We are now in a position to determine the currents and voltages that arise when energy is either released or acquired by an inductor or capacitor in response to an abrupt change in a dc
Trang 1t = 0 the inductor was switched instantaneously to
position b where it remained for 1.6 s before returning
instantaneously to position a The d'Arsonval
volt-meter has a full-scale reading of 20 V and a sensitivity
of 1000 ft/V What will the reading of the voltmeter
be at the instant the switch returns to position a if the
inertia of the d'Arsonval movement is negligible?
Figure P6.13
PSPICE
HULTISIH
Section 6.2
6.14 The current shown in Fig P6.14 is applied to a
0.25 /xF capacitor The initial voltage on the
capaci-tor is zero
a) Find the charge on the capacitor at t = 15 /xs
b) Find the voltage on the capacitor at t = 30 /xs
c) How much energy is stored in the capacitor by
this current?
Figure P6.14
i (mA)
t (fis)
6.15 The initial voltage on the 0.5 /xF capacitor shown in
PSPICE Fig P6.15(a) is - 2 0 V The capacitor current has
m n s m the waveform shown in Fig P6.15(b)
a) How much energy, in microjoules, is stored in
the capacitor at t = 500 /xs?
b) Repeat (a) for t = oo
6.16 The rectangular-shaped current pulse shown in
PSPICE pig P6.16 is applied to a 0.1 fxF capacitor The
ini-' tial voltage on the capacitor is a 15 V drop in the reference direction of the current Derive the expression for the capacitor voltage for the time intervals in (a)-(d)
a) 0 < t < 10 /xs;
b) 10/AS < t =5 20/xs;
c) 20 /xs < t < 40 /xs d) 40 /xs < t < oo c) Sketch v{t) over the interval - 1 0 /xs ^ t ^ 50 ju.s
Figure P6.16
i (mA)
160
PSPICE MULTISIM
100
0
50
6.17 A 20 fxF capacitor is subjected to a voltage pulse
having a duration of 1 s The pulse is described by the following equations:
30t 2 V,
< 3 0 ( f - 1)2V,
0
0 < t < 0.5 s;
0.5 s < t < 1 s;
elsewhere
Sketch the current pulse that exists in the capacitor during the 1 s interval
6.18 The voltage across the terminals of a 0.2 /xF
capaci-tor is
v =
150 V, t < 0;
(Ate**** + A*T saiDt )V, t>0
Figure P6.15
0.5 /xF
-20 V
v
(a)
i (n
50
25
0
lA)
I
100
50e- 20()() 'mA,/>0
1 1 1 l
(b)
Trang 26.19
PSPICE
MULTISIM
V =
The initial current in the capacitor is 250 mA
Assume the passive sign convention
a) What is the initial energy stored in the capacitor?
b) Evaluate the coefficients A { and A 2
c) What is the expression for the capacitor current?
The voltage at the terminals of the capacitor in
Fig 6.10 is known to be
- 2 0 V, / < 0;
100 - 40^200()'(3 cos 1000? + sin 1000/) V t > 0
Assume C = 4 /xF
a) Find the current in the capacitor for t < 0
b) Find the current in the capacitor for t > 0
c) Is there an instantaneous change in the voltage
across the capacitor at t = 0?
d) Is there an instantaneous change in the current
in the capacitor at t = 0?
e) How much energy (in millijoules) is stored in
the capacitor at t = oo?
Section 6.3
6.20 Assume that the initial energy stored in the
induc-PSPICE tors of Fig P6.20 is zero Find the equivalent
induc-tance with respect to the terminals a,b
Figure P6.20
30FH20H
6.21 Assume that the initial energy stored in the
tors of Fig P6.21 is zero Find the equivalent
induc-tance with respect to the terminals a,b
Figure P6.21
3H
8H
6.22 Use realistic inductor values from Appendix H to
con-struct series and parallel combinations of inductors to
yield the equivalent inductances specified below Try
to minimize the number of inductors used Assume
that no initial energy is stored in any of the inductors
a) 3 m H
b) 250/xH
c) 6QfxH
6.23 The three inductors in the circuit in Fig P6.23 are
con-PSPICE nected across the terminals of a black box at t = 0 The resulting voltage for t > 0 is known to be
v a = 2000e"100' V
If /,(0) = - 6 A a n d /2( 0 ) = 1 A, find
a) U0);
b) U0, t > 0;
c) ii(f)»* s 0;
d) /2(/), t > 0:
e) the initial energy stored in the three inductors; f) the total energy delivered to the black box; and g) the energy trapped in the ideal inductors
Figure P6.23
6.24 For the circuit shown in Fig P6.23, how many
milli-seconds after the switch is opened is the energy delivered to the black box 80% of the total energy delivered?
