6.19 < Combining inductors in series If the original inductors carry an initial current of if0, the equivalent inductor carries the same initial current.. In the equivalent circuit, the
Trang 1186 Inductance, Capacitance, and Mutual Inductance
(Note that 5 V is the voltage on the capacitor at
the end of the preceding interval.) Then,
v = (10*7 - 12.5 X 1 0 ¥ - 10) V,
p = t»,
= (62.5 X 1012/3 - 7.5 x 10V + 2.5 X 105/ - 2) W,
w = —Cv~,
2
0
V ( V )
t(fis)
= (15.625 X 10 u r - 2.5 X l ( ) r + 0.125 X 10(V
For/ > 40^ts,
- 2/ + 10~3) J
v = 10 V,
p = vi = 0,
w = —Cv 2 = 10/xJ
b) The excitation current and the resulting voltage,
power, and energy are plotted in Fig 6.12
c) Note that the power is always positive for the
duration of the current pulse, which means that
energy is continuously being stored in the
capac-itor When the current returns to zero, the stored
energy is trapped because the ideal capacitor
offers no means for dissipating energy Thus a
voltage remains on the capacitor after i returns
to zero
10
5
0
— n
10
1
20
l
30
i
40
1
50
I
60 t((XS)
p ( m w
500
400
300
200
100
0
)
^1
10
1
20
1
30 40
1
50
J
60 tips) w(
10
8
6
4
2
0
nJ)
10
i
20
I
30
1
40
l
50
I
60
Figure 6.12 • The variables /', v, p, and to versus
Example 6.5
t(fJ!S)
t for
^ / A S S E S S M E N T PROBLEMS
Objective 2—Know and be able to use the equations for voltage, current, power, and energy in a capacitor
6.2 The voltage at the terminals of the 0.6 /xF
capacitor shown in the figure is 0 for t < 0 and
4Qe -ism)t s i n 30,000/ V for t > 0 Find (a) /'(0);
(b) the power delivered to the capacitor at
/ = 77-/8O ms; and (c) the energy stored in the
capacitor at t = TT/80 ms
0.6 ^F
NOTE: Also try Chapter Problems 6.16 and 6.17
Answer: (a) 0.72 A;
(b) -649.2 mW;
(c) 126.13 AtJ
6.3 The current in the capacitor of Assessment
Problem 6.2 is 0 for t < 0 and 3 cos 50,000/ A for t a 0 Find (a) v{t)\ (b) the maximum power
delivered to the capacitor at any one instant of time; and (c) the maximum energy stored in the capacitor at any one instant of time
Answer: (a) 100 sin 50,000/ V, / > 0;
( b ) 1 5 0 W ; ( c ) 3 m J
Trang 26.3 Series-Parallel Combinations
of Inductance and Capacitance
Just as series-parallel combinations of resistors can be reduced to a single
equivalent resistor, series-parallel combinations of inductors or capacitors
can be reduced to a single inductor or capacitor Figure 6.13 shows
induc-tors in series Here, the inducinduc-tors are forced to carry the same current; thus
we define only one current for the series combination The voltage drops
across the individual inductors are Figure 6.13 A Inductors in series
di di di
„ = £ , - , v 2 = L 1Jt , and v } = L,-
The voltage across the series connection is
v = v\ + v 2 + v 3 = (L { + L 2 + L?)—
at
from which it should be apparent that the equivalent inductance of
series-connected inductors is the sum of the individual inductances For n
induc-tors in series,
LeQ = Li + L 2 + L 3 + ••• + L„ (6.19) < Combining inductors in series
If the original inductors carry an initial current of i(f0), the equivalent
inductor carries the same initial current Figure 6.14 shows the equivalent
circuit for series inductors carrying an initial current
Inductors in parallel have the same terminal voltage In the equivalent
circuit, the current in each inductor is a function of the terminal voltage
and the initial current in the inductor For the three inductors in parallel
shown in Fig 6.15, the currents for the individual inductors are
v
*0b)
•^cu — L\+ L 2 + LT,
i\=— I v dr + /JOO),
'2 v dr + /2(?oX
'2 A t
Figure 6.14 A An equivalent circuit for inductors in
series carrying an initial current i(t () )
h -~T I vdT + i 3 (t {) )
^3 At
(6.20) v LiiliM L 2 i\i 2 (t n ) ^ J j ' a f t ) )
