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다. , [Electrical Engineering Principles and Applications Allan Hambley 3rd Ed Pearson] electrical eng 2기타솔루션 , Electrical Engineering Principles Applications Allan Hambley 3rd Ed Pearson

CHAPTER 1
Exercises
E1.1 Charge = Current × Time = (2 A) × (10 s) = 20 C
E1.2 ( ) ( ) (0.01sin(200t) 0.01 200cos(200t ) 2cos(200t ) A
dt
d
dt
i t = dq t = = × =
E1.3 Because i2 has a positive value, positive charge moves in the same
direction as the reference. Thus positive charge moves downward in
element C.
Because i3 has a negative value, positive charge moves in the opposite
direction to the reference. Thus positive charge moves upward in
element E.
E1.4 Energy = Charge × Voltage = (2 C) × (20 V) = 40 J
Because vab is positive, the positive terminal is a and the negative
terminal is b. Thus the charge moves from the negative terminal to the
positive terminal, and energy is removed from the circuit element.
E1.5 iab enters terminal a. Furthermore, vab is positive at terminal a. Thus
the current enters the positive reference, and we have the passive
reference configuration.
E1.6 (a) p (t ) v (t )i (t ) 20t 2 a = a a =
6667 J
3
20
3
( ) 20 20
10 3
0
10 3
…(투비컨티뉴드 )

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[Electrical Engineering Principles and Applications Allan Hambley 3rd Ed Pearson] electrical eng 2

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