CHAPTER 3
Exercises
E3.1
v (t ) = q (t ) / C = 10 −6 sin(10 5t ) /(2 × 10 −6 ) = 0.5 sin(10 5t ) V
dv
i (t ) = C
= (2 × 10 −6 )(0.5 × 10 5 ) cos(10 5t ) = 0.1 cos(10 5t ) A
dt
E3.2
Because the capacitor voltage is zero at t = 0, the charge on the
capacitor is zero at t = 0.
t
q (t ) = ∫ i (x )dx + 0
0
t
= ∫ 10 −3 dx = 10 −3t for 0 ≤ t ≤ 2 ms
0
=
2E −3
∫ 10
−3
dx +
0
t
∫ − 10
−3
dx = 4 × 10 -6 − 10 −3t for 2 ms ≤ t ≤ 4 ms
2E −3
v (t ) = q (t ) / C
= 10 4t for 0 ≤ t ≤ 2 ms
= 40 − 10 4t for 2 ms ≤ t ≤ 4 ms
p (t ) = i (t )v (t )
= 10t for 0 ≤ t ≤ 2 ms
= −40 × 10 −3 + 10t for 2 ms ≤ t ≤ 4 ms
w (t ) = Cv 2 (t ) / 2
= 5t 2 for 0 ≤ t ≤ 2 ms
= 0.5 × 10 −7 (40 − 10 4t ) 2 for 2 ms ≤ t ≤ 4 ms
in which the units of charge, electrical potential, power, and energy are
coulombs, volts, watts and joules, respectively. Plots of these quantities
are shown in Figure 3.8 in the book.
E3.3
Refer to Figure 3.10 in the book. Applying KVL, we have
v = v1 + v2 + v3
Then using Equation 3.8 to substitute for the voltages we have
72
v (t ) =
1
C1
t
1
t
∫ i (t )dt + v (0) + C ∫ i (t )dt + v
1
0
2 0
2
(0) +
1
C3
t
∫ i (t )dt + v
This can be written as
t
 1
1
1 
 ∫ i (t )dt + v 1 ( 0 ) + v 2 ( 0 ) + v 3 ( 0 )
v (t ) = 
+
+
 C1 C2 C3 0
Now if we define
 1
1
1
1 
 and v (0) = v 1 (0) + v 2 (0) + v 3 (0)
=  +
+
C eq  C 1 C 2 C 3 
we can write Equation (1) as
t
1
v (t ) =
∫ i (t )dt + v (0)
C eq
3
(0)
0
(1)
0
Thus the three capacitances in series have an equivalent capacitance
given by Equation 3.25 in the book.
E3.4
(a) For series capacitances:
1
1
=
= 2 / 3 µF
C eq =
1 / C1 + 1 / C2 1 / 2 + 1 / 1
(b) For parallel capacitances:
C eq = C 1 + C 2 = 1 + 2 = 3 µF
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:
10 −6 × 15 × 10 −6
Cd
Cd
=
=
= 0.4985 m 2
A=
ε
ε r ε 0 3.4 × 8.85 × 10 −12
Then the length of the strip is
L = A /W = 0.4985 /(2 × 10 −2 ) = 24.93 m
E3.6
v (t ) = L
di (t )
d
[
= (10 × 10 −3 )
0.1 cos(10 4t )] = −10 sin(10 4t ) V
dt
dt
w (t ) = 21 Li 2 (t ) = 5 × 10 −3 × [0.1 cos(10 4t )] = 50 × 10 −6 cos2 (10 4t ) J
2
73
E3.7
1
t
t
1
i (t ) = ∫ v (x )dx + i (0) =
v (x )dx
L0
150 × 10 −6 ∫0
t
= 6667 ∫ 7.5 × 10 6 xdx = 25 × 10 9t 2 V for 0 ≤ t ≤ 2 µs
0
= 6667
2E-6
∫ 7.5 × 10
6
xdx = 0.1 V for 2µs ≤ t ≤ 4 µs
0
2E-6
t


= 6667  ∫ 7.5 × 10 6 xdx + ∫ (− 15)dx  = 0.5 − 10 5t V for 4 µs ≤ t ≤ 5 µs
4 E-6
 0

A plot of i(t) versus t is shown in Figure 3.19b in the book.
E3.8
Refer to Figure 3.20a in the book. Using KVL we can write:
v (t ) = v 1 (t ) + v 2 (t ) + v 3 (t )
Using Equation 3.28 to substitute, this becomes
di (t )
di (t )
di (t )
v (t ) = L1
+ L2
+ L3
(1)
dt
dt
Then if we define Leq = L1 + L2 + L3 , Equation (1) becomes:
dt
v (t ) = Leq
di (t )
dt
which shows that the series combination of the three inductances has
the same terminal equation as the equivalent inductance.