The current at the terminals of the three parallel inductors is the sum of Fi 9"re 6.15 A Three inductors in parallel,
the inductor currents:
i = ii + /'2 + /3
Substituting Eq 6.20 into Eq 6.21 yields
(6.21)
I — + — + — I I 1 : ;
Trang 3188 Inductance, Capacitance, and Mutual Inductance
Now we can interpret Eq 6.22 in terms of a single inductor; that is,
1 f
^eq J la
Comparing Eq 6.23 with (6.22) yields
— = - + — + — (6.24)
^ e q -^1 *->2 JUT,
*(%)! 3 ^cq Figure 6.16 shows the equivalent circuit for the three parallel inductors in
Fig 6.15
The results expressed in Eqs 6.24 and 6.25 can be extended to Figure 6.16 • An equivalent circuit for three inductors n i n d u c t o r s m parallel:
in parallel
Combining inductors in parallel • = — + — + + — (6.26)
Equivalent inductance initial current • /(r0) = / ^ ) + j 2 (t 0 ) + • • + i n (t 0 ) (6.27)
Capacitors connected in series can be reduced to a single equivalent capacitor The reciprocal of the equivalent capacitance is equal to the sum
of the reciprocals of the individual capacitances If each capacitor carries its own initial voltage, the initial voltage on the equivalent capacitor is the algebraic sum of the initial voltages on the individual capacitors Figure 6.17 and the following equations summarize these observations:
Combining capacitors in series • = 1 h • • • -\ , (6.28)
Equivalent capacitance initial voltage • v(t 0 ) = vi(t 0 ) + v 2 (t Q ) + • • • + v n (t 0 ) (6.29)
We leave the derivation of the equivalent circuit for series-connected capacitors as an exercise (See Problem 6.32.)
The equivalent capacitance of capacitors connected in parallel is sim-ply the sum of the capacitances of the individual capacitors, as Fig 6.18 and the following equation show:
Capacitors connected in parallel must carry the same voltage Therefore, if there is an initial voltage across the original parallel capacitors, this same initial voltage appears across the equivalent capacitance Ceq The deriva-tion of the equivalent circuit for parallel capacitors is left as an exercise
(See Problem 6.33.)
We say more about series-parallel equivalent circuits of inductors and capacitors in Chapter 7, where we interpret results based on their use
Trang 4+
v
i
Ci?
C 2
T-Cj
(a)
+
^Wi(fe)
+
=: «20»)
+
~v„(ta)
+
V
1
c
c d
W q y
+
^v(t u )
1 + L 4- +
«i('o) + 1¾¾)) + (b)
+
V
e
—^-c,;fc
o 4)
c„
(a)
C + «„('<))
c, cq
c, + c? + - + c„
Figure 6.17 • An equivalent circuit for capacitors connected in
series, (a) The series capacitors, (b) The equivalent circuit
(b)
Figure 6.18 • An equivalent circuit for capacitors connected in
parallel, (a) Capacitors in parallel, (b) The equivalent circuit
/ A S S E S S M E N T PROBLEMS
Objective 3—Be able to combine inductors or capacitors in series and in parallel to form a single equivalent inductor
6.4 The initial values of i[ and /2 in the circuit
shown are + 3 A and - 5 A, respectively The
voltage at the terminals of the parallel
induc-tors for t > 0 is -30e"5' mV
a) If the parallel inductors are replaced by a
single inductor, what is its inductance?
b) What is the initial current and its reference
direction in the equivalent inductor?
c) Use the equivalent inductor to find /(f)
d) Find /t(f) and /2(f) Verify that the solutions
for *']_(*), /2(f), and /(f) satisfy Kirchhoff s
current law
I'W
Answer: (a) 48 mH;
(b) 2 A, up;
(c) 0.125e~5' - 2.125 A, f > 0;
(d)/!(f) = O.le-5' + 2.9 A, t > 0,
kit) = 0.025e~5' - 5.025 A, f ; 0
6.5 The current at the terminals of the two capaci-tors shown is 240e~ 1()'/xA for f > 0 The initial
values of v^ and v 2 are - 1 0 V and - 5 V, respectively Calculate the total energy trapped
in the capacitors as f —* oo (Hint: Don't
com-bine the capacitors in series—find the energy trapped in each, and then add.)
