Electrical Engineering Concepts and Applications Ch2-7, S.A Reza Zekavat

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Electrical Engineering Concepts and Applications Ch2-7, S.A Reza Zekavat에 대한 보고서 자료입니다.

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2.1 Current source
2.2 Voltage source
2.3 Resistor
2.4 Capacitor
2.5 Inductor
Section 2.2 Charge and Current
2.6 b)
The current direction is designated as the direction of the movement of positive
charges.
2.7 The relationship of charge and current is
( ) ( ) ( ) 0
0
q t i t dt q t
t
t
= ∫ +
so
( ) ( ) ( ) 0
0
q t 2sin 10 t dt q t
t
t
= ∫ π +
( ) ( ) ( ) 0 cos 10
10
2
0
q t t q t
t
t
+ ⎥ ⎦
⎤
⎢ ⎣
⎡ −
= π
π
2.8 The coulomb of one electron is denoted by e and
( ) ( ) ( ) 0
0
q t i t dt q t
t
t
= ∫ +
So
( ) ( ) 0
0
( ) / 1 12t dt q t
e
n t q t e
t
t
= = ∫ +
If t0 = 0 and ( ) 0 q t0 = ,
( ) 6 t2
e
n t =
2.9
q(t)= ∫idt
q(t) dt
t
= ∫
0
5
q(t)= 5t
2.10
( ) 5[5 ] 5(5) 5(0) 25
q t =0 t = − = Coulombs
2.11 Using the definition of current-charge relationship, the equation can be rewritten
as:
e
t
n
dt
i dq
Δ
Δ
= =
Thus, the current flow within t1 and t2 time interval is,
i ( 1.6 10 ) 3A
2
(5.75 2) 10 19
19
− × = −
− ×
= −
The negative sign shows the current flow in the opposite direction with respect to the
electric charge.
2.12 Assuming the area of the metal surface is S, The mass of the nickel with depth d
= 0.15mm is
m = ρ × d × S
Meanwhile, using the electro-chemical equivalent, the mass of the nickel can be
expressed as
m = k × I × t
where I = σ × S.
Equating the two expressions of the mass, the coating time is found:
t = ρ × d / σ = 1.24 × 105 s ≈ 34.4 hour
Section 2.3 Voltage
2.13 By the definition of voltage, when a positive charge moves from high voltage to
low voltage, its potential energy decreases.
So a is “+”, b is “-”. In other words, uab=1V.
2.14 The current i(t) is defined as:
( )
⎭ ⎬ ⎫
⎩ ⎨ ⎧
< ≤
=
elsewhere
t
i t
0
3 0 1
Therefore, the charge is
3 3
1
0 ∫
g = dt = C
The energy in Joules is given by:
J =V ×C = 5×3 =15J
2.15
1 electron = 1.6 10−19 − × Coulombs.
Therefore, there are 6.25×1018 electrons in a coulomb.
Coulombs of5×1016 electrons = 3
18
16
8 10
6.25 10
5 10 − = ×
×
×
Therefore, the voltage is
1875
8 10
15
3 =
×
= = C −
V J V
2.16
( ) dt t t q ⎟

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