E3.9
Refer to Figure 3.20b in the book. Using KCL we can write:
i (t ) = i1 (t ) + i2 (t ) + i3 (t )
Using Equation 3.32 to substitute, this becomes
t
t
t
1
1
1
i (t ) = ∫ v (t )dt + i1 (0) + ∫ v (t )dt + i2 (0) + ∫ v (t )dt + i3 (0)
L1
0
L2
L3
0
0
This can be written as
t
1
1
1


v (t ) =  + +  ∫ v (t )dt + i1 (0) + i2 (0) + i3 (0)
 L1 L2 L3  0
Now if we define
1
1
1
1
=  + +  and i (0) = i1 (0) + i2 (0) + i3 (0)
Leq  L1 L2 L3 
we can write Equation (1) as
74
(1)
i (t ) =
1
Leq
t
∫ v (t )dt + i (0)
0
Thus, the three inductances in parallel have the equivalent inductance
shown in Figure 3.20b in the book.
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 5H inductance, which in turn is in parallel with the other 5-H inductance.
This combination has an equivalent inductance of 1/(1/5 + 1/5) = 2.5 H.
Finally the 1-H inductance is in series with the combination of the other
inductances so the equivalent inductance is 1 + 2.5 = 3.5 H.
(b) The 2-H and 3-H inductances are in series and have an equivalent
inductance of 5 H. This equivalent inductance is in parallel with both the
5-H and 4-H inductances. The equivalent inductance of the parallel
combination is 1/(1/5 + 1/4 + 1/5) = 1.538 H. This combination is in series
with the 1-H and 6-H inductances so the overall equivalent inductance is
1.538 + 1 + 6 = 8.538 H.
Problems
P3.1
A dielectric material is an electrical insulator through which virtually no
current flows, assuming normal operating voltages. Some examples of
dielectrics mentioned in the text are air, Mylar, polyester, polypropylene,
and mica. Some others are porcelain, glass, and certain types of oil.
P3.2
Charges (usually in the form of electrons) flow in and accumulate on one
plate. Meanwhile, an equal amount of charge flows out of the other plate.
Thus, current seems to flow through a capacitor.
P3.3
Because we have i = Cdv / dt for a capacitance, the current is zero if the
voltage is constant. Thus, we say that capacitances act as open circuits
for constant dc voltages.
P3.4
Capacitors consist of two conductors separated by an insulating material.
Frequently, the conductors are sheets of metal that are separated by a
thin layer of the insulating material.
75
P3.5
The net charge on each plate is Q = CV = (10 × 10 −6 ) × 50 = 500 µC. One
plate has a net positive charge and the other has a net negative charge so
the net charge for both plates is zero.
dv
dt
i
163 × 10 −6
dv
=
=
= 0.01607 V/s
dt C 1014 × 10 −6
∆v
179
∆t =
=
= 1114 s
dv dt 0.01607
P3.6*
i =C
P3.7
Q = Cv = 4 × 10 −6 × 1559 = 6.236 mC
1
1
Cv 2 = × 4 × 10 −6 × (1559)2 = 4.861 J
2
2
∆W
4.861
=
= 540.1 kW
P =
∆t
9 × 10 −6
W =
P3.8*
i (t ) = C
dv
dt
d
(100 sin 1000t )
dt
= cos(1000t )
= 10 −5
p (t ) = v (t )i (t )
= 100 cos(1000t ) sin(1000t )
= 50 sin(2000t )
1
C [v (t )]2
2
= 0.05 sin2 (1000t )
w (t ) =
76
P3.9
i (t ) = C
dv
dt
d
(
100e − 100t )
dt
= −0.01e −100t A
= 10 − 6
p (t ) = v (t )i (t )
= −e −200t W
w (t ) =
1
C [v (t )]2
2
= 5 × 10 −3 × e −200t J
3.10
77
P3.11
v (t ) =
1
C
t
∫ i (t )dt + v (0)
0
t
v (t ) = 2 × 10 6 ∫ i (t )dt
0
p (t ) = v (t )i (t )
1
Cv 2 (t )
2
= 0.25 × 10 −6 × v 2 (t )
w (t ) =
P3.12
Because the switch is closed prior to t = 0 , the initial voltage is zero and
we have
v (t ) =
1
C
t
0
0
v (15 × 10 −3 ) = 2.143 V
p = vi = 0.2858t
p (15 × 10 −3 ) = 4.286 mW
w (t ) =
t
∫ i (t )dt + v (0) = 71.43 × 10 ∫ (2 × 10
3
1
Cv 2 (t ) = 0.1429t 2 J
2
w (15 × 10 −3 ) = 32.16 µJ
78
−3
)dt + 0 = 142.9t
P3.13
v (t ) =
1
t
1
t
∫ i (t )dt + v (0) = C ∫ I m cos(ωt )dt
=
C 0
0
I
= m sin(ωt )
ωC
Clearly for ω → ∞ , the voltage becomes zero, so the capacitance
becomes the equivalent of a short circuit.