+ v,
2/x¥
/i(0H60mH /2(0H240mH
Answer: 20 /xJ
NOTE: Also try Chapter Problems 6.21, 6.25, 6.26, and 6.31
6.4 Mutual Inductance
The magnetic field we considered in our study of inductors in Section 6.1
was restricted to a single circuit We said that inductance is the parameter
that relates a voltage to a time-varying current in the same circuit; thus,
inductance is more precisely referred to as self-inductance
We now consider the situation in which two circuits are linked by a
magnetic field In this case, the voltage induced in the second circuit can
be related to the time-varying current in the first circuit by a parameter
Trang 5190 Inductance, Capacitance, and Mutual Inductance
Figure 6.19 • Two magnetically coupled coils
Figure 6.20 • Coil currents i { and i 2 used to describe
the circuit shown in Fig 6.19
Figure 6.21 A The circuit of Fig 6.20 with dots added
to the coils indicating the polarity of the mutually
induced voltages
known as mutual inductance The circuit shown in Fig 6.19 represents two magnetically coupled coils The self-inductances of the two coils arc
labeled L] and L2, and the mutual inductance is labeled M The double-headed arrow adjacent to M indicates the pair of coils with this value of
mutual inductance.This notation is needed particularly in circuits contain-ing more than one pair of magnetically coupled coils
The easiest way to analyze circuits containing mutual inductance is
to use mesh currents.The problem is to write the circuit equations that describe the circuit in terms of the coil currents First, choose the refer-ence direction for each coil current Figure 6.20 shows arbitrarily selected reference currents After choosing the reference directions for /, and /2, sum the voltages around each closed path Because of the
mutual inductance M, there will be two voltages across each coil,
namely, a induced voltage and a mutually induced voltage The self-induced voltage is the product of the self-inductance of the coil and the first derivative of the current in that coil The mutually induced voltage
is the product of the mutual inductance of the coils and the first deriva-tive of the current in the other coil Consider the coil on the left in
Fig 6.20 whose self-inductance has the value L\ The self-induced voltage across this coil is L x (di x fdt) and the mutually induced voltage across this coil is M(di 2 /dt) But what about the polarities of these
two voltages?
Using the passive sign convention, the self-induced voltage is a voltage drop in the direction of the current producing the voltage But the polarity
of the mutually induced voltage depends on the way the coils are wound in relation to the reference direction of coil currents In general, showing the details of mutually coupled windings is very cumbersome Instead, we keep
track of the polarities by a method known as the dot convention, in which a
dot is placed on one terminal of each winding, as shown in Fig 6.21 These dots carry the sign information and allow us to draw the coils schematically rather than showing how they wrap around a core structure
The rule for using the dot convention to determine the polarity of mutually induced voltage can be summarized as follows:
Dot convention for mutually coupled coils •
When the reference direction for a current enters the dotted termi-nal of a coil, the reference polarity of the voltage that it induces in the other coil is positive at its dotted terminal
Or, stated alternatively
Dot convention for mutually coupled coils
(alternate) •
When the reference direction for a current leaves the dotted termi-nal of a coil, the reference polarity of the voltage that it induces in the other coil is negative at its dotted terminal
For the most part, dot markings will be provided for you in the circuit diagrams in this text The important skill is to be able to write the appro-priate circuit equations given your understanding of mutual inductance and the dot convention Figuring out where to place the polarity dots if they are not given may be possible by examining the physical configura-tion of an actual circuit or by testing it in the laboratory We will discuss these procedures after we discuss the use of dot markings
