dielectric:
D
e (farads/meter, F/m)
(10.7)
It is a measure of how easily the dielectric will “permit” the establishment of flux lines within the dielectric. The greater its value, the greater
the amount of charge deposited on the plates, and, consequently, the
greater the flux density for a fixed area.
For a vacuum, the value of e (denoted by eo) is 8.85 1012 F/m.
The ratio of the permittivity of any dielectric to that of a vacuum is
called the relative permittivity, er. It simply compares the permittivity
of the dielectric to that of air. In equation form,
e
er eo
(10.8)
The value of e for any material, therefore, is
e ereo
Note that er is a dimensionless quantity. The relative permittivity, or
dielectric constant, as it is often called, is provided in Table 10.1 for
various dielectric materials.
Substituting for D and in Eq. (10.7), we have
D
w/A
Q/A
Qd
e
V/d
V/d
VA
But
Q
C
V
and, therefore,
Cd
e
A
TABLE 10.1
Relative permittivity (dielectric constant) of various dielectrics.
Dielectric
Vacuum
Air
Teflon
Paper, paraffined
Rubber
Transformer oil
Mica
Porcelain
Bakelite
Glass
Distilled water
Barium-strontium
titanite (ceramic)
er (Average Values)
1.0
1.0006
2.0
2.5
3.0
4.0
5.0
6.0
7.0
7.5
80.0
7500.0

379
380

CAPACITORS
and
or
A
Ce
d
(F)
A
A
C eoer 8.85 1012er d
d
(10.9)
(F)
(10.10)
where A is the area in square meters of the plates, d is the distance in
meters between the plates, and er is the relative permittivity. The capacitance, therefore, will be greater if the area of the plates is increased, or
the distance between the plates is decreased, or the dielectric is changed
so that er is increased.
Solving for the distance d in Eq. (10.9), we have
eA
D
C
and substituting into Eq. (10.6) yields
V
CV
V
eA/C
eA
d
But Q CV, and therefore
Q
eA
(V/m)
(10.11)
which gives the electric field intensity between the plates in terms of the
permittivity e, the charge Q, and the surface area A of the plates. Thus,
we have the ratio
C eA/d
e
er
Co eo A/d
eo
or
C erCo
(10.12)
which, in words, states that for the same set of parallel plates, the
capacitance using a dielectric (of relative permittivity er) is er times that
obtained for a vacuum (or air, approximately) between the plates. This
relationship between er and the capacitances provides an excellent
experimental method for finding the value of er for various dielectrics.
EXAMPLE 10.1 Determine the capacitance of each capacitor on the
right side of Fig. 10.7.
Solutions:
a. C 3(5 mF) 15 mF
1
b. C (0.1 mF) 0.05 mF
2
c. C 2.5(20 mF) 50 mF
4
d. C (5) (1000 pF) (160)(1000 pF) 0.16 mF
(1/8)
CAPACITANCE
C = 5 µF
A

381
C = ?
3A
d
d
(a)
C = 0.1 µF
d
C = ?
2d
(b)
C = 20 µF
C = ?
o
(c)
C = 1000 pF
A
d
r = 2.5
(paraffined
paper)
C = ?
4A
o
1
8d
r = 5 (mica)
(d)
FIG. 10.7
Example 10.1.
EXAMPLE 10.2 For the capacitor of Fig. 10.8:
a. Determine the capacitance.
b. Determine the electric field strength between the plates if 450 V are
applied across the plates.
c. Find the resulting charge on each plate.
Solutions:
eo A
(8.85 1012 F/m)(0.01 m2)
a. Co 59.0 1012 F
d
1.5 103 m
59 pF
V
450 V
b. 1.5 103 m
d
300 103 V/m
Q
c. C V
or
Q CV (59.0 1012 F)(450 V)
26.550 109 C
26.55 nC
EXAMPLE 10.3 A sheet of mica 1.5 mm thick having the same area
as the plates is inserted between the plates of Example 10.2.
a. Find the electric field strength between the plates.
b. Find the charge on each plate.
c. Find the capacitance.
Q(+)
o
A = 0.01 m2
Q(–) FIG. 10.8
Example 10.2.
d = 1.5 mm
382

CAPACITORS
Solutions:
a. is fixed by
V
450 V
1.5 103 m
d
300 103 V/m
Q
b. eA
or
Q eA ereoA
(5)(8.85 1012 F/m)(300 103 V/m)(0.01 m2)
132.75 109 C 132.75 nC
(five times the amount for
air between the plates)
c. C erCo
(5)(59 1012 F) 295 pF
10.4 DIELECTRIC STRENGTH
For every dielectric there is a potential that, if applied across the dielectric, will break the bonds within the dielectric and cause current to flow.
The voltage required per unit length (electric field intensity) to establish
conduction in a dielectric is an indication of its dielectric strength and
is called the breakdown voltage. When breakdown occurs, the capacitor has characteristics very similar to those of a conductor. A typical
example of breakdown is lightning, which occurs when the potential
between the clouds and the earth is so high that charge can pass from
one to the other through the atmosphere, which acts as the dielectric.
The average dielectric strengths for various dielectrics are tabulated
in volts/mil in Table 10.2 (1 mil 0.001 in.). The relative permittivity
appears in parentheses to emphasize the importance of considering both
factors in the design of capacitors. Take particular note of bariumstrontium titanite and mica.
TABLE 10.2
Dielectric strength of some dielectric materials.
Dielectric
Air
Barium-strontium
titanite (ceramic)
Porcelain
Transformer oil
Bakelite
Rubber
Paper, paraffined
Teflon
Glass
Mica
Dielectric Strength
(Average Value), in
Volts/Mil
75
75
200
400
400
700
1300
1500
3000
5000
(er)
(1.0006)
(7500)
(6.0)
(4.0)
(7.0)
(3.0)
(2.5)
(2.0)
(7.5)
(5.0)
TYPES OF CAPACITORS

383
EXAMPLE 10.4 Find the maximum voltage that can be applied across
a 0.2-mF capacitor having a plate area of 0.3 m2. The dielectric is porcelain. Assume a linear relationship between the dielectric strength and
the thickness of the dielectric.
Solution:
A
C 8.85 1012er d
or
(8.85)(6)(0.3 m2)
8.85er A
d
7.965 105 m
12
(1012)(0.2 106 F)
10 C
79.65 mm
Converting micrometers to mils, we have
106 m 39.371 in. 1000 mils
79.65 mm 3.136 mils
1 in.
m
mm
Dielectric strength 200 V/mil
Therefore,
10.5
(3.136 mils) 627.20 V
mil 200 V
LEAKAGE CURRENT
Up to this point, we have assumed that the flow of electrons will occur
in a dielectric only when the breakdown voltage is reached. This is the
ideal case. In actuality, there are free electrons in every dielectric due in
part to impurities in the dielectric and forces within the material itself.
When a voltage is applied across the plates of a capacitor, a leakage
current due to the free electrons flows from one plate to the other. The
current is usually so small, however, that it can be neglected for most
practical applications. This effect is represented by a resistor in parallel
with the capacitor, as shown in Fig. 10.9(a), whose value is typically
more than 100 megohms (M). Some capacitors, however, such as the
electrolytic type, have high leakage currents. When charged and then
disconnected from the charging circuit, these capacitors lose their
charge in a matter of seconds because of the flow of charge (leakage
current) from one plate to the other [Fig. 10.9(b)].
10.6
C
Rleakage
(a)
+
–
(b)
FIG. 10.9
Demonstrating the effect of the leakage
current.
TYPES OF CAPACITORS
Like resistors, all capacitors can be included under either of two general
headings: fixed or variable. The symbol for a fixed capacitor is
, and
for a variable capacitor,
. The curved line represents the plate that
is usually connected to the point of lower potential.
Foil
Mica
Foil
Mica
Fixed Capacitors
Many types of fixed capacitors are available today. Some of the most
common are the mica, ceramic, electrolytic, tantalum, and polyesterfilm capacitors. The typical flat mica capacitor consists basically of
mica sheets separated by sheets of metal foil. The plates are connected
to two electrodes, as shown in Fig. 10.10. The total area is the area of
Foil
Mica
Foil
FIG. 10.10
Basic structure of a stacked mica capacitor.
384

CAPACITORS
FIG. 10.11
Mica capacitors. (Courtesy of Custom
Electronics Inc.)
one sheet times the number of dielectric sheets. The entire system is
encased in a plastic insulating material as shown for the two central
units of Fig. 10.11. The mica capacitor exhibits excellent characteristics
under stress of temperature variations and high voltage applications (its
dielectric strength is 5000 V/mil). Its leakage current is also very small
(Rleakage about 1000 M). Mica capacitors are typically between a few
picofarads and 0.2 mF, with voltages of 100 V or more.
The ability to “roll” the mica to form the cylindrical shapes of Fig.
10.11 is due to a process whereby the soluble contaminants in natural
mica are removed, leaving a paperlike structure resulting from the cohesive forces in natural mica. It is commonly referred to as reconstituted
mica, although the terminology does not mean “recycled” or “secondhand” mica. For some of the units in the photograph, different levels of
capacitance are available between different sets of terminals.
The ceramic capacitor is made in many shapes and sizes, two of
which are shown in Fig. 10.12. The basic construction, however, is
about the same for each, as shown in Fig. 10.13. A ceramic base is
coated on two sides with a metal, such as copper or silver, to act as the
two plates. The leads are then attached through electrodes to the plates.
An insulating coating of ceramic or plastic is then applied over the
plates and dielectric. Ceramic capacitors also have a very low leakage
current (Rleakage about 1000 M) and can be used in both dc and ac networks. They can be found in values ranging from a few picofarads to
perhaps 2 mF, with very high working voltages such as 5000 V or more.
In recent years there has been increasing interest in monolithic
(single-structure) chip capacitors such as those appearing in Fig.
10.14(a) due to their application on hybrid circuitry [networks using
both discrete and integrated circuit (IC) components]. There has also
been increasing use of microstrip (strip-line) circuitry such as the one in
Fig. 10.14(b). Note the small chips in this cutaway section. The L and
H of Fig. 10.14(a) indicate the level of capacitance. For example, if in
black ink, the letter H represents 16 units of capacitance (in picofarads),
or 16 pF. If blue ink is used, a multiplier of 100 is applied, resulting in
1600 pF. Although the size is similar, the type of ceramic material controls the capacitance level.
The electrolytic capacitor is used most commonly in situations
where capacitances of the order of one to several thousand microfarads
Lead wire soldered
to silver electrode
Dipped phenolic coating
Solder
Silver electrodes deposited on
top and bottom of ceramic disc
Ceramic dielectric
(a)
(b)
FIG. 10.12
Ceramic disc capacitors: (a) photograph; (b) construction.
TYPES OF CAPACITORS
Dipped
phenolic coating

385
Lead wire
soldered to
electrode pickup
Solder
Ceramic
dielectric
Metal
electrodes
Electrode
pickup
(Alternately deposited layers of
ceramic dielectric material and
metal electrodes fired into a
single homogeneous block)
FIG. 10.13
Multilayer, radial-lead ceramic capacitor.
are required. They are designed primarily for use in networks where
only dc voltages will be applied across the capacitor because they have
good insulating characteristics (high leakage current) between the
plates in one direction but take on the characteristics of a conductor in
the other direction. Electrolytic capacitors are available that can be used
in ac circuits (for starting motors) and in cases where the polarity of the
dc voltage will reverse across the capacitor for short periods of time.
The basic construction of the electrolytic capacitor consists of a roll
of aluminum foil coated on one side with an aluminum oxide, the aluminum being the positive plate and the oxide the dielectric. A layer of
paper or gauze saturated with an electrolyte is placed over the aluminum oxide on the positive plate. Another layer of aluminum without
the oxide coating is then placed over this layer to assume the role of
the negative plate. In most cases the negative plate is connected
directly to the aluminum container, which then serves as the negative
terminal for external connections. Because of the size of the roll of
aluminum foil, the overall area of this capacitor is large; and due to
the use of an oxide as the dielectric, the distance between the plates is
extremely small. The negative terminal of the electrolytic capacitor is
usually the one with no visible identification on the casing. The positive is usually indicated by such designs as , , , and so on. Due
to the polarity requirement, the symbol for an electrolytic capacitor
.
will normally appear as
Associated with each electrolytic capacitor are the dc working voltage and the surge voltage. The working voltage is the voltage that can
be applied across the capacitor for long periods of time without breakdown. The surge voltage is the maximum dc voltage that can be applied for a short period of time. Electrolytic capacitors are characterized
as having low breakdown voltages and high leakage currents (Rleakage
about 1 M). Various types of electrolytic capacitors are shown in Fig.
10.15. They can be found in values extending from a few microfarads
to several thousand microfarads and working voltages as high as 500 V.
However, increased levels of voltage are normally associated with
lower values of available capacitance.
FIG. 10.14
Monolithic chip capacitors. (Courtesy of
Vitramon, Inc.)
386

CAPACITORS
=
There are fundamentally two types of tantalum capacitors: the solid
and the wet-slug. In each case, tantalum powder of high purity is
pressed into a rectangular or cylindrical shape, as shown in Fig. 10.16.
Next the anode () connection is simply pressed into the resulting
structures, as shown in the figure. The resulting unit is then sintered
(baked) in a vacuum at very high temperatures to establish a very
porous material. The result is a structure with a very large surface area
in a limited volume. Through immersion in an acid solution, a very thin
manganese dioxide (MnO2) coating is established on the large, porous
surface area. An electrolyte is then added to establish contact between
the surface area and the cathode, producing a solid tantalum capacitor.
If an appropriate “wet” acid is introduced, it is called a wet-slug tantalum capacitor.
Cathode ( – )
MnO2 coat
Carbon
Solder
Solder
Tantalum
>
Lead wire
FIG. 10.15
Electrolytic capacitors: (a) Radial lead with
extended endurance rating of 2000 h at 85°C.
Capacitance range: 0.1–15,000 mF with a
voltage range of 6.3 to 250 WV dc (Courtesy
of Illinois Capacitor, Inc.). (b) Solid aluminum
electrolytic capacitors available in axial,
resin-dipped, and surface-mount configurations to withstand harsh environmental
conditions (Courtesy of Philips Components,
Inc.).
Tantalum wire
Anode ( + )
FIG. 10.16
Tantalum capacitor. (Courtesy of Union Carbide Corp.)
The last type of fixed capacitor to be introduced is the polyester-film
capacitor, the basic construction of which is shown in Fig. 10.17. It
consists simply of two metal foils separated by a strip of polyester
material such as Mylar®. The outside layer of polyester is applied to act
as an insulating jacket. Each metal foil is connected to a lead that
extends either axially or radially from the capacitor. The rolled construction results in a large surface area, and the use of the plastic dielectric results in a very thin layer between the conducting surfaces.
Polyester (plastic) film
Metal foils
FIG. 10.17
Polyester-film capacitor.
Data such as capacitance and working voltage are printed on the
outer wrapping if the polyester capacitor is large enough. Color coding
is used on smaller devices (see Appendix D). A band (usually black) is
sometimes printed near the lead that is connected to the outer metal foil.
The lead nearest this band should always be connected to the point of
TYPES OF CAPACITORS
lower potential. This capacitor can be used for both dc and ac networks.
Its leakage resistance is of the order of 100 M. An axial lead and
radial lead polyester-film capacitor appear in Fig. 10.18. The axial lead
variety is available with capacitance levels of 0.1 mF to 18 mF, with
working voltages extending to 630 V. The radial lead variety has a
capacitance range of 0.01 mF to 10 mF, with working voltages extending to 1000 V.
(a)
(b)
FIG. 10.18
Polyester-film capacitors: (a) axial lead; (b) radial lead. (Courtesy of Illinois
Capacitor, Inc.)
Variable Capacitors
The most common of the variable-type capacitors is shown in Fig.
10.19. The dielectric for each capacitor is air. The capacitance in Fig.
10.19(a) is changed by turning the shaft at one end to vary the common
area of the movable and fixed plates. The greater the common area, the
larger the capacitance, as determined by Eq. (10.10). The capacitance of
the trimmer capacitor in Fig. 10.19(b) is changed by turning the screw,
which will vary the distance between the plates (the common area is
fixed) and thereby the capacitance.
(a)
(b)
FIG. 10.19
Variable air capacitors. [Part (a) courtesy of James Millen Manufacturing Co.;
part (b) courtesy of Johnson Manufacturing Co.]

387
388

CAPACITORS
Measurement and Testing
FIG. 10.20
Digital reading capacitance meter. (Courtesy
of BK PRECISION, Maxtec International
Corp.)
++
+
+ –
–
––
FIG. 10.21
Checking the dielectric of an electrolytic
capacitor.
A digital reading capacitance meter appears in Fig. 10.20. Simply place
the capacitor between the provided clips with the proper polarity, and
the meter will display the level of capacitance.
The best check of a capacitor is to use a meter designed to perform
the necessary tests. However, an ohmmeter can identify those in which
the dielectric has deteriorated (especially in paper and electrolytic
capacitors). As the dielectric breaks down, the insulating qualities
decrease to a point where the resistance between the plates drops to a
relatively low level. After ensuring that the capacitor is fully discharged, place an ohmmeter across the capacitor, as shown in Fig.
10.21. In a polarized capacitor, the polarities of the meter should match
those of the capacitor. A low-resistance reading (zero ohms to a few
hundred ohms) normally indicates a defective capacitor.
The above test of leakage is not all-inclusive, since some capacitors
will break down only when higher voltages are applied. The test, however, does identify those capacitors that have lost the insulating quality
of the dielectric between the plates.
Standard Values and Recognition Factor
The standard values for capacitors employ the same numerical multipliers encountered for resistors. The most common have the same
numerical multipliers as the most common resistors, that is, those available with the full range of tolerances (5%, 10%, and 20%) as shown in
Table 3.8. They include 0.1 mF, 0.15 mF, 0.22 mF, 0.33 mF, 0.47 mF, and
0.68 mF, and then 1 mF, 1.5 mF, 2.2 mF, 3.3 mF, 4.7 mF, and so on.
Figure 10.22 was developed to establish a recognition factor when it
comes to the various types of capacitors. In other words, it will help
you to develop the skills to identify types of capacitors, their typical
range of values, and some of the most common applications. The figure
is certainly not all-inclusive, but it does offer a first step in establishing
a sense for what to expect for various applications.
Marking Schemes
Due to the small size of some capacitors, various marking schemes
have been adopted to provide the capacitance level, the tolerance, and,
if possible, the maximum working voltage. In general, however, the size
of the capacitor is the first indicator of its value. The smaller units are
typically in picofarads (pF) and the larger units in microfarads (mF).
Keeping this simple fact in mind will usually provide an immediate
indication of the expected capacitance level. On larger mF units, the
value can usually be printed on the jacket with the tolerance and maximum working voltage. However, smaller units need to use some form
of abbreviation as shown in Fig. 10.23. For very small units such as
appearing in Fig. 10.23(a), the value is recognized immediately as in
pF, with the K an indicator of a 10% tolerance level. Too often the K
is read as a multiplier of 103, and the capacitance read as 20,000 pF or
20 nF. For the unit of Fig. 10.23(b), there was room for a lowercase “n”
to represent a multiplier of 109. The presence of the lowercase “n” in
combination with the small size is clear indication that this is a 200-nF
capacitor. To avoid unnecessary confusion, the letters used for tolerance
do not include N or U or P, so any form of these letters will usually sug-
TYPES OF CAPACITORS
Type: Miniature Axial Electrolytic
Typical Values: 0.1 µF to 15,000 µF
Typical Voltage Range: 5 V to 450 V
Capacitor tolerance: ±20%
Applications: Polarized, used in DC
power supplies, bypass filters, DC
blocking.
Type: Silver Mica
Typical Value: 10 pF to 0.001 µ F
Typical Voltage Range: 50 V to 500 V
Capacitor tolerance: ±5%
Applications: Non-polarized, used in
oscillators, in circuits that require a
stable component over a range of
temperatures and voltages.
+
+
+
2
50 2 F
V
µ
Type: Miniature Radial Electrolyte
Typical Values: 0.1 µF to 15,000 µF
Typical Voltage Range: 5 V to 450 V
Capacitor tolerance: ±20%
Applications: Polarized, used in DC
power supplies, bypass filters, DC
blocking.
.33
10
ER 0V
8107 IE
2
10 00
V F
µ
+
+
+
Type: Mylar Paper
Typical Value: 0.001 µF to 0.68 µ F
Typical Voltage Range: 50 V to 600 V
Capacitor tolerance: ±22%
Applications: Non-polarized, used in
all types of circuits, moisture resistant.
0K
47 5P
Y
Type: Ceramic Disc
Typical Values: 10 pF to 0.047 µ F
Typical Voltage Range: 100 V to 6 kV
Capacitor tolerance: ±5%, ±10%
Applications: Non-polarized, NPO
type, stable for a wide range of
temperatures. Used in oscillators, noise
filters, circuit coupling, tank circuits.
Type: Dipped Tantalum (solid and wet)
Typical Values: 0.047 µ F to 470 µ F
Typical Voltage Range: 6.3 V to 50 V
Capacitor tolerance: ±10%, ±20%
Applications: Polarized, low leakage
current, used in power supplies, high
frequency noise filters, bypass filter.
60 .1 ±
0W 10
V %
EM
.0RM
005 Z C
Z51V
V
Type: AC/DC Motor Run
Typical Value: 0.25 µ F to 1200 µ F
Typical Voltage Range: 240 V to 660 V
Capacitor tolerance: ±10%
Applications: Non-polarized, used in
motor run-start, high-intensity lighting
supplies, AC noise filtering.
35
+
+
Type: Surface Mount Type (SMT)
Typical Values: 10 pF to 10 µ F
Typical Voltage Range: 6.3 V to 16 V
Capacitor tolerance: ±10%
Applications: Polarized and nonpolarized, used in all types of circuits,
requires a minimum amount of PC
board real estate.
Type: Trimmer Variable
Typical Value: 1.5 pF to 600 pF
Typical Voltage Range: 5 V to 100 V
Capacitor tolerance: ±10%
Applications: Non-polarized, used in
oscillators, tuning circuits, AC filters.
Type: Tuning variable
Typical Value: 10 pF to 600 pF
Typical Voltage Range: 5 V to 100 V
Capacitor tolerance: ±10%
Applications: Non-polarized, used in
oscillators, radio tuning circuit.
FIG. 10.22
Summary of capacitive elements.
20
K
200n
J
223F
339M
(a)
(b)
(c)
(d)
FIG. 10.23
Various marking schemes for small capacitors.

84-44
640
DIELEKTROL
No PCB'S
CAPACITOR
MADE IN USA
61L1019
10 µ f 90C
400VAC 60HZ
PROTECTED P922
B 10000AFC
16-10000
389
390

CAPACITORS
+ vR –
1
R
iC
+
2
+
E
C
–
gest the multiplier level. The J represents a 5% tolerance level. For
capacitors such as appearing in Fig. 10.23(c), the first two numbers are
actual digits of the value, while the third number is the power of a multiplier (or number of zeros to be added). The F represents a 1% tolerance level. Multipliers of 0.01 use an 8, while 9 is used for 0.1 as shown
for the capacitor of Fig. 10.23(d) where the M represents a 20% tolerance level.
–
vC
e
FIG. 10.24
Basic charging network.
E
R
10.7 TRANSIENTS IN CAPACITIVE NETWORKS:
CHARGING PHASE
iC
Rapid decay
Small change in iC
0
t
FIG. 10.25
iC during the charging phase.
E
vC
Small increase in vC
Rapid increase
0
t
FIG. 10.26
vC during the charging phase.
vR = 0 V
–
+
iC = 0 A
R
+
E
+
VC = E
–
–
FIG. 10.27
Open-circuit equivalent for a capacitor
following the charging phase.
R
+v = E–
R
+
E
–
iC = iR = E
R
+
vC = 0 V
–
FIG. 10.28
Short-circuit equivalent for a capacitor
(switch closed, t 0).
Section 10.3 described how a capacitor acquires its charge. Let us now
extend this discussion to include the potentials and current developed
within the network of Fig. 10.24 following the closing of the switch (to
position 1).
You will recall that the instant the switch is closed, electrons are
drawn from the top plate and deposited on the bottom plate by the battery, resulting in a net positive charge on the top plate and a negative
charge on the bottom plate. The transfer of electrons is very rapid at
first, slowing down as the potential across the capacitor approaches the
applied voltage of the battery. When the voltage across the capacitor
equals the battery voltage, the transfer of electrons will cease and the
plates will have a net charge determined by Q CVC CE.
Plots of the changing current and voltage appear in Figs. 10.25 and
10.26, respectively. When the switch is closed at t 0 s, the current
jumps to a value limited only by the resistance of the network and then
decays to zero as the plates are charged. Note the rapid decay in current
level, revealing that the amount of charge deposited on the plates per
unit time is rapidly decaying also. Since the voltage across the plates is
directly related to the charge on the plates by vC q/C, the rapid rate
with which charge is initially deposited on the plates will result in a
rapid increase in vC. Obviously, as the rate of flow of charge (I)
decreases, the rate of change in voltage will follow suit. Eventually, the
flow of charge will stop, the current I will be zero, and the voltage will
cease to change in magnitude—the charging phase has passed. At this
point the capacitor takes on the characteristics of an open circuit: a voltage drop across the plates without a flow of charge “between” the
plates. As demonstrated in Fig. 10.27, the voltage across the capacitor
is the source voltage since i iC iR 0 A and vR iRR (0)R 0 V. For all future analysis:
A capacitor can be replaced by an open-circuit equivalent once the
charging phase in a dc network has passed.
Looking back at the instant the switch is closed, we can also surmise
that a capacitor behaves as a short circuit the moment the switch is
closed in a dc charging network, as shown in Fig. 10.28. The current
i iC iR E/R, and the voltage vC E vR E iRR E (E/R)R E E 0 V at t 0 s.
Through the use of calculus, the following mathematical equation
for the charging current iC can be obtained:
E
iC et/RC
R
(10.13)
TRANSIENTS IN CAPACITIVE NETWORKS: CHARGING PHASE
The factor et/RC is an exponential function of the form ex, where
x t/RC and e 2.71828 . . . . A plot of ex for x ≥ 0 appears in
Fig. 10.29. Exponentials are mathematical functions that all students
of electrical, electronic, or computer systems must become very familiar with. They will appear throughout the analysis to follow in this
course, and in succeeding courses.
Our current interest in the function ex is limited to values of x
greater than zero, as noted by the curve of Fig. 10.25. All modern-day
scientific calculators have the function ex. To obtain ex, the sign of x
must be changed using the sign key before the exponential function is
keyed in. The magnitude of ex has been listed in Table 10.3 for a range
of values of x. Note the rapidly decreasing magnitude of ex with
increasing value of x.

391
1
e–x
0.3679
0.1353
0.0497
0
1
2
0.0183 0.0067
4
3
5
x
FIG. 10.29
The ex function (x ≥ 0).
TABLE 10.3
Selected values of ex.
x1
x2
x5
x 10
x 100
1
1
ex e0 0 1
e
1
1
1
e1 0.3679
e
2.71828 . . .
1
e2 2 0.1353
e
1
e5 5 0.00674
e
1
e10 0.0000454
e10
1
e100 3.72 1044
e100
The factor RC in Eq. (10.13) is called the time constant of the system and has the units of time as follows:
V Q
Q
V
RC t
I V
Q/t V
TABLE 10.4
iC versus t (charging phase).
Its symbol is the Greek letter t (tau), and its unit of measure is the second. Thus,
t RC
(seconds, s)
(10.14)
If we substitute t RC into the exponential function et/RC, we
obtain et/t. In one time constant, et/t et/t e1 0.3679, or the
function equals 36.79% of its maximum value of 1. At t 2t, et/t e2t/t e2 0.1353, and the function has decayed to only 13.53% of
its maximum value.
The magnitude of et/t and the percentage change between time constants have been tabulated in Tables 10.4 and 10.5, respectively. Note
that the current has dropped 63.2% (100% 36.8%) in the first time
constant but only 0.4% between the fifth and sixth time constants. The
rate of change of iC is therefore quite sensitive to the time constant
determined by the network parameters R and C. For this reason, the universal time constant chart of Fig. 10.30 is provided to permit a more
accurate estimate of the value of the function ex for specific time intervals related to the time constant. The term universal is used because the
axes are not scaled to specific values.
t
Magnitude
0
1t
2t
3t
4t
5t
6t
100%
36.8%
13.5%
5.0%
1.8%
Less than
0.67% ←
1% of maximum
0.24%





x0
TABLE 10.5
Change in iC between time constants.
(0 → 1)t
(1 → 2)t
(2 → 3)t
(3 → 4)t
(4
5)t
(5 → 6)t
63.2%
23.3%
8.6%
3.0%
1.2%
0.4% ← Less than 1%
392

CAPACITORS
y
1.0
0.9
0.8
y
=
1 – e –t/
0.7
0.632 (close to 2 3)
0.6
0.5
0.4
0.368 (close to 1 3)
0.3
y
0.2
=
e–t/
0.1
1
0
2
3
4
5
6
t
1
FIG. 10.30
Universal time constant chart.
Returning to Eq. (10.13), we find that the multiplying factor E/R is
the maximum value that the current iC can attain, as shown in Fig.
10.25. Substituting t 0 s into Eq. (10.13) yields
E
E
E
iC et/RC e0 R
R
R
E
R
verifying our earlier conclusion.
For increasing values of t, the magnitude of et/t, and therefore the
value of iC, will decrease, as shown in Fig. 10.31. Since the magnitude
of iC is less than 1% of its maximum after five time constants, we will
assume the following for future analysis:
iC
iC = E e–t/ R
36.8%
The current iC of a capacitive network is essentially zero after five
time constants of the charging phase have passed in a dc network.
13.5%
5%
0
1
2
3
1.8%
0.67%
4
5
FIG. 10.31
iC versus t during the charging phase.
t
Since C is usually found in microfarads or picofarads, the time constant t RC will never be greater than a few seconds unless R is very
large.
Let us now turn our attention to the charging voltage across the
capacitor. Through further mathematical analysis, the following equation for the voltage across the capacitor can be determined:
vC E(1 et/RC)
(10.15)
Note the presence of the same factor et/RC and the function (1 et/RC) appearing in Fig. 10.30. Since et/t is a decaying function, the
factor (1 et/t) will grow toward a maximum value of 1 with time, as
shown in Fig. 10.30. In addition, since E is the multiplying factor, we
can conclude that, for all practical purposes, the voltage vC is E volts
TRANSIENTS IN CAPACITIVE NETWORKS: CHARGING PHASE
after five time constants of the charging phase. A plot of vC versus t is
provided in Fig. 10.32.
If we keep R constant and reduce C, the product RC will decrease,
and the rise time of five time constants will decrease. The change in
transient behavior of the voltage vC is plotted in Fig. 10.33 for various
values of C. The product RC will always have some numerical value,
even though it may be very small in some cases. For this reason:
vC = E(1 – e–t/ )
63.2%
1
0
2
3
4
charging
5
t
FIG. 10.32
vC versus t during the charging phase.
vC
E
C1
C2
C3 > C2 > C1,
R fixed
C3
q CvC CE(1 et/t)
t
0
(10.16)
FIG. 10.33
Effect of C on the charging phase.
indicating that the charging rate is very high during the first few time
constants and less than 1% after five time constants.
The voltage across the resistor is determined by Ohm’s law:
vR
E
vR iRR RiC Ret/t
R
E
vR Eet/t
or
99.3%
95% 98.2%
86.5%
q
vC C
and
393
vC
E
The voltage across a capacitor cannot change instantaneously.
In fact, the capacitance of a network is also a measure of how much it
will oppose a change in voltage across the network. The larger the
capacitance, the larger the time constant, and the longer it takes to
charge up to its final value (curve of C3 in Fig. 10.33). A lesser capacitance would permit the voltage to build up more quickly since the time
constant is less (curve of C1 in Fig. 10.33).
The rate at which charge is deposited on the plates during the charging phase can be found by substituting the following for vC in Eq.
(10.15):

(10.17)
vR = Ee–t/t
36.8%
A plot of vR appears in Fig. 10.34.
Applying Kirchhoff’s voltage law to the circuit of Fig. 10.24 will
result in
13.5%
5%
vC E vR
0
Substituting Eq. (10.17):
1t
2t
1.8% 0.67%
3t
4t
t
5t
FIG. 10.34
vR versus t during the charging phase.
t/t
vC E Ee
Factoring gives vC E(1 et/t), as obtained earlier.
EXAMPLE 10.5
a. Find the mathematical expressions for the transient behavior of vC,
iC, and vR for the circuit of Fig. 10.35 when the switch is moved to
position 1. Plot the curves of vC, iC, and vR.
b. How much time must pass before it can be assumed, for all practical
purposes, that iC 0 A and vC E volts?
+ vR –
(t = 0)
1
R
2
+
–
Solutions:
a. t RC (8 103 )(4 106 F) 32 103 s 32 ms
By Eq. (10.15),
vC E(1 et/t) 40(1 et/(3210 ))
3
+
C
40 V
E
iC
8 k
4 mF vC
–
FIG. 10.35
Example 10.5.
394