P3.14
v (t ) =
1
C
t
∫ i (t )dt + v (0)
0
t
v (t ) = 0.333 × 10 6 ∫ i (t )dt + 10
0
p (t ) = v (t )i (t )
1
Cv 2 (t )
2
= 1.5 × 10 −6 × v 2 (t )
w (t ) =
P3.15
Im
[sin(ωt ) − sin(0)]
ωC
v (t ) =
1
C
t
∫ i (t )dt + v (0)
0
v (t ) = 2 × 10
4
t
∫ 3 × 10
−3
dt − 20
0
v (t ) = 60t − 20 V
79
p (t ) = i (t )v (t )
= 3 × 10 −3 (60t − 20 ) W
Evaluating at t = 0 , we have p (0 ) = −60 mW . Because the power has a
negative value, the capacitor is delivering energy.
At t = 1 s , we have p (1) = 120 mW . Because the power is positive, we
know that the capacitor is absorbing energy.
P3.16
P3.17
We can write
1
w = Cv 2 = 1.2 × 10 −5v 2 = 119
2
Solving, we find v = 3149 V. Because the stored energy is increasing, the
power is positive. Thus, we have
p
362
i = =
= 115.0 mA
v 3149
Because the stored energy, voltage, and stored charge are increasing,
current flows into of the positive terminal of the capacitor.
We can write v (t 0 ) = Vf =
1
C
t0 + ∆t
∫ i (t )dt . The integral represents the area of
t0
the current pulse, which has units of ampere seconds or coulombs and
must equal Vf C . The pulse area represents the net charge transfered by
the current pulse. Because a constant-amplitude pulse has the largest
area for a given peak amplitude, we can say that the peak amplitude of
the current pulse must be at least as large as Vf C / ∆t .
We conclude that the area of the pulse remains constant and that
the peak amplitude approaches infinity as ∆t approaches zero. In the
limit, this type of pulse is called an impulse.
P3.18
P3.19
dq (t ) d
[C (t )v (t )] = d [43 × 10 −6 + 43 × 10 −6 cos(1098t )]
=
dt
dt
dt
i (t ) = −47.21 sin(1098t ) mA
i (t ) =
By definition, the voltage across a short circuit must be zero. Since we
have v = Ri for a resistor, zero resistance corresponds to a short circuit.
80
For an initially uncharged capacitance, we have
t
1
v (t ) = ∫ i (t )dt
C
0
For the voltage to be zero for all values of current and time, the
capacitance must be infinite. Thus, an infinite initially uncharged
capacitance is equivalent to a short circuit.
For an open circuit, the current must be zero. This requires infinite
resistance. However for a capacitance, we have
i (t ) = C
dv (t )
dt
Thus, a capacitance of zero is equivalent to an open circuit.
P3.20
A capacitance initially charged to 10 V has
t
1
v (t ) = ∫ i (t )dt + 10
C
0
However, if the capacitance is infinite, this becomes v (t ) = 10 V which
describes a 10-V voltage source. Thus, a very large capacitance initially
charged to 10 V is an approximate 10-V voltage source.
P3.21*
W = power × time
= 5 hp × 746 W / hp × 3600 s
= 13.4 × 10 6 J
V =
2W
C
2 × 13.4 × 10 6
=
= 51.8 kV
0.01
It turns out that a 0.01-F capacitor rated for this voltage would be much
too large and massive for powering an automobile. Besides, to have
reasonable performance, an automobile would need much more than 5 hp
for an hour.
P3.22
We assume that i (t ) is referenced into the positive reference for v (t ) ,
and we have
i (t ) = C
dv (t )
dt
d
[10 − 14 exp(−4t )]
dt
= 7.56 × 10 −3 exp( −4t ) A
= 135 × 10 −6
81
p (t ) = v (t )i (t )
[
= [10 − 14 exp( −4t )] 7.65 × 10 −3 exp( −4t )
]
Evaluating at t = 0 , we have p (0 ) = −30.24 mW and the capacitor is
delivering energy. At 0.3 s, we find p (0.3) = 13.33 mW. Because the
power is positive, we know that the capacitor is absorbing energy.
P3.23
Capacitances in parallel are combined by adding their values. Thus,
capacitances in parallel are combined as are resistances in series.
Capacitances in series are combined by taking the reciprocal of the sum
of the reciprocals of the individual capacitances. Thus, capacitances in
series are combined as are resistances in parallel.
P3.24*
(a) C eq = 3 +
1
= 4.5 µF
1 3 + 1/3
(b) The two 2- µ F capacitances are in series and have an equivalent
1
= 1 µ F . This combination is a parallel with the 2capacitance of
1 2+1 2
µ F capacitance, giving an equivalent of 3 µ F. Then the 18 µ F is in series,
1
= 2.571 µF . Finally, the 7 µ F is in
giving a capacitance of
1 18 + 1 3
parallel, giving an equivalent capacitance of C eq = 2.571 + 7 = 9.571 µF .