6.25 The two parallel inductors in Fig P6.25 are
con-nected across the terminals of a black box at t = 0 The resulting voltage v for t > 0 is known to be
64e-4' V It is also known that /,(0) = - 1 0 A and /2(0) = 5 A
a) Replace the original inductors with an equiva-lent inductor and find /(f) for f > 0
b) Find *',(*) for t > 0
c) Find i2 (t) for t > 0
d) How much energy is delivered to the black box
in the time interval 0 < f < oo?
e) How much energy was initially stored in the par-allel inductors?
f) How much energy is trapped in the ideal inductors? g) Show that your solutions for /, and /2 agree with the answer obtained in (f)
Figure P6.25
Trang 36.26 Find the equivalent capacitance with respect to the
terminals a,b for the circuit shown in Fig P6.26
6.27 Find the equivalent capacitance with respect to the
terminals a,b for the circuit shown in Fig P6.27
Figure P6.27
5 V
6.28 Use realistic capacitor values from Appendix H to
construct series and parallel combinations of
capac-itors to yield the equivalent capacitances specified
below Try to minimize the number of capacitors
used Assume that no initial energy is stored in any
of the capacitors
a) 330/AF
b) 750 nF
c) 150 pF
6.29 The four capacitors in the circuit in Fig P6.29 are
con-nected across the terminals of a black box at t — 0
The resulting current i b for t > 0 is known to be
•- 5 0 ' m A
h -5e
If v u (0) = - 2 0 V, v c (0) = - 3 0 V, and
v ( ,(0) = 250 V, find the following for / ^ 0: (a) v h (t),
(b) vM (<0 »<#) (d) * X 0 (e) h(t\ and (f) i 2 (t)
Figure P6.29
200 nF:
6.30 For the circuit in Fig P6.29, calculate a) the initial energy stored in the capacitors; b) the final energy stored in the capacitors;
c) the total energy delivered to the black box; d) the percentage of the initial energy stored that is delivered to the black box; and
e) the time, in milliseconds, it takes to deliver 7.5 mJ to the black box
6.31 The two series-connected capacitors in Fig P6.31
are connected to the terminals of a black box at
t = 0 The resulting current i(t) for t > 0 is known
to be 800<T25' (xA
a) Replace the original capacitors with an
equiva-lent capacitor and find va (t) for t > 0
b) Find »!(0 for/ > 0
c) Find v2 (t) for / > 0
d) How much energy is delivered to the black box
in the time interval 0 < t < oo?
e) How much energy was initially stored in the series capacitors?
f) How much energy is trapped in the ideal capacitors?
g) Show that the solutions for V\ and v 2 agree with the answer obtained in (f)
Figure P6.31
5 V -+ +
25 V ^
-
—»-^2fxF »i
+
^ 8 / x F v 2
+
>
\ / = 0
-*\,
1
Black box
1.25/tF
6.32 Derive the equivalent circuit for a series connection
of ideal capacitors Assume that each capacitor has its own initial voltage Denote these initial voltages
as V\(t[)), ^2(/0)1 and so on (Hint: Sum the voltages
across the string of capacitors, recognizing that the series connection forces the current in each capaci-tor to be the same.)
6.33 Derive the equivalent circuit for a parallel connec-tion of ideal capacitors Assume that the initial
volt-age across the paralleled capacitors is v(t {) ) (Hint:
Sum the currents into the string of capacitors, rec-ognizing that the parallel connection forces the voltage across each capacitor to be the same.)
