R. Allan Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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
Because the capacitor voltage is zero at t = 0, the charge on the capacitor is zero at t = 0.
E3.2
q (t ) = i (x )dx + 0
0
t
= 10 -3 dx = 10 -3t for 0 t 2 ms
0
t
=
2E -3 0
10
-3
dx +
t
2E -3
- 10
-3
dx = 4 10 -6 - 10 -3t for 2 ms t 4 ms
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
61
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v (t ) =
C1
1
t
i (t )dt + v (0) + C i (t )dt + v
1 0 2 0
1
t
2
(0) +
C3
1
t
i (t )dt + v
0
3
(0)
This can be written as t 1 1 1 v (t ) = C + C + C i (t )dt + v 1 ( 0 ) + v 2 ( 0 ) + v 3 ( 0 ) 2 3 0 1 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)
(1)
C eq
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 F C eq = 1 / C1 + 1 / C2 1 / 2 + 1 / 1
(b) For parallel capacitances: C eq = C 1 + C 2 = 1 + 2 = 3 F
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= r 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 [0.1 cos(10 4t )] = -10 sin(10 4t ) V = (10 10 -3 ) dt dt
2
1 w (t ) = 2 Li 2 (t ) = 5 10 -3 [0.1 cos(10 4t )] = 50 10 -6 cos2 (10 4t ) J
62
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E3.7
1 i (t ) = v (x )dx + i (0) = v (x )dx L0 150 10 -6 0 = 6667 7.5 10 6 xdx = 25 10 9t 2 V for 0 t 2 s
0 2E-6 0
1
t
t
t
= 6667
7.5 10
2E-6
6
xdx = 0.1 V for 2s t 4 s
t = 6667 7.5 10 6 xdx + (- 15)dx = 0.5 - 10 5t V for 4 s t 5 s 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
0
L3
0
This can be written as t 1 1 1 + v (t )dt + i1 (0) + i2 (0) + i3 (0) v (t ) = + 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
(1)
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Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
i (t ) =
Leq
1
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
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. 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. 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.
P3.2
P3.3
64
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P3.4
The net charge on each plate is Q = CV = (100 10 -6 ) 200 = 0.02 C. One plate has a net positive charge and the other has a net negative charge so the net charge for both plates is zero.
P3.5*
i =C
dv dt dv i 100 10 -6 = = = 0.05 V/s dt C 2000 10 -6 v 100 t = = = 2000 s dv dt 0.05
P3.6
Q = Cv = 5 10 -6 10 3 = 5 mC W =
2 1 1 Cv 2 = 5 10 -6 (103 ) = 2.5 J 2 2 2. 5 W = -6 = 2.5 MW P = t 10
P3.7*
i (t ) = C
d (100 sin 1000t ) dt = cos(1000t )
= 10 -5
dv dt
p (t ) = v (t )i (t )
= 100 cos(1000t ) sin(1000t ) = 50 sin(2000t )
w (t ) =
1 C [v (t )]2 2 = 0.05 sin2 (1000t )
65
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P3.8
i (t ) = C
= 10 - 6
d (100e -100t ) dt = -0.01e -100t A
dv dt
p (t ) = v (t )i (t )
= -e -200t W
w (t ) =
1 C [v (t )]2 2
= 5 10 -3 e -200t J
P3.9
66
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P3.10
v (t ) =
C
1
t
i (t )dt + v (0)
0
v (t ) = 2 10 6 i (t )dt
0
t
p (t ) = v (t )i (t )
w (t ) =
1 Cv 2 (t ) 2 = 0.25 10 -6 v 2 (t )
P3.11
v (t ) =
C
1
t
i (t )dt + v (0)
0
v (t ) = 0.333 10 6 i (t )dt + 10
0
t
p (t ) = v (t )i (t )
67
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w (t ) =
1 Cv 2 (t ) 2 = 1.5 10 -6 v 2 (t )
P3.12
v (t ) =
C
1
t
i (t )dt + v (0) = 20 10 (5 10
3 0 0
t
-3
)dt + 0 = 100t
v (20 10 -3 ) = 2 V p = vi = 0.5t p (20 10 -3 ) = 10 mW
1 Cv 2 (t ) = 0.25t 2 2 w (20 10 -3 ) = 100 J
w (t ) =
P3.13
We can write 1 w = Cv 2 = 5 10 -6v 2 = 200 2 Solving, we find v = 6325 V. Then because the stored energy is decreasing, the power is negative. Thus, we have p - 500 i = = = - 79.06 mA 6325 v The minus sign shows that the current actually flows opposite to the passive convention. Thus, current flows out of the positive terminal of the capacitor.