P3.25
(a) C eq = 15.34 µF
(b) C eq = 7.35 µF
82
P3.26
C eq =
P3.27
1
= 4 µF
1 / 5 + 1 /(12 + 8)
We obtain the maximum capacitance of 8 µ F by connecting all four 2- µ F
capacitors in parallel. We obtain the minimum capacitance of 1/2 µ F by
connecting all four 2- µ F capacitors in series.
P3.28
C eq =
1
= 6.667 µF
1 C1 + 1 C2
The charges stored on each capacitor and on the equivalent capacitance
are equal because the current through each is the same.
Q = C eq × 13 V = 86.67 µC
Q
= 5.778 V
C1
Q
v2 =
= 7.222 V
C2
v1 =
As a check, we verify that v 1 + v 2 = 13 V .
P3.29*
As shown below, the two capacitors are placed in series with the heart to
produce the output pulse.
While the capacitors are connected, the average voltage supplied to the
heart is 4.95 V. Thus, the average current is I pulse = 4.95 500 = 9.9 mA .
The charge removed from each capacitor during the pulse is
∆Q = 9.9 mA × 1 ms = 9.9 µC . This results in a 0.l V change in voltage, so
83
∆Q 9.9 × 10 −6
=
= 99 µF . Thus, each capacitor has a
2
0.1
∆v
capacitance of C = 198 µF . Then as shown below, the capacitors are
we have C eq =
C
=
placed in parallel with the 2.5-V battery to recharge them.
The battery must supply 9.9 µ C to each battery. Thus, the average
2 × 9.9 µC
= 19.8 µA . The
current supplied by the battery is I battery =
1s
ampere-hour rating of the battery is
19.8 × 10 −6 × 5 × 365 × 24 = 0.867 Ampere hours .
P3.30
The equivalent capacitance is C eq = 50 µF and its initial voltage is 150 V.
The energies are w 1 =
1
2
CV 1 2 = 21 100 × 10 − 6 × 50 2 = 0 . 125 J and
w 2 = 0.5 J. Thus, the total energy stored in the two capacitors is
w total = 0.625 J.
On the other hand, the energy stored in the equivalent capacitance is
w eq = 21 C eqVeq2 = 21 50 × 10 −6 × 150 2 = 0.5625 J.
The reason for the discrepancy is that all of the energy stored in the
original capacitances cannot be accessed as long as they are connected in
series. Net charge is trapped on the plates that are connected together.
P3.31*
C =
P3.32
C =
εr ε 0A
=
εr ε 0A
=
d
d
15 × 8.85 × 10 −12 × 10 × 10 −2 × 30 × 10 −2
= 0.398 µF
0.01 × 10 −3
εr ε 0WL
d
(a) Thus if W and L are both doubled, the capacitance is increased by a
factor of four resulting in C = 832 pF.
(b) If d is doubled, the capacitance is cut in half resulting in C = 104 pF.
(c) The relative dielectric constant of air is approximately unity. Thus,
replacing air with oil increases εr by a factor of 25 increasing the
capacitance to 5200 pF.
84
P3.33
Using C =
εr ε 0A
d
=
εr ε 0WL
d
and Vmax = Kd to substitute into
1
1 εr ε 0WL 2 2 1
2
CVmax
, we have Wmax =
K d = εr ε 0K 2WLd .
2
2 d
2
However, the volume of the dielectric is Vol = WLd, so we have
1
Wmax = εr ε 0K 2 (Vol)
2
Thus, we conclude that the maximum energy stored is independent of W,
L, and d if the volume is constant and if both W and L are much larger
than d. To achieve large energy storage per unit volume, we should look
for a dielectric having a large value for εr K 2 . The dielectric should have
Wmax =
high relative dielectric constant and high breakdown strength.
P3.34*
The charge Q remains constant because the terminals of the capacitor
are open-circuited.
Q = C 1V1 = 1000 × 10 −12 × 1000 = 1 µC
W1 = (1 2)C 1 (V1 )2 = 500 µJ
After the distance between the plates is doubled, the capacitance
becomes C 2 = 500 pF .
10 −6
Q
=
= 2000 V and the stored
The voltage increases to V2 =
C 2 500 × 10 − 12
energy is W2 = (1 2)C 2 (V2 ) = 1000 µJ . The force needed to pull the plates
2
apart supplies the additional energy.
P3.35
Before the switch closes, the energies are
W1 = (1 2)C 1 (V1 )2 = 65.71 mJ
W2 = (1 2)C 2 (V2 )2 = 65.71 mJ
Thus, the total stored energy is 131.4 mJ. The charge on the top plate of
C 1 is Q1 = C 1V1 = +888 µC . The charge on the top plate of C 2
is Q2 = C 2V2 = −888 µC . Thus, the total charge on the top plates is zero.
