CHAPTER 3
Exercises
E3.1
V )10sin(5.0)102/()10sin(10/)()( 5656
ttCtqtv
E3.2 Because the capacitor voltage is zero at
t
= 0, the charge on the
capacitor is zero at
t
= 0.
/)()(
Ctqtv
)()()(
tvtitp
in which the units of charge, electrical potential, power, and energy are
E3.3 Refer to Figure 3.10 in the book. Applying KVL, we have
This can be written as
t
E3.4 (a) For series capacitances:
E3.5 From Table 3.1 we find that the relative dielectric constant of polyester
is 3.4. We solve Equation 3.26 for the area of each sheet:
E3.6
 
 
 
V 10sin1010cos1.0)1010(
)(
)( 443
tt
dt
d
dt
tdi
Ltv
E3.7
)(
10150
1
)0()(
1
)(
6
dxxvidxxv
L
ti
tt
E3.8 Refer to Figure 3.20a in the book. Using KVL we can write:
Using Equation 3.28 to substitute, this becomes
E3.9 Refer to Figure 3.20b in the book. Using KCL we can write:
Using Equation 3.32 to substitute, this becomes
ttt
we can write Equation (1) as
E3.10 Refer to Figure 3.21 in the book.
(a) The 2-H and 3-H inductances are in series and are equivalent to a 5-
H inductance, which in turn is in parallel with the other 5-H inductance.
E3.11 The MATLAB commands including some explanatory comments are:
% We avoid using i alone as a symbol for current because
iC = 0.5*sin((1e4)*t);
ezplot(iC, [0 3*pi*1e-4])
Problems
P3.1 Capacitors consist of two conductors separated by an insulating material.
P3.2 Because we have
dtCdvi
/
for a capacitance, the current is zero if the
P3.3 A dielectric material is an electrical insulator through which virtually no
P3.4 Charge (usually in the form of electrons) flows in and accumulates on one
P3.5*
 
dt
dv
Cti
 
t
dt
d
1000sin10010 5
 
t
1000cos
P3.6*
dv
Ci
P3.7*
   
 
t
vdtti
C
tv
0
0
1
 
t
dttv
0
34 20103102
P3.8*
time power
W
s 3600hp/W 746hp 5
P3.9 The net charge on each plate is
500100)105( 6
CVQ
C. One
P3.10
   
 
 
)0sin()sin()cos(
1
0
1
00
t
C
I
dttI
C
vdtti
C
tv m
t
m
t
P3.11
 
dt
dv
Cti
 
t
e
dt
d
1006 10010
P3.12
mC 12001056
CvQ
P3.13
P3.14
   
 
t
vdtti
C
tv
0
0
1
t
P3.15 Because the switch is closed prior to
0
t
, the initial voltage is zero, and
we have
   
 
tt
tdtvdtti
C
tv
0
35
0
6000)103(1020
1
P3.16
   
 
t
vdtti
C
tv
0
0
1
 
t
dtti
0
61010333.0
P3.17 We can write
200105
2
1262
vCvw
P3.18 A capacitance initially charged to 10 V has
P3.19
   
 
 
)200sin(1010010200)()( 66
t
d
tvtC
d
tdq
ti
P3.20 We can write
 
 
tt
t
fdtti
C
Vtv
0
0
1
0
. The integral represents the area of
P3.21 By definition, the voltage across a short circuit must be zero. Since we
have
Riv
for a resistor, zero resistance corresponds to a short circuit.
For an initially uncharged capacitance, we have
P3.22
   
dt
tdv
Cti
   
 
tt
dt
d
556 10sin210cos31020
P3.23 Capacitances in parallel are combined by adding their values. Thus,
capacitances in parallel are combined as resistances in series are.
P3.24* (a)
F 2
2121
1μ
eq
C
(b) The two 4-
μ
F capacitances are in series and have an equivalent
eq
P3.25* As shown below, the two capacitors are placed in series with the heart to
produce the output pulse.
μ
P3.26 (a)
 
F 667.4
11121
1
1121
1
3μ
eq
C
P3.27 We obtain the maximum capacitance of 2
μ
F by connecting the 1-
μ
F
μ
P3.28
F 6
11
1
21
CC
Ceq
The charges stored on each capacitor and on the equivalent capacitance
P3.29 The equivalent capacitance is
100
eq
C
F and its initial voltage is 150 V.
P3.30
FCeq
4
3/16/1
1
2
P3.32
d
WL
d
A
Crr
00 εεεε
P3.33* The charge
Q
remains constant because the terminals of the capacitor
are open-circuited.
P3.34 Using
WL
A
Crr
00 εεεε
and
KdV
max
to substitute into
P3.35 The capacitance of the microphone is
10101085.8
412
0
A
P3.36 Referring to Figure P3.36 in the book, we see that the transducer
P3.37 With the tank full, we have
25.11000102500 212
1
2
1
mJ
The added energy is supplied from the gravitational potential energy of
P3.38
   
 
 
 
tt
dt
d
dt
tdv
Cti c
c
100sin10100cos1010 47
P3.39 The required volume is
P3.40 Before the switch closes, the energies are
zero. When the switch closes, the charges cancel, the voltage becomes
zero, and the stored energy becomes zero.
P3.41 A fluid-flow analogy for an inductor consists of an incompressible fluid
flowing through a frictionless pipe of constant diameter. The pressure
P3.43*
H 2
L
     
titvtp LL
P3.45*