CAPACITORS
By Eq. (10.13),
vC (V)
40
E
40 V
iC et/t et/(3210
8 k
R
t = 32 ms
3
)
(5 103)et/(3210
3
0
1t
2t
3t
4t
t
5t
)
By Eq. (10.17),
vR Eet/t 40et/(3210
3
iC (mA)
5.0
The curves appear in Fig. 10.36.
b. 5t 5(32 ms) 160 ms
t = 32 ms
0
1t
2t
3t
4t
5t
t
5t
t
Once the voltage across the capacitor has reached the input voltage
E, the capacitor is fully charged and will remain in this state if no further changes are made in the circuit.
If the switch of Fig. 10.24 is opened, as shown in Fig. 10.37(a), the
capacitor will retain its charge for a period of time determined by its
leakage current. For capacitors such as the mica and ceramic, the leakage current (ileakage vC /Rleakage) is very small, enabling the capacitor
to retain its charge, and hence the potential difference across its plates,
for a long time. For electrolytic capacitors, which have very high leakage currents, the capacitor will discharge more rapidly, as shown in Fig.
10.37(b). In any event, to ensure that they are completely discharged,
capacitors should be shorted by a lead or a screwdriver before they are
handled.
vR (V)
40
t = 32 ms
0
1t
2t
3t
)
4t
FIG. 10.36
Waveforms for the network of Fig. 10.35.
vR = 0 V
1
vR = 0 V
i = 0
i = 0
R
2
vC
E
2
+
E
+
vC
Rleakage
–
–
ileakage
(b)
(a)
FIG. 10.37
Effect of the leakage current on the steady-state behavior of a capacitor.
10.8
– vR +
R
2
+
vC = E
–
iC = iR = idischarge
FIG. 10.38
Demonstrating the discharge behavior of a
capacitive network.
DISCHARGE PHASE
The network of Fig. 10.24 is designed to both charge and discharge the
capacitor. When the switch is placed in position 1, the capacitor will
charge toward the supply voltage, as described in the last section. At
any point in the charging process, if the switch is moved to position 2,
the capacitor will begin to discharge at a rate sensitive to the same time
constant t RC. The established voltage across the capacitor will create a flow of charge in the closed path that will eventually discharge the
capacitor completely. In essence, the capacitor functions like a battery
with a decreasing terminal voltage. Note in particular that the current iC
has reversed direction, changing the polarity of the voltage across R.
If the capacitor had charged to the full battery voltage as indicated in
Fig. 10.38, the equation for the decaying voltage across the capacitor
would be the following:
vC Eet/RC
discharging
(10.18)
DISCHARGE PHASE

395
which employs the function ex and the same time constant used above.
The resulting curve will have the same shape as the curve for iC and vR
in the last section. During the discharge phase, the current iC will also
decrease with time, as defined by the following equation:
E
iC et/RC
R
(10.19)
discharging
The voltage vR vC, and
vR Eet/RC
(10.20)
discharging
The complete discharge will occur, for all practical purposes, in five
time constants. If the switch is moved between terminals 1 and 2 every
five time constants, the wave shapes of Fig. 10.39 will result for vC, iC,
and vR. For each curve, the current direction and voltage polarities were
defined by Fig. 10.24. Since the polarity of vC is the same for both the
charging and the discharging phases, the entire curve lies above the
axis. The current iC reverses direction during the charging and discharging phases, producing a negative pulse for both the current and the
voltage vR. Note that the voltage vC never changes magnitude instantaneously but that the current iC has the ability to change instantaneously,
as demonstrated by its vertical rises and drops to maximum values.
Charging Discharging
vC
E
0
Pos. 1 5t Pos. 2 10t Pos. 1 15t Pos. 2
t
E iC
R
5t
0
15t
10t
Pos. 1
t
Pos. 1
Pos. 2
–E
Pos. 2
R
E
vR
5t
0
15t
10t
Pos. 1
t
Pos. 1
Pos. 2
Pos. 2
–E
– vR +
iC
R
FIG. 10.39
The charging and discharging cycles for the
network of Fig. 10.24.
8 k
2
+
C
4 F vC = E = 40 V
–
EXAMPLE 10.6 After vC in Example 10.5 has reached its final value
of 40 V, the switch is thrown into position 2, as shown in Fig. 10.40.
Find the mathematical expressions for the transient behavior of vC, iC,
FIG. 10.40
Example 10.6.
396

CAPACITORS
and vR after the closing of the switch. Plot the curves for vC, iC, and vR
using the defined directions and polarities of Fig. 10.35. Assume that
t 0 when the switch is moved to position 2.
vC (V)
40
t = 32 ms
Solution:
0
0
1t
2t
3t
4t
5t
t
By Eq. (10.18),
iC (mA)
1t
t 32 ms
2t
3t
4t
t = 32 ms
5t
vC Eet/t 40et/(3210
3
t
)
By Eq. (10.19),
5.0
0
E
iC et/t (5 103)et/(3210
R
3
vR (V)
)
By Eq. (10.20),
1t
4t
3t
t = 32 ms
2t
5t
t
vR Eet/t 40et/(3210
3
)
The curves appear in Fig. 10.41.
40
FIG. 10.41
The waveforms for the network of Fig. 10.40.
The preceding discussion and examples apply to situations in which
the capacitor charges to the battery voltage. If the charging phase is disrupted before reaching the supply voltage, the capacitive voltage will be
less, and the equation for the discharging voltage vC will take on the
form
vC Viet/RC
(10.21)
where Vi is the starting or initial voltage for the discharge phase. The
equation for the decaying current is also modified by simply substituting Vi for E; that is,
Vi
iC et/t Ii et/t
R
(10.22)
Use of the above equations will be demonstrated in Examples 10.7 and
10.8.
R1
1
2
3
100 k
E
10 V
+
vC C
–
iC
R2
0.05 mF
FIG. 10.42
Example 10.7.
200 k
EXAMPLE 10.7
a. Find the mathematical expression for the transient behavior of the
voltage across the capacitor of Fig. 10.42 if the switch is thrown into
position 1 at t 0 s.
b. Repeat part (a) for iC.
c. Find the mathematical expressions for the response of vC and iC if
the switch is thrown into position 2 at 30 ms (assuming that the
leakage resistance of the capacitor is infinite ohms).
d. Find the mathematical expressions for the voltage vC and current iC
if the switch is thrown into position 3 at t 48 ms.
e. Plot the waveforms obtained in parts (a) through (d) on the same
time axis for the voltage vC and the current iC using the defined
polarity and current direction of Fig. 10.42.
DISCHARGE PHASE
Solutions:
a. Charging phase:
vC E(1 et/t)
t R1C (100 103 )(0.05 106 F) 5 103 s
5 ms
vC 10(1 et/(510 ))
3
E
b. iC et/t
R1
10 V
et/(510 )
100 103 iC (0.1 103)et/(510 )
3
3
c. Storage phase:
vC E 10 V
iC 0 A
d. Discharge phase (starting at 48 ms with t 0 s for the following
equations):
vC Eet/t′
t′ R2C (200 103 )(0.05 106 F) 10 103 s
10 ms
vC 10et/(1010 )
E
iC et/t′
R2
3
10 V
et/(1010
200 103 3
)
iC (0.05 103)et/(1010
3
)
e. See Fig. 10.43.
vC (V)
10 V
98
48
0
10
5t
20 25 30
40
50
60
70
80
5t'
(t' = 2t)
90
48
50
60
70
90
100 t (ms)
iC (mA)
0.1
0
10
5t
20 25 30
80
40
98
100
t (ms)
5t'
(t' = 2t)
–0.05
FIG. 10.43
The waveforms for the network of Fig. 10.42.

397
398

CAPACITORS
EXAMPLE 10.8
a. Find the mathematical expression for the transient behavior of the
voltage across the capacitor of Fig. 10.44 if the switch is thrown into
position 1 at t 0 s.
iC
1
4 mA
2
I
+
5 k
R1
C
1 k
R2
10 mF vC
–
R3
3 k
FIG. 10.44
Example 10.8.
b. Repeat part (a) for iC.
c. Find the mathematical expression for the response of vC and iC if the
switch is thrown into position 2 at t 1t of the charging phase.
d. Plot the waveforms obtained in parts (a) through (c) on the same
time axis for the voltage vC and the current iC using the defined
polarity and current direction of Fig. 10.44.
Solutions:
a. Charging phase: Converting the current source to a voltage source
will result in the network of Fig. 10.45.
R1
E
iC
1
5 k
2
+
20 V
C
R2
1 k
10 mF vC
–
R3
3 k
FIG. 10.45
The charging phase for the network of Fig. 10.44.
vC E(1 et/t )
t1 (R1 R3)C (5 k 3 k)(10 106 F)
80 ms
t/(8010 )
vC 20(1 e
)
1
iC
2
+
vC C
R2
–
1 k
3
+
10 mF 12.64 V
–
R3
3 k
FIG. 10.46
Network of Fig. 10.45 when the switch is
moved to position 2 at t 1t1.
E
b. iC et/t
R1 R3
1
20 V
et/(8010
8 k
3
)
iC (2.5 103)et/(8010
3
)
c. At t 1t1, vC 0.632E 0.632(20 V) 12.64 V, resulting in the
network of Fig. 10.46. Then vC Viet/t with
2
INITIAL VALUES
t2 (R2 R3)C (1 k 3 k)(10 106 F)
40 ms
vC 12.64et/(4010
3
and
)
At t 1t1, iC drops to (0.368)(2.5 mA) 0.92 mA. Then it
switches to
iC Iiet/t
2
Vi
12.64 V
et/t et/(4010
1 k 3 k
R2 R3
3
2
iC 3.16 103et/(4010
)
3
)
d. See Fig. 10.47.
vC (V)
20 V
12.64 V
0
80
1
160
240
320
400
t (ms)
240
320
400
t (ms)
5 2
iC (mA)
2.5
0.92
0
1
5 2
–3.16
FIG. 10.47
The waveforms for the network of Fig. 10.44.
10.9
INITIAL VALUES
In all the examples examined in the previous sections, the capacitor was
uncharged before the switch was thrown. We will now examine the
effect of a charge, and therefore a voltage (V Q/C), on the plates at
the instant the switching action takes place. The voltage across the
capacitor at this instant is called the initial value, as shown for the general waveform of Fig. 10.48. Once the switch is thrown, the transient

399
400

CAPACITORS
vC
Vi
Initial
conditions
0
Vf
Transient
response
Steady-state
region
t
FIG. 10.48
Defining the regions associated with a transient response.
phase will commence until a leveling off occurs after five time constants.
This region of relatively fixed value that follows the transient response
is called the steady-state region, and the resulting value is called the
steady-state or final value. The steady-state value is found by simply
substituting the open-circuit equivalent for the capacitor and finding the
voltage across the plates. Using the transient equation developed in the
previous section, an equation for the voltage vC can be written for the
entire time interval of Fig. 10.48; that is,
vC Vi (Vf Vi)(1 et/t)
However, by multiplying through and rearranging terms:
vC Vi Vf Vf et/t Vi Vi et/t
Vf Vf et/t Vi et/t
we find
vC Vf (Vi Vf)et/t
(10.23)
If you are required to draw the waveform for the voltage vC from the
initial value to the final value, start by drawing a line at the initial and
steady-state levels, and then add the transient response (sensitive to the
time constant) between the two levels. The example to follow will clarify the procedure.
EXAMPLE 10.9 The capacitor of Fig. 10.49 has an initial voltage
of 4 V.
R1
iC
2.2 k
+
E
vC C
24 V
–
R2
1.2 k
FIG. 10.49
Example 10.9.
+
3.3 F 4 V
–
INITIAL VALUES
a. Find the mathematical expression for the voltage across the capacitor once the switch is closed.
b. Find the mathematical expression for the current during the transient
period.
c. Sketch the waveform for each from initial value to final value.
Solutions:
a. Substituting the open-circuit equivalent for the capacitor will result
in a final or steady-state voltage vC of 24 V.
The time constant is determined by
with
t (R1 R2)C
(2.2 k 1.2 k)(3.3 mF)
11.22 ms
5t 56.1 ms
Applying Eq. (10.23):
and
vC Vf (Vi Vf)et/t
24 V (4 V 24 V)et/11.22ms
vC 24 V 20 Vet/11.22ms
b. Since the voltage across the capacitor is constant at 4 V prior to the
closing of the switch, the current (whose level is sensitive only to
changes in voltage across the capacitor) must have an initial value of
0 mA. At the instant the switch is closed, the voltage across the
capacitor cannot change instantaneously, so the voltage across the
resistive elements at this instant is the applied voltage less the initial
voltage across the capacitor. The resulting peak current is
20 V
24 V 4 V
E VC
Im 5.88 mA
3.4 k
2.2 k 1.2 k
R1 R2
The current will then decay (with the same time constant as the
voltage vC) to zero because the capacitor is approaching its opencircuit equivalence.
The equation for iC is therefore:
iC 5.88 mAet/11.22ms
c. See Fig. 10.50.
vC
24 V
5
4V
0
56.1 ms
t
56.1 ms
t
iC
5.88 mA
0
FIG. 10.50
vC and iC for the network of Fig. 10.49.

401
402

CAPACITORS
The initial and final values of the voltage were drawn first, and
then the transient response was included between these levels. For
the current, the waveform begins and ends at zero, with the peak
value having a sign sensitive to the defined direction of iC in Fig.
10.49.
Let us now test the validity of the equation for vC by substituting
t 0 s to reflect the instant the switch is closed.
et/t e0 1
and
vC 24 V 20 Vet/t 24 V 20 V 4 V
When t 5t,
et/t 0
and
vC 24 V 20 Vet/t 24 V 0 V 24 V
10.10 INSTANTANEOUS VALUES
On occasion it will be necessary to determine the voltage or current at
a particular instant of time that is not an integral multiple of t, as in the
previous sections. For example, if
vC 20(1 et/(210 ))
3
the voltage vC may be required at t 5 ms, which does not correspond
to a particular value of t. Figure 10.30 reveals that (1 et/t) is
approximately 0.93 at t 5 ms 2.5t, resulting in vC 20(0.93) 18.6 V. Additional accuracy can be obtained simply by substituting t 5 ms into the equation and solving for vC using a calculator or table to
determine e2.5. Thus,
vC 20(1 e5ms/2ms)
20(1 e2.5)
20(1 0.082)
20(0.918)
18.36 V
The results are close, but accuracy beyond the tenths place is suspect
using Fig. 10.30. The above procedure can also be applied to any other
equation introduced in this chapter for currents or other voltages.
There are also occasions when the time to reach a particular voltage
or current is required. The procedure is complicated somewhat by the
use of natural logs (loge, or ln), but today’s calculators are equipped to
handle the operation with ease. There are two forms that require some
development. First, consider the following sequence:
vC E(1 et/t)
vC
1 et/t
E
vC
1 et/t
E
vC
loge 1 logeet/t
E
vC
t
loge 1 t
E
INSTANTANEOUS VALUES
and
but
Therefore,
vC
t t loge 1 E
x
y
loge loge
y
x
E
t t loge E vC
(10.24)
The second form is as follows:
and
vC Eet/t
vC
et/t
E
vC
loge
logeet/t
E
vC
t
loge
tv
E
C
t t loge
E
or
E
t t loge
vC
(10.25)
E
t t loge
iCR
(10.26)
For iC (E/R)et/t:
For example, suppose that
vC 20(1 et/(210 ))
3
and the time to reach 10 V is required. Substituting into Eq. (10.24), we
have
20 V
t (2 ms)loge 20 V 10 V
(2 ms)loge2
(2 ms)(0.693)
1.386 ms
F
IN key on calculator
Using Fig. 10.30, we find at (1 et/t) vC /E 0.5 that t 0.7t 0.7(2 ms) 1.4 ms, which is relatively close to the above.
Mathcad
It is time to see how Mathcad can be applied to the transient analysis
described in this chapter. For the first equation described in Section
10.10,
3
vC 20(1 et/(210 ))
the value of t must be defined before the expression is written, or the
value can simply be inserted in the equation. The former approach is

403
404

CAPACITORS
often better because changing the defined value of t will result in an
immediate change in the result. In other words, the value can be used
for further calculations. In Fig. 10.51 the value of t was defined as 5 ms.
The equation was then entered using the e function from the Calculator palette obtained from View-Toolbars-Calculator. Be sure to insert
a multiplication operator between the initial 20 and the main left
bracket. Also, be careful that the control bracket is in the correct place
before placing the right bracket to enclose the equation. It will take
some practice to ensure that the insertion bracket is in the correct place
before entering a parameter, but in time you will find that it is a fairly
direct procedure. The 3 is placed using the shift operator over the
number 6 on the standard keyboard. The result is displayed by simply
entering v again, followed by an equal sign. The result for t 1 ms can
now be obtained by simply changing the defined value for t. The result
of 7.869 V will appear immediately.
FIG. 10.51
Applying Mathcad to the transient R-C equations.
For the second equation of Section 10.10,
vC 20(1 e5ms/2ms)
the equation for t can be entered directly as shown in the bottom of Fig.
10.51. The ln from the Calculator is for a base e calculation, while log
is for a base 10 calculation. The result will appear the instant the equal
sign is placed after the t on the bottom line.
The text you see on the screen to define each operation is obtained
by clicking on Insert-Text Region and then simply typing in the text
material. The boldface was obtained by simply clicking on the text
material and swiping the text to establish a black background. Then
select B from the toolbar, and the boldface will appear.
THÉVENIN EQUIVALENT: t RThC

405
10.11 THÉVENIN EQUIVALENT: t RThC
Occasions will arise in which the network does not have the simple
series form of Fig. 10.24. It will then be necessary first to find the
Thévenin equivalent circuit for the network external to the capacitive
element. ETh will then be the source voltage E of Eqs. (10.15) through
(10.20), and RTh will be the resistance R. The time constant is then t RThC.
EXAMPLE 10.10 For the network of Fig. 10.52:
60 k
10 k
R1
R3
+
E
21 V
–
30 k
R2
1
C = 0.2 mF
iC 2
+
vC R4
10 k
–
FIG. 10.52
Example 10.10.
a. Find the mathematical expression for the transient behavior of the
voltage vC and the current iC following the closing of the switch
(position 1 at t 0 s).
b. Find the mathematical expression for the voltage vC and current iC
as a function of time if the switch is thrown into position 2 at t 9 ms.
c. Draw the resultant waveforms of parts (a) and (b) on the same time
axis.
Solutions:
a. Applying Thévenin’s theorem to the 0.2-mF capacitor, we obtain
Fig. 10.53:
(60 k)(30 k)
RTh R1 R2 R3 10 k
90 k
20 k 10 k
RTh 30 k
1
R2E
(30 k)(21 V)
ETh (21 V) 7 V
3
R2 R1
30 k 60 k
The resultant Thévenin equivalent circuit with the capacitor
replaced is shown in Fig. 10.54. Using Eq. (10.23) with Vf ETh
and Vi 0 V, we find that
becomes
or
with
so that
vC Vf (Vi Vf )et/t
vC ETh (0 V ETh)et/t
vC ETh(1 et/t)
t RC (30 k)(0.2 mF) 6 ms
vC 7(1 et/6ms)
For the current:
E t/RC
iC Th
e
R
7V
et/6ms
30 k
iC (0.233 103)et/6ms
RTh:
10 k
60 k
R3
R1
R2
ETh:
60 k
–
10 k
R3
R1
+
E
21 V
RTh
30 k
R2
ETh
30 k
FIG. 10.53
Applying Thévenin’s theorem to the network of
Fig. 10.52.
RTh = 30 k
ETh = 7 V
C = 0.2 mF
iC
+
vC
–
FIG. 10.54
Substituting the Thévenin equivalent for the
network of Fig. 10.52.
406

CAPACITORS
b. At t 9 ms,
vC ETh(1 et/t) 7(1 e(910 )/(610 ))
7(1 e1.5) 7(1 0.223)
vC 7(0.777) 5.44 V
E t/t
iC Th
e
(0.233 103)e1.5
R
(0.233 103)(0.223)
iC 0.052 103 0.052 mA
3
and
3
Using Eq. (10.23) with Vf 0 V and Vi 5.44 V, we find that
vC Vf (Vi Vf )et/t′
vC 0 V (5.44 V 0 V)et/t′
5.44et/t′
t′ R4C (10 k)(0.2 mF) 2 ms
vC 5.44et/2ms
becomes
with
and
By Eq. (10.22),
5.44 V
Ii 0.054 mA
10 k
iC Iiet/t (0.54 103)et/2ms
and
c. See Fig. 10.55.
vC (V)
ETh = 7
Vi = 5.44 V
5 '
0
5
10
15
5
20
25
30
35
t (ms)
20
25
30
35
t (ms)
iC (mA)
0.233
5 '
0.052
0
10
15
5
5
R1
7 k
R2
18 k
5 k R3
– 0.54
+ vC –
C
E
120 V
40 F
+ 40 V –
FIG. 10.56
Example 10.11.
R4
2 k
FIG. 10.55
The resulting waveforms for the network of Fig. 10.52.
EXAMPLE 10.11 The capacitor of Fig. 10.56 is initially charged to
40 V. Find the mathematical expression for vC after the closing of the
switch.
THÉVENIN EQUIVALENT: t RThC
Solution:
The network is redrawn in Fig. 10.57.
R2
R1
Thévenin 5 k
7 k
+
40 mF
40 V C
–
R4
R3
2 k
18 k
E
120 V
FIG. 10.57
Network of Fig. 10.56 redrawn.
ETh:
R3E
18 k(120 V)
ETh R3 R1 R4
18 k 7 k 2 k
80 V
RTh:
RTh 5 k 18 k (7 k 2 k)
5 k 6 k
11 k
Vi 40 V and Vf 80 V
t RThC (11 k)(40 mF) 0.44 s
vC Vf (Vi Vf)et/t
80 V (40 V 80 V)et/0.44s
vC 80 V 40 Vet/0.44s
Therefore,
and
Eq. (10.23):
and
EXAMPLE 10.12 For the network of Fig. 10.58, find the mathematical
expression for the voltage vC after the closing of the switch (at t 0).
R2
10 I
20 mA
R1 = 6 +
C
500 mF vC
–
FIG. 10.58
Example 10.12.
Solution:
RTh R1 R2 6 10 16 ETh V1 V2 IR1 0
(20 103 A)(6 ) 120 103 V 0.12 V
and
so that
t RThC (16 )(500 106 F) 8 ms
vC 0.12(1 et/8ms)

407
408

CAPACITORS
10.12 THE CURRENT iC
The current iC associated with a capacitance C is related to the voltage
across the capacitor by
dvC
iC C
dt
(10.27)
where dvC /dt is a measure of the change in vC in a vanishingly small
period of time. The function dvC /dt is called the derivative of the voltage vC with respect to time t.
If the voltage fails to change at a particular instant, then
dvC 0
and
dvC
iC C 0
dt
In other words, if the voltage across a capacitor fails to change with
time, the current iC associated with the capacitor is zero. To take this a
step further, the equation also states that the more rapid the change in
voltage across the capacitor, the greater the resulting current.
In an effort to develop a clearer understanding of Eq. (10.27), let us
calculate the average current associated with a capacitor for various
voltages impressed across the capacitor. The average current is defined
by the equation
vC
iCav C t
(10.28)
where indicates a finite (measurable) change in charge, voltage, or
time. The instantaneous current can be derived from Eq. (10.28) by letting t become vanishingly small; that is,
vC
dvC
iCinst lim C C
∆t→0
t
dt
In the following example, the change in voltage vC will be considered for each slope of the voltage waveform. If the voltage increases
with time, the average current is the change in voltage divided by the
change in time, with a positive sign. If the voltage decreases with time,
the average current is again the change in voltage divided by the change
in time, but with a negative sign.
EXAMPLE 10.13 Find the waveform for the average current if the
voltage across a 2-mF capacitor is as shown in Fig. 10.59.
Solutions:
a. From 0 ms to 2 ms, the voltage increases linearly from 0 V to 4 V,
the change in voltage v 4 V 0 4 V (with a positive sign
since the voltage increases with time). The change in time t 2 ms 0 2 ms, and
vC
4V
iCav C
(2 106 F) 2 103 s
t
4 103 A 4 mA
THE CURRENT iC
vC (V)
4
v vv3
2
0
1
2
t1
t2 t 3
t
3
4
5
6
7
8
9
10
11
12
t (ms)
FIG. 10.59
Example 10.13.
b. From 2 ms to 5 ms, the voltage remains constant at 4 V; the change
in voltage v 0. The change in time t 3 ms, and
vC
0
iCav C
C 0
t
t
c. From 5 ms to 11 ms, the voltage decreases from 4 V to 0 V. The change
in voltage v is, therefore, 4 V 0 4 V (with a negative sign since
the voltage is decreasing with time). The change in time t 11 ms 5 ms 6 ms, and
vC
4V
iCav C (2 106 F) 6 103 s
t
1.33 103 A 1.33 mA
d. From 11 ms on, the voltage remains constant at 0 and v 0, so
iCav 0. The waveform for the average current for the impressed
voltage is as shown in Fig. 10.60.
iC (mA)
4
0
1
2
3
4
5
6
7
8
9
10
–1.33
FIG. 10.60
The resulting current iC for the applied voltage of Fig. 10.59.
Note in Example 10.13 that, in general, the steeper the slope, the
greater the current, and when the voltage fails to change, the current is
zero. In addition, the average value is the same as the instantaneous
value at any point along the slope over which the average value was
found. For example, if the interval t is reduced from 0 → t1 to t2 t3,
as noted in Fig. 10.59, v/t is still the same. In fact, no matter how
small the interval t, the slope will be the same, and therefore the current iCav will be the same. If we consider the limit as t → 0, the slope
will still remain the same, and therefore iCav iCinst at any instant of
11
12
t (ms)

409
410

CAPACITORS
time between 0 and t1. The same can be said about any portion of the
voltage waveform that has a constant slope.
An important point to be gained from this discussion is that it is not
the magnitude of the voltage across a capacitor that determines the current but rather how quickly the voltage changes across the capacitor. An
applied steady dc voltage of 10,000 V would (ideally) not create any
flow of charge (current), but a change in voltage of 1 V in a very brief
period of time could create a significant current.
The method described above is only for waveforms with straight-line
(linear) segments. For nonlinear (curved) waveforms, a method of calculus (differentiation) must be employed.
10.13 CAPACITORS IN SERIES AND PARALLEL
Q1
–
+
+
V1
Q3
Q2
–
+
V2
QT
E
–
–
+
V3
Capacitors, like resistors, can be placed in series and in parallel.
Increasing levels of capacitance can be obtained by placing capacitors
in parallel, while decreasing levels can be obtained by placing capacitors in series.
For capacitors in series, the charge is the same on each capacitor
(Fig. 10.61):
QT Q1 Q2 Q3
FIG. 10.61
Series capacitors.
(10.29)
Applying Kirchhoff’s voltage law around the closed loop gives
E V1 V2 V3
However,
so that
Q
V C
QT
Q
Q
Q
1 2 3
CT
C1
C2
C3
Using Eq. (10.29) and dividing both sides by Q yields
1
1
1
1
CT
C1
C2
C3
(10.30)
which is similar to the manner in which we found the total resistance of
a parallel resistive circuit. The total capacitance of two capacitors in
series is
C C2
CT 1
C1 C2
(10.31)
The voltage across each capacitor of Fig. 10.61 can be found by first
recognizing that
QT Q1
or
CT E C1V1
Solving for V1:
CT E
V1 C1
CAPACITORS IN SERIES AND PARALLEL

411
and substituting for CT:
1/C1
V1 E
1/C1 1/C2 1/C3
(10.32)
A similar equation will result for each capacitor of the network.
For capacitors in parallel, as shown in Fig. 10.62, the voltage is the
same across each capacitor, and the total charge is the sum of that on
each capacitor:
QT
+
+ Q1
E
–
QT Q1 Q2 Q3
However,
Q CV
Therefore,
CT E C1V1 C2V2 C3V3
V1
–
+ Q2
–
+ Q3
V2
–
(10.33)
FIG. 10.62
Parallel capacitors.
E V1 V2 V3
but
Thus,
CT C1 C2 C3
(10.34)
which is similar to the manner in which the total resistance of a series
circuit is found.
EXAMPLE 10.14 For the circuit of Fig. 10.63:
a. Find the total capacitance.
b. Determine the charge on each plate.
c. Find the voltage across each capacitor.
Solutions:
1
1
1
1
a. CT
C1
C2
C3
1
1
1
200 106 F
50 106 F
10 106 F
0.005 106 0.02 106 0.1 106
0.125 106
and
1
CT 6 8 mF
0.125 10
b. QT Q1 Q2 Q3
QT CT E (8 106 F)(60 V) 480 mC
Q
480 106 C
c. V1 1 2.4 V
C1
200 106 F
Q
480 106 C
9.6 V
V2 2 C2
50 106 F
Q
480 106 C
V3 3 48.0 V
C3
10 106 F
and
E V1 V2 V3 2.4 V 9.6 V 48 V
60 V (checks)
+
CT
C1
C2
C3
200 mF
50 mF
10 mF
60 V
E
–
FIG. 10.63
Example 10.14.
V3

412
CAPACITORS
CT
E
C2
C1
800 mF
48 V
C3
1200 mF
60 mF
Solutions:
a. CT C1 C2 C3 800 mF 60 mF 1200 mF
2060 mF
b. Q1 C1E (800 106 F)(48 V) 38.4 mC
Q2 C2E (60 106 F)(48 V) 2.88 mC
Q3 C3E (1200 106 F)(48 V) 57.6 mC
c. QT Q1 Q2 Q3 38.4 mC 2.88 mC 57.6 mC
98.88 mC
QT
FIG. 10.64
Example 10.15.
+
E = 120 V
EXAMPLE 10.16 Find the voltage across and charge on each capacitor for the network of Fig. 10.65.
3 mF
C1
C2
Solution:
C3
–
EXAMPLE 10.15 For the network of Fig. 10.64:
a. Find the total capacitance.
b. Determine the charge on each plate.
c. Find the total charge.
4 mF
C′T C2 C3 4 mF 2 mF 6 mF
(3 mF)(6 mF)
C1C′T
CT 2 mF
C1 C′T
3 mF 6 mF
QT CT E (2 106 F)(120 V)
240 mC
2 mF
FIG. 10.65
Example 10.16.
An equivalent circuit (Fig. 10.66) has
C1
3 mF
QT Q1 Q′T
Q
+
E = 120 V
+ V –1
1
C'T
+
V'T
–
Q'T 6 mF –
Q2
C2
Q3 +
V'T
C3 –
FIG. 10.66
Reduced equivalent for the network of
Fig. 10.65.
Q1 240 mC
and, therefore,
Q
240 106 C
V1 1 80 V
C1
3 106 F
Q′T 240 mC
and
and, therefore,
Q′T
240 106 C
V′T 40 V
C′T
6 106 F
and
Q2 C2V′T (4 106 F)(40 V) 160 mC
Q3 C3V′T (2 106 F)(40 V) 80 mC
EXAMPLE 10.17 Find the voltage across and charge on capacitor C1
of Fig. 10.67 after it has charged up to its final value.
R1
+
E = 24 V
–
4
R2
8
+
Q1
C1 = 20 mF VC
–
FIG. 10.67
Example 10.17.
ENERGY STORED BY A CAPACITOR
Solution: As previously discussed, the capacitor is effectively an
open circuit for dc after charging up to its final value (Fig. 10.68).
Therefore,
(8 )(24 V)
VC 16 V
48
Q1 C1VC (20 106 F)(16 V)
320 mC

4
413
+
8
E = 24 V
VC
–
FIG. 10.68
Determining the final (steady-state) value
for vC.
EXAMPLE 10.18 Find the voltage across and charge on each capacitor of the network of Fig. 10.69 after each has charged up to its final
value.
C1 = 2 mF
+ VC1 –
C2 = 3 mF
R1
2
2
+
E = 72 V
–
+ VC2 –
R1
I = 0
+
R2
7
R3
8
–
E = 72 V
R2
7
8
R3
FIG. 10.69
Example 10.18.
Solution:
(7 )(72 V)
VC2 56 V
72
(2 )(72 V)
VC1 16 V
27
Q1 C1VC1 (2 106 F)(16 V) 32 mC
Q2 C2VC2 (3 106 F)(56 V) 168 mC
10.14
ENERGY STORED BY A CAPACITOR
The ideal capacitor does not dissipate any of the energy supplied to it.
It stores the energy in the form of an electric field between the conducting surfaces. A plot of the voltage, current, and power to a capacitor during the charging phase is shown in Fig. 10.70. The power curve
can be obtained by finding the product of the voltage and current at
selected instants of time and connecting the points obtained. The energy
stored is represented by the shaded area under the power curve. Using
calculus, we can determine the area under the curve:
1
WC CE2
2
In general,
1
WC CV 2
2
(J)
(10.35)
v, i, p
E
E
R
vC
p = vC iC
iC
0
FIG. 10.70
Plotting the power to a capacitive element
during the transient phase.
t
414

CAPACITORS
where V is the steady-state voltage across the capacitor. In terms of Q
and C,
1 Q 2
WC C C
2
Q2
WC 2C
or
(J)
(10.36)
EXAMPLE 10.19 For the network of Fig. 10.69, determine the energy
stored by each capacitor.
Solution:
For C1,
1
WC CV 2
2
1
(2 106 F)(16 V)2 (1 106)(256)
2
256 mJ
For C2,
1
WC CV 2
2
1
(3 106 F)(56 V)2 (1.5 106)(3136)
2
4704 mJ
Due to the squared term, note the difference in energy stored
because of a higher voltage.
Conductors
10.15 STRAY CAPACITANCES
(a)
Cbe
E
P
Cbc
N
C
P
In addition to the capacitors discussed so far in this chapter, there are
stray capacitances that exist not through design but simply because
two conducting surfaces are relatively close to each other. Two conducting wires in the same network will have a capacitive effect between
them, as shown in Fig. 10.71(a). In electronic circuits, capacitance levels exist between conducting surfaces of the transistor, as shown in Fig.
10.71(b). As mentioned earlier, in Chapter 12 we will discuss another
element called the inductor, which will have capacitive effects between
the windings [Fig. 10.71(c)]. Stray capacitances can often lead to serious errors in system design if they are not considered carefully.
B Cce
(b)
(c)
FIG. 10.71
Examples of stray capacitance.
10.16
APPLICATIONS
This Applications section for capacitors includes both a description of
the operation of one of the less expensive, throwaway cameras that have
become so popular today and a discussion of the use of capacitors in the
line conditioners (surge protectors) that have found their way into most
homes and throughout the business world. Additional examples of the
use of capacitors will appear throughout the chapter to follow.
APPLICATIONS

