Engineering Circuit Analysis 6th Edition Hayt Solutions Manual – Complete Step-by-Step Problem Solutions | Verified Answers | Updated : Latest

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CHAPTER TWO SOLUTIONS


1. (a) 12 µs (d) 3.5 Gbits (g) 39 pA
(b) 750 mJ (e) 6.5 nm (h) 49 kΩ
(c) 1.13 kΩ (f) 13.56 MHz (i) 11.73 pA




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CHAPTER TWO SOLUTIONS



2. (a) 1 MW (e) 33 µJ (i) 32 mm
(b) 12.35 mm (f) 5.33 nW
(c) 47. kW (g) 1 ns
(d) 5.46 mA (h) 5.555 MW




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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.




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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 average power is

Pavg = (20 pulses)(400 mJ/pulse)/(1 s) = 8 W.




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