Sections 6.1-6.3 6.34 The current in the circuit in Fig P6.34 is known to be
i o = 5<r2()0,)f(2 cos 4000/ + sin 4000/) A
for t > 0 Find z;(0) and v2 (0 )
Trang 4Figure P6.34
40 ft
v 2 < 10 mH
6.35 At t = 0, a series-connected capacitor and
induc-tor are placed across the terminals of a black box,
as shown in Fig P6.35 For t > 0, it is known that
L = 1 5e-] 6'0 0 0' - 0 5 e -4 0 0 0' A
If v c (0) = - 5 0 V find v () for t
Figure P6.35
0
+
25 mH
~625nF
Section 6.4
X
/ - 0
>,>
+
Black box
6.36 a) Show that the differential equations derived in
(a) of Example 6.6 can be rearranged as follows:
4-^- + 25/T - 8 - - 20/2 = 5i g - 8 - ^ ;
dL dii dig
8 — 20/, + 16—^ + 80/2 = 1 6 r
-dt -dt -dt
b) Show that the solutions for i h and i 2 given in
(b) of Example 6.6 satisfy the differential
equations given in part (a) of this problem
6.37 Let v a represent the voltage across the 16 H
inductor in the circuit in Fig 6.25 Assume v0 is
positive at the dot As in Example 6.6,
i g = 1 6 - 16e_ 5'A
a) Can you find v„ without having to
differenti-ate the expressions for the currents? Explain
b) Derive the expression for v0
c) Check your answer in (b) using the
appropri-ate current derivatives and inductances
6.38 Let v R represent the voltage across the current
source in the circuit in Fig 6.25 The reference for
v g is positive at the upper terminal of the current
source
a) Find vg as a function of time when
i K = 16 - 16<T5' A
b) What is the initial value of vj
c) Find the expression for the power developed by the current source
d) How much power is the current source
develop-ing when t is infinite?
e) Calculate the power dissipated in each resistor
when t is infinite
6.39 There is no energy stored in the circuit in Fig P6.39
at the time the switch is opened
a) Derive the differential equation that governs the behavior of /2 if Lj = 4 H, L2 = 16 H,
M = 2 H, and R 0 = 32 H
b) Show that when ig = 8 - 8e"'A, t > 0, the
dif-ferential equation derived in (a) is satisfied
when i2 = e~ l - e~ 2t A, t > 0
c) Find the expression for the voltage V\ across the
current source
d) What is the initial value of v{t Does this make
sense in terms of known circuit behavior?
Figure P6.39
6.40 a) Show that the two coupled coils in Fig P6.40 can
be replaced by a single coil having an inductance
of La b = L\ + L 2 + 2M, (Hint: Express v a ^ as a
function of /ab.) b) Show that if the connections to the terminals
of the coil labeled L2 are reversed,
La b = Lx + L 2 - 2M
Figure P6.40
6.41 a) Show that the two magnetically coupled coils in
Fig P6.41 (see page 210) can be replaced by a single coil having an inductance of
^ab
-L X L 2 M l
L { + U 2M (Hint: Let i\ and i 2 be clockwise mesh currents in
the left and right "windows" of Fig P6.41, respec-tively Sum the voltages around the two meshes
In mesh 1 let v ah be the unspecified applied
volt-age Solve for dijdt as a function of vab.) b) Show that if the magnetic polarity of coil 2 is reversed, then
• ^ a b —
L { L 2 - M 2
L x + L 2 + 2M'
Trang 5Figure P6.41
a«
M
6.42 The polarity markings on two coils are to be
deter-mined experimentally The experimental setup is
shown in Fig P6.42 Assume that the terminal
con-nected to the negative terminal of the battery has
been given a polarity mark as shown When the
switch is opened, the dc voltmeter kicks upscale
Where should the polarity mark be placed on the
coil connected to the voltmeter?
Figure P6.42
/ = 0
r - ^ > \
VBK-voltmeter
6.43 The physical construction of four pairs of
magneti-cally coupled coils is shown in Fig P6.43 (See
page 211.) Assume that the magnetic flux is confined
to the core material in each structure Show two
possi-ble locations for the dot markings on each pair of coils
Section 6.5
6.44 The self-inductances of the coils in Fig 6.30 are
L\ = 18 mH and L 2 = 32 mH If the coefficient of
coupling is 0.85, calculate the energy stored in the
system in millijoules when (a) ix = 6 A, i 2 = 9 A;
(b) i{ = - 6 A, i2 = -9 A; (c) i } = - 6 A, i2 = 9 A;
and (d) 2, = 6 A, i2 = - 9 A
6.45 The coefficient of coupling in Problem 6.44 is
increased to 1.0
a) If /j equals 6 A, what value of i 2 results in zero
stored energy?
b) Is there any physically realizable value of i2 that
can make the stored energy negative?