415
Flash Lamp
The basic circuitry for the flash lamp of the popular, inexpensive,
throwaway camera of Fig. 10.72(a) is provided in Fig. 10.72(b), with
the physical circuitry appearing in Fig. 10.72(c). The labels added to
Fig. 10.72(c) identify broad areas of the design and some individual
components. The major components of the electronic circuitry include
a large 160-mF, 330-V, polarized electrolytic capacitor to store the necessary charge for the flash lamp, a flash lamp to generate the required
light, a dc battery of 1.5 V, a chopper network to generate a dc voltage
in excess of 300 V, and a trigger network to establish a few thousand
volts for a very short period of time to fire the flash lamp. There are
both a 22-nF capacitor in the trigger network as shown in Fig. 10.72(b)
and (c) and a third capacitor of 470 pF in the high-frequency oscillator
of the chopper network. In particular, note that the size of each capacitor is directly related to its capacitance level. It should certainly be of
some interest that a single source of energy of only 1.5 V dc can be converted to one of a few thousand volts (albeit for a very short period of
time) to fire the flash lamp. In fact, that single, small battery has sufficient power for the entire run of film through the camera. Always keep
in mind that energy is related to power and time by W Pt (VI)t.
=
FIG. 10.72(a)
Flash camera: general appearance.
V
V
V
V
V
300 V
1.5 V
0
0
t
0
t
0
t
0
t
Flash button
(face of camera)
On/Off
300 V dc
Rs
Sense
+
1.5 V
Flash
switch
(top of
camera)
A
G
160 µF
4000-V spikes
Vt
Trigger
network
Lamp flashes
Flash unit
0
Rn
60-V
neon
light
–
SCR
K
R2
+
300 V
Highfrequency
transformer
Auto transformer
22 nF
R1
Rr
Highfrequency
oscillator
1.5 V
t
(b)
FIG. 10.72(b)
Flash camera: basic circuitry.
Flash
lamp
t
416

CAPACITORS
#
*
!
"
+
"! $ %!
!"
)) &'( "
# FIG. 10.72(c)
Flash camera: internal construction.
That is, a high level of voltage can be generated for a defined energy
level so long as the factors I and t are sufficiently small.
When you first use the camera, you are directed to press the flash
button on the face of the camera and wait for the flash-ready light to
come on. As soon as the flash button is depressed, the full 1.5 V of the
dc battery are applied to an electronic network (a variety of networks
can perform the same function) that will generate an oscillating waveform of very high frequency (with a high repetitive rate) as shown in
Fig. 10.72(b). The high-frequency transformer will then significantly
increase the magnitude of the generated voltage and will pass it on to a
half-wave rectification system (introduced in earlier chapters), resulting
in a dc voltage of about 300 V across the 160-mF capacitor to charge the
capacitor (as determined by Q CV). Once the 300-V level is reached,
the lead marked “sense” in Fig. 10.72(b) will feed the information back
to the oscillator and will turn it off until the output dc voltage drops to
a low threshold level. When the capacitor is fully charged, the neon
APPLICATIONS
light in parallel with the capacitor will turn on (labeled “flash-ready
lamp” on the camera) to let you know that the camera is ready to use.
The entire network from the 1.5-V dc level to the final 300-V level is
called a dc-dc converter. The terminology chopper network comes from
the fact that the applied dc voltage of 1.5 V was chopped up into one
that changes level at a very high frequency so that the transformer can
perform its function.
Even though the camera may use a 60-V neon light, the neon light
and series resistor Rn must have a full 300 V across the branch before
the neon light will turn on. Neon lights are simply bulbs with a neon
gas that will support conduction when the voltage across the terminals
reaches a sufficiently high level. There is no filament, or hot wire as in
a light bulb, but simply conduction through the gaseous medium. For
new cameras the first charging sequence may take 12 s to 15 s. Succeeding charging cycles may only take some 7 s or 8 s because the
capacitor will still have some residual charge on its plates. If the flash
unit is not used, the neon light will begin to drain the 300-V dc supply
with a drain current in microamperes. As the terminal voltage drops,
there will come a point where the neon light will turn off. For the unit
of Fig. 10.72, it takes about 15 min before the light turns off. Once off,
the neon light will no longer drain the capacitor, and the terminal voltage of the capacitor will remain fairly constant. Eventually, however,
the capacitor will discharge due to its own leakage current, and the terminal voltage will drop to very low levels. The discharge process is
very rapid when the flash unit is used, causing the terminal voltage to
drop very quickly (V Q/C) and, through the feedback-sense connection signal, causing the oscillator to start up again and recharge the
capacitor. You may have noticed when using a camera of this type that
once the camera has its initial charge, there is no need to press the
charge button between pictures—it is done automatically. However, if
the camera sits for a long period of time, the charge button will have to
be depressed again; but you will find that the charge time is only 3 s or
4 s due to the residual charge on the plates of the capacitor.
The 300 V across the capacitor are insufficient to fire the flash lamp.
Additional circuitry, called the trigger network, must be incorporated to
generate the few thousand volts necessary to fire the flash lamp. The
resulting high voltage is one reason that there is a CAUTION note on
each camera regarding the high internal voltages generated and the possibility of electrical shock if the camera is opened.
The thousands of volts required to fire the flash lamp require a discussion that introduces elements and concepts beyond the current level
of the text. However, this description is sensitive to this fact and should
be looked upon as simply a first exposure to some of the interesting
possibilities available from the right mix of elements. When the flash
switch at the bottom left of Fig. 10.72(a) is closed, it will establish a
connection between the resistors R1 and R2. Through a voltage divider
action, a dc voltage will appear at the gate (G) terminal of the SCR
(silicon-controlled rectifier—a device whose state is controlled by the
voltage at the gate terminal). This dc voltage will turn the SCR “on”
and will establish a very low resistance path (like a short circuit)
between its anode (A) and cathode (K) terminals. At this point the trigger capacitor, which is connected directly to the 300 V sitting across the
capacitor, will rapidly charge to 300 V because it now has a direct, lowresistance path to ground through the SCR. Once it reaches 300 V, the
charging current in this part of the network will drop to 0 A, and the

417
418

CAPACITORS
SCR will open up again since it is a device that needs a steady current
in the anode circuit to stay on. The capacitor then sits across the parallel coil (with no connection to ground through the SCR) with its full
300 V and begins to quickly discharge through the coil because the only
resistance in the circuit affecting the time constant is the resistance of
the parallel coil. As a result, a rapidly changing current through the coil
will generate a high voltage across the coil for reasons to be introduced
in Chapter 12.
When the capacitor decays to zero volts, the current through the coil
will be zero amperes, but a strong magnetic field has been established
around the coil. This strong magnetic field will then quickly collapse,
establishing a current in the parallel network that will recharge the
capacitor again. This continual exchange between the two storage elements will continue for a period of time, depending on the resistance in
the ciruit. The more the resistance, the shorter the “ringing” of the voltage at the output. This action of the energy “flying back” to the other
element is the basis for the “flyback” effect that is frequently used to
generate high dc voltages such as needed in TVs. In Fig. 10.72(b), you
will find that the trigger coil is connected directly to a second coil to
form an autotransformer (a transformer with one end connected).
Through transformer action the high voltage generated across the trigger coil will be increased further, resulting in the 4000 V necessary to
fire the flash lamp. Note in Fig. 10.72(c) that the 4000 V are applied to
a grid that actually lies on the surface of the glass tube of the flash lamp
(not internally connected or in contact with the gases). When the trigger voltage is applied, it will excite the gases in the lamp, causing a
very high current to develop in the bulb for a very short period of time
and producing the desired bright light. The current in the lamp is supported by the charge on the 160-mF capacitor which will be dissipated
very quickly. The capacitor voltage will drop very quickly, the photo
lamp will shut down, and the charging process will begin again. If the
entire process didn’t occur as quickly as it does, the lamp would burn
out after a single use.
Line Conditioner (Surge Protector)
=
FIG. 10.73(a)
Line conditioner: general appearance.
In recent years we have all become familiar with the line conditioner as
a safety measure for our computers, TVs, CD players, and other sensitive instrumentation. In addition to protecting equipment from unexpected surges in voltage and current, most quality units will also filter
out (remove) electromagnetic interference (EMI) and radio-frequency
interference (RFI). EMI encompasses any unwanted disturbances down
the power line established by any combination of electromagnetic
effects such as those generated by motors on the line, power equipment
in the area emitting signals picked up by the power line acting as an
antenna, and so on. RFI includes all signals in the air in the audio range
and beyond which may also be picked up by power lines inside or outside the house.
The unit of Fig. 10.73 has all the design features expected in a good
line conditioner. Figure 10.73(a) reveals that it can handle the power
drawn by six outlets and that it is set up for FAX/MODEM protection.
Also note that it has both LED (light-emitting diode) displays which
reveal whether there is fault on the line or whether the line is OK and
an external circuit breaker to reset the system. In addition, when the
surge protector is on, a red light will be visible at the power switch.
APPLICATIONS
Line conditioner casing

419
On/Off switch
228 µH
Circuit
breaker
On/Off lamp
Black
(feed)
2 nF
180 V
MOV
Switch assembly
Black
2 nF
2 nF
White
(return)
Black (feed)
228 µH
White Ground
(bare or
green)
White (return)
2 nF
2 nF
Ground
Black (feed)
Green LED
Protection present
Outlets
Red LED
etc.
Line fault
Green LED
Line on
Voltage monitor network
White (return)
Bare or green
(ground)
(b)
!
"# $
$ FIG. 10.73 (cont.)
Line conditioner: (b) electrical schematic; (c) internal construction.
420

CAPACITORS
The schematic of Fig. 10.73(b) does not include all the details of the
design, but it does include the major components that appear in most
good line conditioners. First note in the photograph of Fig. 10.73(c) that
the outlets are all connected in parallel, with a ground bar used to establish a ground connection for each outlet. The circuit board had to be
flipped over to show the components, so it will take some adjustment to
relate the position of the elements on the board to the casing. The feed
line or hot lead wire (black in the actual unit) is connected directly from
the line to the circuit breaker. The other end of the circuit breaker is connected to the other side of the circuit board. All the large discs that you
see are 2-nF/73 capacitors [not all have been included in Fig 10.73(b) for
clarity]. There are quite a few capacitors to handle all the possibilities.
For instance, there are capacitors from line to return (black wire to white
wire), from line to ground (black to green), and from return to ground
(white to ground). Each has two functions. The first and most obvious
function is to prevent any spikes in voltage that may come down the line
because of external effects such as lightning from reaching the equipment plugged into the unit. Recall from this chapter that the voltage
across capacitors cannot change instantaneously and in fact will act to
squelch any rapid change in voltage across its terminals. The capacitor,
therefore, will prevent the line to neutral voltage from changing too
quickly, and any spike that tries to come down the line will have to find
another point in the feed circuit to fall across. In this way the appliances
to the surge protector are well protected.
The second function requires some knowledge of the reaction of
capacitors to different frequencies and will be discussed in more detail
in later chapters. For the moment, let it suffice to say that the capacitor will have a different impedance to different frequencies, thereby
preventing undesired frequencies, such as those associated with EMI
and RFI disturbances, from affecting the operation of units connected to
the line conditioner. The rectangular-shaped capacitor of 1 mF near the
center of the board is connected directly across the line to take the brunt
of a strong voltage spike down the line. Its larger size is clear evidence
that it is designed to absorb a fairly high energy level that may be established by a large voltage—significant current over a period of time that
might exceed a few milliseconds.
The large, toroidal-shaped structure in the center of the circuit board
of Fig. 10.73(c) has two coils (Chapter 12) of 228 mH that appear in the
line and neutral of Fig. 10.73(b). Their purpose, like that of the capacitors, is twofold: to block spikes in current from coming down the line
and to block unwanted EMI and RFI frequencies from getting to the
connected systems. In the next chapter you will find that coils act as
“chokes” to quick changes in current; that is, the current through a coil
cannot change instantaneously. For increasing frequencies, such as
those associated with EMI and RFI disturbances, the reactance of a coil
increases and will absorb the undesired signal rather than let it pass
down the line. Using a choke in both the line and the neutral makes the
conditioner network balanced to ground. In total, capacitors in a line
conditioner have the effect of bypassing the disturbances, whereas
inductors block the disturbance.
The smaller disc (blue) between two capacitors and near the circuit
breaker is an MOV (metal-oxide varistor) which is the heart of most
line conditioners. It is an electronic device whose terminal characteristics will change with the voltage applied across its terminals. For the
normal range of voltages down the line, its terminal resistance will be
COMPUTER ANALYSIS

421
sufficiently large to be considered an open circuit, and its presence can
be ignored. However, if the voltage is too large, its terminal characteristics will change from a very large resistance to a very small resistance
that can essentially be considered a short circuit. This variation in resistance with applied voltage is the reason for the name varistor. For
MOVs in North America where the line voltage is 120 V, the MOVs are
180 V or more. The reason for the 60-V difference is that the 120-V rating is an effective value related to dc voltage levels, whereas the waveform for the voltage at any 120-V outlet has a peak value of about
170 V. A great deal more will be said about this topic in Chapter 13.
Taking a look at the symbol for an MOV in Fig. 10.73(b), you will
note that it has an arrow in each direction, revealing that the MOV is
bidirectional and will block voltages with either polarity. In general,
therefore, for normal operating conditions, the presence of the MOV
can be ignored; but, if a large spike should appear down the line,
exceeding the MOV rating, it will act as a short across the line to protect the connected circuitry. It is a significant improvement to simply
putting a fuse in the line because it is voltage sensitive, can react much
quicker than a fuse, and will display its low-resistance characteristics
for only a short period of time. When the spike has passed, it will return
to its normal open-circuit characteristic. If you’re wondering where the
spike will go if the load is protected by a short circuit, remember that
all sources of disturbance, such as lightning, generators, inductive
motors (such as in air conditioners, dishwashers, power saws, and so
on), have their own “source resistance,” and there is always some resistance down the line to absorb the disturbance.
Most line conditioners, as part of their advertising, like to mention
their energy absorption level. The rating of the unit of Fig. 10.73 is
1200 J which is actually higher than most. Remembering that W Pt EIt from the earlier discussion of cameras, we now realize that if
a 5000-V spike came down the line, we would be left with the product
It W/E 1200 J/5000 V 240 mAs. Assuming a linear relationship
between all quantities, the rated energy level is revealing that a current
of 100 A could be sustained for t 240 mAs/100 A 2.4 ms, a current of 1000 A for 240 ms, and a current of 10,000 A for 24 ms. Obviously, the higher the power product of E and I, the less the time element.
The technical specifications of the unit of Fig. 10.73 include an
instantaneous response time of 0 ns (questionable), with a phone line
protection of 5 ns. The unit is rated to dissipate surges up to 6000 V and
current spikes up to 96,000 A. It has a very high noise suppression ratio
(80 dB; see Chapter 23) at frequencies from 50 kHz to 1000 MHz, and
(a credit to the company) it has a lifetime warranty.
R
10.17 COMPUTER ANALYSIS
PSpice
Transient RC Response PSpice will now investigate the transient
response for the voltage across the capacitor of Fig. 10.74. In all the
examples in the text involving a transient response, a switch appeared
in series with the source as shown in Fig. 10.75(a). When applying
PSpice, we establish this instantaneous change in voltage level by
iC
5 k
E
20 V
+
C
8 µF vC
–
FIG. 10.74
Circuit to be analyzed using PSpice.
422

CAPACITORS
+
E
20 V
–
applying a pulse waveform as shown in Fig. 10.75(b) with a pulse
width (PW) longer than the period (5t) of interest for the network.
A pulse source is obtained through the sequence Place part keyLibraries-SOURCE-VPULSE-OK. Once in place, the label and all
the parameters can be set by simply double-clicking on each to obtain
the Display Properties dialog box. As you scroll down the list of attributes, you will see the following parameters defined by Fig. 10.76:
e
a
20 V
0
t
b
(a)
a
20 V
+
–
V1 is the initial value.
V2 is the pulse level.
TD is the delay time.
TR is the rise time.
TF is the fall time.
PW is the pulse width at the V2 level.
PER is the period of the waveform.
e
0
t
b
(b)
FIG. 10.75
Establishing a switching dc voltage level:
(a) series dc voltage-switch combination;
(b) PSpice pulse option.
TR
TF
V2
PW
V1
0
t
TD
PER
FIG. 10.76
The defining parameters of PSpice VPULSE.
All the parameters have been set as shown on the schematic of Fig.
10.77 for the network of Fig. 10.74. Since a rise and fall time of 0 s is
unrealistic from a practical standpoint, 0.1 ms was chosen for each in
this example. Further, since t RC (5 k) (8 mF) 20 ms and
5t 200 ms, a pulse width of 500 ms was selected. The period was
simply chosen as twice the pulse width.
Now for the simulation process. First the New Simulation Profile
key is selected to obtain the New Simulation dialog box in which
TransientRC is inserted for the Name and Create is chosen to leave
the dialog box. The Simulation Settings-Transient RC dialog box will
result, and under Analysis, the Time Domain (Transient) option is
chosen under Analysis type. The Run to time is set at 200 ms so that
only the first five time constants will be plotted. The Start saving data
after option will be 0 s to ensure that the data are collected immediately. The Maximum step size is 1 ms to provide sufficient data points
for a good plot. Click OK, and we are ready to select the Run PSpice
key. The result will be a graph without a plot (since it has not been
defined yet) and an x-axis that extends from 0 s to 200 ms as defined
above. To obtain a plot of the voltage across the capacitor versus time,
the following sequence is applied: Add Trace key-Add Traces dialog
box-V1(C)-OK, and the plot of Fig. 10.78 will result. The color and
thickness of the plot and the axis can be changed by placing the cursor
on the plot line and performing a right click. A list will appear in which
Properties should be selected; then a Trace Properties dialog box will
appear in which the color and thickness of the line can be changed.
Since the plot is against a black background, a better printout occurred
when yellow was selected and the line was made thicker as shown in
Fig. 10.78. Next, the cursor can be put on the axis, and another right
click will allow you to make the axis yellow and thicker for a better
printout. For comparison it seemed appropriate to plot the applied pulse
signal also. This is accomplished by going back to Trace and selecting
Add Trace followed by V(Vpulse:) and OK. Now both waveforms
appear on the same screen as shown in Fig. 10.78. In this case, the plot
was left a greenish tint so it could be distinguished from the axis and
the other plot. Note that it follows the left axis to the top and travels
across the screen at 20 V.
If you want the magnitude of either plot at any instant, simply select
the Toggle cursor key. Then click on V1(C) at the bottom left of the
screen. A box will appear around V1(C) that will reveal the spacing
between the dots of the cursor on the screen. This is important when
COMPUTER ANALYSIS
FIG. 10.77
Using PSpice to investigate the transient response of the series R-C circuit
of Fig. 10.74.
FIG. 10.78
Transient response for the voltage across the capacitor of Fig. 10.74 when
VPulse is applied.

423
424

CAPACITORS
more than one cursor is used. By moving the cursor to 200 ms, we find
that the magnitude (A1) is 19.865 V (in the Probe Cursor dialog box),
clearly showing how close it is to the final value of 20 V. A second cursor can be placed on the screen with a right click and then a click on the
same V1(C) on the bottom of the screen. The box around V1(C) cannot
show two boxes, but the spacing and the width of the lines of the box
have definitely changed. There is no box around the Pulse symbol since
it was not selected—although it could have been selected by either cursor. If we now move the second cursor to one time constant of 40 ms,
we find that the voltage is 12.633 V as shown in the Probe Cursor dialog box. This confirms the fact that the voltage should be 63.2% of its
final value of 20 V in one time constant (0.632 20 V 12.4 V). Two
separate plots could have been obtained by going to Plot-Add Plot to
Window and then using the trace sequence again.
Average Capacitive Current As an exercise in using the pulse
source and to verify our analysis of the average current for a purely
capacitive network, the description to follow will verify the results of
Example 10.13. For the pulse waveform of Fig. 10.59, the parameters
of the pulse supply appear in Fig. 10.79. Note that the rise time is now
2 ms, starting at 0 s, and the fall time is 6 ms. The period was set at
15 ms to permit monitoring the current after the pulse had passed.
Simulation is initiated by first selecting the New Simulation Profile
key to obtain the New Simulation dialog box in which AverageIC is
entered as the Name. Create is then chosen to obtain the Simulation
FIG. 10.79
Using PSpice to verify the results of Example 10.13.
COMPUTER ANALYSIS
Settings-AverageIC dialog box. Analysis is selected, and Time
Domain(Transient) is chosen under the Analysis type options. The
Run to time is set to 15 ms to encompass the period of interest, and the
Start saving data after is set at 0 s to ensure data points starting at t 0 s. The Maximum step size is selected from 15 ms/1000 15 ms to
ensure 1000 data points for the plot. Click OK, and the Run PSpice
key is selected. A window will appear with a horizontal scale that
extends from 0 to 15 ms as defined above. Then the Add Trace key is
selected, and I(C) is chosen to appear in the Trace Expression below.
Click OK, and the plot of I(C) appears in the bottom of Fig. 10.80. This
time it would be nice to see the pulse waveform in the same window but
as a separate plot. Therefore, continue with Plot-Add Plot to WindowTrace-Add Trade-V(Vpulse:)-OK, and both plots appear as shown
in Fig. 10.80.
FIG. 10.80
The applied pulse and resulting current for the 2-mF capacitor of Fig. 10.79.
The cursors can now be used to measure the resulting average current levels. First, select the I(C) plot to move the SEL>> notation to the
lower plot. The SEL>> defines which plot for multiplot screens is
active. Then select the Toggle cursor key, and left-click on the I(C)
plot to establish the crosshairs of the cursor. Set the value at 1 ms, and
the magnitude A1 is displayed as 4 mA. Right-click on the same plot,
and a second cursor will result that can be placed at 6 ms to get a
response of 1.33 mA (A2) as expected from Example 10.13. Both
plots were again placed in the yellow color with a wider line by rightclicking on the curve and choosing Properties.

425
426

CAPACITORS
PROBLEMS
SECTION 10.2 The Electric Field
1. Find the electric field strength at a point 2 m from a
charge of 4 mC.
2. The electric field strength is 36 newtons/coulomb (N/C)
at a point r meters from a charge of 0.064 mC. Find the
distance r.
SECTION 10.3 Capacitance
3. Find the capacitance of a parallel plate capacitor if
1400 mC of charge are deposited on its plates when 20 V
are applied across the plates.
4. How much charge is deposited on the plates of a 0.05-mF
capacitor if 45 V are applied across the capacitor?
5. Find the electric field strength between the plates of a
parallel plate capacitor if 100 mV are applied across the
plates and the plates are 2 mm apart.
6. Repeat Problem 5 if the plates are separated by 4 mils.
7. A 4-mF parallel plate capacitor has 160 mC of charge on
its plates. If the plates are 5 mm apart, find the electric
field strength between the plates.
8. Find the capacitance of a parallel plate capacitor if the
area of each plate is 0.075 m2 and the distance between
the plates is 1.77 mm. The dielectric is air.
9. Repeat Problem 8 if the dielectric is paraffin-coated paper.
10. Find the distance in mils between the plates of a 2-mF
capacitor if the area of each plate is 0.09 m2 and the
dielectric is transformer oil.
11. The capacitance of a capacitor with a dielectric of air is
1200 pF. When a dielectric is inserted between the plates,
the capacitance increases to 0.006 mF. Of what material
is the dielectric made?
12. The plates of a parallel plate air capacitor are 0.2 mm
apart and have an area of 0.08 m2, and 200 V are applied
across the plates.
a. Determine the capacitance.
b. Find the electric field intensity between the plates.
c. Find the charge on each plate if the dielectric is air.
13. A sheet of Bakelite 0.2 mm thick having an area of
0.08 m2 is inserted between the plates of Problem 12.
a. Find the electric field strength between the plates.
b. Determine the charge on each plate.
c. Determine the capacitance.
SECTION 10.4 Dielectric Strength
14. Find the maximum voltage ratings of the capacitors of
Problems 12 and 13 assuming a linear relationship
between the breakdown voltage and the thickness of the
dielectric.
15. Find the maximum voltage that can be applied across a
parallel plate capacitor of 0.006 mF. The area of one plate
is 0.02 m2 and the dielectric is mica. Assume a linear
relationship between the dielectric strength and the thickness of the dielectric.
PROBLEMS

16. Find the distance in millimeters between the plates of a
parallel plate capacitor if the maximum voltage that can
be applied across the capacitor is 1250 V. The dielectric
is mica. Assume a linear relationship between the breakdown strength and the thickness of the dielectric.
R
iC
SECTION 10.7 Transients in Capacitive Networks:
Charging Phase
17. For the circuit of Fig. 10.81:
a. Determine the time constant of the circuit.
b. Write the mathematical equation for the voltage vC
following the closing of the switch.
c. Determine the voltage vC after one, three, and five
time constants.
d. Write the equations for the current iC and the voltage vR.
e. Sketch the waveforms for vC and iC.
E
100 k
+ vR –
20 V
+
–
FIG. 10.81
Problems 17 and 18.
R1
iC
18. Repeat Problem 17 for R 1 M, and compare the
results.
19. For the circuit of Fig. 10.82:
a. Determine the time constant of the circuit.
b. Write the mathematical equation for the voltage vC
following the closing of the switch.
c. Determine vC after one, three, and five time constants.
d. Write the equations for the current iC and the voltage vR.
e. Sketch the waveforms for vC and iC.
5 mF vC
C
2.2 k
E
100 V
+
1 mF vC
C
–
R2
3.3 k
– vR +
FIG. 10.82
Problem 19.
20. For the circuit of Fig. 10.83:
a. Determine the time constant of the circuit.
b. Write the mathematical equation for the voltage vC
following the closing of the switch.
c. Write the mathematical expression for the current iC
following the closing of the switch.
d. Sketch the waveforms of vC and iC.
+15 V
(t = 0 s)
R
56 k
iC
C
+
0.1 µ
µF vC
–
–10 V
FIG. 10.83
Problem 20.
SECTION 10.8 Discharge Phase
21. For the circuit of Fig. 10.84:
a. Determine the time constant of the circuit when the
switch is thrown into position 1.
b. Find the mathematical expression for the voltage
across the capacitor after the switch is thrown into
position 1.
c. Determine the mathematical expression for the current following the closing of the switch (position 1).
d. Determine the voltage vC and the current iC if the
switch is thrown into position 2 at t 100 ms.
e. Determine the mathematical expressions for the voltage vC and the current iC if the switch is thrown into
position 3 at t 200 ms.
f. Plot the waveforms of vC and iC for a period of time
extending from t 0 to t 300 ms.
+ vC –
C
R1
3 k
50 V
1
iC
2 mF
2
3
FIG. 10.84
Problems 21 and 22.
R2
2 k
427
428

CAPACITORS
E
22. Repeat Problem 21 for a capacitance of 20 mF.
2
1
+
iC
10 pF vC R2
80 V C
390 k
–
100 k
R1
FIG. 10.85
Problem 23.
– vR +
+ 40 V –
iC
*23. For the network of Fig. 10.85:
a. Find the mathematical expression for the voltage
across the capacitor after the switch is thrown into
position 1.
b. Repeat part (a) for the current iC.
c. Find the mathematical expressions for the voltage vC
and current iC if the switch is thrown into position 2
at a time equal to five time constants of the charging
circuit.
d. Plot the waveforms of vC and iC for a period of time
extending from t 0 to t 30 ms.
24. The capacitor of Fig. 10.86 is initially charged to 40 V
before the switch is closed. Write the expressions for the
voltages vC and vR and the current iC for the decay phase.
R
2.2 k
C = 2000 mF
+ vC –
FIG. 10.86
Problem 24.
C = 1000 µ
µF
+
6V
25. The 1000-mF capacitor of Fig. 10.87 is charged to 6 V. To
discharge the capacitor before further use, a wire with a
resistance of 0.002 is placed across the capacitor.
a. How long will it take to discharge the capacitor?
b. What is the peak value of the current?
c. Based on the answer to part (b), is a spark expected
when contact is made with both ends of the capacitor?
–
FIG. 10.87
Problems 25 and 29.
R1
SECTION 10.9 Initial Values
iC
4.7 k
E
10 V
+
vC
–
10 F 3 V
C
+
–
26. The capacitor in Fig. 10.88 is initially charged to 3 V
with the polarity shown.
a. Find the mathematical expressions for the voltage vC
and the current iC when the switch is closed.
b. Sketch the waveforms for vC and iC.
*27. The capacitor of Fig. 10.89 is initially charged to 12 V
with the polarity shown.
a. Find the mathematical expressions for the voltage vC
and the current iC when the switch is closed.
b. Sketch the waveforms for vC and iC.
FIG. 10.88
Problem 26.
+ vC –
40 V
R1
R2
10 k
8.2 k
FIG. 10.89
Problem 27.
iC
C
–12 V
6.8 F
+ 12 V –
SECTION 10.10 Instantaneous Values
28. Given the expression vC 8(1 et/(2010 )):
a. Determine vC after five time constants.
b. Determine vC after 10 time constants.
c. Determine vC at t 5 ms.
6

PROBLEMS
429
29. For the situation of Problem 25, determine when the discharge current is one-half its maximum value if contact is
made at t 0 s.
30. For the network of Fig. 10.90, VL must be 8 V before the
system is activated. If the switch is closed at t 0 s, how
long will it take for the system to be activated?
(t = 0 s)
(t = 0 s)
R
E
33 k
R
12 V
C
20 mF
System
R = ∞
+
VL = 12 V to turn on
20 V
E
System
R = ∞
200 mF
C
VL
–
FIG. 10.91
Problem 31.
FIG. 10.90
Problem 30.
*31. Design the network of Fig. 10.91 such that the system
will turn on 10 s after the switch is closed.
R1
32. For the circuit of Fig. 10.92:
a. Find the time required for vC to reach 60 V following
the closing of the switch.
b. Calculate the current iC at the instant vC 60 V.
c. Determine the power delivered by the source at the
instant t 2t.
iC
8 k
+
E
80 V
C
6 mF vC
–
R2
12 k
FIG. 10.92
Problem 32.
+ vR1 –
R1
*33. For the network of Fig. 10.93:
a. Calculate vC, iC, and vR at 0.5 s and 1 s after the
switch makes contact with position 1.
b. The network sits in position 1 10 min before the
switch is moved to position 2. How long after making
contact with position 2 will it take for the current iC to
drop to 8 mA? How much longer will it take for vC to
drop to 10 V?
1
1 M
1
iC
2
E
60 V
+
C
4 M
R2
0.2 mF vC
–
FIG. 10.93
Problem 33.
+ vC –
C
34. For the system of Fig. 10.94, using a DMM with a
10-M internal resistance in the voltmeter mode:
a. Determine the voltmeter reading one time constant
after the switch is closed.
b. Find the current iC two time constants after the switch
is closed.
c. Calculate the time that must pass after the closing of
the switch for the voltage vC to be 50 V.
iC
E
DMM
0.2 mF
60 V
+ –
FIG. 10.94
Problem 34.
430

CAPACITORS
SECTION 10.11 Thévenin Equivalent: t RThC
R
24 V
E
35. For the system of Fig. 10.95, using a DMM with a
10-M internal resistance in the voltmeter mode:
a. Determine the voltmeter reading four time constants
after the switch is closed.
b. Find the time that must pass before iC drops to 3 mA.
c. Find the time that must pass after the closing of the
switch for the voltage across the meter to reach 10 V.
iC
2 M
1 mF
C
+ –
FIG. 10.95
Problem 35.
36. For the circuit of Fig. 10.96:
a. Find the mathematical expressions for the transient
behavior of the voltage vC and the current iC following the closing of the switch.
b. Sketch the waveforms of vC and iC.
+ vC –
C
R1
8 k
E
iC
15 mF
24 k
20 V R2
4 k
R3
FIG. 10.96
Problem 36.
*37. Repeat Problem 36 for the circuit of Fig. 10.97.
5 mA
3.9 k
R1
6.8 k
+4V
iC
R2
R3
20 mF +
vC
C
0.56 k
–
FIG. 10.97
Problems 37 and 55.
38. The capacitor of Fig. 10.98 is initially charged to 4 V
with the polarity shown.
a. Write the mathematical expressions for the voltage vC
and the current iC when the switch is closed.
b. Sketch the waveforms of vC and iC.
R2
3.9 k
iC
+
R1
–
20 F 4 V
vC C
1.8 k
–
E
+
36 V
FIG. 10.98
Problem 38.
R2
1.5 k
iC
+
I = 4 mA
R1
6.8 k
vC C
–
FIG. 10.99
Problem 39.
+
2.2 F 2 V
–
39. The capacitor of Fig. 10.99 is initially charged to 2 V
with the polarity shown.
a. Write the mathematical expressions for the voltage vC
and the current iC when the switch is closed.
b. Sketch the waveforms of vC and iC.
PROBLEMS
*40. The capacitor of Fig. 10.100 is initially charged to 3 V
with the polarity shown.
a. Write the mathematical expressions for the voltage vC
and the current iC when the switch is closed.
b. Sketch the waveforms of vC and iC.