P3.14
dq (t ) d [C (t )v (t )] = d [200 10 -6 + 100 10 -6 sin(200t )] = dt dt dt i (t ) = 20 cos(200t ) mA i (t ) =
By definition, the voltage across a short circuit must be zero. For an initially uncharged capacitance, we have t 1 v (t ) = i (t )dt
P3.15
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. 68
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For an open circuit, the current must be zero. For a capacitance, we have
i (t ) = C
Thus, a capacitance of zero is equivalent to an open circuit.
dv (t ) dt
P3.16
A capacitance initially charged to 10 V has t 1 v (t ) = i (t )dt + 10
C
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.
P3.17*
0
W = power time
= 13.4 10 J
6
= 5 hp 746 W / hp 3600 s
V =
2 W
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.
P3.18
i (t ) = C
d [3 cos(105t ) + 2 sin(105t )] dt (105t ) + 4 cos(105t ) = -6 sin
= 20 10 -6
dv (t ) dt
p (t ) = v (t )i (t )
= 3 cos(10 5t ) + 2 sin(10 5t ) - 6 sin(10 5t ) + 4 cos(10 5t )
[
][
]
Evaluating at t = 0 , we have p (0 ) = 12 W . Because p (0 ) is positive, we know that the capacitor is absorbing energy at t = 0 . t Evaluating at t2 = 2 10 -5 , we have p (t2 ) = -12 W . Because p ( 2 ) is
negative, we know that the capacitor is supplying energy at t = t2 .
69
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P3.19
v (t ) =
C
1
t
i (t )dt + v (0)
0
v (t ) = 10 4 5 10 - 3 dt - 20
v (t ) = 50t - 20 V p (t ) = i (t )v (t )
0
t
= 5 10 -3 (50t - 20 ) W
Evaluating at t = 0 , we have p (0 ) = -0.1 W . Because the power has a negative value, the capacitor is delivering energy. At t = 1 s , we have p (1) = 0.15 W . Because the power is positive, we know that the capacitor is absorbing energy.
P3.20
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.
P3.21*
(a) C eq = 1 +
1 = 2 F 1 2+1 2
(b) The two 4- F capacitances are in series and have an equivalent 1 capacitance of = 2 F . This combination is a parallel with the 1 4+1 4 2- F capacitance, giving an equivalent of 4 F. Then the 6 F is in 1 = 2.4 F . Finally, the 3 F is in series, giving a capacitance of 1 6+1 4 parallel, giving an equivalent capacitance of C eq = 3 + 2.4 = 5.4 F .
P3.22
(a) C eq = 3 + (b) C eq
1 1 + = 4.667 F 1 2 + 1 1 1 2 + 1 (1 + 1) 1 = = 2 F 1 (2 + 1) + 1 (4 + 2)
70
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P3.23
C eq =
P3.24
1 = 1.5 F 1 / 3 + 1 /(1 + 2)
We obtain the maximum capacitance of 4 F by connecting all four 1- F capacitors in parallel. We obtain the minimum capacitance of 1/4 F by connecting all four 1- F capacitors in series.
P3.25
C eq =
1 = 2 3 F 1 C1 + 1 C2
The charges stored on each capacitor and on the equivalent capacitance are equal. Q = C eq 12 V = 8 C
As a check, we verify that v 1 + v 2 = 12 V .
P3.26*
Q = 8V C1 Q v2 = = 4V C2 v1 =
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 C . This results in a 0.l V change in voltage, so we have C eq =
C
2
=
Q 9.9 10 -6 = = 99 F . Thus, each capacitor has a 0.1 v
71
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capacitance of C = 198 F . Then as shown below, the capacitors are placed in parallel with the 2.5 V battery to recharge them.
The battery must supply 9.9 C to each battery. Thus, the average 2 9.9 C = 19.8 A . 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 .
P3.27
C =
r 0A
(a) Thus if W and L are both doubled, the capacitance is increased by a factor of four resulting in C = 400 pF. (b) If d is doubled the capacitance is cut in half resulting in C = 50 pF. (c) The relative dielectric constant of air is approximately unity. Thus replacing with air oil increases r by a factor of 25 increasing the capacitance to 2500 pF.