In Fig 6.21, the dot convention rule indicates that the reference
polar-ity for the voltage induced in coil 1 by the current i 2 is negative at the
dot-ted terminal of coil l.This voltage (Mdi 2 /dt) is a voltage rise with respect
to /]_ The voltage induced in coil 2 by the current /| is Mdi\jdt, and its
ref-erence polarity is positive at the dotted terminal of coil 2 This voltage is a
voltage rise in the direction of i 2 Figure 6.22 shows the self- and mutually
induced voltages across coils 1 and 2 along with their polarity marks
Trang 6Figure 6.22 • The self- and mutually induced voltages appearing
across the coils shown in Fig 6.21
Now let's look at the sum of the voltages around each closed loop In
Eqs 6.31 and 6.32, voltage rises in the reference direction of a current
are negative:
dii du ioRo + L 2 -f - M -1
dt dt 0
(6.31)
(6.32)
The Procedure for Determining Dot Markings
We shift now to two methods of determining dot markings The first
assumes that we know the physical arrangement of the two coils and the
mode of each winding in a magnetically coupled circuit The following six
steps, applied here to Fig 6.23, determine a set of dot markings:
a) Arbitrarily select one terminal—say, the D terminal—of one coil and
mark it with a dot
b) Assign a current into the dotted terminal and label it /D
c) Use the right-hand rule1 to determine the direction of the magnetic
field established by /D inside the coupled coils and label this field <j6D
d) Arbitrarily pick one terminal of the second coil—say, terminal A—and
assign a current into this terminal, showing the current as /A
e) Use the right-hand rule to determine the direction of the flux
estab-lished by /A inside the coupled coils and label this flux <£A
f) Compare the directions of the two fluxes <£D and <£A If the fluxes
have the same reference direction, place a dot on the terminal of the
second coil where the test current (/A) enters (In Fig 6.23, the fluxes
<£D and c/>A have the same reference direction, and therefore a dot
goes on terminal A.) If the fluxes have different reference
direc-tions, place a dot on the terminal of the second coil where the test
current leaves
The relative polarities of magnetically coupled coils can also be
deter-mined experimentally.This capability is important because in some situations,
determining how the coils are wound on the core is impossible One
experi-mental method is to connect a dc voltage source, a resistor, a switch, and a dc
voltmeter to the pair of coils, as shown in Fig 6.24 The shaded box covering
the coils implies that physical inspection of the coils is not possible The
resis-tor R limits the magnitude of the current supplied by the dc voltage source
The coil terminal connected to the positive terminal of the dc source
via the switch and limiting resistor receives a polarity mark, as shown in
Fig 6.24 When the switch is closed, the voltmeter deflection is observed If
the momentary deflection is upscale, the coil terminal connected to the
positive terminal of the voltmeter receives the polarity mark If the
eP2)
Arbitrarily dotted
D terminal ( S t e p l )
Figure 6.23 • A set of coils showing a method for
determining a set of dot markings
R
voltmeter
Figure 6.24 • An experimental setup for determining
polarity marks
See discussion of Faraday's law on page 193
Trang 7192 Inductance, Capacitance, and Mutual Inductance
deflection is downscale, the coil terminal connected to the negative termi-nal of the voltmeter receives the polarity mark
Example 6.6 shows how to use the dot markings to formulate a set of circuit equations in a circuit containing magnetically coupled coils
a) Write a set of mesh-current equations that
describe the circuit in Fig 6.25 in terms of the
currents /j and i 2
b) Verify that if there is no energy stored in the
cir-cuit at t = 0 and if L = 16 - 16tf_5t A, the
solu-tions for i\ and i 2 are
b) To check the validity of /j and i 2 , we begin by testing the initial and final values of i L and i 2 We know by hypothesis that /^(0) = i 2 (0) = 0 From
the given solutions we have
/j(0) = 4 + 64 - 68 = 0,
4;
/ , = 4 + 64e_:* - 6&T" A,
/2(0) = 1 - 52 + 51 = 0
/2 = 1 - 52«?"* + 51e"4' A