431
+ vC –
2 k
+10 V
C
6.8 k
iC
–20 V
39 F
+ 3V –
FIG. 10.100
Problem 40.
SECTION 10.12 The Current iC
41. Find the waveform for the average current if the voltage
across a 0.06-mF capacitor is as shown in Fig. 10.101.
v (V)
100
80
60
40
20
0
– 20
– 40
– 60
1
2
3
4
5
6
7
8
9 10 11 12 13 14 15
t (ms)
FIG. 10.101
Problem 41.
42. Repeat Problem 41 for the waveform of Fig. 10.102.
3
v (V)
2
1
0
–1
–2
–3
1
2
3
4
5
6
7
8
9 10 11 12 13 14 15 16 t (ms)
FIG. 10.102
Problem 42.
*43. Given the waveform of Fig. 10.103 for the current of a
20-mF capacitor, sketch the waveform of the voltage vC
across the capacitor if vC 0 V at t 0 s.
iC
+40 mA
0
4
6
16 18
20
–80 mA
–120 mA
FIG. 10.103
Problem 43.
25
t (ms)
432

CAPACITORS
SECTION 10.13 Capacitors in Series and Parallel
44. Find the total capacitance CT between points a and b of
the circuits of Fig. 10.104.
3 mF
6 mF
0.2 mF
60 pF
7 mF
a
a
CT
b
CT
b
(a)
30 pF
10 pF
20 pF
(b)
FIG. 10.104
Problem 44.
45. Find the voltage across and charge on each capacitor for
the circuits of Fig. 10.105.
200 pF
C2
C1
6 mF
C2
1200 pF
40 V
6 mF
C1
10 V
C3
400 pF
C4
600 pF
12 mF
C3
(a)
(b)
FIG. 10.105
Problem 45.
*46. For each configuration of Fig. 10.106, determine the
voltage across each capacitor and the charge on each
capacitor.
9 mF
C1
E
24 V
C3
9 mF
10 mF
C2
E
C4
7 mF
42 mF
C1
C2
16 V
C3
72 mF
(a)
(b)
FIG. 10.106
Problem 46.
C4
14 mF
C5
35 mF
2 mF
PROBLEMS
*47. For the network of Fig. 10.107, determine the following
100 ms after the switch is closed:
a. Vab
b. Vac
c. Vcb
d. Vda
e. If the switch is moved to position 2 one hour later,
find the time required for vR2 to drop to 20 V.

R1
d
a
1
20 k
8 mF
C1
2
c
E
100 V
C2
R2
48. For the circuits of Fig. 10.108, find the voltage across and
charge on each capacitor after each capacitor has charged
to its final value.
40 k
12 mF C3
b
FIG. 10.107
Problem 47.
2 k
48 V
C1
C2
0.04 mF
4 k
0.08 mF
(a)
C2
60 mF
40 mF
6 k
C1
80 V
4 k
5 k
(b)
FIG. 10.108
Problem 48.
SECTION 10.14 Energy Stored by a Capacitor
49. Find the energy stored by a 120-pF capacitor with 12 V
across its plates.
50. If the energy stored by a 6-mF capacitor is 1200 J, find
the charge Q on each plate of the capacitor.
*51. An electronic flashgun has a 1000-mF capacitor that is
charged to 100 V.
a. How much energy is stored by the capacitor?
b. What is the charge on the capacitor?
c. When the photographer takes a picture, the flash fires
for 1/2000 s. What is the average current through the
flashtube?
d. Find the power delivered to the flashtube.
e. After a picture is taken, the capacitor has to be
recharged by a power supply that delivers a maximum
current of 10 mA. How long will it take to charge the
capacitor?
52. For the network of Fig. 10.109:
a. Determine the energy stored by each capacitor under
steady-state conditions.
b. Repeat part (a) if the capacitors are in series.
433
6 k
24 V
4 k
3 k
FIG. 10.109
Problem 52.
6 mF
12 mF
12 mF
434

CAPACITORS
SECTION 10.17 Computer Analysis
PSpice or Electronics Workbench
53. Using schematics:
a. Obtain the waveforms for vC and iC versus time for
the network of Fig. 10.35.
b. Obtain the power curve (representing the energy
stored by the capacitor over the same time interval),
and compare it to the plot of Fig. 10.70.
*54. Using schematics, obtain the waveforms of vC and iC versus time for the network of Fig. 10.49 using the IC option.
55. Verify your solution to Problem 37 (Fig. 10.97) using
schematics.
Programming Language (C, QBASIC, Pascal, etc.)
56. Write a QBASIC program to tabulate the voltage vC and
current iC for the network of Fig. 10.44 for five time constants after the switch is moved to position 1 at t 0 s.
Use an increment of (1/5)t.
*57. Write a program to write the mathematical expression for
the voltage vC for the network of Fig. 10.52 for any element values when the switch is moved to position 1.
*58. Given three capacitors in any series-parallel arrangement,
write a program to determine the total capacitance. That
is, determine the total number of possibilities, and ask the
user to identify the configuration and provide the capacitor values. Then calculate the total capacitance.
GLOSSARY
Breakdown voltage Another term for dielectric strength,
listed below.
Capacitance A measure of a capacitor’s ability to store
charge; measured in farads (F).
Capacitive time constant The product of resistance and
capacitance that establishes the required time for the charging and discharging phases of a capacitive transient.
Capacitive transient The waveforms for the voltage and
current of a capacitor that result during the charging and
discharging phases.
Capacitor A fundamental electrical element having two conducting surfaces separated by an insulating material and
having the capacity to store charge on its plates.
Coulomb’s law An equation relating the force between two
like or unlike charges.
Dielectric The insulating material between the plates of a
capacitor that can have a pronounced effect on the charge
stored on the plates of a capacitor.
Dielectric constant Another term for relative permittivity,
listed below.
Dielectric strength An indication of the voltage required for
unit length to establish conduction in a dielectric.
Electric field strength The force acting on a unit positive
charge in the region of interest.
Electric flux lines Lines drawn to indicate the strength and
direction of an electric field in a particular region.
Fringing An effect established by flux lines that do not pass
directly from one conducting surface to another.
Leakage current The current that will result in the total discharge of a capacitor if the capacitor is disconnected from
the charging network for a sufficient length of time.
Permittivity A measure of how well a dielectric will permit
the establishment of flux lines within the dielectric.
Relative permittivity The permittivity of a material compared to that of air.
Stray capacitance Capacitances that exist not through
design but simply because two conducting surfaces are relatively close to each other.
Surge voltage The maximum voltage that can be applied
across a capacitor for very short periods of time.
Working voltage The voltage that can be applied across a
capacitor for long periods of time without concern for
dielectric breakdown.
11
Magnetic Circuits
11.1
INTRODUCTION
Magnetism plays an integral part in almost every electrical device used
today in industry, research, or the home. Generators, motors, transformers, circuit breakers, televisions, computers, tape recorders, and
telephones all employ magnetic effects to perform a variety of important tasks.
The compass, used by Chinese sailors as early as the second century
A.D., relies on a permanent magnet for indicating direction. The permanent magnet is made of a material, such as steel or iron, that will
remain magnetized for long periods of time without the need for an
external source of energy.
In 1820, the Danish physicist Hans Christian Oersted discovered
that the needle of a compass would deflect if brought near a currentcarrying conductor. For the first time it was demonstrated that electricity and magnetism were related, and in the same year the French
physicist André-Marie Ampère performed experiments in this area and
developed what is presently known as Ampère’s circuital law. In subsequent years, men such as Michael Faraday, Karl Friedrich Gauss,
and James Clerk Maxwell continued to experiment in this area and
developed many of the basic concepts of electromagnetism—magnetic effects induced by the flow of charge, or current.
There is a great deal of similarity between the analyses of electric
circuits and magnetic circuits. This will be demonstrated later in this
chapter when we compare the basic equations and methods used to
solve magnetic circuits with those used for electric circuits.
Difficulty in understanding methods used with magnetic circuits will
often arise in simply learning to use the proper set of units, not because
of the equations themselves. The problem exists because three different
systems of units are still used in the industry. To the extent practical, SI
will be used throughout this chapter. For the CGS and English systems,
a conversion table is provided in Appendix F.
436

MAGNETIC CIRCUITS
11.2
Same area
Flux lines
b
S
a
N
FIG. 11.1
Flux distribution for a permanent magnet.
N
S
N
S
FIG. 11.2
Flux distribution for two adjacent, opposite
poles.
MAGNETIC FIELDS
In the region surrounding a permanent magnet there exists a magnetic
field, which can be represented by magnetic flux lines similar to electric flux lines. Magnetic flux lines, however, do not have origins or terminating points as do electric flux lines but exist in continuous loops, as
shown in Fig. 11.1. The symbol for magnetic flux is the Greek letter (phi).
The magnetic flux lines radiate from the north pole to the south pole,
returning to the north pole through the metallic bar. Note the equal
spacing between the flux lines within the core and the symmetric distribution outside the magnetic material. These are additional properties of
magnetic flux lines in homogeneous materials (that is, materials having
uniform structure or composition throughout). It is also important to
realize that the continuous magnetic flux line will strive to occupy as
small an area as possible. This will result in magnetic flux lines of minimum length between the like poles, as shown in Fig. 11.2. The strength
of a magnetic field in a particular region is directly related to the density of flux lines in that region. In Fig. 11.1, for example, the magnetic
field strength at a is twice that at b since twice as many magnetic flux
lines are associated with the perpendicular plane at a than at b. Recall
from childhood experiments that the strength of permanent magnets
was always stronger near the poles.
If unlike poles of two permanent magnets are brought together, the
magnets will attract, and the flux distribution will be as shown in Fig.
11.2. If like poles are brought together, the magnets will repel, and the
flux distribution will be as shown in Fig. 11.3.
Flux lines
S
Soft iron
N
N
N
S
S
Glass
FIG. 11.3
Flux distribution for two adjacent, like poles.
FIG. 11.4
Effect of a ferromagnetic sample on the flux
distribution of a permanent magnet.
Soft iron
Sensitive
instrument
FIG. 11.5
Effect of a magnetic shield on the flux
distribution.
If a nonmagnetic material, such as glass or copper, is placed in the
flux paths surrounding a permanent magnet, there will be an almost
unnoticeable change in the flux distribution (Fig. 11.4). However, if a
magnetic material, such as soft iron, is placed in the flux path, the flux
lines will pass through the soft iron rather than the surrounding air
because flux lines pass with greater ease through magnetic materials
than through air. This principle is put to use in the shielding of sensitive
electrical elements and instruments that can be affected by stray magnetic fields (Fig. 11.5).
As indicated in the introduction, a magnetic field (represented by
concentric magnetic flux lines, as in Fig. 11.6) is present around every
wire that carries an electric current. The direction of the magnetic flux
lines can be found simply by placing the thumb of the right hand in the
direction of conventional current flow and noting the direction of the
MAGNETIC FIELDS
Magnetic flux lines

437
Conductor
I
FIG. 11.6
Magnetic flux lines around a current-carrying
conductor.
fingers. (This method is commonly called the right-hand rule.) If the
conductor is wound in a single-turn coil (Fig. 11.7), the resulting flux
will flow in a common direction through the center of the coil. A coil of
more than one turn would produce a magnetic field that would exist in
a continuous path through and around the coil (Fig. 11.8).
I
I
FIG. 11.7
Flux distribution of a single-turn coil.
N
S
I
I
N
FIG. 11.8
Flux distribution of a current-carrying coil.
The flux distribution of the coil is quite similar to that of the permanent magnet. The flux lines leaving the coil from the left and entering
to the right simulate a north and a south pole, respectively. The principal difference between the two flux distributions is that the flux lines
are more concentrated for the permanent magnet than for the coil. Also,
since the strength of a magnetic field is determined by the density of the
flux lines, the coil has a weaker field strength. The field strength of the
coil can be effectively increased by placing certain materials, such as
iron, steel, or cobalt, within the coil to increase the flux density (defined
in the next section) within the coil. By increasing the field strength with
the addition of the core, we have devised an electromagnet (Fig. 11.9)
that, in addition to having all the properties of a permanent magnet, also
has a field strength that can be varied by changing one of the component values (current, turns, and so on). Of course, current must pass
through the coil of the electromagnet in order for magnetic flux to be
developed, whereas there is no need for the coil or current in the permanent magnet. The direction of flux lines can be determined for the
electromagnet (or in any core with a wrapping of turns) by placing the
fingers of the right hand in the direction of current flow around the core.
The thumb will then point in the direction of the north pole of the
induced magnetic flux, as demonstrated in Fig. 11.10(a). A cross section of the same electromagnet is included as Fig. 11.10(b) to introduce
the convention for directions perpendicular to the page. The cross and
dot refer to the tail and head of the arrow, respectively.
S
I
Steel
I
FIG. 11.9
Electromagnet.
I
N
S
I
(a)
N
S
(b)
FIG. 11.10
Determining the direction of flux for an
electromagnet: (a) method; (b) notation.
438

MAGNETIC CIRCUITS
Other areas of application for electromagnetic effects are shown in
Fig. 11.11. The flux path for each is indicated in each figure.
Cutaway section
Laminated
sheets of steel
Φ
Flux path
N
Air
gap
Secondary
Primary
Φ
Transformer
S
Loudspeaker
Generator
Air gap
Air gap Φ
N
Φ
Φ
S
Relay
Medical Applications: Magnetic
resonance imaging.
Meter movement
FIG. 11.11
Some areas of application of magnetic effects.
11.3 FLUX DENSITY
German (Wittenberg,
Göttingen)
(1804–91)
Physicist
Professor of Physics,
University of
Göttingen
In the SI system of units, magnetic flux is measured in webers (note
Fig. 11.12) and has the symbol . The number of flux lines per unit
area is called the flux density, is denoted by the capital letter B, and is
measured in teslas (note Fig. 11.15). Its magnitude is determined by the
following equation:
Courtesy of the
Smithsonian Institution
Photo No. 52,604
An important contributor to the establishment of a
system of absolute units for the electrical sciences,
which was beginnning to become a very active area of
research and development. Established a definition
of electric current in an electromagnetic system
based on the magnetic field produced by the current.
He was politically active and, in fact, was dismissed
from the faculty of the Universiity of Göttingen for
protesting the suppression of the constitution by the
King of Hanover in 1837. However, he found other
faculty positions and eventually returned to Göttingen as director of the astronomical observatory.
Received honors from England, France, and Germany, including the Copley Medal of the Royal Society.
FIG. 11.12
Wilhelm Eduard Weber.
B
A
B teslas (T)
webers (Wb)
A square meters (m2)
(11.1)
where is the number of flux lines passing through the area A (Fig.
11.13). The flux density at position a in Fig. 11.1 is twice that at b
because twice as many flux lines are passing through the same area.
By definition,
1 T 1 Wb/m2
A
Φ
FIG. 11.13
Defining the flux density B.
PERMEABILITY
EXAMPLE 11.1 For the core of Fig. 11.14, determine the flux density
B in teslas.
Solution:
6 105 Wb
B 5 102 T
1.2 103 m2
A
EXAMPLE 11.2 In Fig. 11.14, if the flux density is 1.2 T and the area
is 0.25 in.2, determine the flux through the core.
Solution:
By Eq. (11.1),
BA
However, converting 0.25 in.2 to metric units,
1m
1m
A 0.25 in.2 1.613 104 m2
39.37 in. 39.37 in.
and
(1.2 T)(1.613 104 m2)
1.936 104 Wb
An instrument designed to measure flux density in gauss (CGS system) appears in Fig. 11.16. Appendix F reveals that 1 T 104 gauss.
The magnitude of the reading appearing on the face of the meter in Fig.
11.16 is therefore
1T
1.964 gauss 1.964 104 T
4
10 gauss
11.4
PERMEABILITY
If cores of different materials with the same physical dimensions are
used in the electromagnet described in Section 11.2, the strength of the
magnet will vary in accordance with the core used. This variation in
strength is due to the greater or lesser number of flux lines passing
through the core. Materials in which flux lines can readily be set up are
said to be magnetic and to have high permeability. The permeability
(m) of a material, therefore, is a measure of the ease with which magnetic flux lines can be established in the material. It is similar in many
respects to conductivity in electric circuits. The permeability of free
space mo (vacuum) is

439
A
= 6 10–5 Wb
A = 1.2 10–3 m2
FIG. 11.14
Example 11.1.
Croatian-American
(Smiljan, Paris,
Colorado Springs,
New York City)
(1856–1943)
Electrical Engineer
and Inventor
Recipient of the
Edison Medal in
1917
Courtesy of the
Smithsonian Institution
Photo No. 52,223
Often regarded as one of the most innovative and
inventive individuals in the history of the sciences.
He was the first to introduce the alternating-current
machine, removing the need for commutator bars of
dc machines. After emigrating to the United States
in 1884, he sold a number of his patents on ac
machines, transformers, and induction coils (including the Tesla coil as we know it today) to the Westinghouse Electric Company. Some say that his most
important discovery was made at his laboratory in
Colorado Springs, where in 1900 he discovered terrestrial stationary waves. The range of his discoveries and inventions is too extensive to list here but
extends from lighting systems to polyphase power
systems to a wireless world broadcasting system.
FIG. 11.15
Nikola Tesla.
Wb
A· m
mo 4p 107 As indicated above, m has the units of Wb/A· m. Practically speaking, the permeability of all nonmagnetic materials, such as copper, aluminum, wood, glass, and air, is the same as that for free space. Materials that have permeabilities slightly less than that of free space are said
to be diamagnetic, and those with permeabilities slightly greater than
that of free space are said to be paramagnetic. Magnetic materials,
such as iron, nickel, steel, cobalt, and alloys of these metals, have permeabilities hundreds and even thousands of times that of free space.
Materials with these very high permeabilities are referred to as ferromagnetic.
FIG. 11.16
Digital display gaussmeter. (Courtesy of LDJ
Electronics, Inc.)
440

MAGNETIC CIRCUITS
The ratio of the permeability of a material to that of free space is
called its relative permeability; that is,
m
mr mo
(11.2)
In general, for ferromagnetic materials, mr ≥ 100, and for nonmagnetic
materials, mr 1.
Since mr is a variable, dependent on other quantities of the magnetic
circuit, values of mr are not tabulated. Methods of calculating mr from
the data supplied by manufacturers will be considered in a later section.
11.5
RELUCTANCE
The resistance of a material to the flow of charge (current) is determined for electric circuits by the equation
l
R r
A
(ohms, )
The reluctance of a material to the setting up of magnetic flux lines
in the material is determined by the following equation:
l
mA
(rels, or At/Wb)
(11.3)
where is the reluctance, l is the length of the magnetic path, and A is
the cross-sectional area. The t in the units At/Wb is the number of turns
of the applied winding. More is said about ampere-turns (At) in the next
section. Note that the resistance and reluctance are inversely proportional to the area, indicating that an increase in area will result in a
reduction in each and an increase in the desired result: current and flux.
For an increase in length the opposite is true, and the desired effect is
reduced. The reluctance, however, is inversely proportional to the permeability, while the resistance is directly proportional to the resistivity.
The larger the m or the smaller the r, the smaller the reluctance and
resistance, respectively. Obviously, therefore, materials with high permeability, such as the ferromagnetics, have very small reluctances and
will result in an increased measure of flux through the core. There is no
widely accepted unit for reluctance, although the rel and the At/Wb are
usually applied.
11.6 OHM’S LAW FOR MAGNETIC CIRCUITS
Recall the equation
cause
Effect opposition
appearing in Chapter 4 to introduce Ohm’s law for electric circuits. For
magnetic circuits, the effect desired is the flux . The cause is the magnetomotive force (mmf) , which is the external force (or “pressure”)
required to set up the magnetic flux lines within the magnetic material.
The opposition to the setting up of the flux is the reluctance .
MAGNETIZING FORCE

441
Substituting, we have
(11.4)
The magnetomotive force is proportional to the product of the
number of turns around the core (in which the flux is to be established) and the current through the turns of wire (Fig. 11.17). In equation form,
I
N turns
NI
(ampere-turns, At)
(11.5)
I
This equation clearly indicates that an increase in the number of turns
or the current through the wire will result in an increased “pressure” on
the system to establish flux lines through the core.
Although there is a great deal of similarity between electric and
magnetic circuits, one must continue to realize that the flux is not a
“flow” variable such as current in an electric circuit. Magnetic flux is
established in the core through the alteration of the atomic structure of
the core due to external pressure and is not a measure of the flow of
some charged particles through the core.
11.7
FIG. 11.17
Defining the components of a magnetomotive
force.
MAGNETIZING FORCE
The magnetomotive force per unit length is called the magnetizing
force (H). In equation form,
H l
(At/m)
(11.6)
Substituting for the magnetomotive force will result in
NI
H l
(At/m)
(11.7)
For the magnetic circuit of Fig. 11.18, if NI 40 At and l 0.2 m,
then
NI
40 At
H 200 At/m
0.2 m
l
In words, the result indicates that there are 200 At of “pressure” per
meter to establish flux in the core.
Note in Fig. 11.18 that the direction of the flux can be determined
by placing the fingers of the right hand in the direction of current
around the core and noting the direction of the thumb. It is interesting
to realize that the magnetizing force is independent of the type of core
material—it is determined solely by the number of turns, the current,
and the length of the core.
The applied magnetizing force has a pronounced effect on the resulting permeability of a magnetic material. As the magnetizing force
increases, the permeability rises to a maximum and then drops to a minimum, as shown in Fig. 11.19 for three commonly employed magnetic
materials.
I
N turns
I
Mean length l = 0.2 m
FIG. 11.18
Defining the magnetizing force of a magnetic
circuit.
442

MAGNETIC CIRCUITS
µ (permeability) × 10–3
10
9
8
7
6
5
4
3
2
1
0
Cast steel
Sheet steel
Cast iron
300 600 900 1200 1500 1800 2100 2400 2700 3000 3300 3600 3900 4200 4500
H (At/m)
FIG. 11.19
Variation of m with the magnetizing force.
The flux density and the magnetizing force are related by the following equation:
B mH
(11.8)
This equation indicates that for a particular magnetizing force, the
greater the permeability, the greater will be the induced flux density.
Since henries (H) and the magnetizing force (H) use the same capital letter, it must be pointed out that all units of measurement in the text,
such as henries, use roman letters, such as H, whereas variables such as
the magnetizing force use italic letters, such as H.
11.8 HYSTERESIS
I
N turns
I
A
Steel
FIG. 11.20
Series magnetic circuit used to define the
hysteresis curve.
A curve of the flux density B versus the magnetizing force H of a material is of particular importance to the engineer. Curves of this type can
usually be found in manuals, descriptive pamphlets, and brochures published by manufacturers of magnetic materials. A typical B-H curve for
a ferromagnetic material such as steel can be derived using the setup of
Fig. 11.20.
The core is initially unmagnetized and the current I 0. If the current I is increased to some value above zero, the magnetizing force H
will increase to a value determined by
H
NI D
l
The flux and the flux density B (B /A) will also increase with the
current I (or H). If the material has no residual magnetism, and the
magnetizing force H is increased from zero to some value Ha, the B-H
curve will follow the path shown in Fig. 11.21 between o and a. If the
HYSTERESIS
b
B (T)
a
c
BR
Saturation
Bmax
– Hs
d
– Hd
o
– Bmax
e
f
Ha
Hs
H (NI/l)
– BR
Saturation
FIG. 11.21
Hysteresis curve.
magnetizing force H is increased until saturation (Hs) occurs, the curve
will continue as shown in the figure to point b. When saturation occurs,
the flux density has, for all practical purposes, reached its maximum
value. Any further increase in current through the coil increasing H NI/l will result in a very small increase in flux density B.
If the magnetizing force is reduced to zero by letting I decrease to
zero, the curve will follow the path of the curve between b and c. The
flux density BR, which remains when the magnetizing force is zero, is
called the residual flux density. It is this residual flux density that makes
it possible to create permanent magnets. If the coil is now removed
from the core of Fig. 11.20, the core will still have the magnetic properties determined by the residual flux density, a measure of its “retentivity.” If the current I is reversed, developing a magnetizing force, H,
the flux density B will decrease with an increase in I. Eventually, the
flux density will be zero when Hd (the portion of curve from c to d)
is reached. The magnetizing force Hd required to “coerce” the flux
density to reduce its level to zero is called the coercive force, a measure
of the coercivity of the magnetic sample. As the force H is increased
until saturation again occurs and is then reversed and brought back to
zero, the path def will result. If the magnetizing force is increased in the
positive direction (H), the curve will trace the path shown from f to b.
The entire curve represented by bcdefb is called the hysteresis curve for
the ferromagnetic material, from the Greek hysterein, meaning “to lag
behind.” The flux density B lagged behind the magnetizing force H during the entire plotting of the curve. When H was zero at c, B was not
zero but had only begun to decline. Long after H had passed through
zero and had become equal to Hd did the flux density B finally
become equal to zero.
If the entire cycle is repeated, the curve obtained for the same core
will be determined by the maximum H applied. Three hysteresis loops
for the same material for maximum values of H less than the saturation
value are shown in Fig. 11.22. In addition, the saturation curve is
repeated for comparison purposes.
Note from the various curves that for a particular value of H, say, Hx,
the value of B can vary widely, as determined by the history of the core.
In an effort to assign a particular value of B to each value of H, we
compromise by connecting the tips of the hysteresis loops. The resulting curve, shown by the heavy, solid line in Fig. 11.22 and for various

443
444

MAGNETIC CIRCUITS
B (T )
H1 H2
H3
H (At/m)
HS
Hx
FIG. 11.22
Defining the normal magnetization curve.
B (T)
2.0
1.8
Sheet steel
1.6
Cast steel
1.4
1.2
1.0
0.8
0.6
Cast iron
0.4
0.2
0
300
600
900
1200
1500
1800
2100
2400
2700
3000
3300
3600
3900
4200
4500
H(At/m)
FIG. 11.23
Normal magnetization curve for three ferromagnetic materials.
materials in Fig. 11.23, is called the normal magnetization curve. An
expanded view of one region appears in Fig. 11.24.
A comparison of Figs. 11.19 and 11.23 shows that for the same value
of H, the value of B is higher in Fig. 11.23 for the materials with the
higher m in Fig. 11.19. This is particularly obvious for low values of H.
This correspondence between the two figures must exist since B mH.
In fact, if in Fig. 11.23 we find m for each value of H using the equation m B/H, we will obtain the curves of Fig. 11.19. An instrument
that will provide a plot of the B-H curve for a magnetic sample appears
in Fig. 11.25.
It is interesting to note that the hysteresis curves of Fig. 11.22 have
a point symmetry about the origin; that is, the inverted pattern to the left
of the vertical axis is the same as that appearing to the right of the ver-
HYSTERESIS
B (T)
1.4
1.3
Sheet steel
1.2
1.1
1.0
0.9
0.8
Cast steel
0.7
0.6
0.5
0.4
0.3
Cast iron
0.2
0.1
0
100
200
300
400
500
600
700
FIG. 11.24
Expanded view of Fig. 11.23 for the low magnetizing force region.
H (At/m)

445
446

MAGNETIC CIRCUITS
FIG. 11.25
Model 9600 vibrating sample magnetometer. (Courtesy of LDJ Electronics, Inc.)
tical axis. In addition, you will find that a further application of the
same magnetizing forces to the sample will result in the same plot. For
a current I in H NI/l that will move between positive and negative
maximums at a fixed rate, the same B-H curve will result during each
cycle. Such will be the case when we examine ac (sinusoidal) networks
in the later chapters. The reversal of the field () due to the changing
current direction will result in a loss of energy that can best be
described by first introducing the domain theory of magnetism.
Within each atom, the orbiting electrons (described in Chapter 2) are
also spinning as they revolve around the nucleus. The atom, due to its
spinning electrons, has a magnetic field associated with it. In nonmagnetic materials, the net magnetic field is effectively zero since the magnetic fields due to the atoms of the material oppose each other. In magnetic materials such as iron and steel, however, the magnetic fields of
groups of atoms numbering in the order of 1012 are aligned, forming
very small bar magnets. This group of magnetically aligned atoms is
called a domain. Each domain is a separate entity; that is, each domain
is independent of the surrounding domains. For an unmagnetized sample of magnetic material, these domains appear in a random manner,
such as shown in Fig. 11.26(a). The net magnetic field in any one direction is zero.
S
(a)
N
(b)
FIG. 11.26
Demonstrating the domain theory of magnetism.
When an external magnetizing force is applied, the domains that are
nearly aligned with the applied field will grow at the expense of the less
favorably oriented domains, such as shown in Fig. 11.26(b). Eventually,
if a sufficiently strong field is applied, all of the domains will have the
orientation of the applied magnetizing force, and any further increase in
external field will not increase the strength of the magnetic flux through
the core—a condition referred to as saturation. The elasticity of the
AMPÈRE’S CIRCUITAL LAW
above is evidenced by the fact that when the magnetizing force is
removed, the alignment will be lost to some measure, and the flux density will drop to BR. In other words, the removal of the magnetizing
force will result in the return of a number of misaligned domains within
the core. The continued alignment of a number of the domains, however, accounts for our ability to create permanent magnets.
At a point just before saturation, the opposing unaligned domains are
reduced to small cylinders of various shapes referred to as bubbles.
These bubbles can be moved within the magnetic sample through the
application of a controlling magnetic field. These magnetic bubbles
form the basis of the recently designed bubble memory system for computers.
11.9
AMPÈRE’S CIRCUITAL LAW
As mentioned in the introduction to this chapter, there is a broad similarity between the analyses of electric and magnetic circuits. This
has already been demonstrated to some extent for the quantities in
Table 11.1.
TABLE 11.1
Electric Circuits
Magnetic Circuits
E
I
R
Cause
Effect
Opposition
If we apply the “cause” analogy to Kirchhoff’s voltage law ( V 0), we obtain the following:
0
(for magnetic circuits)
(11.9)
which, in words, states that the algebraic sum of the rises and drops of
the mmf around a closed loop of a magnetic circuit is equal to zero; that
is, the sum of the rises in mmf equals the sum of the drops in mmf
around a closed loop.
Equation (11.9) is referred to as Ampère’s circuital law. When it is
applied to magnetic circuits, sources of mmf are expressed by the equation
NI
(At)
(11.10)
The equation for the mmf drop across a portion of a magnetic circuit
can be found by applying the relationships listed in Table 11.1; that is,
for electric circuits,
V IR
resulting in the following for magnetic circuits:
(At)
(11.11)
where is the flux passing through a section of the magnetic circuit
and is the reluctance of that section. The reluctance, however, is sel-

447

448
MAGNETIC CIRCUITS
dom calculated in the analysis of magnetic circuits. A more practical
equation for the mmf drop is
Hl
Steel
Iron
I
b
N turns
I
(11.12)
as derived from Eq. (11.6), where H is the magnetizing force on a section of a magnetic circuit and l is the length of the section.
As an example of Eq. (11.9), consider the magnetic circuit appearing
in Fig. 11.27 constructed of three different ferromagnetic materials.
Applying Ampère’s circuital law, we have
a
(At)
0
NI Hablab Hbclbc Hcalca 0
Cobalt
c
Rise
Drop
Drop
Drop
NI Hablab Hbclbc Hcalca
FIG. 11.27
Series magnetic circuit of three different
materials.
Impressed
mmf
mmf drops
All the terms of the equation are known except the magnetizing force
for each portion of the magnetic circuit, which can be found by using
the B-H curve if the flux density B is known.
11.10 THE FLUX If we continue to apply the relationships described in the previous section to Kirchhoff’s current law, we will find that the sum of the fluxes
entering a junction is equal to the sum of the fluxes leaving a junction;
that is, for the circuit of Fig. 11.28,
a
a
I
c
b
or
N
I
a
c
a b c
(at junction a)
b c a
(at junction b)
both of which are equivalent.
b
FIG. 11.28
Flux distribution of a series-parallel magnetic
network.
11.11 SERIES MAGNETIC CIRCUITS:
DETERMINING NI
We are now in a position to solve a few magnetic circuit problems, which
are basically of two types. In one type, is given, and the impressed
mmf NI must be computed. This is the type of problem encountered in
the design of motors, generators, and transformers. In the other type, NI
is given, and the flux of the magnetic circuit must be found. This type
of problem is encountered primarily in the design of magnetic amplifiers
and is more difficult since the approach is “hit or miss.”
As indicated in earlier discussions, the value of m will vary from
point to point along the magnetization curve. This eliminates the possibility of finding the reluctance of each “branch” or the “total reluctance” of a network, as was done for electric circuits where r had a
fixed value for any applied current or voltage. If the total reluctance
could be determined, could then be determined using the Ohm’s law
analogy for magnetic circuits.
For magnetic circuits, the level of B or H is determined from the
other using the B-H curve, and m is seldom calculated unless asked for.
SERIES MAGNETIC CIRCUITS: DETERMINING NI

449
An approach frequently employed in the analysis of magnetic circuits is the table method. Before a problem is analyzed in detail, a table
is prepared listing in the extreme left-hand column the various sections
of the magnetic circuit. The columns on the right are reserved for the
quantities to be found for each section. In this way, the individual doing
the problem can keep track of what is required to complete the problem
and also of what the next step should be. After a few examples, the usefulness of this method should become clear.
This section will consider only series magnetic circuits in which the
flux is the same throughout. In each example, the magnitude of the
magnetomotive force is to be determined.
EXAMPLE 11.3 For the series magnetic circuit of Fig. 11.29:
a. Find the value of I required to develop a magnetic flux of 4 104 Wb.
b. Determine m and mr for the material under these conditions.
Solutions: The magnetic circuit can be represented by the system
shown in Fig. 11.30(a). The electric circuit analogy is shown in Fig.
11.30(b). Analogies of this type can be very helpful in the solution of
magnetic circuits. Table 11.2 is for part (a) of this problem. The table is
fairly trivial for this example, but it does define the quantities to be
found.
I
A = 2 10–3 m2
N = 400 turns
Cast-steel core
I
l = 0.16 m
(mean length)
FIG. 11.29
Example 11.3.
(a)
I
E
R
(b)
FIG. 11.30
(a) Magnetic circuit equivalent and
(b) electric circuit analogy.
TABLE 11.2
Section
(Wb)
A (m2)
One continuous section
4 104
2 103
a. The flux density B is
4 104 Wb
B 2 101 T 0.2 T
A
2 103 m2
B (T)
H (At/m)
l (m)
0.16
Hl (At)
450