When the switch closes, the charges cancel, the voltage becomes zero,
and the stored energy becomes zero.
Where did the energy go? Usually, the resistance of the wires absorbs
it. If the superconductors are used so that the resistance is zero, the
energy can be accounted for by considering the inductance of the circuit.
(It is not possible to have a real circuit that is precisely modeled by
85
Figure P3.35; there is always resistance and inductance associated with
the wires that connect the capacitances.)
P3.36
Refering to Figure P3.36 in the book, we see that the transducer consists
of two capacitors in parallel: one above the surface of the liquid and one
below. Furthermore, the capacitance of each portion is proprotional to
its length and the relative dielectric constant of the material between
the plates. Thus for the portion above the liquid, the capacitance in pF is
100 − x
100
in which x is the height of the liquid in cm. For the portion of the plates
below the surface of the liquid:
C above = 200
C below = 200(25)
x
100
Then the total capacitance is:
C = C above + C below
C = 200 + 48x
P3.37
pF
With the tank full, we have
2
w full = 21 C fullVfull
= 21 × 2000 × 10 −12 × 1000 2 = 1 mJ
Q = C fullVfull = 2000 × 10 −12 × 1000 = 2 µC
The charge cannot change when the tank is drained, so we have
Q
2 × 10 −6
Vempty =
=
= 10 4 V
−12
C empty 200 × 10
2
w empty = 21 C emptyVempty
= 21 × 200 × 10 −12 × 10 8 = 10 mJ
The added energy is supplied from the gravitational potential energy of
the insulating fluid. When there is liquid between the plates, the charge
separation of the dielectric partly cancels the electrical forces of the
charges on the plates. When the dielectric fluid drains, this cancellation
effect is lost, which is why the voltage increases. The charge on the
plates creates a small force of attraction on the fluid, and it is the force
of gravity acting against this force of attraction as the fluid drains that
accounts for the added energy.
86
P3.38
The capacitance of the microphone is
ε A 8.85 × 10 −12 × 1.43 × 10 −2
C = 0 =
d
[1 + 0.01 sin(200t )]10 −3
≅ 12.66 × 10 −11 [1 − 0.01 sin(200t )]
The current flowing through the microphone is
dq (t ) d [Cv ]
i (t ) =
=
≅ −50.6 × 10 −9 cos(200t )
dt
P3.39
dt
A
dv c (t )
d
[10 cos(100t )] = −10 − 4 sin(100t )
= 10 − 7
dt
dt
−3
v r (t ) = Ric (t ) = −10 sin(100t )
v (t ) = v c (t ) + v r (t )
ic (t ) = C
v (t ) = 10 cos(100t ) − 10 − 3 sin(100t )
Thus, v (t ) = v c (t ) to within 1% accuracy, and the resistance can be
neglected.
Repeating for v c (t ) = 0.1 cos(10 7t ) , we find
ic (t ) = −0.1 sin(10 7t )
v r (t ) = − sin(10 7 )
v (t ) = v c (t ) + v r (t ) = 0.1 cos(10 7t ) − sin(10 7t )
Thus, in this case, the voltage across the parasitic resistance is larger in
magnitude than the voltage across the capacitance.
P3.40*
For square plates, we have L = W , the plate area is A = L2 , and the
volume of the dielectric is
Vol = L2d .
The minimum thickness of the dielectric is
V
1000
d min = max =
= 0.3125 mm
K
32 × 10 5
The required volume is
2Wmax
2 × 10 −3
Vol = Ad =
=
= 22.07 × 10 −6 m 3
2
2
−
12
5
ε 0 εr K
8.85 × 10 (32 × 10 )
and the area is
A = Vol d = 0.07062 m 2
The length of each side of the square plate is
L = A = 0.2657 m
87
P3.41
P3.42
P3.43
Inductors consist of coils of wire wound on coil forms such as toriods.
1 2
Li , the energy stored
2
in the inductor increases when the current magnitude increases.
Therefore, energy is flowing into the inductor.
Because the energy stored in an inductor is w =
Because we have v L (t ) = L
diL (t )
, the voltage is zero when the current is
dt
constant. Thus, we say that inductors act as short circuits for steady dc
currents.
P3.44
A fluid flow analogy for an inductor consists of an incompressible fluid
flowing through a frictionless pipe of constant diameter. The pressure
differential between the ends of the pipe is analogous to the voltage
across the inductor and the flow rate of the liquid is proportional to the
current. (If the pipe had friction, the electrical analog would have series
resistance. If the ends of the pipe had different diameters, a pressure
differential would exist for constant flow rate, whereas an inductance
has zero voltage for constant flow rate.)
