Baixe Solutions do Rashid 1 e 2 e outras Exercícios em PDF para Eletrônica de Potência, somente na Docsity!
I RMS
= 50
I AVG
I p
pi
:= I AVG
=31.
t 1
:= 0 t 1
= 0
t 2
T
2
:= t 2
8.335 10
− 3 = ×
T o
:= k T⋅ T o
8.335 10
− 3 = ×
θ 1
2 pi⋅
t 1
T
:= ⋅ θ 1
= 0
θ 2
2 pi⋅
t 2
T
:= ⋅ θ 2
=3.
I AVG
I p
2 pi⋅ θ 1
θ 2
sin x( ) x
⌠
⎮
⌡
:= ⋅ d
I AVG
=31.
Chapter 1 - Introduction
pi := 4 atan 1( ) pi =3. Problem 1.
I p
:= 100 T o
8.3 10
− 3 := ⋅ T 16.67 10
− 3 := ⋅
I RMS
I p
2
:= I RMS
=70.
I AVG
2
I p
pi
:= ⋅ I AVG
=63.
Problem 1.
I p
:= 100 k := 0.5 T 16.67 10
− 3 := ⋅
I RMS
I p
k
2
:= ⋅
Chapter 1- Introduction
https://www.book4me.xyz/solution-manual-for-power-electronics-handbook-rashid/
Access Full Complete Solution Manual Here
Problem 1.
I p
:= 100 k := 0.4 T 1 10
− 3 := ⋅
T o
:= k T⋅ T o
4 10
− 4 = ×
I RMS
I p
:= ⋅ k I RMS
=63.
I AVG
I p
:= ⋅k
I AVG
= 40
Problem 1.
I a
:= 80 I b
:= 100 k := 0.4 T 1 10
− 3 := ⋅
T o
:= k T⋅ T o
4 10
− 4 = ×
Problem 1.
I p
:= 100 k := 0.8 T 16.67 10
− 3 := ⋅
t 1
:= T ⋅( 1 −k) t 1
3.334 10
− 3 = ×
t 2
T
2
:= t 2
8.335 10
− 3 = ×
T o
:= k T⋅ T o
=0.
I RMS
I p
k
2
sin T o
⋅( 1 −k) ⎡ ⎣
cos pi[ ⋅( 1 −k)]
2 pi⋅
:= ⋅ + I RMS
=63.
θ 1
2 pi⋅
t 1
T
:= ⋅ θ 1
=1.
θ 2
2 pi⋅
t 2
T
:= ⋅ θ 2
=3.
I AVG
I p
2 pi⋅ θ 1
θ 2
sin x( ) x
⌠
⎮
⌡
:= ⋅ d
I AVG
=20.
Chapter 1- Introduction
Chapter 2 – Diodes Circuits
Prob 2.
t rr
− 6 := ⋅ di_dt 80 10
6 := ⋅
( ) a Eq. (2.10)
Q
RR
0.5 di_dt⋅ t rr
2 := ⋅ Q RR
6 ⋅ 1 10
3 = × μC
( )b Eq. (2-11)
I
RR
2 Q
RR
:= ⋅ ⋅di_dt I RR
= 400 A
Prob 2.
Chapter 2-Diodes Circuits
Qୖୖ ൌ 10000μC and Iୖୖ ൌ 4000A
Qୖୖ ൌ 0.5 ൬
di
dt
൰ t (^) ୰୰
ଶ
Iୖୖ ൌ ඨ2Qୖୖ ൬
di
dt
൰
Solving both equations,
We get, t ୰୰
ൌ 5μs andୢ
୧ୢ
୲
ൌ 800A/μs
Prob 2.
t rr
− 6 := ⋅ SF :=0.
t a
t rr
1 +SF
t a
− 6 = ×
t b
SF t a
:= ⋅ t b
− 6 = ×
m
t a
⋅ t rr
m 8.333 10
− 12 = ×
Q
RR
( )x :=m x⋅
2
. 10
8 4
. 10
8 6
. 10
8 8
. 10
8 1
. 10
9
0
Plot of the charge storage verus di/dt
di/dt
Charge storage
Q RR
( )x
x
Chapter 2-Diodes Circuits
I
I
T
R
− 3 R = × 2
V
D
V
D
I
R
− 3 R = × 1
V
D
V
D
I
V
D
V
D
I
I
I
I
I = 150
I
T
V
D
I :=2.
T
Prob 2.
R
5 R = × 2
V
D
I
R
I
R
− 3 = ×
I
R
I
S
I
R
+ I
S
( )b := −
Using Eq. (2-13),
I
R
I =0.
R
V
D
R
( ) a
I
S
− 3 I := ⋅ S
− 3 := ⋅
R
3 := ⋅
V
D
V := 2200
D
Prob 2-
Chapter 2-Diodes Circuits
I
peak
I = 500
peak
I
p
I
RMS
I =20.
RMS
I
p
T
0
t 1
sin 2(^ ⋅π ⋅ f⋅t) t
2
:= ⋅ d
I
AVG
I =3.