MAGNETIC CIRCUITS
Using the B-H curves of Fig. 11.24, we can determine the magnetizing force H:
H (cast steel) 170 At/m
Applying Ampère’s circuital law yields
NI Hl
and
Hl
(170 At/m)(0.16 m)
I 68 mA
N
400 t
(Recall that t represents turns.)
b. The permeability of the material can be found using Eq. (11.8):
B
H
0.2 T
170 At/m
m 1.176 103 Wb/A • m
and the relative permeability is
m
1.176 103
mr 935.83
4p 107
mo
EXAMPLE 11.4 The electromagnet of Fig. 11.31 has picked up a section of cast iron. Determine the current I required to establish the indicated flux in the core.
N = 50 turns
I
Sheet steel
I
f
Solution: To be able to use Figs. 11.23 and 11.24, we must first convert to the metric system. However, since the area is the same throughout, we can determine the length for each material rather than work
with the individual sections:
a
e
b
d
c
lefab 4 in. 4 in. 4 in. 12 in.
lbcde 0.5 in. 4 in. 0.5 in. 5 in.
1m
12 in. 304.8 103 m
39.37 in.
Cast iron
1m
5 in. 127 10 m
39.37 in.
1m
1m
1 in. 6.452 10 m
39.37 in. 39.37 in.
lab = lcd = lef = lfa = 4 in.
lbc = lde = 0.5 in.
Area (throughout) = 1 in.2
= 3.5 × 10–4 Wb
3
FIG. 11.31
Electromagnet for Example 11.4.
4
2
2
The information available from the specifications of the problem has
been inserted in Table 11.3. When the problem has been completed,
each space will contain some information. Sufficient data to complete
the problem can be found if we fill in each column from left to right. As
the various quantities are calculated, they will be placed in a similar
table found at the end of the example.
TABLE 11.3
Section
efab
bcde
(Wb)
4
3.5 10
3.5 104
A (m2)
B (T)
H (At/m)
4
l (m)
Hl (At)
3
6.452 10
6.452 104
304.8 10
127 103
The flux density for each section is
3.5 104 Wb
B 0.542 T
A
6.452 104 m2
SERIES MAGNETIC CIRCUITS: DETERMINING NI

451
and the magnetizing force is
H (sheet steel, Fig. 11.24) 70 At/m
H (cast iron, Fig. 11.23) 1600 At/m
Note the extreme difference in magnetizing force for each material for
the required flux density. In fact, when we apply Ampère’s circuital law,
we will find that the sheet steel section could be ignored with a minimal error in the solution.
Determining Hl for each section yields
Hefablefab (70 At/m)(304.8 103 m) 21.34 At
Hbcdelbcde (1600 At/m)(127 103 m) 203.2 At
Inserting the above data in Table 11.3 will result in Table 11.4.
TABLE 11.4
Section
(Wb)
A (m2)
B (T)
H (At/m)
l (m)
Hl (At)
efab
bcde
3.5 104
3.5 104
6.452 104
6.452 104
0.542
0.542
70
1600
304.8 103
127 103
21.34
203.2
The magnetic circuit equivalent and the electric circuit analogy for
the system of Fig. 11.31 appear in Fig. 11.32.
Applying Ampère’s circuital law,
NI Hefablefab Hbcdelbcde
21.34 At 203.2 At 224.54 At
efab
(50 t)I 224.54 At
and
224.54 At
I 4.49 A
50 t
so that
EXAMPLE 11.5 Determine the secondary current I2 for the transformer of Fig. 11.33 if the resultant clockwise flux in the core is 1.5 105 Wb.
a
I1 (2 A)
N1 = 60 turns
I1
d
Sheet steel
b
I2
N2 = 30 turns
I2
c
Area (throughout) = 0.15 × 10–3 m2
labcda = 0.16 m
FIG. 11.33
Transformer for Example 11.5.
Solution: This is the first example with two magnetizing forces to
consider. In the analogies of Fig. 11.34 you will note that the resulting
flux of each is opposing, just as the two sources of voltage are opposing in the electric circuit analogy.
The structural data appear in Table 11.5.
bcde
(a)
E
–
+
Refab
Rbcde
(b)
FIG. 11.32
(a) Magnetic circuit equivalent and
(b) electric circuit analogy for the
electromagnet of Fig. 11.31.
452

MAGNETIC CIRCUITS
abcda
Rabcda
I
2
1
E2
E1
(a)
(b)
FIG. 11.34
(a) Magnetic circuit equivalent and (b) electric circuit analogy for the
transformer of Fig. 11.33.
TABLE 11.5
Section
(Wb)
A (m2)
abcda
1.5 105
0.15 103
B (T)
H (At/m)
l (m)
Hl (At)
0.16
The flux density throughout is
1.5 105 Wb
B 10 102 T 0.10 T
0.15 103 m2
A
and
1
H (from Fig. 11.24) (100 At/m) 20 At/m
5
Applying Ampère’s circuital law,
N1I1 N2I2 Habcdalabcda
(60 t)(2 A) (30 t)(I2) (20 At/m)(0.16 m)
120 At (30 t)I2 3.2 At
and
or
c
Air gap
c
(a)
c
c
c
(b)
FIG. 11.35
Air gaps: (a) with fringing; (b) ideal.
(30 t)I2 120 At 3.2 At
116.8 At
I2 3.89 A
30 t
For the analysis of most transformer systems, the equation N1I1 N2I2 is employed. This would result in 4 A versus 3.89 A above. This
difference is normally ignored, however, and the equation N1I1 N2I2
considered exact.
Because of the nonlinearity of the B-H curve, it is not possible to
apply superposition to magnetic circuits; that is, in Example 11.5, we
cannot consider the effects of each source independently and then find
the total effects by using superposition.
11.12 AIR GAPS
Before continuing with the illustrative examples, let us consider the
effects that an air gap has on a magnetic circuit. Note the presence of
air gaps in the magnetic circuits of the motor and meter of Fig. 11.11.
The spreading of the flux lines outside the common area of the core for
the air gap in Fig. 11.35(a) is known as fringing. For our purposes, we
shall neglect this effect and assume the flux distribution to be as in Fig.
11.35(b).
AIR GAPS

453
The flux density of the air gap in Fig. 11.35(b) is given by
g
Bg Ag
(11.13)
where, for our purposes,
g core
Ag Acore
and
For most practical applications, the permeability of air is taken to be
equal to that of free space. The magnetizing force of the air gap is then
determined by
Bg
Hg (11.14)
mo
and the mmf drop across the air gap is equal to Hglg. An equation for
Hg is as follows:
Bg
Bg
Hg 4p 107
mo
Hg (7.96 105)Bg
and
(At/m)
(11.15)
EXAMPLE 11.6 Find the value of I required to establish a magnetic
flux of 0.75 104 Wb in the series magnetic circuit of Fig.
11.36.
All cast steel
core
Area (throughout)
= 1.5 × 10–4 m2
f
I
N = 200 turns
a
b
c
Air gap
gap
= 0.75 × 10–4 Wb
I
(a)
e
d
Rcdefab
lcdefab = 100 × 10–3 m
lbc = 2 × 10–3 m
FIG. 11.36
Relay for Example 11.6.
Solution: An equivalent magnetic circuit and its electric circuit
analogy are shown in Fig. 11.37.
The flux density for each section is
0.75 104 Wb
B 0.5 T
A
1.5 104 m2
I
Rbc
E
(b)
FIG. 11.37
(a) Magnetic circuit equivalent and
(b) electric circuit analogy for the relay of
Fig. 11.36.
454

MAGNETIC CIRCUITS
From the B-H curves of Fig. 11.24,
H (cast steel) 280 At/m
Applying Eq. (11.15),
Hg (7.96 105)Bg (7.96 105)(0.5 T) 3.98 105 At/m
The mmf drops are
Hcorelcore (280 At/m)(100 103 m) 28 At
Hglg (3.98 105 At/m)(2 103 m) 796 At
Applying Ampère’s circuital law,
NI Hcorelcore Hglg
28 At 796 At
(200 t)I 824 At
I 4.12 A
Note from the above that the air gap requires the biggest share (by
far) of the impressed NI due to the fact that air is nonmagnetic.
11.13 SERIES-PARALLEL MAGNETIC CIRCUITS
As one might expect, the close analogies between electric and magnetic
circuits will eventually lead to series-parallel magnetic circuits similar
in many respects to those encountered in Chapter 7. In fact, the electric
circuit analogy will prove helpful in defining the procedure to follow
toward a solution.
EXAMPLE 11.7 Determine the current I required to establish a flux of
1.5 104 Wb in the section of the core indicated in Fig. 11.38.
efab
a
T
1
be
1
2
2
bcde
I
b
T
1
Sheet steel
2 = 1.5 × 10–4 Wb
2
1
N = 50 turns
c
I
f
d
lbcde = lefab = 0.2 m
lbe = 0.05 m
Cross-sectional area = 6 × 10–4 m2 throughout
(a)
FIG. 11.38
Example 11.7.
Refab
IT
E
e
I1
1
Rbe
I2
2
Rbcde
Solution: The equivalent magnetic circuit and the electric circuit
analogy appear in Fig. 11.39. We have
1.5 104 Wb
B2 2 0.25 T
A
6 104 m2
(b)
FIG. 11.39
(a) Magnetic circuit equivalent and
(b) electric circuit analogy for the seriesparallel system of Fig. 11.38.
From Fig. 11.24,
Hbcde 40 At/m
Applying Ampère’s circuital law around loop 2 of Figs. 11.38 and
11.39,
SERIES-PARALLEL MAGNETIC CIRCUITS

455
0
Hbelbe Hbcdelbcde 0
Hbe(0.05 m) (40 At/m)(0.2 m) 0
8 At
Hbe 160 At/m
0.05 m
From Fig. 11.24,
B1 0.97 T
and
1 B1A (0.97 T)(6 104 m2) 5.82 104 Wb
The results are entered in Table 11.6.
TABLE 11.6
Section
(Wb)
A (m2)
bcde
be
efab
1.5 104
5.82 104
6 104
6 104
6 104
B (T)
0.25
0.97
The table reveals that we must now turn our attention to section
efab:
T 1 2 5.82 104 Wb 1.5 104 Wb
7.32 104 Wb
T
7.32 104 Wb
B A
6 104 m2
1.22 T
From Fig. 11.23,
Hefab 400 At
Applying Ampère’s circuital law,
NI Hefablefab Hbelbe 0
NI (400 At/m)(0.2 m) (160 At/m)(0.05 m)
(50 t)I 80 At 8 At
88 At
I 1.76 A
50 t
To demonstrate that m is sensitive to the magnetizing force H, the
permeability of each section is determined as follows. For section bcde,
B
H
0.25 T
40 At/m
m 6.25 103
and
m
6.25 103
mr 4972.2
mo 12.57 107
For section be,
B
H
0.97 T
160 At/m
m 6.06 103
H (At/m)
40
160
l (m)
0.2
0.05
0.2
Hl (At)
8
8
456

MAGNETIC CIRCUITS
m
6.06 103
mr 4821
mo 12.57 107
and
For section efab,
B
H
1.22 T
400 At/m
m 3.05 103
m
3.05 103
mr 2426.41
mo 12.57 107
and
11.14
DETERMINING The examples of this section are of the second type, where NI is given
and the flux must be found. This is a relatively straightforward problem if only one magnetic section is involved. Then
NI
H l
H→B
(B-H curve)
BA
and
For magnetic circuits with more than one section, there is no set order
of steps that will lead to an exact solution for every problem on the first
attempt. In general, however, we proceed as follows. We must find the
impressed mmf for a calculated guess of the flux and then compare
this with the specified value of mmf. We can then make adjustments to
our guess to bring it closer to the actual value. For most applications, a
value within 5% of the actual or specified NI is acceptable.
We can make a reasonable guess at the value of if we realize that the
maximum mmf drop appears across the material with the smallest permeability if the length and area of each material are the same. As shown
in Example 11.6, if there is an air gap in the magnetic circuit, there will
be a considerable drop in mmf across the gap. As a starting point for
problems of this type, therefore, we shall assume that the total mmf (NI)
is across the section with the lowest m or greatest (if the other physical dimensions are relatively similar). This assumption gives a value of
that will produce a calculated NI greater than the specified value. Then,
after considering the results of our original assumption very carefully, we
shall cut and NI by introducing the effects (reluctance) of the other portions of the magnetic circuit and try the new solution. For obvious reasons,
this approach is frequently called the cut and try method.
A (throughout) = 2 ×
I = 5A
a
10–4
m2
EXAMPLE 11.8 Calculate the magnetic flux for the magnetic circuit of Fig. 11.40.
Solution:
b
NI Habcdalabcda
N = 60 turns
NI
(60 t)(5 A)
Habcda 0.3 m
la bc da
300 At
1000 At/m
0.3 m
or
I
c
d
labcda = 0.3 m
FIG. 11.40
Example 11.8.
By Ampère’s circuital law,
Cast iron
and
Babcda (from Fig. 11.23) 0.39 T
Since B /A, we have
BA (0.39 T)(2 104 m2) 0.78 104 Wb
DETERMINING Φ
EXAMPLE 11.9 Find the magnetic flux for the series magnetic circuit of Fig. 11.41 for the specified impressed mmf.
Solution:
gap,
457
Cast iron
Assuming that the total impressed mmf NI is across the air
Φ
NI Hglg
Air gap
0.001 m
I = 4A
NI
400 At
Hg 4 105 At/m
0.001 m
lg
or
and

7
Area = 0.003 m2
N = 100 turns lcore = 0.16 m
Bg moHg (4p 10 )(4 10 At/m)
0.503 T
5
FIG. 11.41
Example 11.9.
The flux
g core Bg A
(0.503 T)(0.003 m2)
core 1.51 103 Wb
Using this value of , we can find NI. The data are inserted in Table
11.7.
TABLE 11.7
(Wb)
Section
A (m2)
B (T)
H (At/m)
l (m)
1500
(B-H curve)
4 105
0.16
3
Core
1.51 10
0.003
0.503
Gap
1.51 103
0.003
0.503
0.001
Hl (At)
400
Hcorelcore (1500 At/m)(0.16 m) 240 At
Applying Ampère’s circuital law results in
NI Hcorelcore Hglg
240 At 400 At
NI 640 At > 400 At
Since we neglected the reluctance of all the magnetic paths but the
air gap, the calculated value is greater than the specified value. We must
therefore reduce this value by including the effect of these reluctances.
Since approximately (640 At 400 At)/640 At 240 At/640 At 37.5% of our calculated value is above the desired value, let us reduce
by 30% and see how close we come to the impressed mmf of 400 At:
(1 0.3)(1.51 103 Wb)
1.057 103 Wb
See Table 11.8.
TABLE 11.8
Section
Core
Gap
(Wb)
A (m2)
3
1.057 10
1.057 103
0.003
0.003
B (T)
H (At/m)
l (m)
0.16
0.001
Hl (At)
458

MAGNETIC CIRCUITS
1.057 103 Wb
B 0.352 T
A
0.003 m2
Hglg (7.96 105)Bglg
(7.96 105)(0.352 T)(0.001 m)
280.19 At
From the B-H curves,
Hcore 850 At/m
Hcorelcore (850 At/m)(0.16 m) 136 At
Applying Ampère’s circuital law yields
NI Hcorelcore Hglg
136 At 280.19 At
NI 416.19 At > 400 At (but within 5%
and therefore acceptable)
The solution is, therefore,
1.057 103 Wb
11.15
APPLICATIONS
Recording Systems
The most common application of magnetic material is probably in the
increasing number of recording instruments used every day in the office
and the home. For instance, the VHS tape and the 8-mm cassette of Fig.
11.42 are used almost daily by every family with a VCR or cassette
player. The basic recording process is not that difficult to understand
and will be described in detail in the section to follow on computer hard
disks.
=
FIG. 11.42
Magnetic tape: (a) VHS and 8-mm cassette (Courtesy of Maxell Corporation of
America); (b) manufacturing process (Courtesy of Ampex Corporation).
>
APPLICATIONS

459
Speakers and Microphones
Electromagnetic effects are the moving force in the design of speakers
such as the one shown in Fig. 11.43. The shape of the pulsating waveform
of the input current is determined by the sound to be reproduced by the
speaker at a high audio level. As the current peaks and returns to the valleys of the sound pattern, the strength of the electromagnet varies in
exactly the same manner. This causes the cone of the speaker to vibrate at
a frequency directly proportional to the pulsating input. The higher the
pitch of the sound pattern, the higher the oscillating frequency between the
peaks and valleys and the higher the frequency of vibration of the cone.
A second design used more frequently in more expensive speaker
systems appears in Fig. 11.44. In this case the permanent magnet is
fixed and the input is applied to a movable core within the magnet, as
shown in the figure. High peaking currents at the input produce a strong
flux pattern in the voice coil, causing it to be drawn well into the flux
pattern of the permanent magnet. As occurred for the speaker of Fig.
11.43, the core then vibrates at a rate determined by the input and provides the audible sound.
Flexible cone
Electromagnet
i
i
Sound
i
Magnetic sample
(free to move)
FIG. 11.43
Speaker.
Magnetized
ferromagnetic
material
Lead terminal
Magnet
Magnetic gap
Cone
i
i
Voice coil
Magnet
(a)
(b)
FIG. 11.44
Coaxial high-fidelity loudspeaker: (a) construction; (b) basic operation;
(c) cross section of actual unit. (Courtesy of Electro-Voice, Inc.)
Microphones such as those in Fig. 11.45 also employ electromagnetic effects. The sound to be reproduced at a higher audio level causes
the core and attached moving coil to move within the magnetic field of
the permanent magnet. Through Faraday’s law (e N df/dt), a voltage
is induced across the movable coil proportional to the speed with which
it is moving through the magnetic field. The resulting induced voltage
pattern can then be amplified and reproduced at a much higher audio
level through the use of speakers, as described earlier. Microphones of
this type are the most frequently employed, although other types that
use capacitive, carbon granular, and piezoelectric* effects are available.
This particular design is commercially referred to as a dynamic microphone.
*Piezoelectricity is the generation of a small voltage by exerting pressure across certain
crystals.
(c)
460

MAGNETIC CIRCUITS
FIG. 11.45
Dynamic microphone. (Courtesy of Electro-Voice, Inc.)
Computer Hard Disks
The computer hard disk is a sealed unit in a computer that stores data
on a magnetic coating applied to the surface of circular platters that
spin like a record. The platters are constructed on a base of aluminum
or glass (both nonferromagnetic), which makes them rigid—hence the
term hard disk. Since the unit is sealed, the internal platters and components are inaccessible, and a “crash” (a term applied to the loss of
data from a disk or the malfunction thereof) usually requires that the
entire unit be replaced. Hard disks are currently available with diameters from less than 1 in. to 51⁄4 in., with the 31⁄ 2 in. the most popular for
today’s desktop units. Lap-top units typically use 21⁄ 2 in. All hard disk
drives are often referred to as Winchester drives, a term first applied in
the 1960s to an IBM drive that had 30 MB [a byte is a series of binary
bits (0s and 1s) representing a number, letter, or symbol] of fixed
(nonaccessible) data storage and 30 MB of accessible data storage. The
term Winchester was applied because the 30-30 data capacity matched
the name of the popular 30-30 Winchester rifle.
The magnetic coating on the platters is called the media and is of
either the oxide or the thin-film variety. The oxide coating is formed by
first coating the platter with a gel containing iron-oxide (ferromagnetic)
particles. The disk is then spun at a very high speed to spread the material evenly across the surface of the platter. The resulting surface is then
covered with a protective coating that is made as smooth as possible.
The thin-film coating is very thin, but durable, with a surface that is
smooth and consistent throughout the disk area. In recent years the
trend has been toward the thin-film coating because the read/write
heads (to be described shortly) must travel closer to the surface of the
platter, requiring a consistent coating thickness. Recent techniques have
resulted in thin-film magnetic coatings as thin as one-millionth of an
inch.
The information on a disk is stored around the disk in circular paths
called tracks or cylinders, with each track containing so many bits of
information per inch. The product of the number of bits per inch and the
number of tracks per inch is the Areal density of the disk, which provides an excellent quantity for comparison with early systems and
reveals how far the field has progressed in recent years. In the 1950s the
first drives had an Areal density of about 2 kbits/in.2 compared to
today’s typical 4 Gbits/in.2, an incredible achievement; consider
APPLICATIONS
4,000,000,000,000 bits of information on an area the size of the face of
your watch. Electromagnetism is the key element in the writing of
information on the disk and the reading of information off the disk.
In its simplest form the write/read head of a hard disk (or floppy
disk) is a U-shaped electromagnet with an air gap that rides just above
the surface of the disk, as shown in Fig. 11.46. As the disk rotates,
information in the form of a voltage with changing polarities is applied
to the winding of the electromagnet. For our purposes we will associate
a positive voltage level with a 1 level (of binary arithmetic) and a negative voltage level with a 0 level. Combinations of these 0 and 1 levels
can be used to represent letters, numbers, or symbols. If energized as
shown in Fig. 11.46 with a 1 level (positive voltage), the resulting magnetic flux pattern will have the direction shown in the core. When the
flux pattern encounters the air gap of the core, it jumps to the magnetic
material (since magnetic flux always seeks the path of least reluctance
and air has a high reluctance) and establishes a flux pattern, as shown
on the disk, until it reaches the other end of the core air gap, where it
returns to the electromagnet and completes the path. As the head then
moves to the next bit sector, it leaves behind the magnetic flux pattern
just established from the left to the right. The next bit sector has a 0
level input (negative voltage) that reverses the polarity of the applied
voltage and the direction of the magnetic flux in the core of the head.
The result is a flux pattern in the disk opposite that associated with a 1
level. The next bit of information is also a 0 level, resulting in the same
pattern just generated. In total, therefore, information is stored on the
disk in the form of small magnets whose polarity defines whether they
are representing a 0 or a 1.
+
V
–
I
Write
head
Φ
S
N
Track width
Ferromagnetic surface
Air gap
S
S
S N
N N
1
0
0
Disk motion
FIG. 11.46
Hard disk storage using a U-shaped electromagnet write head.
Now that the data have been stored, we must have some method to
retrieve the information when desired. The first few hard disks used the
same head for both the write and the read functions. In Fig. 11.47(a),
the U-shaped electromagnet in the read mode simply picks up the flux
pattern of the current bit of information. Faraday’s law of electromagnet induction states that a voltage is induced across a coil if exposed to
a changing magnetic field. The change in flux for the core in Fig.
11.47(a) is minimal as it passes over the induced bar magnet on the surface of the disk. A flux pattern is established in the core because of the

461
462

MAGNETIC CIRCUITS
∆Φ ≅ 0 Wb
V≅0V
+
V
–
0
1
0V
0V
a
N
S S
0
a
N N
1
c
t
b
S
0
c
1
b
(a)
0V
0
(b)
FIG. 11.47
Reading the information off a hard disk using a U-shaped electromagnet.
bar magnet on the disk, but the lack of a significant change in flux level
results in an induced voltage at the output terminals of the pickup of
approximately 0 V, as shown in Fig. 11.47(b) for the readout waveform.
A significant change in flux occurs when the head passes over the transition region so marked in Fig. 11.47(a). In region a the flux pattern
changes from one direction to the other—a significant change in flux
occurs in the core as it reverses direction, causing a measurable voltage
to be generated across the terminals of the pickup coil as dictated by
Faraday’s law and indicated in Fig. 11.47(b). In region b there is no significant change in the flux pattern from one bit area to the next, and a
voltage is not generated, as also revealed in Fig. 11.47(b). However,
when region c is reached, the change in flux is significant but opposite
that occurring in region a, resulting in another pulse but of opposite
polarity. In total, therefore, the output bits of information are in the
form of pulses that have a shape totally different from the read signals
but that are certainly representative of the information being stored. In
addition, note that the output is generated at the transition regions and
not in the constant flux region of the bit storage.
In the early years, the use of the same head for the read and write
functions was acceptable; but as the tracks became narrower and the seek
time (the average time required to move from one track to another a random distance away) had to be reduced, it became increasingly difficult to
construct the coil or core configuration in a manner that was sufficiently
thin with minimum weight. In the late 1970s IBM introduced the thinfilm inductive head, which was manufactured in much the same way as
the small integrated circuits of today. The result is a head having a length
typically less than 1⁄10 in., a height less than 1⁄50 in., and minimum mass
and high durability. The average seek time has dropped from a few hundred milliseconds to 6 ms to 8 ms for very fast units and 8 ms to 10 ms
for average units. In addition, production methods have improved to the
point that the head can “float” above the surface (to minimize damage to
the disk) at a height of only 5 microinches or 0.000005 in. Using a typical lap-top hard disk speed of 3600 rpm (as high as 7200 rpm for desktops) and an average diameter of 1.75 in. for a 3.5-in. disk, the speed of
the head over the track is about 38 mph. Scaling the floating height up to
1
⁄4 in. (multiplying by a factor of 50,000), the speed would increase to
about 1.9 106 mph. In other words, the speed of the head over the surface of the platter is analogous to a mass traveling 1⁄4 in. above a surface
at 1.9 million miles per hour, all the while ensuring that the head never
touches the surface of the disk—quite a technical achievement and amaz-
APPLICATIONS
ingly enough one that perhaps will be improved by a factor of 10 in the
next decade. Incidentally, the speed of rotation of floppy disks is about
1
⁄10 that of the hard disk, or 360 rpm. In addition, the head touches the
magnetic surface of the floppy disk, limiting the storage life of the unit.
The typical magnetizing force needed to lay down the magnetic orientation is 400 mA-turn (peak-to-peak). The result is a write current of only
40 mA for a 10-turn, thin-film inductive head.
Although the thin-film inductive head could also be used as a read
head, the magnetoresistive (MR) head has improved reading characteristics. The MR head depends on the fact that the resistance of a soft ferromagnetic conductor such as permolloy is sensitive to changes in
external magnetic fields. As the hard disk rotates, the changes in magnetic flux from the induced magnetized regions of the platter change the
terminal resistance of the head. A constant current passed through the
sensor displays a terminal voltage sensitive to the magnitude of the
resistance. The result is output voltages with peak values in excess of
300 V, which exceeds that of typical inductive read heads by a factor of
2 or 31.
Further investigation will reveal that the best write head is of the
thin-film inductive variety and that the optimum read head is of the MR
variety. Each has particular design criteria for maximum performance,
resulting in the increasingly common dual-element head, with each
head containing separate conductive paths and different gap widths.
The Areal density of the new hard disks will essentially require the
dual-head assembly for optimum performance.
As the density of the disk increases, the width of the tracks or
cylinders will decrease accordingly. The net result will be smaller
heads for the read/write function, an arm supporting the head that
must be able to move into and out of the rotating disk in smaller
increments, and an increased sensitivity to temperature effects which
can cause the disk itself to contract or expand. At one time the
mechanical system with its gears and pulleys was sensitive enough to
perform the task. However, today’s density requires a system with less
play and with less sensitivity to outside factors such as temperature
and vibration. A number of modern drives use a voice coil and ferromagnetic arm as shown in Fig. 11.48. The current through the coil will
determine the magnetic field strength within the coil and will cause
Voice coil
Read/Write
head
FIG. 11.48
Disk drive with voice coil and ferromagnetic arm.
Ferromagnetic
shaft
Control

463
464

MAGNETIC CIRCUITS
the supporting arm for the head to move in and out, thereby establishing a rough setting for the extension of the arm over the disk. It
would certainly be possible to relate the position of the arm to the
applied voltage to the coil, but this would lack the level of accuracy
required for high-density disks. For the desired accuracy, a laser beam
has been added as an integral part of the head. Circular strips placed
around the disk (called track indicators) ensure that the laser beam
homes in and keeps the head in the right position. Assuming that the
track is a smooth surface and the surrounding area a rough texture, a
laser beam will be reflected back to the head if it’s on the track,
whereas the beam will be scattered if it hits the adjoining areas. This
type of system permits continuous recalibration of the arm by simply
comparing its position with the desired location—a maneuver referred
to as “recalibration on the fly.”
As with everything, there are limits to any design. However, in this
case, it is not because larger disks cannot be made or that more tracks
cannot be put on the disk. The limit to the size of hard drives in PCs
is set by the BIOS (Basic Input Output System) drive that is built into
all PCs. When first developed years ago, it was designed around a
maximum storage possibility of 8.4 gigabytes. At that time this number seemed sufficiently large to withstand any new developments for
many years to come. However, 8-gigabyte drives and larger are now
becoming commonplace, with lap-tops averaging 20 gigabytes and
desktops averaging 40 gigabytes. The result is that mathematical methods had to be developed to circumvent the designed maximum for
each component of the BIOS system. Fundamentally, the maximum
values for the BIOS drive are the following:
Cylinders (or tracks)
Heads
Sectors
Bytes per sector
FIG. 11.49
A 3.5-in. hard disk drive with a capacity of
17.2 GB and an average search time of 9 ms.
(Courtesy of Seagate Corporation.)
1024
128
128
512
Multiplying through all the factors results in a maximum of 8.59 gigabytes, but the colloquial reference is normally 8.4 gigabytes.
Most modern drives use a BIOS translation technique whereby they
play a mathematical game in which they make the drive appear different
to the BIOS system than it actually is. For instance, the drive may have
2048 tracks and 16 heads, but through the mathematical link with the
BIOS system it will appear to have 1024 tracks and 32 heads. In other
words, there was a trade-off between numbers in the official maximum
listing. This is okay for certain combinations, but the total combination
of figures for the design still cannot exceed 8.4 gigabytes. Also be aware
that this mathematical manipulation is possible only if the operating system has BIOS translation built in. By implementing new enchanced IDE
controllers, BIOS can have access drives greater than 8.4 gigabytes.
The above is clear evidence of the importance of magnetic effects in
today’s growing industrial, computer-oriented society. Although
research continues to maximize the Areal density, it appears certain that
the storage will remain magnetic for the write/read process and will not
be replaced by any of the growing alternatives such as the optic laser
variety used so commonly in CD-ROMs.
A 3.5-in. full-height disk drive, which is manufactured by the
Seagate Corporation and has a formatted capacity of 17.2 gigabytes
(GB) with an average search time of 9 ms, appears in Fig. 11.49.
APPLICATIONS

465
Hall Effect Sensor
The Hall effect sensor is a semiconductor device that generates an output voltage when exposed to a magnetic field. The basic construction
consists of a slab of semiconductor material through which a current
is passed, as shown in Fig. 11.50(a). If a magnetic field is applied as
shown in the figure perpendicular to the direction of the current, a
voltage VH will be generated between the two terminals, as indicated
in Fig. 11.50(a). The difference in potential is due to the separation of
charge established by the Lorentz force first studied by Professor Hendrick Lorentz in the early eighteenth century. He found that electrons
in a magnetic field are subjected to a force proportional to the velocity of the electrons through the field and the strength of the magnetic
field. The direction of the force is determined by the left-hand rule.
Simply place the index finger of the left hand in the direction of the
magnetic field, with the second finger at right angles to the index finger in the direction of conventional current through the semiconductor
material, as shown in Fig. 11.50(b). The thumb, if placed at right
angles to the index finger, will indicate the direction of the force on
the electrons. In Fig. 11.50(b), the force causes the electrons to accumulate in the bottom region of the semiconductor (connected to the
negative terminal of the voltage VH), leaving a net positive charge in
the upper region of the material (connected to the positive terminal of
VH). The stronger the current or strength of the magnetic field, the
greater the induced voltage VH.
In essence, therefore, the Hall effect sensor can reveal the strength of
a magnetic field or the level of current through a device if the other
determining factor is held fixed. Two applications of the sensor are
therefore apparent—to measure the strength of a magnetic field in the
vicinity of a sensor (for an applied fixed current) and to measure the
level of current through a sensor (with knowledge of the strength of the
magnetic field linking the sensor). The gaussmeter in Fig. 11.16
employs a Hall effect sensor. Internal to the meter, a fixed current is
passed through the sensor with the voltage VH indicating the relative
strength of the field. Through amplification, calibration, and proper
scaling, the meter can display the relative strength in gauss.
The Hall effect sensor has a broad range of applications that are often
quite interesting and innovative. The most widespread is as a trigger for
an alarm system in large department stores, where theft is often a difficult problem. A magnetic strip attached to the merchandise sounds an
alarm when a customer passes through the exit gates without paying for
the product. The sensor, control current, and monitoring system are
housed in the exit fence and react to the presence of the magnetic field as
the product leaves the store. When the product is paid for, the cashier
removes the strip or demagnetizes the strip by applying a magnetizing
force that reduces the residual magnetism in the strip to essentially zero.
The Hall effect sensor is also used to indicate the speed of a bicycle
on a digital display conveniently mounted on the handlebars. As shown
in Fig. 11.51(a), the sensor is mounted on the frame of the bike, and a
small permanent magnet is mounted on a spoke of the front wheel. The
magnet must be carefully mounted to be sure that it passes over the
proper region of the sensor. When the magnet passes over the sensor,
the flux pattern in Fig. 11.51(b) results, and a voltage with a sharp peak
is developed by the sensor. Assuming a bicycle with a 26-in.-diameter
B
I
(conventional
flow)
+
VH
–
(a)
Magnetic field
into page
I
e–
++++++++++++++++
e–
e–
e–
+
I
VH
––––––––––––––––
–
(b)
FIG. 11.50
Hall effect sensor: (a) orientation of
controlling parameters; (b) effect on electron
flow.
466

MAGNETIC CIRCUITS
+
VH
–
I
I (from battery)
I
I
Hall effect sensor
Permanent
magnet
+
VH
–
Hall
effect
sensor
Time
for one
rotation
B
N
S
Motion
Spoke
(a)
(b)
FIG. 11.51
Obtaining a speed indication for a bicycle using a Hall effect sensor:
(a) mounting the components; (b) Hall effect response.
wheel, the circumference will be about 82 in. Over 1 mi, the number of
rotations is
12 in. 1 rotation
5280 ft 773 rotations
1 ft
82 in.
Reeds
Embedded
permanent
magnet
Plastic
housing
N
S
If the bicycle is traveling at 20 mph, an output pulse will occur at
a rate of 4.29 per second. It is interesting to note that at a speed of
20 mph, the wheel is rotating at more than 4 revolutions per second,
and the total number of rotations over 20 mi is 15,460.
Magnetic Reed Switch
Sealed
capsule
FIG. 11.52
Magnetic reed switch.
Permanent
magnet
Reed switch
Control
FIG. 11.53
Using a magnetic reed switch to monitor the
state of a window.
One of the most frequently employed switches in alarm systems is the
magnetic reed switch shown in Fig. 11.52. As shown by the figure, there
are two components of the reed switch—a permanent magnet embedded
in one unit that is normally connected to the movable element (door, window, and so on) and a reed switch in the other unit that is connected to
the electrical control circuit. The reed switch is constructed of two ironalloy (ferromagnetic) reeds in a hermetically sealed capsule. The cantilevered ends of the two reeds do not touch but are in very close proximity to one another. In the absence of a magnetic field the reeds remain
separated. However, if a magnetic field is introduced, the reeds will be
drawn to each other because flux lines seek the path of least reluctance
and, if possible, exercise every alternative to establish the path of least
reluctance. It is similar to placing a ferromagnetic bar close to the ends
of a U-shaped magnet. The bar is drawn to the poles of the magnet, establishing a magnetic flux path without air gaps and with minimum reluctance. In the open-circuit state the resistance between reeds is in excess
of 100 M, while in the on state it drops to less than 1 .
In Fig. 11.53 a reed switch has been placed on the fixed frame of a
window and a magnet on the movable window unit. When the window is
closed as shown in Fig. 11.53, the magnet and reed switch are suffi-
APPLICATIONS
ciently close to establish contact between the reeds, and a current is
established through the reed switch to the control panel. In the armed
state the alarm system accepts the resulting current flow as a normal
secure response. If the window is opened, the magnet will leave the
vicinity of the reed switch, and the switch will open. The current
through the switch will be interrupted, and the alarm will react appropriately.
One of the distinct advantages of the magnetic reed switch is that the
proper operation of any switch can be checked with a portable magnetic
element. Simply bring the magnet to the switch and note the output
response. There is no need to continually open and close windows and
doors. In addition, the reed switch is hermetically enclosed so that oxidation and foreign objects cannot damage it, and the result is a unit that
can last indefinitely. Magnetic reed switches are also available in other
shapes and sizes, allowing them to be concealed from obvious view.
One is a circular variety that can be set into the edge of a door and door
jam, resulting in only two small visible disks when the door is open.

