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Tipologia: Exercícios
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Engineering Circuit Analysis, 7 Edition Chapter Two Solutions 10 March 2006
(b) 750 mJ (e) 6.5 nm (h) 49 kΩ
(c) 1.13 kΩ (f) 13.56 MHz (i) 11.73 pA
PROPRIETARY MATERIAL. © 2007 Th e McGraw-Hill Companies, Inc. Lim ited distr ibution perm itted onl y to
Engineering Circuit Analysis, 7 Edition Chapter Two Solutions 10 March 2006
(b) 12.35 mm (f) 5.33 nW
(c) 47. kW (g) 1 ns
(d) 5.46 mA (h) 5.555 MW
PROPRIETARY MATERIAL. © 2007 Th e McGraw-Hill Companies, Inc. Lim ited distr ibution perm itted onl y to
Engineering Circuit Analysis, 7 Edition Chapter Two Solutions 10 March 2006
3 hrs running at this power level equates to a transfer of energy equal to
(1.5 J/s)(3 hr)(60 min/ hr)(60 s/ min) = 16.2 kJ
PROPRIETARY MATERIAL. © 2007 Th e McGraw-Hill Companies, Inc. Lim ited distr ibution perm itted onl y to
Engineering Circuit Analysis, 7 Edition Chapter Two Solutions 10 March 2006
(a) With 100% efficient mechanical to electrical power conversion,
(175 Hp)[1 W/ (1/745.7 Hp)] = 130.5 kW
(b) Running for 3 hours,
Energy = (130.5× 10
3 W)(3 hr)(60 min/hr)(60 s/min) = 1.409 GJ
(c) A single battery has 430 kW-hr capacity. We require
(130.5 kW)(3 hr) = 391.5 kW-hr therefore one battery is sufficient.
PROPRIETARY MATERIAL. © 2007 Th e McGraw-Hill Companies, Inc. Lim ited distr ibution perm itted onl y to
Engineering Circuit Analysis, 7 Edition Chapter Two Solutions 10 March 2006
(a) To compute the peak power, we assume the pulse shape is square:
1
Energy (mJ)
t (fs) 75
Then P = 1× 10
(b) At 100 pulses per second, the average power is
P avg = (100 pulses)(1 mJ/pulse)/(1 s) = 100 mW.
PROPRIETARY MATERIAL. © 2007 Th e McGraw-Hill Companies, Inc. Lim ited distr ibution perm itted onl y to
Engineering Circuit Analysis, 7 Edition Chapter Two Solutions 10 March 2006
5 7 17 24
P (W)
10
6
t (min)
(a) Total energy (in J) expended is
[6(5) + 0(2) + 0.5(10)(10) + 0.5(10)(7)]60 = 6.9 kJ.
(b) The average power in Btu/hr is
(6900 J/24 min)(60 min/1 hr)(1 Btu/1055 J) = 16.35 Btu/hr.
PROPRIETARY MATERIAL. © 2007 Th e McGraw-Hill Companies, Inc. Lim ited distr ibution perm itted onl y to
Engineering Circuit Analysis, 7 Edition Chapter Two Solutions 10 March 2006
2
4 C.
(a) q (2 s) = 40 C.
(b) To find the m aximum charge within 0≤ t ≤ 3 s, we need to take the f irst and
second derivitives:
dq / dt = 36 t – 8 t
3 = 0, leading to roots at 0, ± 2.121 s
d
2 q / dt
2 = 36 – 24 t
2
substituting t = 2.121 s into the expression for d
2 q / dt
2 , we obtain a value of
–14.9, so that this root represents a maximum.
Thus, we find a maximum charge q = 40.5 C at t = 2.121 s.
(c) The rate of charge accumulation at t = 8 s is
dq/dt |t = 0.8 = 36(0.8) – 8(0.8)
3 = 24.7 C/s.
(d) See Fig. (a) and (b).
0 0.5 1 1.5 2 2.5 3
0
10
20
30
40
50
60
70
tim e (s )
q (C)
0 0.5 1 1.5 2 2.5 3
0
50
i (A )
tim e (t)
(a) (b)
PROPRIETARY MATERIAL. © 2007 Th e McGraw-Hill Companies, Inc. Lim ited distr ibution perm itted onl y to
Engineering Circuit Analysis, 7 Edition Chapter Two Solutions 10 March 2006
3
5
1 ⎩
−
e t
e t i t t
t
Thus,
(a) i 1 (-0.2) = 6.155 A
(b) i 1 (0.2) = 3.466 A
(c) To determine the instants at which i 1 = 0, we must consider t < 0 and t > 0 separately:
for t < 0, - 2 + 3 e
-5 t = 0 leads to t = -0.2 ln (2/3) = +0.0811 s (impossible)
for t > 0, -2 + 3 e
3 t = 0 leads to t = (1/3) ln (2/3) = –0.135 s (impossible)
Therefore, the current is never negative.