P3.45*
L = 2H
v L (t ) = L
diL (t )
dt
p (t ) = v L (t )iL (t )
88
w (t ) =
P3.46
1
L[iL (t )] 2
2
L = 0.1 H
iL (t ) = 0.5 sin(1000t ) A
v L (t ) = L
diL (t )
dt
= 50 cos(1000t ) V
p (t ) = v L (t )iL (t )
= 25 cos(1000t ) sin(1000t )
= 12.5 sin(2000t ) W
1
L[iL (t )] 2
2
= 0.0125 sin2 (1000t ) J
w (t ) =
P3.47
L = 2H
iL (t ) = 5e −20t
89
diL (t )
dt
= −200e − 20t V
v L (t ) = L
p (t ) = v L (t )iL (t )
= (− 200e − 20t )(5e − 20t )
= −1000e − 40t W
1
L[iL (t )] 2
2
= 25e − 40t J
w (t ) =
P3.48
L = 2H
iL (t ) =
=
1
t
L ∫0
v L (t )dt + iL (0 )
t
1
v (t )dt
2 ∫0 L
p (t ) = v L (t )iL (t )
90
w (t ) =
1
L[iL (t )]2
2
= [iL (t )] 2
P3.49
L = 10 µH
v L (t ) = 5 sin(10 6t )
iL (t ) =
1
t
L ∫0
v L (t )dt + iL (0 )
t
= 10 5 ∫ 5 sin(10 6t )dt − 0.5
0
= −0.5 cos(10 6t ) A
p (t ) = v L (t )iL (t )
= −2.5 sin(10 6t )cos(10 6t )
= −1.25 sin(2 × 10 6t )
1
L[iL (t )]2
2
= 1.25 cos2 (10 6t ) µJ
w (t ) =
P3.50
Using either tables or integration by parts, we have
1
t
1
t
v (t )dt + i L (0 ) = ∫t exp( −t )dt
L∫ L
L
i L (t ) = [1 − t exp( −t ) − exp( −t )]
i L (t ) =
0
0
=
1
L
[− exp(−t )(t + 1)]t0
1
2
A sequence of MATLAB commands to obtain the desired plots is
L = 2;
t = 0:0.01:10;
v = t.*exp(-t);
91
i = (1/2)*(1 -t.*exp(-t) - exp(-t));
plot(t, v)
hold
plot(t, i, ':')
The resulting plot is:
i (t )
v (t )
t
P3.51*
i L (t ) =
1
L
t
∫ v L (t )dt + iL (0)
0
= 16.39 × 10
3
t
∫ 28dt − 0.164
0
= 4.590 × 10 t − 0.164 A
5
Solving for the time that the current reaches + 164 mA , we have
i L (t x ) = 0.164 = 4.590 × 10 5t 0 − 0.164
t x = 0.7146 µs
P3.52*
vL = L
8
di
= 0.6
= 12 V
0. 4
dt
92
P3.53
vL = L
di
dt
d
[exp(−4t ) sin(19t )]
dt
= 5.225 exp( −4t ) cos(19t ) − 1.1 exp( −4t ) sin(19t )
= 0.275
A sequence of MATLAB commands that produces the desired plots is
t = 0:0.01:3;
v = exp(-4*t).*sin(19*t);
i = 5.225*exp(-4*t).*cos(19*t) - 1.1*exp(-4*t).*sin(19*t);
plot(t, v)
hold
plot(t, i)
The resulting plot is:
v (t )
i (t )
93
P3.54
P3.55
i L (t 0 ) =
1
L
t0
∫v L (t )dt + iL (0 )
0
i L ( 4) =
p (4 ) = v L (4 )i L (4 ) = 21.33 W
P3.56
We can write i L (t 0 + ∆t ) = If =
1
3
4
∫ 4dt + 0 = 5.333 A
0
w (4) = Li L2 (4) = 42.67 J
1
2
1
L
t0 + ∆t
∫v L (t )dt . The integral represents the
t0
area of the voltage pulse, which has units of volt seconds and must equal
If L . Because a constant-amplitude pulse has the largest area for a given
peak amplitude, we can say that the peak amplitude of the voltage pulse
must be at least as large as If L / ∆t .
We conclude that the area of the pulse remains constant and that
the peak amplitude approaches infinity as ∆t approaches zero. In the
limit, this type of pulse is called an impulse.
P3.57
w (3) = 21 Li L2 (3) = 295 J ⇒ i L (3) = 12.14 A
Since a reference is not specified, we can choose i L (3) = +12.14 A. Also,
because the stored energy is increasing, the power for the inductor
carries a plus sign. Thus p (3) = +162 = v L (3)i L (3) and we have
v L (3) = +13.34 V. Finally, because the current and voltage have the same
algebraic signs, the current flows into the positive polarity.