P3.28*
d
=
r 0WL
d
C =
r 0A
d
=
15 8.85 10 -12 10 10 -2 30 10 -2 = 0.398 F 0.01 10 -3
=
P3.29
Using C =
r 0A
Wmax =
1 1 r 0WL 2 2 1 2 CVmax , we have Wmax = K d = r 0K 2WLd . 2 2 d 2 However, the volume of the dielectric is Vol = WLd, so we have 1 Wmax = r 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 r K 2 . The dielectric should have high relative dielectric constant and high breakdown strength.
d
r 0WL
d
and Vmax = Kd to substitute into
72
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P3.30*
The charge Q remains constant because the terminals of the capacitor are open-circuited. Q = C 1V1 = 1000 10 -12 1000 = 1 C
W1 = (1 2)C 1 ( 1 )2 = 500 J V
After the distance between the plates is doubled, the capacitance becomes C 2 = 500 pF . The voltage increases to V2 =
2
V energy is W2 = (1 2)C 2 ( 2 ) = 1000 J . The additional energy is supplied
by the force needed to pull the plates apart.
P3.31
10 -6 Q = = 2000 V and the stored C 2 500 10 - 12
Before the switch closes, the energies are W1 = (1 2)C 1 ( 1 )2 = 5000 J V Thus, the total stored energy is 10,000 J. The charge on the top plate is Q2 = C 2 2 = -100 C . Thus, the total charge on the top plates is zero. V 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 Figure P3.31; there is always resistance and inductance associated with the wires that connect the capacitances.) of C 1 is Q1 = C 1V1 = +100 C . The charge on the top plate of C 2
W2 = (1 2)C 2 ( 2 )2 = 5000 J V
P3.32
Refering to Figure P3.32 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
C above = 200
100 - x 100
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in which x is the height of the liquid in cm. For the portion of the plates below the surface of the liquid: 100 Then the total capacitance is: C = C above + C below
C below = 200(25)
x
C = 200 + 48x
pF
P3.33
The capacitance of the microphone is A 8.85 10 -12 10 -2 C = 0 = d [1 + 0.01 sin(200t )]10 -3 8.85 10 -11 [1 - 0.01 sin(200t )] The current flowing through the microphone is dq (t ) d [Cv ] i (t ) = = -35.4 10 -9 cos(200t )
A
dt
dt
P3.34
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
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 r (t ) = - sin(10 7 )
ic (t ) = -0.1 sin(10 7t )
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.
P3.35*
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 74
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The required volume is 2 max W 2 10 -3 Vol = Ad = = = 22.07 10 -6 m 3 0 r K 2 8.85 10 -12 (32 10 5 )2 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
P3.36 P3.37
Inductors consist of coils of wire wound on coil forms such as toriods. 1 2 Li , the energy stored 2 in the inductor decreases when the current magnitude decreases. Therefore, energy is flowing out of the inductor. Because the energy stored in an inductor is w = Because we have v L (t ) = L
P3.38
constant. Thus we say that inductors act as short circuits for steady dc currents.
P3.39
diL (t ) , the voltage is zero when the current is dt
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.)