(d) The total charge passed left to right in the interval –0. 8 < t < 0.1 s is
q ( t ) =
1
i ( ) t dt ∫ −
0 0. 5 3 0.8 0
t t e dt e d
− −
∫ ⎣ ⎦ ∫ ⎣
⎤ t ⎦
= (^) ( )
0
5 3
0 -0.
t t t e t e
PROPRIETARY MATERIAL. © 2007 Th e McGraw-Hill Companies, Inc. Lim ited distr ibution perm itted onl y to
Engineering Circuit Analysis, 7 Edition Chapter Two Solutions 10 March 2006
2 pJ
(b) VED =
(c) VDC = –
3 pJ
(d) It takes – 3 pJ to move +1.602x
It takes 2 pJ to move –1.602x
+1.602x
Thus, it requires –3 pJ + 2 pJ = –1 pJ to m ove +1.602x
B.
Hence, (^) DB = V
1 pJ
PROPRIETARY MATERIAL. © 2007 Th e McGraw-Hill Companies, Inc. Lim ited distr ibution perm itted onl y to
Engineering Circuit Analysis, 7 Edition Chapter Two Solutions 10 March 2006
Voltmeter
Voltmeter
From the diagram, we see that V 2 = –V 1 = +2.86 V.
PROPRIETARY MATERIAL. © 2007 Th e McGraw-Hill Companies, Inc. Lim ited distr ibution perm itted onl y to
Engineering Circuit Analysis, 7 Edition Chapter Two Solutions 10 March 2006
-100 t mA and v = [6 – 600 t ] e
-100 t mV
(a) The power absorbed at t = 5 ms is
100 100
t t t e te =
− − − ⋅
= 0.01655W = μ16.55 nW
(b) The energy delivered over the interval 0 < t < ∞ is
∞ ∞ − = − 0 0
200
t abs
Making use of the relationship
0 1
∞ − =
n ax
a
n x e dx where n is a positive integer and a > 0,
we find the energy delivered to be
2
3
PROPRIETARY MATERIAL. © 2007 Th e McGraw-Hill Companies, Inc. Lim ited distr ibution perm itted onl y to
Engineering Circuit Analysis, 7 Edition Chapter Two Solutions 10 March 2006
-100 t
2 8
100 (^360) t ms
t e (^) =
− = 72.68 W
2 8
100 ⎟ = = ⎠
=
− t ms
t i e dt
di
t ms
t t idt e 8
100 0
=
− ⎟ ⎠
t ms
t t t t e e dt e 8
100 0
100 100 90 3 60 =
− − ′ − ⎟ ⎠
PROPRIETARY MATERIAL. © 2007 Th e McGraw-Hill Companies, Inc. Lim ited distr ibution perm itted onl y to
Engineering Circuit Analysis, 7 Edition Chapter Two Solutions 10 March 2006
first 2 hours P = 5 V 0.001 A 5 mW
(b) Energy = (5 V)(0.001 A)(2 hr)(60 min/ hr)(60 s/ min) = 36 J
(c) 36 – (2)(0.001)(60)(60) = 21.6 J
PROPRIETARY MATERIAL. © 2007 Th e McGraw-Hill Companies, Inc. Lim ited distr ibution perm itted onl y to
Engineering Circuit Analysis, 7 Edition Chapter Two Solutions 10 March 2006
“absorbing” power; the remaining sources are supplying power to the circuit.
Source Absorbed Power Absorbed Power
2 V source (2 V)(-2 A) - 4 W
8 V source (8 V)(-2 A) - 16 W
-4 A source (10 V)[-(-4 A)] 40 W
10 V source (10 V)(-5 A) - 50 W
-3 A source (10 V)[-(-3 A)] 30 W
The 5 powe r quantities sum to –4 – 16 + 40 – 50 + 30 = 0, as demanded from
conservation of energy.
PROPRIETARY MATERIAL. © 2007 Th e McGraw-Hill Companies, Inc. Lim ited distr ibution perm itted onl y to