94
P3.58
Because we the current through an open circuit is zero by definition and
t
1
we have i (t ) = ∫ v (t )dt (assuming zero initial current), infinite
Lt
0
inductance corresponds to an open circuit.
Because the voltage across a short circuit is zero by definition and
di (t )
v L = L L , we see that L = 0 corresponds to a short circuit.
dt
P3.59
For an inductor with an initial current of 10 A, we have
t
1
i (t ) = ∫ v (t )dt + 10 . However for an infinite inductance. this becomes
Lt
0
i (t ) = +10 which is the specification for a 10-A current source. Thus, a
very large inductance with an initial current of 10 A is an approximation
to a 10-A current source.
P3.60
i (t ) =
1
t
∫v (t )dt + i (0) =
1
t
∫Vm cos(ωt )dt =
Vm
[sin(ωt ) − sin(0)]
ωL
L0
L0
V
= m sin(ωt )
ωL
Clearly for ω → ∞ , the current becomes zero, so the inductance becomes
the equivalent of an open circuit.
P3.61
Inductances are combined in the same way as resistances. Inductances in
series are added. Inductances in parallel are combined by taking the
reciprocal of the sum of the reciprocals of the several inductances.
P3.62*
(a) Leq = 2 +
1
= 5.077 H
1 5 + 1 (4 + 4 )
(b) 7 H in parallel with 28 H is equivalent to 5.6 H. Also, 29 H in parallel
with 5 H is equivalent to 4.265 H. Finally, we have
1
Leq =
= 6.242 H
1 17 + 1 (5.6 + 4.265)
95
P3.63
(a) The 1-H, 6-H and 0.6-H inductors have no effect because they are in
parallel with a short circuit. Thus, Leq = 8 H .
(b) Leq = 5.063 H .
P3.64
If all four inductors are connected in series, we obtain the maximum
inductance:
Lmax = 4 H
By connecting all four inductors in parallel, we obtain the minimum
inductance:
1
Lmin =
= 0.25 H
1 1+1 1+1 1+1 1
P3.65
Ordinarily negative inductance is not practical. Thus, adding inductance
in series always increases the equivalent inductance and placing
inductance in parallel results in smaller inductance. Thus, we need to
consider a parallel inductance such that
1
=2
1/ L + 1/6
Solving, we find that L = 3 H.
P3.66
In this case we need to place 2 H in series with the original 6-H
inductance.
P3.67*
L1 + L2 t
i (t ) =
v (t )dt =
v (t )dt
Leq ∫0
L1 L2 ∫0
1
i1 (t ) =
1
L1
t
t
∫ v (t )dt
0
Thus, we can write i1 (t ) =
L2
i (t ) =
2
i (t ) .
3
L1 + L2
L1
1
Similarly, we have i2 (t ) =
i (t ) = i (t ) .
3
L1 + L2
This is similar to the current-division principle for resistances. Keep in
mind that these formulas assume that the initial currents are zero.
96
P3.68
(a) v R (t ) = Ri (t ) = 0.1 cos (105 )
di (t )
= −100 sin (10 5 )
dt
v (t ) = v R (t ) + v L (t )
v L (t ) = L
= 0.1 cos (105t ) − 100 sin(105t )
In this case to obtain 1% accuracy, the resistance can be neglected.
(b) For i (t ) = 0.1 cos(10t ) , we have
v R (t ) = 0.1 cos(10t )
v L (t ) = −0.01 sin(10t )
v (t ) = 0.1 cos(10t ) − 0.01 sin(10t )
Thus in this case, the parasitic resistance cannot be neglected.
P3.69
See Figure 3.22 in the book.
di L (t )
= 0 for currents that are constant in time, we
dt
P3.70
Because v L = L
P3.71
conclude that the inductance behaves as a short-circuit for dc currents.
Thus, the circuit simplifies the single resistance Rs , which is given
500 mV
by Rs =
= 5 Ω.
100 mA
dv (t )
i (t ) = C C
= 500 × 10 −6 × 10 × 1000 cos(1000t ) = 5 cos(1000t ) A
dt
di (t )
v L (t ) = L
= −10 sin(1000t ) V
dt
v (t ) = v C (t ) + v L (t ) = 0
w C (t ) = 21 Cv C2 (t ) = 25 sin 2 (1000t ) = 12.5 − 12.5 cos(2000t ) mJ
w L (t ) = 21 Li 2 (t ) = 25 cos 2 (1000t ) = 12.5 + 12.5 cos(2000t ) mJ
w (t ) = w C (t ) + w L (t ) = 25 mJ
The values in this circuit have been carefully selected so the source
voltage is zero. Because the source voltage is zero and there are no
resistances, there is no source or sink for energy in the circuit. Thus, we
would expect the total energy to be constant, as the equations show. The
total energy surges back and forth between the capacitance and the
inductance. In a real circuit, the parasitic resistances would eventually
97
absorb the energy. The circuit is analogous to a swinging pendulum or a
ringing bell.