P3.40*
L = 2H
75
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v L (t ) = L
diL (t ) dt
p (t ) = v L (t )iL (t )
w (t ) =
1 L[iL (t )] 2 2
P3.41
L = 0.1 H iL (t ) = 0.5 sin(1000t ) A
v L (t ) = L
= 50 cos(1000t ) V
diL (t ) dt
p (t ) = v L (t )iL (t )
= 25 cos(1000t ) sin(1000t ) = 12.5 sin(2000t ) W
76
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w (t ) =
1 L[iL (t )] 2 2 = 0.0125 sin2 (1000t ) J
P3.42
L = 2H iL (t ) = 5e -20t
v L (t ) = L
diL (t ) dt = -200e - 20t V
p (t ) = v L (t )iL (t )
= (- 200e - 20t )(5e - 20t ) = -1000e - 40t W
w (t ) =
1 L[i (t )] 2 2 L = 25e - 40t J
P3.43
L = 2H
77
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iL (t ) =
=
L 0
t
1
t
v L (t )dt + iL (0 )
1 v (t )dt 2 L 0
p (t ) = v L (t )iL (t )
w (t ) =
= [iL (t )] 2
1 L[iL (t )]2 2
P3.44
L = 10 H
v L (t ) = 5 sin(10 6t )
iL (t ) =
L 0
1
t
v L (t )dt + iL (0 )
t
= 10 5 5 sin(10 6t )dt - 0.5 = -0.5 cos(10 6t ) A
0
p (t ) = v L (t )iL (t )
= -2.5 sin(10 6t )cos(10 6t ) = -1.25 sin(2 10 6t )
78
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w (t ) =
1 L[iL (t )]2 2 = 1.25 cos2 (10 6t ) J
P3.45*
iL (t ) =
L 0
1
t
v L (t )dt + iL (0 )
t
= 20 10 3 10dt - 0.1
0
= 2 10 t - 0.1 A
5
Solving for the time that the current reaches + 100 mA , we have
iL (tx ) = 0.1 = 2 10 5t0 - 0.1 tx = 1 s
P3.46* P3.47
vL = L
di 4 = 0.5 = 10 V dt 0. 2
79
Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
P3.48
i L (t 0 ) =
L
1
t0
0
v L (t )dt + iL (0 )
iL (2) =
1 3 1 2
5dt + 0 = 3.333 A
0
2
p (2) = v L (2)iL (2) = 16.67 W
P3.49
w (2) = LiL2 (2) = 16.67 J
1 w (5) = 2 LiL2 (5) = 200 J i L (5) = 14.14 A
Since a reference is not specified, we can choose i L (5) = +14.14 A. Also, because the stored energy is increasing, the power for the inductor carries a plus sign. Thus p (5) = +100 = v L (5)i L (5) and we have
v L (5) = +7.071 V. Finally, because the current and voltage have the same
algebraic signs, the current flows into the positive polarity.
P3.50
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
P3.51
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.
P3.52
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. (a) Leq = 1 + 1 = 2.333 H 1 2 + 1 (1 + 3)
P3.53*
(b) 2 H in parallel with 2 H is equivalent to 1H. Also, 2 H in parallel with 6 1 H is equivalent to 1.5 H. Finally, we have Leq = = 1.538 H 1 4 + 1 (1 + 1.5)
80
Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
P3.54
(a) The 2 H inductors and 0.5 H inductor have no effect because they are in parallel with a short circuit. Thus, Leq = 1 H . (b) The two 2-H inductances in parallel are equivalent to 1 H. Also, the 1 H in parallel with the 3 H inductance is equivalent to 0.75 H. Thus, 1 Leq = 1 + = 2.158 H . 1 (1 + 1) + 1 (2 + 0.75)
P3.55
If all four inductors are connected in series, we obtain the maximum inductance: Lmax = 8 H By connecting all four inductors in parallel, we obtain the minimum inductance: 1 Lmin = = 0. 5 H 1 2+1 2+1 2+1 2
P3.56
Ordinarilly 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 + 1/5 Solving we find that L = 3.333 H. In this case we need to place 3 H in series with the original 5-H inductance.
P3.57
P3.58*
i (t ) =
Leq
1
t
v (t )dt =
0
L1 + L2 t v (t )dt L1 L2 0
i1 (t ) =
L1
1
t
v (t )dt
0
Thus, we can write i1 (t ) =
L1 + L2 L1 1 Similarly, we have i2 (t ) = i (t ) = i (t ) . 3 L1 + L2
L2
i (t ) =
2 i (t ) . 3
This is similar to the current-division principle for resistances. Keep in mind that these formulas assume that the initial currents are zero.
81
Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
P3.59
(a) v R (t ) = Ri (t ) = 0.1 cos (105 )
v L (t ) = L
di (t ) = -100 sin (10 5 ) dt v (t ) = v R (t ) + v L (t )
= 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.
P3.60 P3.61
See Figure 3.22 in the book. 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. Refer to Figures 3.23 and P3.62. For the dots as shown in Figure P3.62, we have
P3.62
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
P3.63*
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
82
Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
P3.64*
(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 v 1 (t ) = (L1 + M )
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 .
(b) Similarly, for the dot at the bottom end of L2 , we have
Leq = L1 - 2M + L2
P3.65
In general, we have
Substituting the given information, we have
di di1 M 2 dt dt di di v 2 (t ) = M 1 + L2 2 dt dt v 1 (t ) = L1
83
Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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.
84