6.46 Two magnetically coupled coils have self-inductances
of 60 mH and 9.6 mH, respectively.The mutual
induc-tance between the coils is 22.8 mH
a) What is the coefficient of coupling?
b) For these two coils, what is the largest value that
M can have?
c) Assume that the physical structure of these
cou-pled coils is such that 2?*j = 9>2- What is the turns
ratio N\/N2 if N\ is the number of turns on the
60 mH coil?
6.47 The self-inductances of two magnetically coupled
coils are 72 mH and 40.5 mH, respectively The 72 mH
coil has 250 turns, and the coefficient of coupling
between the coils is %.The coupling medium is non-magnetic When coil 1 is excited with coil
2 open, the flux linking only coil 1 is 0.2 as large as the flux linking coil 2
a) How many turns does coil 2 have?
b) What is the value of <& 2 in nanowebers per ampere?
c) What is the value of S^n in nanowebers per ampere?
d) What is the ratio (^22/^12)?
6.48 Two magnetically coupled coils are wound on a
nonmagnetic core The self-inductance of coil 1 is
288 mH, the mutual inductance is 90 mH, the coeffi-cient of coupling is 0.75, and the physical structure
of the coils is such that SPn =
9^2-a) Find L2 and the turns ratio Ni/N2
b) If N { = 1200, what is the value of S^ and 2P2?
6.49 The self-inductances of two magnetically coupled coils
are L{ = 180/xH and L2 = 500 /xH The coupling
medium is nonmagnetic If coil 1 has 300 turns and
coil 2 has 500 turns, find <?fin and 5P2i (in nanowebers per ampere) when the coefficient of coupling is 0.6
6.50 a) Starting with Eq 6.59, show that the coefficient
of coupling can also be expressed as
¢1
¢12
02
b) On the basis of the fractions 4>2 \f4>\ and <f>\ 2 /(f> 2 ,
explain why k is less than 1.0
Sections 6.1-6.5 6.51 Rework the Practical Perspective example, except
PERSSE t n a t this time, put the button on the bottom of the divider circuit, as shown in Fig P6.51 Calculate the
output voltage v(t) when a finger is present
Figure P6.51
>M&)
Fixed capacitor' 25 pF
Button
+
( A ) » ( 0
6.52 Some lamps are made to turn on or off when the
PERSPECTIVEbase *s touched These use a one-terminal variation
of the capacitive switch circuit discussed in the Practical Perspective Figure P6.52 shows a circuit model of such a lamp Calculate the change in the
voltage v(t) when a person touches the lamp
Assume all capacitors are initially discharged
Trang 6Figure P6.43
10 pF Lamp Person 10 pF
«*>© 10 pF!
t •
+ tt
loo pF;
6.53 In the Practical Perspective example, we calculated
PRACTICAL the output voltage when the elevator button is the
upper capacitor in a voltage divider In
Problem 6.51, we calculated the voltage when the
button is the bottom capacitor in the divider, and we
got the same result! You may wonder if this will be
true for all such voltage dividers Calculate the
volt-age difference (finger versus no finger) for the
cir-cuits in Figs P6.53 (a) and (b), which use two
identical voltage sources
— ' Fixed
Button
25 pF capacitor
25 pF +
"•* v{t) No finger
(a)
25 pF 25 pF
Button
25 pF F i x c d
-" capacitor
25 pF +
"* v(t) Finger
(b)
Trang 77.1 The Natural Response of an RL Circuit p 214
7.2 The Natural Response of an RCCircuit p 220
7.3 The Step Response of RL and
RC Circuits p 224
7.4 A General Solution for Step and Natural
Responses p 231
7.5 Sequential Switching p 235
7.6 Unbounded Response p 240
7.7 The Integrating Amplifier p 242
1 Be able to determine the natural response of
both RL and RC circuits
2 Be able to determine the step response of both
RL and RC circuits
3 Know how to analyze circuits with sequential
switching
4 Be able to analyze op amp circuits containing
resistors and a single capacitor
212
Response of First-Order
RL and RC Circuits
In Chapter 6, we noted that an important attribute of inductors
and capacitors is their ability to store energy We are now in a position to determine the currents and voltages that arise when energy is either released or acquired by an inductor or capacitor
in response to an abrupt change in a dc voltage or current source
In this chapter, we will focus on circuits that consist only of sources, resistors, and either (but not both) inductors or
capaci-tors For brevity, such configurations are called RL