5 0
4H
8H 20 n
1
16 H f i2 \ 60 H
Figure 6.25 • The circuit for Example 6.6
Now we observe that as t approaches infinity the
source current (/<,) approaches a constant value
of 16 A, and therefore the magnetically coupled
coils behave as short circuits Hence at t = oo
the circuit reduces to that shown in Fig 6.26
From Fig 6.26 we see that at t = oo the three
resistors are in parallel across the 16 A source The equivalent resistance is 3.75 fl and thus the voltage across the 16 A current source is 60 V It follows that
z,(oo) = — + — = 4 A ,
u } 20 60
Solution
a) Summing the voltages around the i x mesh yields
4 - ^ + 8 - ¾ - i 2 ) + 20(/, - i 2 ) + 5(i t - i g ) = 0
,2 (o o ) = | | = l A
These values agree with the final values
pre-dicted by the solutions for i x and i 2
Finally we check the solutions by seeing if they satisfy the differential equations derived in (a) We will leave this final check to the reader via Problem 6.37
The i 2 mesh equation is
20(/2 - h) + 60/2 + 16-7-(¾ - **) ~ 8 - r = 0
Note that the voltage across the 4 H coil due to
the current (i g - i 2 ), that is, 8d(i g - i 2 )/dt, is a
voltage drop in the direction of i x The voltage
induced in the 16 Ff coil by the current /t, that is,
8di\/dt, is a voltage rise in the direction of i 2
16A
20 n
V A r
-:60O
Figure 6.26 • The circuit of Example 6.6 when t — oo
Trang 8• A S S E S S M E N T PROBLEM
Objective 4—Use the dot convention to write mesh-current equations for mutually coupled coils
6.6 a) Write a set of mesh-current equations for
the circuit in Example 6.6 if the dot on the
4 H inductor is at the right-hand terminal,
the reference direction of/^, is reversed, and
the 60 ft resistor is increased to 780 ft
b) Verify that if there is no energy stored in the
circuit at t = 0,andififf = 1.96 - 1.96e~* A,
the solutions to the differential equations
derived in (a) of this Assessment Problem are
Answer: (a) A{dijdt) + 25i x + 8(di 2 /dt) - 20i 2
= -5i g - $(dig/dt)
and
%{di<Jdt) - 20¾ + 16(di 2 /dt) + 800/2
= -16(dig/dt);
(b) verification
0.4 0.01
11.6e"4' + 12e_ 5'A,
,-4/
0.99<T" + e~™ A -St NOTE: Also try Chapter Problem 6.39
6.5 A Closer Look at Mutual Inductance
In order to fully explain the circuit parameter mutual inductance, and to
examine the limitations and assumptions made in the qualitative discussion
presented in Section 6.4, we begin with a more quantitative description of
self-inductance than was previously provided
A Review of Self-Inductance
The concept of inductance can be traced to Michael Faraday, who did
pio-neering work in this area in the early 1800s Faraday postulated that a
magnetic field consists of lines of force surrounding the current-carrying
conductor Visualize these lines of force as energy-storing elastic bands
that close on themselves As the current increases and decreases, the
elas-tic bands (that is, the lines of force) spread and collapse about the
conduc-tor The voltage induced in the conductor is proportional to the number of
lines that collapse into, or cut, the conductor This image of induced
volt-age is expressed by what is called Faraday's law; that is,
dX_
where A is referred to as the flux linkage and is measured in weber-turns
How do we get from Faraday's law to the definition of inductance
pre-sented in Section 6.1? We can begin to draw this connection using Fig 6.27
as a reference
The lines threading the N turns and labeled 4> represent the magnetic
lines of force that make up the magnetic field The strength of the
mag-netic field depends on the strength of the current, and the spatial
orienta-tion of the field depends on the direcorienta-tion of the current The right-hand
rule relates the orientation of the field to the direction of the current:
When the fingers of the right hand are wrapped around the coil so that the
fingers point in the direction of the current, the thumb points in the
direc-tion of that pordirec-tion of the magnetic field inside the coil The flux linkage is
the product of the magnetic field (<£), measured in webers (Wb), and the
number of turns linked by the field (N):
N turns
Figure 6.27 • Representation of a magnetic field link-ing an vV-turn coil
A = N<f> (6.34)