467
FIG. 11.54
Magnetic resonance imaging equipment.
(Courtesy of Siemens Medical Systems, Inc.)
Magnetic Resonance Imaging
Magnetic resonance imaging [MRI, also called nuclear magnetic resonance (NMR)] is receiving more and more attention as we strive to
improve the quality of the cross-sectional images of the body so useful
in medical diagnosis and treatment. MRI does not expose the patient to
potentially hazardous X rays or injected contrast materials such as those
employed to obtain computerized axial tomography (CAT) scans.
The three major components of an MRI system are a huge magnet
that can weigh up to 100 tons, a table for transporting the patient into
the circular hole in the magnet, and a control center, as shown in Fig.
11.54. The image is obtained by placing the patient in the tube to a precise depth depending on the cross section to be obtained and applying
a strong magnetic field that causes the nuclei of certain atoms in the
body to line up. Radio waves of different frequencies are then applied
to the patient in the region of interest, and if the frequency of the wave
matches the natural frequency of the atom, the nuclei will be set into a
state of resonance and will absorb energy from the applied signal.
When the signal is removed, the nuclei release the acquired energy in
the form of weak but detectable signals. The strength and duration of
the energy emission vary from one tissue of the body to another. The
weak signals are then amplified, digitized, and translated to provide a
cross-sectional image such as the one shown in Fig. 11.55.
MRI units are very expensive and therefore are not available at all
locations. In recent years, however, their numbers have grown, and one
is available in almost every major community. For some patients the
claustrophobic feeling they experience while in the circular tube is difficult to contend with. Today, however, a more open unit has been
developed, as shown in Fig. 11.56, that has removed most of this discomfort.
Patients who have metallic implants or pacemakers or those who
have worked in industrial environments where minute ferromagnetic
particles may have become lodged in open, sensitive areas such as the
eyes, nose, and so on, may have to use a CAT scan system because it
does not employ magnetic effects. The attending physician is well
trained in such areas of concern and will remove any unfounded fears
or suggest alternative methods.
FIG. 11.55
Magnetic resonance image. (Courtesy
of Siemens Medical Systems, Inc.)
FIG. 11.56
Magnetic resonance imaging equipment (open
variety). (Courtesy of Siemens Medical
Systems, Inc.)
468

MAGNETIC CIRCUITS
PROBLEMS
SECTION 11.3 Flux Density
1. Using Appendix F, fill in the blanks in the following
table. Indicate the units for each quantity.
SI
CGS
English
B
5 104 Wb
_________
_________
8 104 T
_________
_________
2. Repeat Problem 1 for the following table if area 2 in.2:
SI
CGS
English
_________
60,000 maxwells
_________
Φ = 4 × 10–4 Wb
N turns
SECTION 11.5 Reluctance
4. Which section of Fig. 11.58—(a), (b), or (c)—has the
largest reluctance to the setting up of flux lines through
its longest dimension?
FIG. 11.57
Problem 3.
3 in.
0.01 m
1 cm
2 cm
Iron
6 cm
0.01 m
Iron
Iron
0.1 m
1 in.
2
(a)
_________
_________
_________
3. For the electromagnet of Fig. 11.57:
a. Find the flux density in the core.
b. Sketch the magnetic flux lines and indicate their
direction.
c. Indicate the north and south poles of the magnet.
Area = 0.01 m2
I
B
(b)
(c)
FIG. 11.58
Problem 4.
SECTION 11.6 Ohm’s Law for Magnetic Circuits
5. Find the reluctance of a magnetic circuit if a magnetic
flux 4.2 104 Wb is established by an impressed
mmf of 400 At.
6. Repeat Problem 5 for 72,000 maxwells and an
impressed mmf of 120 gilberts.
SECTION 11.7 Magnetizing Force
7. Find the magnetizing force H for Problem 5 in SI units if
the magnetic circuit is 6 in. long.
8. If a magnetizing force H of 600 At/m is applied to a magnetic circuit, a flux density B of 1200 104 Wb/m2 is
established. Find the permeability m of a material that
will produce twice the original flux density for the same
magnetizing force.
PROBLEMS

469
SECTION 11.8 Hysteresis
9. For the series magnetic circuit of Fig. 11.59, determine
the current I necessary to establish the indicated flux.
10. Find the current necessary to establish a flux of 3 104 Wb in the series magnetic circuit of Fig. 11.60.
Area (throughout)
= 3 × 10–3 m2
Cast iron
Sheet steel
I
Φ
N
I
N = 75 turns
Cast iron
Φ
I
liron core = lsteel core = 0.3 m
Area (throughout) = 5 10–4 m2
N = 100 turns
Φ = 10 × 10–4 Wb
Mean length = 0.2 m
FIG. 11.60
Problem 10.
FIG. 11.59
Problem 9.
11. a. Find the number of turns N1 required to establish a
flux 12 104 Wb in the magnetic circuit of
Fig. 11.61.
b. Find the permeability m of the material.
12. a. Find the mmf (NI) required to establish a flux 80,000 lines in the magnetic circuit of Fig. 11.62.
b. Find the permeability of each material.
Cast steel
Cast steel
Φ
Sheet steel
NI
I = 1A
I =
2A
N2 = 30 turns
N1
lcast steel = 5.5 in.
lsheet steel = 0.5 in.
lm
2
Area = 0.0012 m
lm (mean length) = 0.2 m
Uniform area
(throughout)
= 1 in.2
FIG. 11.62
Problem 12.
FIG. 11.61
Problem 11.
Cast steel
*13. For the series magnetic circuit of Fig. 11.63 with two
impressed sources of magnetic “pressure,” determine the
current I. Each applied mmf establishes a flux pattern in
the clockwise direction.
I
Φ = 0.8 10
–4 Wb
I
N1 = 20 turns I N2 = 30 turns
lcast steel = 5.5 in.
lcast iron = 2.5 in.
Cast iron
Area (throughout) = 0.25 in.2
FIG. 11.63
Problem 13.
470

MAGNETIC CIRCUITS
SECTION 11.12 Air Gaps
Sheet steel
Φ
a
N =
100
turns
b
I
0.003 m
c
d
I
f
14. a. Find the current I required to establish a flux 2.4 104 Wb in the magnetic circuit of Fig. 11.64.
b. Compare the mmf drop across the air gap to that
across the rest of the magnetic circuit. Discuss your
results using the value of m for each material.
e
Area (throughout) = 2 × 10–4 m2
lab = lef = 0.05 m
laf = lbe = 0.02 m
lbc = lde
FIG. 11.64
Problem 14.
*15. The force carried by the plunger of the door chime of
Fig. 11.65 is determined by
4 cm
Chime
f
Plunger
FIG. 11.65
Door chime for Problem 15.
I1
Sheet steel
0.002 m
N1 = 200 turns
0.3
I1
Φ
(newtons)
where df/dx is the rate of change of flux linking the coil
as the core is drawn into the coil. The greatest rate of
change of flux will occur when the core is 1⁄4 to 3⁄4 the
way through. In this region, if changes from 0.5 104 Wb to 8 104 Wb, what is the force carried by
the plunger?
I
I = 900 mA
N = 80 turns
1
df
f NI 2
dx
16. Determine the current I1 required to establish a flux of
2 104 Wb in the magnetic circuit of Fig.
11.66.
m
I2 = 0.3 A
N2 = 40 turns
Area (throughout) = 1.3 × 10–4 m2
FIG. 11.66
Problem 16.
Spring
Armature
Air gap = 0.2 cm
Contacts
Coil
N = 200 turns
Diameter of core = 0.01 m
Solenoid
I
FIG. 11.67
Relay for Problem 17.
*17. a. A flux of 0.2 104 Wb will establish sufficient
attractive force for the armature of the relay of Fig.
11.67 to close the contacts. Determine the required
current to establish this flux level if we assume that
the total mmf drop is across the air gap.
b. The force exerted on the armature is determined by
the equation
2
1 Bg A
F (newtons) · 2 mo
where Bg is the flux density within the air gap and A
is the common area of the air gap. Find the force in
newtons exerted when the flux specified in part (a)
is established.
PROBLEMS
*18. For the series-parallel magnetic circuit of Fig. 11.68, find
the value of I required to establish a flux in the gap of
g 2 104 Wb.

471
Sheet steel throughout
T
a
I
N =
200 turns
1
b
1
Area =
2 × 10–4 m2
h
2
2
c
0.002 m
d
e
f
g
Area for sections other than bg = 5 ×
lab = lbg = lgh = lha = 0.2 m
lbc = lfg = 0.1 m, lcd = lef = 0.099 m
10–4
m2
FIG. 11.68
Problem 18.
SECTION 11.14 Determining 19. Find the magnetic flux established in the series magnetic circuit of Fig. 11.69.
Φ
I = 2A
8m
0.0
N = 100 turns
Area =
0.009 m2
Cast steel
FIG. 11.69
Problem 19.
*20. Determine the magnetic flux established in the series
magnetic circuit of Fig. 11.70.
a
Cast steel
I = 2A
Φ
b
c
N = 150 turns
d
f
e
lcd = 8 × 10 – 4 m
lab = lbe = lef = lfa = 0.2 m
Area (throughout) = 2 × 10 – 4 m2
lbc = lde
FIG. 11.70
Problem 20.
*21. Note how closely the B-H curve of cast steel in Fig.
11.23 matches the curve for the voltage across a capacitor as it charges from zero volts to its final value.
a. Using the equation for the charging voltage as a
guide, write an equation for B as a function of H [B f(H)] for cast steel.
b. Test the resulting equation at H 900 At/m, 1800
At/m, and 2700 At/m.
c. Using the equation of part (a), derive an equation for
H in terms of B [H f(B)].
d. Test the resulting equation at B 1 T and B 1.4 T.
e. Using the result of part (c), perform the analysis of
Example 11.3, and compare the results for the current I.
COMPUTER ANALYSIS Programming Language
(C, QBASIC, Pascal, etc.)
*22. Using the results of Problem 21, write a program to perform the analysis of a core such as that shown in Example 11.3; that is, let the dimensions of the core and the
applied turns be input variables requested by the program.
*23. Using the results of Problem 21, develop a program to
perform the analysis appearing in Example 11.9 for cast
steel. A test routine will have to be developed to determine whether the results obtained are sufficiently close to
the applied ampere-turns.
472

MAGNETIC CIRCUITS
GLOSSARY
Ampère’s circuital law A law establishing the fact that the
algebraic sum of the rises and drops of the mmf around a
closed loop of a magnetic circuit is equal to zero.
Diamagnetic materials Materials that have permeabilities
slightly less than that of free space.
Domain A group of magnetically aligned atoms.
Electromagnetism Magnetic effects introduced by the flow
of charge or current.
Ferromagnetic materials Materials having permeabilities
hundreds and thousands of times greater than that of free
space.
Flux density (B) A measure of the flux per unit area perpendicular to a magnetic flux path. It is measured in teslas (T)
or webers per square meter (Wb/m2).
Hysteresis The lagging effect between the flux density of a
material and the magnetizing force applied.
Magnetic flux lines Lines of a continuous nature that reveal
the strength and direction of a magnetic field.
Magnetizing force (H) A measure of the magnetomotive
force per unit length of a magnetic circuit.
Magnetomotive force (mmf) () The “pressure” required to
establish magnetic flux in a ferromagnetic material. It is
measured in ampere-turns (At).
Paramagnetic materials Materials that have permeabilities
slightly greater than that of free space.
Permanent magnet A material such as steel or iron that will
remain magnetized for long periods of time without the aid
of external means.
Permeability (m) A measure of the ease with which magnetic flux can be established in a material. It is measured in
Wb/Am.
Relative permeability (mr) The ratio of the permeability of a
material to that of free space.
Reluctance () A quantity determined by the physical characteristics of a material that will provide an indication of
the “reluctance” of that material to the setting up of magnetic flux lines in the material. It is measured in rels or
At/Wb.
12
Inductors
12.1
INTRODUCTION
We have examined the resistor and the capacitor in detail. In this chapter we shall consider a third element, the inductor, which has a number
of response characteristics similar in many respects to those of the
capacitor. In fact, some sections of this chapter will proceed parallel to
those for the capacitor to emphasize the similarity that exists between
the two elements.
12.2 FARADAY’S LAW OF
ELECTROMAGNETIC INDUCTION
If a conductor is moved through a magnetic field so that it cuts magnetic lines of flux, a voltage will be induced across the conductor, as
shown in Fig. 12.1. The greater the number of flux lines cut per unit
time (by increasing the speed with which the conductor passes through
the field), or the stronger the magnetic field strength (for the same tra-
–
eind
+
V
Φ
r
cto
du
n
Co
S
N
Motion
FIG. 12.1
Generating an induced voltage by moving a
conductor through a magnetic field.
474

INDUCTORS
versing speed), the greater will be the induced voltage across the conductor. If the conductor is held fixed and the magnetic field is moved so
that its flux lines cut the conductor, the same effect will be produced.
If a coil of N turns is placed in the region of a changing flux, as in
Fig. 12.2, a voltage will be induced across the coil as determined by
Faraday’s law:
Changing flux
+
e
–
df
e N dt
dφ
φ
dt
FIG. 12.2
Demonstrating Faraday’s law.
e induced
–
+
I
I
(volts, V)
(12.1)
where N represents the number of turns of the coil and df/dt is the
instantaneous change in flux (in webers) linking the coil. The term linking refers to the flux within the turns of wire. The term changing simply indicates that either the strength of the field linking the coil changes
in magnitude or the coil is moved through the field in such a way that
the number of flux lines through the coil changes with time.
If the flux linking the coil ceases to change, such as when the coil
simply sits still in a magnetic field of fixed strength, df/dt 0, and the
induced voltage e N(df/dt) N(0) 0.
12.3 LENZ’S LAW
FIG. 12.3
Demonstrating the effect of Lenz’s law.
American (Albany,
NY; Princeton, NJ)
(1797–1878)
Physicist and
Mathematician
Professor of Natural
Philosophy,
Princeton
University
Courtesy of the
Smithsonian Institution
Photo No. 59,054
In the early 1800s the title Professor of Natural Philosophy was applied to educators in the sciences. As
a student and teacher at the Albany Academy, Henry
performed extensive research in the area of electromagnetism. He improved the design of electromagnets by insulating the coil wire to permit a tighter
wrap on the core. One of his earlier designs was
capable of lifting 3600 pounds. In 1832 he discovered and delivered a paper on self-induction. This
was followed by the construction of an effective electric telegraph transmitter and receiver and extensive
research on the oscillatory nature of lightning and
discharges from a Leyden jar. In 1845 he was
appointed the first Secretary of the Smithsonian.
FIG. 12.4
Joseph Henry.
In Section 11.2 it was shown that the magnetic flux linking a coil of N
turns with a current I has the distribution of Fig. 12.3.
If the current increases in magnitude, the flux linking the coil also
increases. It was shown in Section 12.2, however, that a changing flux
linking a coil induces a voltage across the coil. For this coil, therefore,
an induced voltage is developed across the coil due to the change in
current through the coil. The polarity of this induced voltage tends to
establish a current in the coil that produces a flux that will oppose any
change in the original flux. In other words, the induced effect (eind) is a
result of the increasing current through the coil. However, the resulting
induced voltage will tend to establish a current that will oppose the
increasing change in current through the coil. Keep in mind that this is
all occurring simultaneously. The instant the current begins to increase
in magnitude, there will be an opposing effect trying to limit the
change. It is “choking” the change in current through the coil. Hence,
the term choke is often applied to the inductor or coil. In fact, we will
find shortly that the current through a coil cannot change instantaneously. A period of time determined by the coil and the resistance of
the circuit is required before the inductor discontinues its opposition to
a momentary change in current. Recall a similar situation for the voltage across a capacitor in Chapter 10. The reaction above is true for
increasing or decreasing levels of current through the coil. This effect is
an example of a general principle known as Lenz’s law, which states
that
an induced effect is always such as to oppose the cause that produced it.
12.4
SELF-INDUCTANCE
The ability of a coil to oppose any change in current is a measure of the
self-inductance L of the coil. For brevity, the prefix self is usually
dropped. Inductance is measured in henries (H), after the American
physicist Joseph Henry (Fig. 12.4).
TYPES OF INDUCTORS
Inductors are coils of various dimensions designed to introduce
specified amounts of inductance into a circuit. The inductance of a coil
varies directly with the magnetic properties of the coil. Ferromagnetic
materials, therefore, are frequently employed to increase the inductance
by increasing the flux linking the coil.
A close approximation, in terms of physical dimensions, for the
inductance of the coils of Fig. 12.5 can be found using the following
equation:
N 2mA
L l

475
N turns
A
l
d
(henries, H)
(12.2)
where N represents the number of turns; m, the permeability of the core
(as introduced in Section 11.4; recall that m is not a constant but
depends on the level of B and H since m B/H); A, the area of the core
in square meters; and l, the mean length of the core in meters.
Substituting m mr mo into Eq. (12.2) yields
Magnetic or nonmagnetic core
Solenoid (for l >>10)
d
(a)
A
N turns
N 2mrmo A
N 2mo A
L mr l
l
L mr Lo
and
Magnetic or
nonmagnetic core
(12.3)
l
where Lo is the inductance of the coil with an air core. In other words,
the inductance of a coil with a ferromagnetic core is the relative permeability of the core times the inductance achieved with an air core.
Equations for the inductance of coils different from those shown
above can be found in reference handbooks. Most of the equations are
more complex than those just described.
Toroid
(b)
FIG. 12.5
Inductor configurations for which Equation
(12.2) is appropriate.
EXAMPLE 12.1 Find the inductance of the air-core coil of Fig. 12.6.
Solution:
d = 4 mm
m mrmo (1)(mo) mo
(p)(4 103 m)2
pd 2
A 12.57 106 m2
4
4
N 2mo A
(100 t)2(4p 107 Wb/A⋅m)(12.57 106 m2)
Lo l
0.1 m
1.58 mH
EXAMPLE 12.2 Repeat Example 12.1, but with an iron core and conditions such that mr 2000.
Solution:
By Eq. (12.3),
L mr Lo (2000)(1.58 106 H) 3.16 mH
12.5
l = 100 mm
TYPES OF INDUCTORS
Practical Equivalent
Inductors, like capacitors, are not ideal. Associated with every inductor
are a resistance equal to the resistance of the turns and a stray capaci-
100 turns
FIG. 12.6
Example 12.1.
Air (
o)
476

INDUCTORS
Inductance of
coil
Resistance of the
turns of wire
Rl
L
C
Stray capacitance
FIG. 12.7
Complete equivalent model for an inductor.
L
Rl
L
FIG. 12.8
Practical equivalent model for an inductor.
tance due to the capacitance between the turns of the coil. To include
these effects, the equivalent circuit for the inductor is as shown in Fig.
12.7. However, for most applications considered in this text, the stray
capacitance appearing in Fig. 12.7 can be ignored, resulting in the
equivalent model of Fig. 12.8. The resistance Rl can play an important
role in the analysis of networks with inductive elements. For most
applications, we have been able to treat the capacitor as an ideal element and maintain a high degree of accuracy. For the inductor, however,
Rl must often be included in the analysis and can have a pronounced
effect on the response of a system (see Chapter 20, “Resonance”). The
level of Rl can extend from a few ohms to a few hundred ohms. Keep
in mind that the longer or thinner the wire used in the construction of
the inductor, the greater will be the dc resistance as determined by R rl / A. Our initial analysis will treat the inductor as an ideal element.
Once a general feeling for the response of the element is established,
the effects of Rl will be included.
Symbols
The primary function of the inductor, however, is to introduce inductance—not resistance or capacitance—into the network. For this reason,
the symbols employed for inductance are as shown in Fig. 12.9.
Air-core
Iron-core
Variable
(permeability-tuned)
FIG. 12.9
Inductor symbols.
Appearance
All inductors, like capacitors, can be listed under two general headings:
fixed and variable. The fixed air-core and iron-core inductors were
described in the last section. The permeability-tuned variable coil has a
ferromagnetic shaft that can be moved within the coil to vary the flux
linkages of the coil and thereby its inductance. Several fixed and variable inductors appear in Fig. 12.10.
Testing
The primary reasons for inductor failure are shorts that develop
between the windings and open circuits in the windings due to factors
such as excessive currents, overheating, and age. The open-circuit condition can be checked easily with an ohmmeter (∞ ohms indication), but
the short-circuit condition is harder to check because the resistance of
many good inductors is relatively small and the shorting of a few windings will not adversely affect the total resistance. Of course, if one is
aware of the typical resistance of the coil, it can be compared to the
TYPES OF INDUCTORS
(a)
(c)
(b)
(e)
(d)
FIG. 12.10
Various types of inductors: (a) toroidal power inductor (1.4 mH to 5.6 mH)
(courtesy of Microtan Co., Inc.); (b) surface-mount inductors on reels (0.1 mH
through 1000 mH on 500-piece reels in 46 values) (courtesy of Bell Industries);
(c) molded inductors (0.1 mH to 10 mH); (d) high-current filter inductors
(24 mH at 60 A to 500 mH at 15 A); (e) toroid filter inductors (40 mH to 5 H);
(f) air-core inductors (1 to 32 turns) for high-frequency applications. [Parts (c)
through (f) courtesy of Dale Electronics, Inc.]
measured value. A short between the windings and the core can be
checked by simply placing one lead of the meter on one wire (terminal)
and the other on the core itself. An indication of zero ohms reflects a
short between the two because the wire that makes up the winding has
an insulation jacket throughout. The universal LCR meter of Fig. 10.20
can be used to check the inductance level.
Standard Values and Recognition Factor
The standard values for inductors employ the same numerical multipliers used with resistors and inductors. Like the capacitor, the most common employ the same numerical multipliers as the most common resistors, that is, those with the full range of tolerances (5%, 10%, and
20%), as appearing in Table 3.8. However, inductors are also readily
available with the multipliers associated with the 5% and 10% resistors
of Table 3.8. In general, therefore, expect to find inductors with the following multipliers: 0.1 mH, 0.12 mH, 0.15 mH, 0.18 mH, 0.22 mH, 0.27
mH, 0.33 mH, 0.39 mH, 0.47 mH, 0.56 mH, 0.68 mH, and 0.82 mH, and
then 1 mH, 1.2 mH, 1.5 mH, 1.8 mH, 2.2 mH, 2.7 mH, and so on.
Figure 12.11 was developed to establish a recognition factor when it
comes to the various types and uses for inductors—in other words, to
help the reader develop the skills to identify types of inductors, their
typical range of values, and some of the most common applications.
Figure 12.11 is certainly not all-inclusive, but it does offer a first step in
establishing a sense for what to expect for various applications.
(f)

477
478

INDUCTORS
Type: Open Core Coil
Typical Values: 3 mH to 40 mH
Applications: Used in low-pass
filter circuits. Found in speaker
crossover networks.
Type: RF Chokes
Typical Values: 10 µ H to 50 µ H
Applications: Used in radio,
television, and communication
circuits. Found in AM, FM, and
UHF circuits.
Type: Toroid Coil
Typical Values: 1 mH to 30 mH
Applications: Used as a choke in AC
power lines circuits to filter transient
and reduce EMI interference. This
coil is found in many electronic
appliances.
Type: Moiled Coils
Typical Values: 0.1 µH to 100 µ H
Applications: Used in a wide variety
of circuit such as oscillators, filters,
pass-band filters, and others.
Type: Hash Choke Coil
Typical Values: 3 µH to 1 mH
Applications: Used in AC supply
lines that deliver high currents.
Plastic tube
Type: Delay Line Coil
Typical Values: 10 µH to 50 µ H
Applications: Used in color
televisions to correct for timing
differences between the color
signal and black and white signal.
3"
Fiber
insulator
Type: Common Mode Choke Coil
Typical Values: 0.6 mH to 50 mH
Applications: Used in AC line filters,
switching power supplies, battery
charges and other electronic equipment.
Coil
Type: Surface Mounted Inductors
Typical Values: 0.01 µH to 100 µ H
Applications: Found in many
electronic circuits that require
miniature components on
multilayered PCB.
Inner
core
Type: Adjustable RF Coil
Typical Values: 1 µH to 100 µ H
Applications: Variable inductor
used in oscillators and various RF
circuits such as CB transceivers,
televisions, and radios.
FIG. 12.11
Typical areas of application for inductive elements.
12.6 INDUCED VOLTAGE
The inductance of a coil is also a measure of the change in flux linking
a coil due to a change in current through the coil; that is,
df
L N di
(H)
(12.4)
where N is the number of turns, f is the flux in webers, and i is the current through the coil. If a change in current through the coil fails to
result in a significant change in the flux linking the coil through its center, the resulting inductance level will be relatively small. For this reason the inductance of a coil is sensitive to the point of operation on the
hysteresis curve (described in detail in Section 11.8). If the coil is operating on the steep slope, the change in flux will be relatively high for a
change in current through the coil. If the coil is operating near or in saturation, the change in flux will be relatively small for the same change
in current, resulting in a reduced level of inductance. This effect is particularly important when we examine ac circuits since a dc level asso-
INDUCED VOLTAGE
ciated with the applied ac signal may put the coil at or near saturation,
and the resulting inductance level for the applied ac signal will be significantly less than expected. You will find that the maximum dc current
is normally provided in supply manuals and data sheets to ensure avoidance of the saturation region.
Equation (12.4) also reveals that the larger the inductance of a coil
(with N fixed), the larger will be the instantaneous change in flux
linking the coil due to an instantaneous change in current through the
coil.
If we write Eq. (12.1) as
df di
df
eL N N dt
di dt
and substitute Eq. (12.4), we then have
di
eL L dt
(V)
(12.5)
revealing that the magnitude of the voltage across an inductor is directly
related to the inductance L and the instantaneous rate of change of current through the coil. Obviously, therefore, the greater the rate of
change of current through the coil, the greater will be the induced voltage. This certainly agrees with our earlier discussion of Lenz’s law.
When induced effects are employed in the generation of voltages
such as those available from dc or ac generators, the symbol e is appropriate for the induced voltage. However, in network analysis the voltage
across an inductor will always have a polarity such as to oppose the
source that produced it, and therefore the following notation will be
used throughout the analysis to come:
di
vL L dt
(12.6)
If the current through the coil fails to change at a particular instant,
the induced voltage across the coil will be zero. For dc applications,
after the transient effect has passed, di/dt 0, and the induced voltage
is
di
vL L L(0) 0 V
dt
Recall that the equation for the current of a capacitor is the following:
dvC
iC C dt
Note the similarity between this equation and Eq. (12.6). In fact, if we
apply the duality v
i (that is, interchange the two) and L
C for
capacitance and inductance, each equation can be derived from the
other.
The average voltage across the coil is defined by the equation
Di
vLav L Dt
(V)
(12.7)

479
480

INDUCTORS
where D signifies finite change (a measurable change). Compare this to
iC C(Dv/Dt), and the meaning of D and application of this equation
should be clarified from Chapter 10. An example follows.
EXAMPLE 12.3 Find the waveform for the average voltage across the
coil if the current through a 4-mH coil is as shown in Fig. 12.12.
iL (mA)
10
5
0
1
2
3
4
5
6
7
8
9
t (ms)
10
FIG. 12.12
Example 12.3.
Solutions:
a. 0 to 2 ms: Since there is no change in current through the coil, there
is no voltage induced across the coil; that is,
0
Di
vL L L 0
Dt
Dt
b. 2 ms to 4 ms:
Di
10 103 A
vL L (4 103 H) 20 103 V
Dt
2 103 s
20 mV
c. 4 ms to 9 ms:
Di
10 103 A
vL L (4 103 H) 8 103 V
Dt
5 103 s
8 mV
d. 9 ms to ∞:
0
Di
vL L L 0
Dt
Dt
The waveform for the average voltage across the coil is shown in
Fig. 12.13. Note from the curve that
vL (mV)
20
10
0
1
2
3
4
5
6
7
8
9
10
t (ms)
–10
FIG. 12.13
Voltage across a 4-mH coil due to the current of Fig. 12.12.
R-L TRANSIENTS: STORAGE CYCLE

481
the voltage across the coil is not determined solely by the magnitude
of the change in current through the coil (Di), but also by the rate of
change of current through the coil (Di/Dt).
A similar statement was made for the current of a capacitor due to a
change in voltage across the capacitor.
A careful examination of Fig. 12.13 will also reveal that the area
under the positive pulse from 2 ms to 4 ms equals the area under the
negative pulse from 4 ms to 9 ms. In Section 12.13, we will find that
the area under the curves represents the energy stored or released by
the inductor. From 2 ms to 4 ms, the inductor is storing energy,
whereas from 4 ms to 9 ms, the inductor is releasing the energy
stored. For the full period zero to 10 ms, energy has simply been
stored and released; there has been no dissipation as experienced for
the resistive elements. Over a full cycle, both the ideal capacitor and
inductor do not consume energy but simply store and release it in their
respective forms.
12.7
R-L TRANSIENTS: STORAGE CYCLE
The changing voltages and current that result during the storing of
energy in the form of a magnetic field by an inductor in a dc circuit can
best be described using the circuit of Fig. 12.14. At the instant the
switch is closed, the inductance of the coil will prevent an instantaneous
change in current through the coil. The potential drop across the coil,
vL, will equal the impressed voltage E as determined by Kirchhoff’s
voltage law since vR iR (0)R 0 V. The current iL will then build
up from zero, establishing a voltage drop across the resistor and a corresponding drop in vL. The current will continue to increase until the
voltage across the inductor drops to zero volts and the full impressed
voltage appears across the resistor. Initially, the current iL increases
quite rapidly, followed by a continually decreasing rate until it reaches
its maximum value of E/R.
You will recall from the discussion of capacitors that a capacitor has
a short-circuit equivalent when the switch is first closed and an opencircuit equivalent when steady-state conditions are established. The
inductor assumes the opposite equivalents for each stage. The instant
the switch of Fig. 12.14 is closed, the equivalent network will appear as
shown in Fig. 12.15. Note the correspondence with the earlier comments regarding the levels of voltage and current. The inductor obviously meets all the requirements for an open-circuit equivalent: vL E
volts, and iL 0 A.
When steady-state conditions have been established and the storage
phase is complete, the “equivalent” network will appear as shown in
Fig. 12.16. The network clearly reveals the following:
An ideal inductor (Rl 0 ) assumes a short-circuit equivalent in a
dc network once steady-state conditions have been established.
Fortunately, the mathematical equations for the voltages and current
for the storage phase are similar in many respects to those encountered
for the R-C network. The experience gained with these equations in
Chapter 10 will undoubtedly make the analysis of R-L networks somewhat easier to understand.
+ vR –
iL
R
+
E
L
vL
–
FIG. 12.14
Basic R-L transient network.
vR = iR = (0)R = 0 V
i = 0
R
iL = 0 A
+
vL = E volts
E
–
FIG. 12.15
Circuit of Fig. 12.14 the instant the switch is
closed.
vR = iR =
i
E
( ER ) R = E volts
R
iL = E
R
+
vL = 0 V
–
FIG. 12.16
Circuit of Fig. 12.14 under steady-state
conditions.
482