P3.72
di L (t )
= 4 × 10 −3 × 0.1 × 5000[− sin(5000t )] = −2 sin(5000t ) V
dt
dv (t )
iC (t ) = C
= −0.1 cos(5000t ) A
dt
i (t ) = iC (t ) + i L (t ) = 0
v (t ) = L
w C (t ) = 21 Cv 2 (t ) = 20 sin 2 (5000t ) = 10 − 10 cos(2000t ) µJ
w L (t ) = 21 LiL2 (t ) = 20 cos 2 (5000t ) = 10 + 10 cos(2000t ) µJ
w (t ) = w C (t ) + w L (t ) = 20 µJ
The values in this circuit have been carefully selected so the source
current is zero. Because the source current is zero and there are no
resistances, there is no source or sink for energy in the circuit. Thus, we
would expect the total energy to be constant, as the equations show. The
total energy surges back and forth between the capacitance and the
inductance. In a real circuit, the parasitic resistances would eventually
absorb the energy. The circuit is analogous to a swinging pendulum or a
ringing bell.
P3.73
When a time-varying current flows in a coil, a time-varying magnetic
field is produced. If some of this field links a second coil, voltage is
induced in it. Thus time-varying current in one coil results in a
contribution to the voltage across a second coil.
P3.74
Refer to Figures 3.23 and P3.74. For the dots as shown in Figure P3.74,
we have
di1 (t )
di (t )
+M 2
= 15 cos(10t )
dt
dt
di (t )
di (t )
v 2 (t ) = M 1 + L2 2
= 20 cos(10t )
dt
dt
v 1 (t ) = L1
P3.75*
With the dot moved to the bottom of L2 , we have
di1 (t )
di (t )
−M 2
= 5 cos(10t )
dt
dt
di (t )
di (t )
v 2 (t ) = −M 1 + L2 2
=0
dt
dt
v 1 (t ) = L1
98
P3.76*
(a)
As in Figure 3.23a, we can write
di1 (t )
di (t )
+M 2
dt
dt
di (t )
di (t )
v 2 (t ) = M 1 + L2 2
dt
dt
However, for the circuit at hand, we have i (t ) = i1 (t ) = i2 (t ) .
v 1 (t ) = L1
Thus,
di (t )
dt
di (t )
v 2 (t ) = (L2 + M )
dt
Also, we have v (t ) = v 1 (t ) + v 2 (t ) .
di (t )
Substituting, we obtain v (t ) = (L1 + 2M + L2 )
.
dt
di (t )
, in which
Thus, we can write v (t ) = Leq
dt
Leq = L1 + 2M + L2 .
v 1 (t ) = (L1 + M )
(b) Similarly, for the dot at the bottom end of L2 , we have
Leq = L1 − 2M + L2
P3.77
Because of the parallel connection, we have v 1 (t ) = v 2 (t ) = v (t ) and the
equations for the mutually coupled inductors become
di1 (t )
di (t )
+M 2
dt
dt
di (t )
di (t )
v (t ) = M 1 + L2 2
dt
dt
v (t ) = L1
Using determinants to solve for the derivatives, we have
99
v (t )
di1 (t ) v (t )
=
L1
dt
M
M
L2
L −M
= 2
v (t )
M
L1 L2 − M 2
L2
L1 v (t )
di2 (t ) M v (t )
L −M
=
= 1
v (t )
L1 M
dt
L1L2 − M 2
M L2
Then, we have
i (t ) = i1 (t ) + i2 (t )
di (t ) di1 (t ) di1 (t ) L1 + L2 − 2M
=
+
=
v (t )
dt
dt
dt
L1L2 − M 2
from which we conclude that
LL − M 2
Leq = 1 2
L1 + L2 − 2M
P3.78
With a short circuit across the terminals of the second coil, we have
di1 (t )
di (t )
+M 2
dt
dt
di (t )
di (t )
v 2 (t ) = 0 = M 1 + L2 2
dt
dt
Solving the second equation for di2 (t ) / dt and substituting into the first
v 1 (t ) = L1
equation, we have
di1 (t ) M 2 di1 (t ) L1 L2 − M 2 di1 (t )
v 1 (t ) = L1
−
=
dt
L2 dt
L2
dt
from which we can see that
LL − M2
Leq = 1 2
L2
P3.79
In general, we have
di
di1
±M 2
dt
dt
di
di
v 2 (t ) = ±M 1 + L2 2
dt
dt
v 1 (t ) = L1
Substituting the given information, we have
v 1 (t ) = −2 × 10 4 sin(1000t )
10 4 sin(1000t ) = mM 10 4 sin(1000t )
We deduce that M = 1 H. Furthermore, because the lower of the two
algebraic signs applies, we know that the currents are referenced into
unlike terminals.
100