Solucionario-analisis de circuitos en ingenieria - 6ta edicion - j.r. hayt
Engineering Circuit Analysis, 6th Edition
Copyright ©2002 McGraw-Hill, Inc. All Rights Reserved.
CHAPTER TWO SOLUTIONS
2.
(a) 1 MW (b) 12.35 mm (c) 47. kW (d) 5.46 mA
(e) 33 µJ (f) 5.33 nW (g) 1 ns (h) 5.555 MW
(i) 32 mm
Engineering CircuitAnalysis, 6th Edition
Copyright ©2002 McGraw-Hill, Inc. All Rights Reserved.
CHAPTER TWO SOLUTIONS
3.
Motor power = 175 Hp (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 = (130.5×103 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.
Engineering Circuit Analysis, 6th Edition
Copyright ©2002 McGraw-Hill, Inc. All Rights Reserved.
CHAPTER TWO SOLUTIONS
4.
The 400-mJ pulse lasts 20 ns. (a) To compute the peak power, we assume the pulse shape is square:
Energy (mJ)
400
t (ns) 20
Then P = 400×10-3/20×10-9 = 20 MW. (b) At 20 pulses per second, the averagepower is Pavg = (20 pulses)(400 mJ/pulse)/(1 s) = 8 W.
Engineering Circuit Analysis, 6th Edition
Copyright ©2002 McGraw-Hill, Inc. All Rights Reserved.
CHAPTER TWO SOLUTIONS
5.
The 1-mJ pulse lasts 75 fs. (c) To compute the peak power, we assume the pulse shape is square:
Energy (mJ)
1
t (fs) 75
Then P = 1×10-3/75×10-15 = 13.33 GW. (d) At 100 pulses per second, the averagepower is Pavg = (100 pulses)(1 mJ/pulse)/(1 s) = 100 mW.
Engineering Circuit Analysis, 6th Edition
Copyright ©2002 McGraw-Hill, Inc. All Rights Reserved.
CHAPTER TWO SOLUTIONS
6.
The power drawn from the battery is (not quite drawn to scale):
P (W)
10
6
t (min) 5 7 17 24
(a) Total energy (in J) expended is [6(5) + 0(2) + 0.5(10)(10) + 0.5(10)(7)]60 = 6.9 kJ. (b) Theaverage power in Btu/hr is (6900 J/24 min)(60 min/1 hr)(1 Btu/1055 J) = 16.35 Btu/hr.
Engineering Circuit Analysis, 6th Edition
Copyright ©2002 McGraw-Hill, Inc. All Rights Reserved.
CHAPTER TWO SOLUTIONS
7.
Total charge q = 18t2 – 2t4 C. (a) q(2 s) = 40 C. (b) To find the maximum charge within 0 ≤ t ≤ 3 s, we need to take the first and second derivitives: dq/dt = 36t – 8t3 = 0,leading to roots at 0, ± 2.121 s d2q/dt2 = 36 – 24t2 substituting t = 2.121 s into the expression for d2q/dt2, 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).
(a)
(b)
Engineering Circuit Analysis,6th Edition
Copyright ©2002 McGraw-Hill, Inc. All Rights Reserved.
CHAPTER TWO SOLUTIONS 8. Referring to Fig. 2.6c, t0
- 2 + 3e −5t A, i1 (t ) = 3t - 2 + 3e A, Thus, (a) i1(-0.2) = 6.155 A (b) i1 (0.2) = 3.466 A
(c) To determine the instants at which i1 = 0, we must consider t < 0 and t > 0 separately: for t < 0, - 2 + 3e-5t = 0 leads to t = -0.2 ln (2/3) = +2.027 s (impossible)for t > 0, -2 + 3e3t = 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 0.08 < t < 0.1 s is q(t) = =
∫
0.1
− 0.08 0
i1 (t )dt
−5t
∫ [− 2 + 3e ]dt
− 0.08 0 -0.08
+
∫ [− 2 + 3e ]dt
0.1 3t 0 0.1 0
= − 2 + 3e −5t
+ − 2 + 3e3t
= 0.1351 + 0.1499 = 285 mCEngineering Circuit Analysis, 6th Edition
Copyright ©2002 McGraw-Hill, Inc. All Rights Reserved.
CHAPTER TWO SOLUTIONS
9.
Referring to Fig. 2.28,
(a) The average current over one period (10 s) is iavg = [-4(2) + 2(2) + 6(2) + 0(4)]/10 = 800 mA
(b) The total charge transferred over the interval 1 < t < 12 s is qtotal =
∫
12
1
i (t )dt = -4(2) + 2(2) + 6(2) + 0(4) –4(2) = 0 C...
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