INDUCTORS
The equation for the current iL during the storage phase is the following:
E
iL Im(1 et/t ) (1 et/(L/R))
R
(12.8)
Note the factor (1 et/t ), which also appeared for the voltage vC of a
capacitor during the charging phase. A plot of the equation is given in
Fig. 12.17, clearly indicating that the maximum steady-state value of iL
is E/R, and that the rate of change in current decreases as time passes.
The abscissa is scaled in time constants, with t for inductive circuits
defined by the following:
L
t R
(seconds, s)
(12.9)
iL = E (1 – e–t/(L/R))
R
iL
Im = E
R
0.865Im
0.951Im 0.981Im 0.993Im
0.632Im
0
1
2
3
4
5
t
FIG. 12.17
Plotting the waveform for iL during the storage cycle.
The fact that t has the units of time can be verified by taking the
equation for the induced voltage
di
vL L dt
and solving for L:
vL
L
di/dt
which leads to the ratio
vL
di/dt
vL
L
—
t R —
di
R
R
dt
V
V
——t
IR
V
t
t
(s)
Our experience with the factor (1 e−t/t) verifies the level of 63.2%
after one time constant, 86.5% after two time constants, and so on. For
convenience, Figure 10.29 is repeated as Fig. 12.18 to evaluate the
functions (1 et/t) and et/t at various values of t.
If we keep R constant and increase L, the ratio L/R increases and the
rise time increases. The change in transient behavior for the current iL
is plotted in Fig. 12.19 for various values of L. Note again the duality
between these curves and those obtained for the R-C network in Fig.
10.32.
R-L TRANSIENTS: STORAGE CYCLE
y
1.0
0.9
0.8
y
=
1 – e – t/
0.7
0.632 (close to 2 3)
0.6
0.5
0.4
0.368 (close to 1 3)
0.3
y
0.2
=
e – t/t
0.1
0
1t
2t
3t
4t
1t
FIG. 12.18
Plotting the functions y 1 et/t and y et/t.
E
R
iL
L1
L2
L3
L3 > L2 > L1
(R fixed)
t (s)
FIG. 12.19
Effect of L on the shape of the iL storage
waveform.
For most practical applications, we will assume that
the storage phase has passed and steady-state conditions have been
established once a period of time equal to five time constants has
occurred.
In addition, since L/R will always have some numerical value, even
though it may be very small, the period 5t will always be greater than
zero, confirming the fact that
the current cannot change instantaneously in an inductive network.
In fact, the larger the inductance, the more the circuit will oppose a
rapid buildup in current level.
Figures 12.15 and 12.16 clearly reveal that the voltage across the
coil jumps to E volts when the switch is closed and decays to zero volts
with time. The decay occurs in an exponential manner, and vL during
5t
6t
t

483
484

INDUCTORS
the storage phase can be described mathematically by the following
equation:
vL Eet/t
(12.10)
A plot of vL appears in Fig. 12.20 with the time axis again divided
into equal increments of t. Obviously, the voltage vL will decrease to
zero volts at the same rate the current presses toward its maximum
value.
vL
E
vL = Ee – t/t
0.368E
0.135E
0
1t
2t
0.049E
0.019E
3t
4t
0.007E
5t
6t
t
FIG. 12.20
Plotting the voltage vR versus time for the network of Fig. 12.14.
R1
iL
2 k
+
50 V
E
L
4 H vL
–
FIG. 12.21
Example 12.4.
iL
In five time constants, iL E/R, vL 0 V, and the inductor can be
replaced by its short-circuit equivalent.
Since
vR iR R iL R
then
E
vR (1 et/t) R
R
and
vR E(1 et/t)
(12.11)
and the curve for vR will have the same shape as obtained for iL.
25 mA
EXAMPLE 12.4 Find the mathematical expressions for the transient
behavior of iL and vL for the circuit of Fig. 12.21 after the closing of the
switch. Sketch the resulting curves.
t = 2 ms
1t
0
2t
3t
4t
5t
t
Solution:
L
4H
t 2 ms
2 k
R1
By Eq. (12.8),
vL
50 V
E
50
Im 25 103 A 25 mA
2 k
R1
t = 2 ms
and
3)
iL (25 103)(1 et/(210
By Eq. (12.10),
0
1t
2t
3t
4t
5t
FIG. 12.22
iL and vL for the network of Fig. 12.21.
t
3)
vL 50et/(210
Both waveforms appear in Fig. 12.22.
)
INITIAL VALUES
12.8

485
INITIAL VALUES
This section will parallel Section 10.9 (Initial Values—Capacitors) on
the effect of initial values on the transient phase. Since the current
through a coil cannot change instantaneously, the current through a
coil will begin the transient phase at the initial value established by
the network (note Fig. 12.23) before the switch was closed. It will
then pass through the transient phase until it reaches the steady-state
(or final) level after about five time constants. The steady-state level of
the inductor current can be found by simply substituting its shortcircuit equivalent (or Rl for the practical equivalent) and finding the
resulting current through the element.
Using the transient equation developed in the previous section, an
equation for the current iL can be written for the entire time interval of
Fig. 12.23; that is,
iL
If
Ii
initial
conditions
0
transient
response
steady-state
region
t
FIG. 12.23
Defining the three phases of a transient waveform.
iL Ii (If Ii)(1 et/t)
with (If Ii) representing the total change during the transient phase.
However, by multiplying through and rearranging terms:
iL Ii If If et/t Ii Ii et/t
If If et/t Ii et/t
we find
iL If (Ii If)et/t
(12.12)
If you are required to draw the waveform for the current iL from initial value to final value, start by drawing a line at the initial value and
steady-state levels, and then add the transient response (sensitive to the
time constant) between the two levels. The following example will clarify the procedure.
EXAMPLE 12.5 The inductor of Fig. 12.24 has an initial current level
of 4 mA in the direction shown. (Specific methods to establish the initial current will be presented in the sections and problems to follow.)
a. Find the mathematical expression for the current through the coil
once the switch is closed.
b. Find the mathematical expression for the voltage across the coil during the same transient period.
c. Sketch the waveform for each from initial value to final value.
Solutions:
a. Substituting the short-circuit equivalent for the inductor will result in
a final or steady-state current determined by Ohm’s law:
E
16 V
16 V
If 1.78 mA
2.2 k 6.8 k
9 k
R1 R2
The time constant is determined by
L
100 mH
100 mH
t 11.11 ms
RT
2.2 k 6.8 k
9 k
R1
2.2 k
iL
4 mA
+
E
16 V
L = 100 mH
R2
6.8 k
FIG. 12.24
Example 12.5.
vL
–
486

INDUCTORS
Applying Eq. (12.12):
iL If (Ii If)et/t
1.78 mA (4 mA 1.78 mA)et/11.11 ms
1.78 mA 2.22 mAet/11.11 ms
b. Since the current through the inductor is constant at 4 mA prior to
the closing of the switch, the voltage (whose level is sensitive only
to changes in current through the coil) must have an initial value of
0 V. At the instant the switch is closed, the current through the coil
cannot change instantaneously, so the current through the resistive
elements will be 4 mA. The resulting peak voltage at t 0 s can
then be found using Kirchhoff’s voltage law as follows:
Vm E VR1 VR2
16 V (4 mA)(2.2 k) (4 mA)(6.8 k)
16 V 8.8 V 27.2 V 16 V 36 V
20 V
Note the minus sign to indicate that the polarity of the voltage vL is
opposite to the defined polarity of Fig. 12.24.
The voltage will then decay (with the same time constant as the
current iL) to zero because the inductor is approaching its shortcircuit equivalence.
The equation for vL is therefore:
vL 20et/11.11 ms
c. See Fig. 12.25. The initial and final values of the current were drawn
first, and then the transient response was included between these
levels. For the voltage, the waveform begins and ends at zero, with
the peak value having a sign sensitive to the defined polarity of vL in
Fig. 12.24.
iL (mA)
4 mA
3
1.78 mA
2
s
τ = 11.11 1
0
τ
1τ
τ
2τ
τ
3τ
τ
4τ
τ
5τ
τ
2τ
τ
3τ
τ
4τ
τ
5τ
t (s)
vL (volts)
0V
0
τ
1τ
–20 V
FIG. 12.25
iL and vL for the network of Fig. 12.24.
0V
R-L TRANSIENTS: DECAY PHASE

487
Let us now test the validity of the equation for iL by substituting
t 0 s to reflect the instant the switch is closed.
et/t e0 1
and
iL 1.78 mA 2.22 mAet/t 1.78 mA 2.22 mA
4 mA
When t > 5t,
et/t 0
and
12.9
iL 1.78 mA 2.22 mAet/t 1.78 mA
R-L TRANSIENTS: DECAY PHASE
In the analysis of R-C circuits, we found that the capacitor could hold
its charge and store energy in the form of an electric field for a period
of time determined by the leakage factors. In R-L circuits, the energy is
stored in the form of a magnetic field established by the current through
the coil. Unlike the capacitor, however, an isolated inductor cannot continue to store energy since the absence of a closed path would cause the
current to drop to zero, releasing the energy stored in the form of a
magnetic field. If the series R-L circuit of Fig. 12.26 had reached
steady-state conditions and the switch were quickly opened, a spark
would probably occur across the contacts due to the rapid change in
current from a maximum of E/R to zero amperes. The change in current
di/dt of the equation vL L(di/dt) would establish a high voltage vL
across the coil that in conjunction with the applied voltage E appears
across the points of the switch. This is the same mechanism as applied
in the ignition system of a car to ignite the fuel in the cylinder. Some
25,000 V are generated by the rapid decrease in ignition coil current
that occurs when the switch in the system is opened. (In older systems,
the “points” in the distributor served as the switch.) This inductive reaction is significant when you consider that the only independent source
in a car is a 12-V battery.
If opening the switch to move it to another position will cause such
a rapid discharge in stored energy, how can the decay phase of an R-L
circuit be analyzed in much the same manner as for the R-C circuit?
The solution is to use a network such as that appearing in Fig.
12.27(a). When the switch is closed, the voltage across the resistor R2
is E volts, and the R-L branch will respond in the same manner as
described above, with the same waveforms and levels. A Thévenin network of E in parallel with R2 would simply result in the source as
+ vR1 –
+
+
–
vR2
–
E
L
R2
Th
+
vL
–
+ vcontact –
vcontact = vL + E
E
–
(a)
(b)
FIG. 12.27
Initiating the storage phase for the inductor L
by closing the switch.
–
+
iL
L
E
–
vL
L
0A
FIG. 12.26
Demonstrating the effect of opening a switch
in series with an inductor with a steady-state
current.
R1
+
+
R
iL
+ vR1 –
iL
R1
vR = iRR = (0 A)R = 0 V
+
vL
–

INDUCTORS
(same polarity)
+ vR1 –
R1
–
(reversed
v
polarity) R2
R2
iL
(same
direction)
+
iL
iL
shown in Fig. 12.27(b) since R2 would be shorted out by the shortcircuit replacement of the voltage source E when the Thévenin resistance is determined.
After the storage phase has passed and steady-state conditions are
established, the switch can be opened without the sparking effect or
rapid discharge due to the resistor R2, which provides a complete path
for the current iL. In fact, for clarity the discharge path is isolated in Fig.
12.28. The voltage vL across the inductor will reverse polarity and have
a magnitude determined by
L
vL (vR1 vR2 )
vL
(12.13)
–
488
+
FIG. 12.28
Network of Fig. 12.27 the instant the switch
is opened.
Recall that the voltage across an inductor can change instantaneously but the current cannot. The result is that the current iL must
maintain the same direction and magnitude as shown in Fig. 12.28.
Therefore, the instant after the switch is opened, iL is still Im E/R1,
and
vL (vR1 vR2) (i1R1 i2R2)
E
R1
R2
iL(R1 R2) (R1 R2) E
R1
R1
R1
and
R2
vL 1 E
R1
(12.14)
which is bigger than E volts by the ratio R2 /R1. In other words, when
the switch is opened, the voltage across the inductor will reverse polarity and drop instantaneously from E to [1 (R2 /R1)]E volts.
As an inductor releases its stored energy, the voltage across the coil
will decay to zero in the following manner:
vL Vi et/t′
(12.15)
with
R2
Vi 1 E
R1
and
L
L
t′ RT
R1 R2
The current will decay from a maximum of Im E/R1 to zero. Using
Eq. (12.20), Ii E/R1 and If 0 A so that
iL If (Ii If)et/t′
E
0 A 0 A et/t′
R1
and
with
E
iL et/t′
R1
L
t′ R1 R2
(12.16)
R-L TRANSIENTS: DECAY PHASE
The mathematical expression for the voltage across either resistor
can then be determined using Ohm’s law:
vR1 iR1R1 iL R1
E
R1et/t′
R1
vR1 Eet/t′
and
(12.17)
The voltage vR1 has the same polarity as during the storage phase since
the current iL has the same direction. The voltage vR2 is expressed as
follows using the defined polarity of Fig. 12.27:
vR2 iR2 R2 iL R2
E
R2et/t′
R1
R2
vR2 Eet/t′
R1
and
(12.18)
EXAMPLE 12.6 The resistor R2 was added to the network of Fig.
12.21, as shown in Fig. 12.29.
a. Find the mathematical expressions for iL, vL, vR1, and vR2 for five
time constants of the storage phase.
b. Find the mathematical expressions for iL, vL, vR1, and vR2 if the
switch is opened after five time constants of the storage phase.
c. Sketch the waveforms for each voltage and current for both phases
covered by this example and Example 12.4 if five time constants
pass between phases. Use the defined polarities of Fig. 12.27.
+ vR1 –
R1
E = 50 V
+
vR2 R2
–
iL
2 k
3 k
L
+
4 H vL
–
FIG. 12.29
Defined polarities for vR1, vR2, vL , and current direction for iL for
Example 12.6.
Solutions:
L
4H
a. t 2 ms
R
2 k
Eq. (12.10):
vL Eet/t
3
vL 50et/210
Eq. (12.8):
iL Im(1 et/t)

489
490

INDUCTORS
50 V
E
Im = 25 mA
R1
2 k
3
iL 25 103(1 et/210 )
Eq. (12.11):
vR1 E(1 et/t)
3
vR1 50(1 et/210 )
vR2 50 V
4H
L
4H
b. t′ 0.8 103 s
5 103 2 k 3 k
R1 R2
0.8 ms
By Eq. (12.15),
3 k
R
Vi 1 2 E 1 (50 V) 125 V
R1
2 k
and
3)
vL Vi et/t′ 125et/(0.810
By Eq. (12.16),
50 V
E
Ii Im 25 mA
2 k
R1
and
3)
iL (25 103)et/(0.810
By Eq. (12.17),
3)
vR1 Eet/t′ 50et/(0.810
By Eq. (12.18),
R2
3 k
3
vR2 Eet/t′ (50 V)et/t′ 75et/(0.810 )
R1
2 k
c. See Fig. 12.30 (opposite page).
In the preceding analysis, it was assumed that steady-state conditions
were established during the charging phase and Im E/R1, with vL 0 V. However, if the switch of Fig. 12.28 is opened before iL reaches its
maximum value, the equation for the decaying current of Fig. 12.28
must change to
iL Ii et/t′
(12.19)
where Ii is the starting or initial current. Equation (12.15) would be
modified as follows:
vL Vi et/t′
with
(12.20)
Vi Ii (R1 R2)
12.10 INSTANTANEOUS VALUES
The development presented in Section 10.10 for capacitive networks
can also be applied to R-L networks to determine instantaneous voltages, currents, and time. The instantaneous values of any voltage or current can be determined by simply inserting t into the equation and using
INSTANTANEOUS VALUES
vL:
+
E
50 V
vL
Switch opened
–
Defined
polarity
5t′ = 5(0.8 ms) = 4 ms
t
0
5t
5(2 ms)
= 10 ms
Switch
closed
Instantaneous
change
–125
iL
iL:
(mA)
No instantaneous
change
25
Defined
direction
0
vR1 :
+ vR1 –
5t
volts
50
Same shape
as iL since
vR1 = iL R1
R1
Defined
polarity
0
vR2 :
5t
volts
R2
–
Defined
polarity
t
5t′
50
+
vR2
t
5t′
5t′
0
t
5t
75
FIG. 12.30
The various voltages and the current for the network of Fig. 12.29.
a calculator or table to determine the magnitude of the exponential
term.
The similarity between the equations vC E(1 et/t) and iL Im(1 et/t) results in a derivation of the following for t that is identical to that used to obtain Eq. (10.24):
Im
t t loge Im iL
(12.21)

491
492

INDUCTORS
For the other form, the equation vC Eet/t is a close match with
vL Eet/t, permitting a derivation similar to that employed for Eq.
(10.25):
E
t t loge
vL
(12.22)
The similarities between the above and the equations in Chapter 10
should make the equation for t fairly easy to obtain.
12.11 THÉVENIN EQUIVALENT: t L/RTh
In Chapter 10 (“Capacitors”), we found that there are occasions when
the circuit does not have the basic form of Fig. 12.14. The same is true
for inductive networks. Again, it is necessary to find the Thévenin
equivalent circuit before proceeding in the manner described in this
chapter. Consider the following example.
E
20 k
4 k
R1
R2
iL
12 V
R3
+
vL
16 k
L = 80 mH
EXAMPLE 12.7 For the network of Fig. 12.31:
a. Find the mathematical expression for the transient behavior of the
current iL and the voltage vL after the closing of the switch (Ii 0 mA).
b. Draw the resultant waveform for each.
Solutions:
–
a. Applying Thévenin’s theorem to the 80-mH inductor (Fig. 12.32)
yields
FIG. 12.31
Example 12.7.
R
20 k
RTh 10 k
N
2
RTh:
20 k
4 k
R1
R2
RTh
R2 + R3 =
R3
16 k
RTh
R1
20 k
4 k + 16 k
= 20 k
FIG. 12.32
Determining RTh for the network of Fig. 12.31.
ETh:
E
20 k
4 k
R1
R2
12 V
ETh
Applying the voltage divider rule (Fig. 12.33),
R3
16 k
FIG. 12.33
Determining ETh for the network of Fig. 12.31.
(R2 R3)E
ETh R1 R2 R3
(20 k)(12 V)
(4 k 16 k)(12 V)
6 V
20 k 4 k 16 k
40 k
The Thévenin equivalent circuit is shown in Fig. 12.34. Using Eq.
(12.8),
THÉVENIN EQUIVALENT: t L/RTh
ETh
iL (1 et/t)
R
493
Thévenin equivalent circuit:
RTh
3
L
80 10 H
t 8 106 s
RTh
10 103 ETh
6V
Im 0.6 103 A
RTh
10 103 10 k
ETh
6V
iL
+
vL
80 mH
–
6)
iL (0.6 103)(1 et/(810
and
)
Using Eq. (12.10),
FIG. 12.34
The resulting Thévenin equivalent circuit for
the network of Fig. 12.31.
vL EThet/t
6)
vL 6et/(810
so that
b. See Fig. 12.35.
vL, iL
ETh = 6 V
Im = ETh
R
= 0.6 mA
vL
iL
t (ms)
5 10 15 20 25 30 35 40 45 50
5t
FIG. 12.35
The resulting waveforms for iL and vL for the network of Fig. 12.31.
EXAMPLE 12.8 The switch S1 of Fig. 12.36 has been closed for a
long time. At t 0 s, S1 is opened at the same instant S2 is closed to
avoid an interruption in current through the coil.
S2
(t = 0 s)
S1
(t = 0 s)
R2
8.2 k
iL
R3
1 k
–
I

12 mA
R1 = 2.2 k
L
FIG. 12.36
Example 12.8.
680 mH
E
+
6V
494

INDUCTORS
a. Find the initial current through the coil. Pay particular attention to its
direction.
b. Find the mathematical expression for the current iL following the
closing of the switch S2.
c. Sketch the waveform for iL.
Solutions:
a. Using Ohm’s law, the initial current through the coil is determined
by
6V
E
Ii 6 mA
R3
1 k
b. Applying Thévenin’s theorem:
RTh R1 R2 2.2 k 8.2 k 10.4 k
ETh IR1 (12 mA)(2.2 k) 26.4 V
RTh
iL
10.4 k
ETh
26.4 V
L
The Thévenin equivalent network appears in Fig. 12.37.
The steady-state current can then be determined by substituting
the short-circuit equivalent for the inductor:
E
26.4 V
If 2.54 mA
RTh
10.4 k
680 mH
6 mA
The time constant:
FIG. 12.37
Thévenin equivalent circuit for the network of
Fig. 12.36 for t ≥ 0 s.
L
680 mH
t 65.39 ms
RTh
10.4 k
Applying Eq. (12.12):
iL If (Ii If)et/t
2.54 mA (6 mA 2.54 mA)et/65.39 ms
2.54 mA 8.54 mAet/(65.39 ms)
c. Note Fig. 12.38.
iL (mA)
2.54 mA
3
2
1
0
–1
–2
1τ
τ
τ
2τ
τ
3τ
τ
4τ
s
τ = 65.39 –3
–4
–5
–6 mA
FIG. 12.38
The current iL for the network of Fig. 12.37.
τ
5τ
t
INDUCTORS IN SERIES AND PARALLEL
12.12

INDUCTORS IN SERIES AND PARALLEL
Inductors, like resistors and capacitors, can be placed in series or parallel. Increasing levels of inductance can be obtained by placing inductors
in series, while decreasing levels can be obtained by placing inductors
in parallel.
For inductors in series, the total inductance is found in the same
manner as the total resistance of resistors in series (Fig. 12.39):
LT L1 L2 L3 • • • LN
L1
L2
L3
(12.23)
LN
LT
FIG. 12.39
Inductors in series.
For inductors in parallel, the total inductance is found in the same
manner as the total resistance of resistors in parallel (Fig. 12.40):
1
1
1
1
1
• • • LT
L1
L2
L3
LN
LT
L1
L2
L3
(12.24)
LN
FIG. 12.40
Inductors in parallel.
For two inductors in parallel,
L 1L 2
LT L1 L2
(12.25)
EXAMPLE 12.9 Reduce the network of Fig. 12.41 to its simplest
form.
Solution: The inductors L2 and L3 are equal in value and they are in
parallel, resulting in an equivalent parallel value of
L
1.2 H
L′T 0.6 H
N
2
L1
L4
0.56 H
L2
1.2 H
R
1.8 H
L3 = 1.2 H
1.2 k
The resulting 0.6 H is then in parallel with the 1.8-H inductor, and
(L′T)(L4)
(0.6 H)(1.8 H)
L″T L′T L4
0.6 H 1.8 H
0.45 H
FIG. 12.41
Example 12.9.
495
496

INDUCTORS
The inductor L1 is then in series with the equivalent parallel value, and
LT
LT L1 L″T 0.56 H 0.45 H
1.01 H
1.01 H
R
The reduced equivalent network appears in Fig. 12.42.
1.2 k
FIG. 12.42
Terminal equivalent of the network of
Fig. 12.41.
12.13 R-L AND R-L-C CIRCUITS
WITH dc INPUTS
We found in Section 12.7 that, for all practical purposes, an inductor
can be replaced by a short circuit in a dc circuit after a period of time
greater than five time constants has passed. If in the following circuits
we assume that all of the currents and voltages have reached their final
values, the current through each inductor can be found by replacing
each inductor with a short circuit. For the circuit of Fig. 12.43, for
example,
E
10 V
I1 5 A
R1
2
R1
R1
I1
2
+
E
10 V
–
L = 2H
I1
2
R2
3
+
E
10 V
–
R2
3
R3
6
FIG. 12.43
Substituting the short-circuit equivalent for the inductor for t > 5t.
For the circuit of Fig. 12.44,
E
21 V
I 10.5 A
2
R2 R3
I
10 mH
I
R1
R1
I1
6 mH
5
+
E
21 V
R3
–
R2
5
I
6
I1
+
E
21 V
–
3
R2
3
2
FIG. 12.44
Establishing the equivalent network for t > 5t.
R-L AND R-L-C CIRCUITS WITH dc INPUTS
Applying the current divider rule,
63 A
R3 I
(6 )(10.5 A)
I1 7 A
9
R3 R2
63
In the following examples we will assume that the voltage across
the capacitors and the current through the inductors have reached their
final values. Under these conditions, the inductors can be replaced
with short circuits, and the capacitors can be replaced with open circuits.
EXAMPLE 12.10 Find the current IL and the voltage VC for the network of Fig. 12.45.
R1
+
E
–
2
IL
+ VC –
R1
C
2
L1
10 V
+
4
R3
R2
+ VC –
I = 0
IL
R3
4
10 V
E
–
3
R2
3
+
V = 0
–
FIG. 12.45
Example 12.10.
Solution:
10 V
E
IL 2 A
5
R1 R2
(3 )(10 V)
R2 E
VC 6 V
R2 R1
32
EXAMPLE 12.11 Find the currents I1 and I2 and the voltages V1 and
V2 for the network of Fig. 12.46.
R1
2
+
E
–
50 V
I1
I2
R3
1
L1
R2
L2
5
R4
+
C1
V1
–
FIG. 12.46
Example 12.11.
4
+
C2
V2
–
R5
7

497
498

INDUCTORS
Solution:
Note Fig. 12.47:
I1
R1
R3
2
+
R2
50 V
E
–
1
5
I2
R4
4
+
+
V1
V2
R5
7
–
–
FIG. 12.47
Substituting the short-circuit equivalents for the inductors and the open-circuit
equivalents for the capacitor for t > 5t.
I1 I2
50 V
E
50 V
I1 5 A
R1 R3 R5
217
10 V2 I2R5 (5 A)(7 ) 35 V
Applying the voltage divider rule,
(8 )(50 V)
(R3 R5)E
(1 7 )(50 V)
V1 40 V
217
10 R1 R3 R5
12.14 ENERGY STORED BY AN INDUCTOR
The ideal inductor, like the ideal capacitor, does not dissipate the electrical energy supplied to it. It stores the energy in the form of a magnetic field. A plot of the voltage, current, and power to an inductor is
shown in Fig. 12.48 during the buildup of the magnetic field surrounding the inductor. The energy stored is represented by the shaded area
under the power curve. Using calculus, we can show that the evaluation
of the area under the curve yields
1
Wstored LIm2
2
(joules, J)
(12.26)
E
Im
iL
pL = vLiL
vL
Energy stored
t
FIG. 12.48
The power curve for an inductive element under transient conditions.
APPLICATIONS
EXAMPLE 12.12 Find the energy stored by the inductor in the circuit
of Fig. 12.49 when the current through it has reached its final value.
R1
3
+
E
R1
3
+
15 V
L
6 mH
–
E
Im
15 V
–
2
2
R2
R2
FIG. 12.49
Example 12.12.
Solution:
E
15 V
15 V
Im 3 A
32
5
R1 R2
1 2
1
54
Wstored LIm (6 103 H)(3 A)2 103 J
2
2
2
27 mJ
12.15
APPLICATIONS
Camera Flash Lamp and Line Conditioner
The inductor (or coil as some prefer to call it) played important roles in
both the camera flash lamp circuitry and the line conditioner (surge protector) described in the Applications section of Chapter 10 on capacitors. For the camera it was the important component that resulted in the
high spike voltage across the trigger coil which was then magnified by
the autotransformer action of the secondary to generate the 4000 V necessary to ignite the flash lamp. Recall that the capacitor in parallel with
the trigger coil charged up to 300 V using the low-resistance path provided by the SCR. However, once the capacitor was fully charged, the
short-circuit path to ground provided by the SCR was removed, and the
capacitor immediately started to discharge through the trigger coil.
Since the only resistance in the time constant for the inductive network
is the relatively low resistance of the coil itself, the current through the
coil grew at a very rapid rate. A significant voltage was then developed
across the coil as defined by Eq. (12.6): vL L(di/dt). This voltage was
in turn increased by transformer action to the secondary coil of the
autotransformer, and the flash lamp was ignited. That high voltage generated across the trigger coil will also appear directly across the capacitor of the trigger network. The result is that it will begin to charge up
again until the generated voltage across the coil drops to zero volts.
However, when it does drop, the capacitor will again discharge through
the coil, establish another charging current through the coil, and again
develop a voltage across the coil. The high-frequency exchange of
energy between the coil and capacitor is called flyback because of the
“flying back” of energy from one storage element to the other. It will
begin to decay with time because of the resistive elements in the loop.

