[전기] Fundamental of Electric Circuits 3e Alexander Sadiku
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작성일 21-10-07 17:49본문
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(d) i=dq/dt = 1200120ππcost pA
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(a) (b) 1710482.6×181024.1×
Chapter 1, Solution 1
PROPRIETARY
(b) q = 1. 24x1018 x [-1.602x10-19 C] = -0.19865 C
[전기] Fundamental of Electric Circuits 3e Alexander Sadiku
순서
Chapter 1, Problem 1
Chapter 1, Problem 2.
(c) ()()nC e5e3tqt2-t−−=
(d) ()pCt sin10 120tqπ=
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(c) (d) 191046.2×2010628.1×
(c) i = dq/dt = (-3e-t + 10e-2t) nA
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[참고자료(data)] Fundamental of Electric Circuits 3e - Alexander Sadiku -
(a) i = dq/dt = 3 mA
Chapter 1, Solution 2
(a) q = 6.482x1017 x [-1.602x10-19 C] = -0.10384 C
[자료(data)범위] Fundamental of Electric Circuits 3e - Alexander Sadiku -
(a) ()()mC 83+=ttq
(e) ()Ct 50cos204μtetq−=
(d) q = 1.628x1020 x [-1.602x10-19 C] = -26.08 C
How many coulombs are represented by these amounts of electrons:
(c) q = 2.46x1019 x [-1.602x10-19 C] = -3.941 C
[이용대상] Fundamental of Electric Circuits 3e - Alexander Sadiku -
(e) i =dq/dt = −+−ettt48050100050(cossin) Aμ
설명
레포트 > 공학,기술계열
(b) i = dq/dt = (16t + 4) A
Determine the current flowing through an element if the charge flow is given by
Fundamental of Electric Circuits 3e - Alexander Sadiku - [참고자료] Fundamental of Electric Circuits 3e - Alexander Sadiku - [자료범위] Fundamental of Electric Circuits 3e - Alexander Sadiku - [이용대상] Fundamental of Electric Circuits 3e - Alexander Sadiku -
(b) ()C 2)482t-t(tq+=
Fundamental of Electric Circuits 3e - Alexander Sadiku -
다.