文档介绍：CHAPTER TWO SOLUTIONS
1. (a) 12 µs (d) Gbits (g) 39 pA
(b) 750 mJ (e) nm (h) 49 kΩ
(c) kΩ(f) MHz (i) pA
Engineering Circuit Analysis, 6th Edition Copyright ©2002 McGraw-Hill, Inc. All Rights Reserved.
CHAPTER TWO SOLUTIONS
2. (a) 1 MW (e) 33 µJ (i) 32 mm
(b) mm (f) nW
(c) 47. kW (g) 1 ns
(d) mA (h) MW
Engineering Circuit Analysis, 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/ Hp)] = kW
(b) Running for 3 hours,
Energy = (×103 W)(3 hr)(60 min/hr)(60 s/min) = GJ
(c) A single battery has 430 kW-hr capacity. We require
( kW)(3 hr) = 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) pute 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 average power 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) pute the peak power, we assume the pulse shape is square:
Energy (mJ)
1
t (fs)
75
Then P = 1×10-3/75×10-15 = GW.
(d) At 100 pulses per second, the average power 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