499
500

INDUCTORS
The more resistance, the more quickly it will die out. If the capacitorinductor pairing were isolated and “tickled” along the way with the
application of a dc voltage, the high frequency-generated voltage across
the coil could be maintained and put to good use. In fact, it is this flyback effect that is used to generate a steady dc voltage (using rectification to convert the oscillating waveform to one of a steady dc nature)
that is commonly used in TVs.
In the line conditioner, the primary purpose of the inductors is to
“choke” out spikes of current that may come down the line using the
effect described under the discussion of Lenz’s law in this chapter.
Inductors are such that a rapidly changing current through a coil will
result in the development of a current in the coil that will oppose the
current that established the induced effect in the first place. This effect
is so strong that it can squelch current spikes of a significant number of
amperes in the line. An undesirable result in line conditioners, however,
is the voltage across the coil that will develop when it “chokes” this
rapidly changing current through the coil. However, as mentioned in
Chapter 10, there are two coils in the system that will generate opposing emf’s so that the net voltage to ground is zero. This is fairly clear
when you carefully examine the two coils on the ferromagnetic core
and note that they are wound in a way to develop opposing fields. The
reaction of the coils in the line conditioner to different frequencies and
their ability to help out with the blocking of EMI and RFI disturbances
will have to wait until we discuss the effect of frequency on the reaction of an inductor in a later chapter.
Household Dimmer Switch
Inductors can be found in a wide variety of common electronic circuits
in the home. The typical household dimmer uses an inductor to protect
the other components and the applied load from “rush” currents—currents that increase at very high rates and often to excessively high levels. This feature is particularly important for dimmers since they are
most commonly used to control the light intensity of an incandescent
lamp. At “turn on,” the resistance of incandescent lamps is typically
very low, and relatively high currents may flow for short periods of time
until the filament of the bulb heats up. The inductor is also effective in
blocking high-frequency noise (RFI) generated by the switching action
of the triac in the dimmer. A capacitor is also normally included from
line to neutral to prevent any voltage spikes from affecting the operation
of the dimmer and the applied load (lamp, etc.) and to assist with the
suppression of RFI disturbances.
A photograph of one of the most common dimmers is provided in
Fig. 12.50(a), with an internal view shown in Fig. 12.50(b). The basic
components of most commercially available dimmers appear in the
schematic of Fig. 12.50(c). In this design, a 14.5-mH inductor is used in
the “choking” capacity described above, with a 0.068-mF capacitor for
the “bypass” operation. Note the size of the inductor with its heavy wire
and large ferromagnetic core and the relatively large size of the two
0.068-mF capacitors. Both suggest that they are designed to absorb
high-energy disturbances.
The general operation of the dimmer is shown in Fig. 12.51. The
controlling network is in series with the lamp and will essentially act as
an impedance that can vary between very low and very high levels: very
low impedance levels resembling a short circuit so that the majority of
APPLICATIONS

501
!" #
=
Feed
+
DIMMER
Dimmer
switch
on/off
120 V ac
A
47 k DIAC
0.068 µF
330-k
rheostat
14.5 µH
G
TRIAC
K
0.068 µF
Lamp
under control
–
Return
(c)
FIG. 12.50
Dimmer control: (a) external appearance; (b) internal construction;
(c) schematic.
the applied voltage appears across the lamp [Fig. 12.51(a)] and very
high impedances approaching open circuit where very little voltage
appears across the lamp [Fig. 12.51(b)]. Intermediate levels of impedance will control the terminal voltage of the bulb accordingly. For
instance, if the controlling network has a very high impedance (opencircuit equivalent) through half the cycle as shown in Fig. 12.51(c), the
brightness of the bulb will be less than full voltage but not 50% due to
the nonlinear relationship between the brightness of a bulb and the
applied voltage. There is also a lagging effect present in the actual operation of the dimmer, but this subject will have to wait until leading and
lagging networks are examined in the ac chapters.
The controlling knob, slide, or whatever other method is used on the
face of the switch to control the light intensity is connected directly to
the rheostat in the branch parallel to the triac. Its setting will determine
502

INDUCTORS
+
+
+
DIMMER
DIMMER
DIMMER
A
Zcontrol
Zlow
Vline
A
Vline
Zcontrol
(high)
Zhigh
A
Vline
K
Zchange
K
K
+
+
+
Vlamp
Vlamp
Vlamp
–
–
–
Vline
–
–
Change state
at 1/4 cycle
–
Vline
Vlamp
Zcontrol
Vline
Vlamp
t
(a)
t
t
Vlamp
(b)
(c)
FIG. 12.51
Basic operation of the dimmer of Fig. 12.50: (a) full voltage to the lamp; (b)
approaching the cutoff point for the bulb; (c) reduced illumination of the lamp.
+
Vline = 120 V
–
+
60 V
– 240 Rheostat
dimmer
1 k in wall
+
60 V
– 240 FIG. 12.52
Direct rheostat control of the brightness of a
60-W bulb.
when the voltage across the capacitor reaches a sufficiently high level
to turn on the diac (a bidirectional diode) and establish a voltage at the
gate (G) of the triac to turn it on. When it does, it will establish a very
low resistance path from the anode (A) to the cathode (K), and the
applied voltage will appear directly across the lamp. A more detailed
explanation of this operation will appear in a later chapter following the
examination of some important concepts for ac networks. During the
period the SCR is off, its terminal resistance between anode and cathode will be very high and can be approximated by an open circuit. During this period the applied voltage will not reach the load (lamp). During such intervals the impedance of the parallel branch containing the
rheostat, fixed resistor, and capacitor is sufficiently high compared to
the load that it can also be ignored, completing the open-circuit equivalent in series with the load. Note the placement of the elements in the
photograph of Fig. 12.50 and the fact that the metal plate to which the
triac is connected is actually a heat sink for the device. The on/off
switch is in the same housing as the rheostat. The total design is certainly well planned to maintain a relatively small size for the dimmer.
Since the effort here is simply to control the amount of power getting to the load, the question is often asked, Why don’t we simply use
a rheostat in series with the lamp? The question is best answered by
examining Fig. 12.52, which shows a rather simple network with a
rheostat in series with the lamp. At full wattage, a 60-W bulb on a 120-V
line theoretically has an internal resistance of R V 2/P (from the equation P V2/R) (120 V)2/60 W 240 . Although the resistance
is sensitive to the applied voltage, we will assume this level for the
following calculations. If we consider the case where the rheostat is set
for the same level as the bulb, as shown in Fig. 12.52, there will be
60 V across the rheostat and the bulb. The power to each element will
then be P V 2/R (60 V)2/240 15 W. The bulb is certainly quite
dim, but the rheostat inside the dimmer switch would be dissipating 15 W
of power on a continuous basis. When you consider the size of a 2-W
potentiometer in your laboratory, you can imagine the size rheostat you
APPLICATIONS
would need for 15 W, not to mention the purchase cost, although the
biggest concern would probably be all the heat developed in the walls
of the house. You would certainly be paying for electric power that
would not be performing a useful function. Also, if you had four dimmers set at the same level, you would actually be wasting sufficient
power to fully light another 60-W bulb.
On occasion, especially when the lights are set very low by the dimmer, a faint “singing” can sometimes be heard from the light bulb. This
effect will sometimes occur when the conduction period of the dimmer
is very small. The short, repetitive voltage pulse applied to the bulb will
set the bulb into a condition that could be likened to a resonance (Chapter 20) state. The short pulses are just enough to heat up the filament
and its supporting structures, and then the pulses are removed to allow
the filament to cool down again for a longer period of time. This repetitive heating and cooling cycle can set the filament in motion, and the
“singing” can be heard in a quiet environment. Incidentally, the longer
the filament, the louder the “singing.” A further condition for this effect
is that the filament be in the shape of a coil and not a straight wire so
that the “slinky” effect can develop.
TV or PC Monitor Yolk
Inductors and capacitors play a multitude of roles in the operation of a
TV or PC monitor. However, the most obvious use of the coil is in the
yolk assembly wrapped around the neck of the tube as shown in Fig.
12.53. As shown in the figure, the tube itself, in addition to providing
the screen for viewing, is actually a large capacitor which plays an integral part in establishing the high dc voltage for the proper operation of
the monitor.
A photograph of the yolk assembly of a black-and-white TV tube is
provided in Fig. 12.54(a). It is constructed of four 28-mH coils with
two sets of two coils connected at one point [Fig. 12.54(b)] so that they
will share the same current and will establish the same magnetic field.
The purpose of the yolk assembly is to control the direction of the electron beam from the cathode to the screen of the tube. When the cathode
DANGER!
Can hold charge
for many years!
+20,000 V
Cathode
(GND)
Control
grids
Vacuum
Base
Neck
of
tube
Yolk
assembly
++
+
+
+
Anode
+
+
e
+
+
+
+
+
+
+
+
+
+
+
+
+
FIG. 12.53
Yolk assembly for a TV or PC tube.
Screen
(picture)
Phosphor coating
(emits light when
struck by electron
beam)

503
504

INDUCTORS
V
28 mH
Vertical output
driver network
H
28 mH
28 mH
28 mH
Horizontal output
driver network
=
(b)
FIG. 12.54
(a) Black-and-white TV yolk assembly; (b) schematic representation.
Vertical
control
coils
e
Electron beam
Φ
N
S
Looking down on
vertical control coils
(a)
F
e
N
B
S
(b)
FIG. 12.55
Deflection coils: (a) vertical control;
(b) right-hand-rule (RHR)
for electron flow.
is heated to a very high temperature by a filament internal to the structure, electrons are emitted into the surrounding media. The placement
of a very high positive potential (10 kV to 25 kV dc) to the conductive
coating on the face of the tube will attract the emitted electrons at a
very high speed and therefore at a high level of kinetic energy. When
the electrons hit the phosphorescent coating (usually white, green, or
amber) on the screen, light will be emitted which can be seen by someone facing the monitor. The beam characteristics (such as intensity,
focus, and shape) are controlled by a series of grids placed relatively
close to the cathode in the neck of the tube. The grid is such that the
negatively charged electrons can easily pass through, but the number
and speed with which they pass can be controlled by a negative potential applied to the grid. The grids cannot have a positive potential
because the negatively charged electrons would be attracted to the grid
structure and would eventually disintegrate from the high rate of conduction. Negative potentials on the grids control the flow of electrons
by repulsion and by masking the attraction for the large positive potential applied to the face of the tube.
Once the beam has been established with the desired intensity and
shape, it must be directed to a particular location on the screen using
the yolk assembly. For vertical control, the two coils on the side establish a magnetic flux pattern as shown in Fig. 12.55(a). The resulting
direction of the magnetic field is from left to right as shown in Figs.
12.55(a) and 12.55(b). Using the right hand, with the index finger pointing in the direction of the magnetic field and the middle finger (at right
angles to the index finger) in the direction of electron flow, will result
in the thumb (also at right angles to the index finger) pointing in the
direction of the force on the electron beam. The result is a bending of
the beam as shown in Fig. 12.53. The stronger the magnetic field of the
coils as determined by the current through the coils, the greater the
deflection of the beam.
Before continuing, it is important to realize that when the electron
beam hits the phosphorescent screen as shown in Fig. 12.56, it is moving with sufficient velocity to cause a secondary emission of X rays that
APPLICATIONS
will scatter to all sides of the monitor. Even though the X rays will die
off exponentially with distance from the source, there is some concern
about safety, and all modern-day monitors have shields all around the
outside surface of the tube as shown in Fig. 12.56. It is therefore interesting that it is not the direct viewing that is of some concern but rather
viewing by individuals to the side, above, or below the screen. Monitors
are currently limited to 25 kV at the anode because the application of
voltages in excess of 25 kV can result in a direct emission of X rays.
Internally, all monitors currently have a safety shutoff to ensure that this
level is never attained in the operating system.
Time and space do not permit a detailed discussion of the full operation of a monitor, but there are some facts about its operation that
reveal the sophistication of the design. When an image is generated on
a screen, it is done one pixel at a time along one horizontal line at a
time. A pixel is one point on the screen. Pixels are black (no signal) or
white (with signal) for black-and-white (monochromic) TVs or black
and white or some color for color TVs. For EGA monitors the resolution is 640 pixels wide and 35 pixels high, whereas VGA monitors are
also 640 pixels wide but 480 pixels high. Obviously the more pixels in
the same area, the sharper the image. A typical scan rate is 31.5 kHz
which means that 31,500 lines can be drawn in 1 s, or one line of 640
pixels can be drawn in about 31.7 ms.
Patterns on the screen are developed by the sequence of lines appearing in Fig. 12.57. Starting at the top left, the image moves across the
screen down to the next line until it ends at the bottom right of the screen,
at which point there is a rapid retrace (invisible) back to the starting
point. Typical scanning rates (full image generated) extend from 60
frames per second to 80 frames per second. The slower the rate, the
higher the possibility of flickering in the images. At 60 frames per second, one entire frame is generated every 1/60 16.67 ms 0.017 s.
Color monitors are particularly interesting because all colors on the
screen are generated by the colors red, blue, and green. The reason is that
the human eye is responding to the wavelengths and energy levels of the
various colors. The absence of any color is black, and the result of full
energy to each of the three colors is white. The color yellow is a combination of red and green with no blue, and pink is primarily red energy with
smaller amounts of blue and green. An in-depth description of this “additive” type of color generation must be left as an exercise for the reader.
The fact that three colors define the resulting color requires that
there be three cathodes in a color monitor to generate three electron
beams. However, the three beams must sweep the screen in the same
relative positions. Each pixel is now made up of three color dots in the
same relative position for each pixel, as shown in Fig. 12.58. Each dot
has a phosphorescent material that will generate the desired color when
hit with an electron beam. For situations where the desired color has no
green, the electron beam associated with the color green will be turned
off. In fact, between each pixel, each beam is shut down to provide definition between the color pixels. The dots within the pixel are so close
that the human eye cannot pick up the individual colors but simply the
color that would result from the “additive” process.
During the entire “on” time of a monitor, a full 10 kV to 25 kV are
applied to the conductor on the screen to attract electrons. Over time
there will naturally be an accumulation of negative charge on the screen
which will remain after the power is turned off—a typical capacitive
storage charge. For a brief period of time, it will sit with 25 kV across

505
Shield
Screen
Electron beam
e
Secondary
emission
Shield
FIG. 12.56
Secondary emission from and protective
measures for a TV or PC monitor.
FIG. 12.57
Pattern generation.
Green
Cyan
Yellow
White
Blue
Magenta
Red
FIG. 12.58
Color television pixels.
506

INDUCTORS
the plates which will drop as the “capacitor” begins to discharge. However, the lack of a low-resistance path will often result in a storage of
the charge for a fairly long period of time. This stored charge and the
associated voltage across the plates are sufficiently high to cause severe
damage. It is therefore paramount that TVs and monitors be repaired or
investigated only by someone who is well versed in how to discharge
the tube. One commonly applied procedure is to attach a long lead from
the metal shaft of a flat-edge screwdriver to a good ground connection.
Then leave the anode connection to the tube in place, and simply insert
the screwdriver under the cap until it touches the metal clip of the cap.
You will probably hear a loud snap when discharge occurs. Because of
the enormous amount of residual charge, it is recommended that the
above procedure be repeated two or three times. Even then, treat the
tube with a great deal of respect. In short, until you become familiar
with the discharge procedure, leave the investigation of TVs and monitors to someone with the necessary experience. A further concern is the
very high pulse voltages generated in an operating system. Be aware
that they are of a magnitude that could destroy standard test equipment.
The capacitive effect of the tube is an integral part of developing the
high dc anode potential. Its filtering action smooths out the repetitive,
high-voltage pulses generated by the flyback action of the TV. Otherwise, the screen would simply be a flickering pattern as the anode
potential switched on and off with the pulsating signal.
12.16 COMPUTER ANALYSIS
PSpice
Transient RL Response The computer analysis will begin with a
transient analysis of the network of parallel inductive elements in Fig.
12.59. The inductors are picked up from the ANALOG library in the
Place Part dialog box. As noted in Fig. 12.59, the inductor is displayed
with its terminal identification which is helpful for identifying nodes
when calling for specific output plots and values. In general, when an
element is first placed on a schematic, the number 1 is assigned to the
left end on a horizontal display and to the top on a vertical display.
Similarly, the number 2 is assigned to the right end of an element in the
horizontal position and to the bottom in the vertical position. Be aware,
however, that the option Rotate rotates the element in the CCW direction, so taking a horizontal resistor to the vertical position requires
three rotations to get the number 1 to the top again. In previous chapters you may have noted that a number of the outputs were taken off
terminal 2 because a single rotation placed this terminal at the top of
the vertical display. Also note in Fig. 12.59 the need for a series resistor Rl within the parallel loop of inductors. In PSpice, inductors must
have a series resistor to reflect real-world conditions. The chosen value
of 1 m is so small, however, that it will not affect the response of the
system. For VPulse, the rise time was selected as 0.01 ms, and the
pulse width was chosen as 10 ms because the time constant of the network is t LT /R (4 H 12 H)/2 k 1.5 ms and 5t 7.5 ms.
The simulation is the same as applied when obtaining the transient
response of capacitive networks. In condensed form, the sequence to
obtain a plot of the voltage across the coils versus time is as follows: New
SimulationProfilekey-TransientRL-Create-TimeDomain(Transient)Run to time:10ms-Start saving data after:0s and Maximum step
size:5ms-OK-Run PSpice key-Add Trace key-V1(L2)-OK. The result-
COMPUTER ANALYSIS
FIG. 12.59
Using PSpice to obtain the transient response of a parallel inductive network due to an applied
pulse of 50 V.
ing trace appears in the bottom of Fig. 12.60. A maximum step size of 5
ms was chosen to ensure that it was less than the rise or fall times of 10
ms. Note that the voltage across the coil jumps to the 50-V level almost
immediately; then it decays to 0 V in about 8 ms. A plot of the total current through the parallel coils can be obtained through Plot-Plot to Window-Add Trace key-I(R)-OK, resulting in the trace appearing at the top
of Fig. 12.60. When the trace first appeared, the vertical scale extended
from 0 A to 40 mA even though the maximum value of iR was 25 mA. To
bring the maximum value to the top of the graph, Plot was selected followed by Axis Settings-Y Axis-User Defined-0A to 25mA-OK.
For values, the voltage plot was selected, SEL, followed by the
Toggle cursor key and a click on the screen to establish the crosshairs.
The left-click cursor was set on one time constant to reveal a value of
18.461 V for A1 (about 36.8% of the maximum as defined by the exponential waveform). The right-click cursor was set at 7.5 ms or five time
constants, resulting in a relatively low 0.338 V for A2.
Transient Response with Initial Conditions The next application will verify the results of Example 12.5 which has an initial condition associated with the inductive element. VPULSE is again employed
with the parameters appearing in Fig. 12.61. Since t L /R 100 mH/
(2.2 k 6.8 k) 100 mH/9 k 11.11 ms and 5t 55.55 ms, the
pulse width (PW) was set to 100 ms. The rise and fall times were set at
100 ms/1000 0.1 ms. Note again that the labels 1 and 2 appear with
the inductive element.
Setting the initial conditions for the inductor requires a procedure
that has not been described as yet. First double-click on the inductor
symbol to obtain the Property Editor dialog box. Then select Parts at

507
508

INDUCTORS
FIG. 12.60
The transient response of vL and iR for the network of Fig. 12.59.
the bottom of the dialog box, and select New Column to obtain the
Add New Column dialog box. Under Name, enter IC (an abbreviation
for “initial condition”—not “capacitive current”) followed by the initial
condition of 4 mA under Value; then click OK. The Property Editor
dialog box will appear again, but now the initial condition appears as a
New Column in the horizontal listing dedicated to the inductive element. Now select Display to obtain the Display Properties dialog box,
and under Display Format choose Name and Value so that both IC
and 4mA will appear. Click OK, and we return to the Property Editor
dialog box. Finally, click on Apply and exit the dialog box (X). The
result is the display of Fig. 12.61 for the inductive element.
Now for the simulation. First select the New Simulation Profile key,
insert the name InitialCond(L), and follow up with Create. Then in the
Simulation Settings dialog box, select Time Domain(Transient) for
the Analysis type and General Settings for the Options. The Run to
time should be 200 ms so that we can see the full effect of the pulse
source on the transient response. The Start saving data after should
remain at 0 s, and the Maximum step size should be 200 ms/1000 200 ns. Click OK and then select the Run PSpice key. The result will
be a screen with an x-axis extending from 0 to 200 ms. Selecting Trace
to get to the Add Traces dialog box and then selecting I(L) followed
by OK will result in the display of Fig. 12.62. The plot for I(L) clearly
starts at the initial value of 4 mA and then decays to 1.78 mA as defined
by the left-click cursor. The right-click cursor reveals that the current
has dropped to 0.222 mA (essentially 0 A) after the pulse source
has dropped to 0 V for 100 ms. The VPulse source was placed in the
same figure through Plot-Add Plot to Window-Trace-Add TraceV(VPulse:)-OK to permit a comparison between the applied voltage
and the resulting inductor current.
COMPUTER ANALYSIS
FIG. 12.61
Using PSpice to determine the transient response for a circuit in which the
inductive element has an initial condition.
FIG. 12.62
A plot of the applied pulse and resulting current for the circuit of Fig. 12.61.

509
510

INDUCTORS
Electronics Workbench
The transient response of an R-L network can also be obtained using
Electronics Workbench. The circuit to be examined appears in Fig.
12.63 with a pulse voltage source to simulate the closing of a switch at
t 0 s. The source, referred to as PULSE–VOLTAGE–SOURCE in
the Source listing, is the near the bottom left of the Sources parts bin.
When selected, it will appear with a label, an initial voltage, a step voltage, and a frequency. All can be changed by simply double-clicking on
the source symbol to obtain the Pulse Voltage dialog box. As shown in
Fig. 12.63, the Pulsed Value will be set at 20 V, and the Delay Time to
0 s. The Rise Time and Fall Time will both remain at the default levels of 1 ns. For our analysis we want a Pulse Width that is at least
twice the 5t transient period of the circuit. For the chosen values of R
and L, t L/R 10 mH/100 0.1 ms 100 ms. The transient
period of 5t is therefore 500 ms or 0.5 ms. Thus, a Pulse Width of 1 ms
would seem appropriate with a Period of 2 ms. The result is a frequency of f 1/T 1/2 ms 500 Hz. When all have been set and
selected, the parameters of the pulse source will appear as shown in Fig.
12.63. Next the resistor, inductor, and ground are placed on the screen
to complete the circuit.
This time we will want to see the node names so that we can call for
them when we set up the simulation process. This is accomplished
through Options-Preferences-Show node names. In this case we have
two—one at the positive terminal of the supply (1) and the other at the
top end of the inductor (2) representing the voltage across the inductor.
The simulation process is initiated by the following sequence: Simulate-Analyses-Transient Analysis. The result is the Transient Analysis dialog box in which Analysis Parameters is chosen first. Under
FIG. 12.63
Using Electronics Workbench to obtain the transient response for an inductive circuit.
PROBLEMS
Parameters, use 0 s as the Start time and 4 ms as the End time so that
we get two full cycles of the applied voltage. After enabling the Maximum time step settings(TMAX), we set the Minimum number of
time points at 1000 to get a reasonably good plot during the rapidly
changing transient period. Next, the Output variables section must be
selected and the program told which voltage and current levels we are
interested in. On the left side of the dialog box is a list of Variables that
have been defined for the circuit. On the right is a list of Selected variables for analysis. In between you see a Plot during simulation or
Remove. To move a variable from the left to the right column, simply
select it in the left column and choose Plot during simulation. It will
then appear in the right column. For our purposes it seems appropriate
that we plot both the applied voltage and the voltage across the coil, so
1 and 2 were moved to the right column. Then Simulate is selected, and
a window titled Analysis Graphs will appear with the selected plots as
shown in Fig. 12.63. Click on the Show/Hide Grid key (a red grid on
a black axis), and the grid lines will appear. Then selecting the
Show/Hide Legend key on the immediate right will result in the small
Transient Anal dialog box that will identify the color that goes with
each nodal voltage. In our case, blue is the color of the applied voltage,
and red is the color of the voltage across the coil.
The source voltage appears as expected with its transition to 20 V,
50% duty cycle, and the period of 2 ms. The voltage across the coil
jumped immediately to the 20-V level and then began its decay to 0 V
in about 0.5 ms as predicted. When the source voltage dropped to
zero, the voltage across the coil reversed polarity to maintain the same
direction of current in the inductive circuit. Remember that for a coil,
the voltage can change instantaneously, but the inductor will “choke”
any instantaneous change in current. By reversing its polarity, the
voltage across the coil ensures the same polarity of voltage across the
resistor and therefore the same direction of current through the coil
and circuit.
PROBLEMS
SECTION 12.2 Faraday’s Law of
Electromagnetic Induction
1. If the flux linking a coil of 50 turns changes at a rate of
0.085 Wb/s, what is the induced voltage across the coil?
2. Determine the rate of change of flux linking a coil if
20 V are induced across a coil of 40 turns.
3. How many turns does a coil have if 42 mV are induced
across the coil by a change of flux of 0.003 Wb/s?
SECTION 12.4 Self-Inductance
4. Find the inductance L in henries of the inductor of Fig.
12.64.
5. Repeat Problem 4 with l 4 in. and d 0.25 in.
l = 0.075 m
d = 0.005 m
200 turns
FIG. 12.64
Problems 4 and 5.
Wood core

511
512

INDUCTORS
300 turns
A = 1.5 × 10–4 m2
Air core
6. a. Find the inductance L in henries of the inductor of
Fig. 12.65.
b. Repeat part (a) if a ferromagnetic core is added having a mr of 2000.
l = 0.1 m
FIG. 12.65
Problem 6.
SECTION 12.6 Induced Voltage
7. Find the voltage induced across a coil of 5 H if the rate
of change of current through the coil is
a. 0.5 A/s
b. 60 mA/s
c. 0.04 A/ms
8. Find the induced voltage across a 50-mH inductor if the
current through the coil changes at a rate of 0.1 mA/ms.
9. Find the waveform for the voltage induced across a 200mH coil if the current through the coil is as shown in Fig.
12.66.
iL
40 mA
0
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17
t (ms)
FIG. 12.66
Problem 9.
10. Sketch the waveform for the voltage induced across a
0.2-H coil if the current through the coil is as shown in
Fig. 12.67.
66 µµA
iL
50 µA
µ
0
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17
– 44 µµA
FIG. 12.67
Problem 10.
t (µ
µs)
PROBLEMS

vL
*11. Find the waveform for the current of a 10-mH coil if the
voltage across the coil follows the pattern of Fig. 12.68.
The current iL is 4 mA at t 0 s.
+60 V
0
5
10
12
16
t (ms)
24
–5 V
–24 V
FIG. 12.68
Problem 11.
SECTION 12.7 R-L Transients: Storage Cycle
+ vR –
R
12. For the circuit of Fig. 12.69:
a. Determine the time constant.
b. Write the mathematical expression for the current iL
after the switch is closed.
c. Repeat part (b) for vL and vR.
d. Determine iL and vL at one, three, and five time constants.
e. Sketch the waveforms of iL, vL, and vR.
iL
20 k
+
E
40 mV
L
250 mH vL
–
FIG. 12.69
Problem 12.
13. For the circuit of Fig. 12.70:
a. Determine t.
b. Write the mathematical expression for the current iL
after the switch is closed at t 0 s.
c. Write the mathematical expressions for vL and vR
after the switch is closed at t 0 s.
d. Determine iL and vL at t 1t, 3t, and 5t.
e. Sketch the waveforms of iL, vL, and vR for the storage
phase.
+12 V
R
L
iL
2.2 k
5 mH
+ vR –
+ vL –
FIG. 12.70
Problem 13.
SECTION 12.8 Initial Values
14. For the network of Fig. 12.71:
a. Write the mathematical expressions for the current iL
and the voltage vL following the closing of the switch.
Note the magnitude and direction of the initial current.
b. Sketch the waveform of iL and vL for the entire period
from initial value to steady-state level.
513
R2
iL
2.2 k
+
I
5 mA
R1 = 1.2 k
L
2 H vL
–
3 mA
FIG. 12.71
Problem 14.
514

INDUCTORS
R1
4.7 k
E
+ vL –
iL
L
120 mH
8 mA
R2
36 V
3.9 k
15. For the network of Fig. 12.72:
a. Write the mathematical expressions for the current iL
and the voltage vL following the closing of the switch.
Note the magnitude and direction of the initial current.
b. Sketch the waveform of iL and vL for the entire period
from initial value to steady-state level.
FIG. 12.72
Problem 15.
+ vL –
L = 200 mH
iL
I
4 mA
3 mA
R1 = 2.2 k
E
16 V
*16. For the network of Fig. 12.73:
a. Write the mathematical expressions for the current iL
and the voltage vL following the closing of the switch.
Note the magnitude and direction of the initial current.
b. Sketch the waveform of iL and vL for the entire period
from initial value to steady-state level.
R2
8.2 k
FIG. 12.73
Problem 16.
SECTION 12.9 R-L Transients: Decay Phase
R1
iL
10 k
+
20 V
R2
10 k
L
10 mH vL
–
17. For the network of Fig. 12.74:
a. Determine the mathematical expressions for the current iL and the voltage vL when the switch is closed.
b. Repeat part (a) if the switch is opened after a period
of five time constants has passed.
c. Sketch the waveforms of parts (a) and (b) on the same
set of axes.
FIG. 12.74
Problems 17, 45, and 46.
E
–6V
iL
R1
6.8 k
L
5 mH vL
–
vR2 R2
+
8.2 k
–
+
iL
FIG. 12.75
Problem 18.
*18. For the network of Fig. 12.75:
a. Write the mathematical expression for the current iL
and the voltage vL following the closing of the switch.
b. Determine the mathematical expressions for iL and vL
if the switch is opened after a period of five time constants has passed.
c. Sketch the waveforms of iL and vL for the time periods defined by parts (a) and (b).
d. Sketch the waveform for the voltage across R2 for the
same period of time encompassed by iL and vL. Take
careful note of the defined polarities and directions of
Fig. 12.75.
PROBLEMS
*19. For the network of Fig. 12.76:
a. Determine the mathematical expressions for the current iL and the voltage vL following the closing of the
switch.
b. Repeat part (a) if the switch is opened at t 1 ms.
c. Sketch the waveforms of parts (a) and (b) on the same
set of axes.
R1

iL
2 k
+
12 V
E
R2
10 k
1 mH vL
L
–
FIG. 12.76
Problem 19.
SECTION 12.10 Instantaneous Values
20. Referring to the solution to Example 12.4, determine the
time when the current iL reaches a level of 10 mA. Then determine the time when the voltage drops to a level of 10 V.
21. Referring to the solution to Example 12.5, determine the
time when the current iL drops to 2 mA.
SECTION 12.11 Thévenin Equivalent: t L/RTh
22. a. Determine the mathematical expressions for iL and vL
following the closing of the switch in Fig. 12.77.
b. Determine iL and vL at t 100 ns.
R2
iL
24 k
I =
4 mA
+
12 k
R1
2 mH vL
L
–
FIG. 12.77
Problem 22.
*23. a. Determine the mathematical expressions for iL and vL
following the closing of the switch in Fig. 12.78.
b. Calculate iL and vL at t 10 ms.
c. Write the mathematical expressions for the current iL
and the voltage vL if the switch is opened at t 10 ms.
d. Sketch the waveforms of iL and vL for parts (a) and (c).
E = +8V
R1
2.2 k
iL
+
R2
4.7 k
L
10 mH vL
–
FIG. 12.78
Problem 23.
515
516

INDUCTORS
+20 V
4 k
R2
*24. a. Determine the mathematical expressions for iL and vL
following the closing of the switch in Fig. 12.79.
b. Determine iL and vL after two time constants of the
storage phase.
c. Write the mathematical expressions for the current iL
and the voltage vL if the switch is opened at the
instant defined by part (b).
d. Sketch the waveforms of iL and vL for parts (a) and (c).
1.5 k
R4
R1
iL
12 k
+
3 k
R3
5 mH vL
L
–
–6V
FIG. 12.79
Problem 24.
+ vR1 –
R1
iL
100 +
36 V
E
R2
470 0.6 H vL
L
–
*25. For the network of Fig. 12.80, the switch is closed at
t 0 s.
a. Determine vL at t 25 ms.
b. Find vL at t 1 ms.
c. Calculate vR1 at t 1t.
d. Find the time required for the current iL to reach
100 mA.
R3
20 – vR 3 +
FIG. 12.80
Problem 25.
2 M
E
iL
24 V
+
vL L
+
5H
–
–
*26. The switch for the network of Fig. 12.81 has been closed
for about 1 h. It is then opened at the time defined as t 0 s.
a. Determine the time required for the current iR to drop
to 1 mA.
b. Find the voltage vL at t 1 ms.
c. Calculate vR3 at t 5t.
FIG. 12.81
Problems 26 and 27.
R1
R2
4.7 k
3.3 k
iL
+
E
(t = 0 s)
16 V
L
2 H vL
–
R3
1 k
FIG. 12.82
Problem 28.
27. The network of Fig. 12.81 employs a DMM with an
internal resistance of 10 M in the voltmeter mode. The
switch is closed at t 0 s.
a. Find the voltage across the coil the instant after the
switch is closed.
b. What is the final value of the current iL?
c. How much time must pass before iL reaches 10 mA?
d. What is the voltmeter reading at t 12 ms?
*28. The switch in Fig. 12.82 has been open for a long time.
It is then closed at t 0 s.
a. Write the mathematical expression for the current iL
and the voltage vL after the switch is closed.
b. Sketch the waveform of iL and vL from the initial
value to the steady-state level.
PROBLEMS
*29. The switch of Fig. 12.83 has been closed for a long time.
It is then opened at t 0 s.
a. Write the mathematical expression for the current iL
and the voltage vL after the switch is opened.
b. Sketch the waveform of iL and vL from initial value to
the steady-state level.

517
(t = 0 s)
R1
R2
2.2 k
4.7 k
iL
+
E
1.2 H vL
24 V
–
FIG. 12.83
Problem 29.
*30. The switch of Fig. 12.84 has been open for a long time.
It is then closed at t 0 s.
a. Write the mathematical expression for the current iL
and the voltage vL after the switch is closed.
b. Sketch the waveform of iL and vL from initial value to
the steady-state level.
iL + vL –
R2
1.2 k
220 mH
R1
1 k
I
4 mA
E
(t = 0 s)
FIG. 12.84
Problems 30 and 43.
SECTION 12.12 Inductors in Series and Parallel
31. Find the total inductance of the circuits of Fig. 12.85.
4H
3.6 H
LT
3H
6H
12 H
LT
2H
6H
(a)
4H
(b)
FIG. 12.85
Problem 31.
32. Reduce the networks of Fig. 12.86 to the fewest elements.
5 mH
42 µ
µF
14 mH
35 mH
9µ
µF
7µ
µF
12 µ
µF
10 µ
µF
90 µ
µF
(a)
20 mH
(b)
FIG. 12.86
Problem 32.
18 V
518

INDUCTORS
4H
E
20 V
33. Reduce the network of Fig. 12.87 to the fewest number
of components.
1H
1 k
9.1 k
6H
2H
4.7 k
FIG. 12.87
Problem 33.
20 V
R1
L1
iL
+
5 k
R2
5H
+
vL
20 k
30 H vL3
6 H L3
L2
*34. For the network of Fig. 12.88:
a. Find the mathematical expressions for the voltage vL
and the current iL following the closing of the switch.
b. Sketch the waveforms of vL and iL obtained in part (a).
c. Determine the mathematical expression for the voltage vL3 following the closing of the switch, and
sketch the waveform.
–
–
FIG. 12.88
Problem 34.
+ V1 –
4 k
I1
SECTION 12.13 R-L and R-L-C Circuits
with dc Inputs
2H
+
2 k
16 V
3 H V2
–
For Problems 35 through 37, assume that the voltage across
each capacitor and the current through each inductor have
reached their final values.
35. Find the voltages V1 and V2 and the current I1 for the circuit of Fig. 12.89.
FIG. 12.89
Problems 35 and 38.
36. Find the current I1 and the voltage V1 for the circuit of
Fig. 12.90.
L = 6H
+
V1
–
C = 5
F
4
I1
6
20 V
FIG. 12.90
Problems 36 and 39.
PROBLEMS
37. Find the voltage V1 and the current through each inductor
in the circuit of Fig. 12.91.

519
*42. Verify the results of Example 12.3 using the VPULSE
function and a PW equal to 1 ns.
*43. Verify the results of Problem 30 using the VPULSE function and the appropriate initial current.
20 50 V
5
3
Programming Language (C, QBASIC, Pascal, etc.)
3
6
+
I1
V1
6
F
I2
4H
–
0.5 H
FIG. 12.91
Problems 37 and 40.
SECTION 12.14 Energy Stored by an Inductor
38. Find the energy stored in each inductor of Problem 35.
39. Find the energy stored in the capacitor and inductor of
Problem 36.
44. Write a program to provide a general solution for the circuit of Fig. 12.14; that is, given the network parameters,
generate the equations for iL, vL, and vR.
45. Write a program that will provide a general solution for
the storage and decay phase of the network of Fig. 12.74;
that is, given the network values, generate the equations
for iL and vL for each phase. In this case, assume that the
storage phase has passed through five time constants
before the decay phase begins.
46. Repeat Problem 45, but assume that the storage phase
was not completed, requiring that the instantaneous values of iL and vL be determined when the switch is
opened.
40. Find the energy stored in each inductor of Problem 37.
SECTION 12.16 Computer Analysis
PSpice or Electronics Workbench
*41. Verify the results of Example 12.6 using the VPULSE
function and a pulse width (PW) equal to five time constants of the charging network.
GLOSSARY
Choke A term often applied to an inductor, due to the ability
of an inductor to resist a change in current through it.
Faraday’s law A law relating the voltage induced across a
coil to the number of turns in the coil and the rate at which
the flux linking the coil is changing.
Inductor A fundamental element of electrical systems constructed of numerous turns of wire around a ferromagnetic
core or an air core.
Lenz’s law A law stating that an induced effect is always
such as to oppose the cause that produced it.
Self-inductance (L) A measure of the ability of a coil to
oppose any change in current through the coil and to store
energy in the form of a magnetic field in the region surrounding the coil.