AVG
I
p
T
0
t 1
sin 2(^ ⋅ π⋅f ⋅t) t
:= ⋅ d
T 10
3 ⋅ = 2
T
f
t 1
− 6 f := 500 := ⋅
I
p
Prob 2-
Chapter 2-Diodes Circuits
R
ଵ
ൌ R
ଶ
ൌ 50kΩ, ܸ ௦
ଵ
ଶ
From Eq. (2-12), ܫ ௦ଵ
ோଵ
௦ଶ
ோଶ
or ܫ௦ଵ
(^) ವభ
ோభ
(^) ವమ
ோమ
ଵ
ଶ
ܴ௦ଵ ଵ
ܴ௦ଶ ଶ
Solving for ܫ௦ଵ and ܫ௦ଶ gives ܫ௦ଵ ܣ݉ൌ 20 and ܫ (^) ௦ଶ ܣ݉ൌ 30.
Prob 2-
t 1
− 6 := ⋅ t 2
− 6 := ⋅ t 3
− 6 := ⋅ t 4
− 6 := ⋅
t 5
− 3 := ⋅ f := 250 I a
:= 150 I
b
:= 100 I
p
( ) a I AVG
I
a
⋅ ft 3
⋅ I
b
⋅ f t 5
t 4
( )
+ ⋅ 2 I
p
I
a
( )
⋅ ⋅f
t 2
t 1
( )
π
I
AVG
( ) b
I
r
I
p
I
a
( )
f
t 2
t 1
( )
I
r
I
r
I
a
f t 3
:= ⋅ ⋅ I
r
I
r
I
b
f t 5
t 4
( )
:= ⋅ ⋅ I
r
I
rms
I
r
2 I r
2
2 := + I rms
Prob 2-
t 1
− 6 := ⋅ t 2
− 6 := ⋅ t 3
− 6 := ⋅ t 4
− 6 := ⋅
t 5
− 3 := ⋅ f := 250 I a
:= 150 I
b
:= 100 I
p
( ) a I AVG
I
a
⋅ ft 3
⋅ I
b
⋅ f t 5
t 4
( )
+ ⋅ 2 I
p
I
a
( )
⋅ ⋅f
t 2
t 1
( )
π
I
AVG
( ) b
I
r
I
p
I
a
( )
f
t 2
t 1
( )
I
r
I
r
I
a
f t 3
:= ⋅ ⋅ I
r
I
r
I
b
f t 5
t 4
( )
:= ⋅ ⋅ I
r
I
rms
I
r
2 I r
2
2 := + I rms
Chapter 2-Diodes Circuits
I
AVG
I
AVG
I
a
⋅ ft 3
⋅ I
b
⋅ f t 5
t 4
( )
+ ⋅ 2 I
p
I
a
( )
⋅ ⋅f
t 2
t 1
( )
π
( )b
I
rms
I = 180
rms
I
r
2 I r
2
2 := +
I
r
I
r
I
p
I
a
( )
f
t 2
t 1
( )
I
p
3 I = × p
I
r
f
t 2
t 1
( )
I
a
( ) a
I
r
I
r
I
rms
2 I r
2 − I r
2 := −
I
r
I
b
f t 5
t 4
( )
I
r
I
r
I =47.
r
I
a
f t 3
I
rms
I
p
I := 150
b
I := 100
a
t f := 250 := 150 5
− 3 := ⋅
t 4
− 6 t := ⋅ 3
− 6 t := ⋅ 2
− 6 t := ⋅ 1
− 6 := ⋅
Prob 2-
Chapter 2-Diodes Circuits
Prob 2-
V
S
:= 110 R := 4.7 L 6.5 10
− 3 := ⋅
τ
R
L
:= τ =723.
Using Eq. (2-25),
( )a I D
V
S
R
I
D
( ) b I O
I
D
W 0.5 L⋅ I
O
2 := ⋅ W =1.
Using Eq. (2-27),
( )c di
V
S
L
:= di 1.692 10
4 = ×
Prob 2-
V
S
:= 220 R := 4.7 C 10 10
− 6 := ⋅ t 2 10
− 6 := ⋅
τ := R C⋅ τ 4.7 10
− 5 = ×
Using Eq. (2-20),
( )a I p
V
S
R
I
p
( ) b V O
V
S
W 0.5 C⋅ V
O
2 := ⋅ W =0.
Using Eq. (2-21),
( )c V c
V
S
1 e
−t
τ −
:= ⋅ V
c
Chapter 2-Diodes Circuits
V
C
V = 220
C
2 V
S
( )c := ⋅
Using Eq. (2-35),
t 1
− 5 t = × 1
( )b :=π ⋅ L C⋅
I
p
I =49.
p
V
S
C
L
( )a := ⋅
Using Eq. (2-32),
L 50 10
− 6 C 10 10 := ⋅
− 6 V := ⋅ S
Prob 2-
di 3.385 10
4 = ×
Prob 2-
V
S
:= 220 R := 4.7 L 6.5 10
− 3 := ⋅
τ
R
L
:= τ =723.
Using Eq. (2-25),
( )a I D
V
S
R
I
D
( ) b I O
I
D
W 0.5 L⋅ I
O
2 := ⋅ W =7.
Using Eq. (2-27),
( )c di
V
S
L
Chapter 2-Diodes Circuits