Trang 9194 Inductance, Capacitance, and Mutual Inductance
The magnitude of the flux, ¢, is related to the magnitude of the coil
current by the relationship
where N is the number of turns on the coil, and SP is the permeance of the
space occupied by the flux Permeance is a quantity that describes the magnetic properties of this space, and as such, a detailed discussion of per-meance is outside the scope of this text Here, we need only observe that, when the space containing the flux is made up of magnetic materials (such
as iron, nickel, and cobalt), the permeance varies with the flux, giving a
nonlinear relationship between 4> and i But when the space containing the
flux is comprised of nonmagnetic materials, the permeance is constant,
giving a linear relationship between cj> and L Note from Eq 6.35 that the
flux is also proportional to the number of turns on the coil
Here, we assume that the core material—the space containing the flux—
is nonmagnetic Then, substituting Eqs 6.34 and 6.35 into Eq 6.33 yields
v =
dX d(N4>)
dt dt
d<b d,
dt dt
dt
di
which shows that self-inductance is proportional to the square of the num-ber of turns on the coil We make use of this observation later
The polarity of the induced voltage in the circuit in Fig 6.27 reflects the
reaction of the field to the current creating the field For example, when i is increasing, di/dt is positive and v is positive Thus energy is required to establish the magnetic field The product vi gives the rate at which energy is stored in the field When the field collapses, di/dt is negative, and again the
polarity of the induced voltage is in opposition to the change As the field collapses about the coil, energy is returned to the circuit
Keeping in mind this further insight into the concept of self-inductance,
we now turn back to mutual inductance
Figure 6.28 A Two magnetically coupled coils
The Concept of Mutual Inductance
Figure 6.28 shows two magnetically coupled coils You should verify that the dot markings on the two coils agree with the direction of the coil
wind-ings and currents shown The number of turns on each coil are A^ and N 2 ,
respectively Coil 1 is energized by a time-varying current source that
establishes the current i\ in the A7] turns Coil 2 is not energized and is open The coils are wound on a nonmagnetic core The flux produced by
the current i\ can be divided into two components, labeled <f> n and 02 J
The flux component 4> X \ is the flux produced by i\ that links only the Ny
turns.The component <£21 is the flux produced by f] that links the N2 turns
and the Ny turns The first digit in the subscript to the flux gives the coil
number, and the second digit refers to the coil current Thus </>n is a flux
linking coil 1 and produced by a current in coil 1, whereas cf) 2 \ is a flux
link-ing coil 2 and produced by a current in coil 1
Trang 1001 = 0 n + <t>2\- (6-37) The flux <f> { and its components 4>n and 4>z\ are related to the coil current
ii as follows:
</>! = ?i>1N1/i, (6.38)
^21 = ^ 2 l W i , (6.40) where flf^ is the permeance of the space occupied by the flux 01,27*11 is the
permeance of the space occupied by the flux 0U, and P?21 is t n e permeance
of the space occupied by the flux 02i- Substituting Eqs 6.38,6.39, and 6.40
into Eq 6.37 yields the relationship between the permeance of the space
occupied by the total flux 4> x and the permeances of the spaces occupied
by its components 4>\\ and 02 I:
PJ>, = 0>ii + 9*21- (6-4l)
We use Faraday's law to derive expressions for ?;t and v2 :
rfA, d(NM d
, di\ » di\ di\
= Aft*,, + 9 >2, ) ^ = N P ^ = L , ^ , (6.42)
and
dk 2 d(N 2 (f>2\) i T d ,™ A, x
The coefficient of d/t/d7 in Eq 6.42 is the self-inductance of coil 1 The
coefficient of di x jdt in Eq 6.43 is the mutual inductance between coils
1 and 2 Thus
The subscript on M specifies an inductance that relates the voltage induced
in coil 2 to the current in coil 1
The coefficient of mutual inductance gives
Note that the dot convention is used to assign the polarity reference to v 2
in Fig 6.28
For the coupled coils in Fig 6.28, exciting coil 2 from a time-varying
cur-rent source (i 2 ) and leaving coil 1 open produces the circuit arrangement