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[전기] Fundamental of Electric Circuits 3e Alexander Sadiku

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작성일 20-04-14 23:51

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(d) i=dq/dt = 1200120ππcost pA

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 -

(a) (b) 1710482.6×181024.1×

(e) ()Ct 50cos204μtetq−=
(c) i = dq/dt = (-3e-t + 10e-2t) nA
Chapter 1, Solution 1

[전기] Fundamental of Electric Circuits 3e Alexander Sadiku

(a) i = dq/dt = 3 mA
레포트 > 공학,기술계열

만약 내용이 다를시 해피래포트에 환불요청하시면 환불됩니다.

(b) q = 1. 24x1018 x [-1.602x10-19 C] = -0.19865 C
Fundamental of Electric Circuits 3e - Alexander Sadiku -


(a) q = 6.482x1017 x [-1.602x10-19 C] = -0.10384 C
[참고data(자료)] Fundamental of Electric Circuits 3e - Alexander Sadiku -

Download : Fundamental of .zip( 60 )



(c) (d) 191046.2×2010628.1×


(b) ()C 2)482t-t(tq+=
(c) q = 2.46x1019 x [-1.602x10-19 C] = -3.941 C
[data(자료)범위] Fundamental of Electric Circuits 3e - Alexander Sadiku -
Chapter 1, Problem 1
(d) ()pCt sin10 120tqπ=
[이용대상] Fundamental of Electric Circuits 3e - Alexander Sadiku -
(d) q = 1.628x1020 x [-1.602x10-19 C] = -26.08 C

Determine the current flowing through an element if the charge flow is given by

How many coulombs are represented by these amounts of electrons:

Chapter 1, Problem 2.
(b) i = dq/dt = (16t + 4) A

(e) i =dq/dt = −+−ettt48050100050(cossin) Aμ


PROPRIETARY
(a) ()()mC 83+=ttq

(c) ()()nC e5e3tqt2-t−−=
Chapter 1, Solution 2

순서
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Fundamental of -2825_01_.gif Fundamental of -2825_02_.gif Fundamental of -2825_03_.gif Fundamental of -2825_04_.gif Fundamental of -2825_05_.gif
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