(resistor-inductor) and RC (resistor-capacitor) circuits
Our analysis of RL and RC circuits will be divided into three
phases In the first phase, we consider the currents and voltages that arise when stored energy in an inductor or capacitor is sud-denly released to a resistive network This happens when the inductor or capacitor is abruptly disconnected from its dc source Thus we can reduce the circuit to one of the two equivalent forms shown in Fig 7.1 on page 214 The currents and voltages that arise
in this configuration are referred to as the natural response of the
circuit, to emphasize that the nature of the circuit itself, not exter-nal sources of excitation, determines its behavior
In the second phase of our analysis, we consider the currents and voltages that arise when energy is being acquired by an induc-tor or capaciinduc-tor due to the sudden application of a dc voltage or
current source This response is referred to as the step response
The process for finding both the natural and step responses is the same; thus, in the third phase of our analysis, we develop a general
method that can be used to find the response of RL and RC
cir-cuits to any abrupt change in a dc voltage or current source
Figure 7.2 on page 214 shows the four possibilities for the
gen-eral configuration of RL and RC circuits Note that when there
are no independent sources in the circuit, the Thevenin voltage or Norton current is zero, and the circuit reduces to one of those shown in Fig 7.1; that is, we have a natural-response problem
RL and RC circuits are also known as first-order circuits,
because their voltages and currents are described by first-order differential equations No matter how complex a circuit may
Trang 8Practical Perspective
A Flashing Light Circuit
You can probably think of many different applications that
require a flashing light A still camera used to take pictures in
low light conditions employs a bright flash of light to
illumi-nate the scene for just long enough to record the image on
film Generally, the camera cannot take another picture until
the circuit that creates the flash of light has "re-charged."
Other applications use flashing lights as warning for
haz-ards, such as tall antenna towers, construction sites, and
secure areas In designing circuits to produce a flash of light
the engineer must know the requirements of the application
For example, the design engineer has to know whether the
flash is controlled manually by operating a switch (as in the
case of a camera) or if the flash is to repeat itself
automati-cally at a predetermined rate The engineer also has to know if
the flashing light is a permanent fixture (as on an antenna) or
+
Vs
1
i
k^i
, c 1
<
+
^^ ""^p
v
=s %
5H rt%
a temporary installation (as at a construction site) Another question that has to be answered is whether a power source is readily available
Many of the circuits that are used today to control flashing lights are based on electronic circuits that are beyond the scope of this text Nevertheless we can get a feel for the thought process involved in designing a flashing light circuit
by analyzing a circuit consisting of a dc voltage source, a resis-tor, a capaciresis-tor, and a lamp that is designed to discharge a flash of light at a critical voltage Such a circuit is shown in the figure We shall discuss this circuit at the end of the chapter
213
Trang 9R eq C, eq
(a) (b)
Figure 7.1 • The two forms of the circuits for natural
response, (a) RL circuit, (b) RC circuit
R rii L i v
(b)
(d)
Figure 7.2 A Four possible first-order circuits
(a) An inductor connected to a Thevem'n equivalent
(b) An inductor connected to a Norton equivalent
(c) A capacitor connected to a Thevem'n equivalent
(d) A capacitor connected to a Norton equivalent
^ V
r = ()
Figure 7.3 • An RL circuit
/(0) = I \ L RZv
appear, if it can be reduced to a Thevenin or Norton equivalent connected
to the terminals of an equivalent inductor or capacitor, it is a first-order circuit (Note that if multiple inductors or capacitors exist in the original circuit, they must be interconnected so that they can be replaced by a sin-gle equivalent element.)
After introducing the techniques for analyzing the natural and step responses of first-order circuits, we discuss some special cases of interest The first is that of sequential switching, involving circuits in which switching can take place at two or more instants in time Next is the unbounded response Finally, we analyze a useful circuit called the integrating amplifier
7.1 The Natural Response
of an RL Circuit
The natural response of an RL circuit can best be described in terms of the
circuit shown in Fig 7.3 We assume that the independent current source
generates a constant current of I s A, and that the switch has been in a
closed position for a long time We define the phrase a long time more
accurately later in this section For now it means that all currents and volt-ages have reached a constant value Thus only constant, or dc, currents can exist in the circuit just prior to the switch's being opened, and therefore
the inductor appears as a short circuit (Ldi/dt = 0) prior to the release of
the stored energy
Because the inductor appears as a short circuit, the voltage across the
inductive branch is zero, and there can be no current in either R() or R
Therefore, all the source current /s appears in the inductive branch Finding the natural response requires finding the voltage and current at the terminals of the resistor after the switch has been opened, that is, after the source has been disconnected and the inductor begins releasing
energy If we let t = 0 denote the instant when the switch is opened, the problem becomes one of finding v(t) and i(t) for f > 0 For t S 0, the
cir-cuit shown in Fig 7.3 reduces to the one shown in Fig 7.4
Deriving the Expression for the Current
To find i(i), we use Kirchhoff s voltage law to obtain an expression involv-ing i, R, and L Summinvolv-ing the voltages around the closed loop gives
r dl
L— + Ri
where we use the passive sign convention Equation 7.1 is known as a first-order ordinary differential equation, because it contains terms involving
the ordinary derivative of the unknown, that is, di/dt The highest order
derivative appearing in the equation is 1; hence the term first-order
We can go one step further in describing this equation The
coeffi-cients in the equation, R and L, are constants; that is, they are not func-tions of either the dependent variable i or the independent variable f.Thus
the equation can also be described as an ordinary differential equation with constant coefficients
To solve Eq 7.1, we divide by L, transpose the term involving i to the right-hand side, and then multiply both sides by a differential time dt The
result is
Trang 10Next, we recognize the left-hand side of Eq 7.2 as a differential change in
the current /, that is, di We now divide through by i, getting
— = —rdt
We obtain an explicit expression for i as a function of f by integrating both
sides of Eq 7.3 Using x and y as variables of integration yields
i(/„) X U <
(7.4)
in which /(f0) is the current corresponding to time fg, and /(f) is the current
corresponding to time f Here, f() = 0 Therefore, carrying out the
indi-cated integration gives
/(f)
In
/(0)
R
= ~1L
Based on the definition of the natural logarithm,
/(f) = i(0)e~ls/L K
(7.5)
(7.6)
Recall from Chapter 6 that an instantaneous change of current cannot
occur in an inductor Therefore, in the first instant after the switch has
been opened, the current in the inductor remains unchanged If we use 0~
to denote the time just prior to switching, and 0+ for the time immediately
following switching, then
,(0-) = /(0 + ) = / 0 , -4 Initial inductor current
where, as in Fig 7.1, /() denotes the initial current in the inductor The initial
current in the inductor is oriented in the same direction as the reference
direction of / Hence Eq 7.6 becomes
/(f) = /0<TW L )', f ^ 0, (7.7) -4 Natural response of an RL circuit
which shows that the current starts from an initial value I() and decreases
exponentially toward zero as f increases Figure 7.5 shows this response
We derive the voltage across the resistor in Fig 7.4 from a direct
appli-cation of Ohm's law:
v = iR = I 0 Re- WL)t , t > 0+ (7.8)
Note that in contrast to the expression for the current shown in Eq 7.7,
the voltage is defined only for t > 0, not at f = 0 The reason is that a step
change occurs in the voltage at zero Note that for t < 0, the derivative of
the current is zero, so the voltage is also zero (This result follows from
v = Ldi/dt = 0.) Thus
?;(CT) = 0,
v(0 + ) = I 0 R,
(7.9) (7.10)
where v(0+ ) is obtained from Eq 7.8 with f = 0"1".1 With this step change at
an instant in time, the value of the voltage at f = 0 is unknown Thus we
use f > 0+ in defining the region of validity for these solutions
Figure 7.5 A The current response for the circuit shown
in Fig 7.4
1 We can define the expressions ()"" and u + more formally The expression x(0~) refers to the
limit of the variable x as / —»0 from the left, or from negative time The expression v(O')
refers to the limit of the variable x as / —*• 0 from the right, or from positive time