351115115-Solutions-Manual-Power-Electronics-Circuits-Devices-Applications-4th-Edition-Muhammad-H-Rashid.pdf

Instant download and all chapters Solutions Manual Power Electronics Circuits, Devices Applications 4th Edition Muhammad H. Rashid https://testbankdata.com/download/solutions-manual-power-electronics-circuits-devices-_applications-4th-edition-muhammad-h-rashid/

Chapter 3-Diodes Rectifiers Prob 3-1

Vm

:= 170

R := 5

f := 60

Using Usin g Eq. (3-11) (3-11) 2⋅ V m Vdc :=

Vdc = 108.23

π

Vdc := 0.6366⋅ Vm

Prob 3-2

Vm

Vdc = 108.22

:= 170

−3

R := 10

Lc := 0.5⋅ 10

f := 60

Using Eq. (3-11) (3-11) Vdc := 0.6366⋅ V m Vdc

Idc :=

Using Eq. (3-83) Vx

:= 2⋅ f ⋅ Lc⋅ Idc

Vo

Prob 3-3

Idc = 10.82

R

:=

Vdc − Vx

:= 170

Vm

Vx = 0.65 Vo = 107.57 R := 5

Using Eq. (3-25) Vdc := Vm⋅

Prob 3-4

Vm

6

π

f := 60

π ⋅ sin⎛ ⎞

Vdc = 162.34

⎝ 6 ⎠

:= 170

R := 5

f := 60

Using Eq. (3-25) Vdc := Vm⋅ Idc :=

6

π

π ⋅ sin⎛ ⎞

⎝ 6 ⎠

Vdc = 162.34

Vdc Idc = 32.47

R

Using Eq. (3-83) Vx := 6⋅ f ⋅ Lc⋅ Idc Vo

:=

Vx = 5.84

Vdc − Vx

Vo = 156.49

Chapter 3-Diodes Rectifiers Page # 3 -1

−3

Lc := 0.5⋅ 10

Prob 3-5

Vs := 280 Vm

R := 5

2

:= 280⋅

f := 60 Vm = 228.62

3

Using Eq. (3-33) 1.6542⋅ V m Vdc :=

Prob 3-6

Vs := 280 Vm

Vdc = 378.18 R := 5

f := 60

2

:= 280⋅

Lc

Vm = 228.62

3

Using Eq. (3-33)

Vdc = 378.18

1.6542⋅ V m Vdc := Vdc

Idc :=

Idc = 75.64

R

Using Eq. (3-83) Vx := 6⋅ f ⋅ Lc⋅ Idc

:=

Vo

Prob 3-7

Vdc := 240

Vm

0.6366

:=

Vdc

Idc

R

Vm

= 24

I p = 37.7

R

Id :=

= 377

Vs = 266.58

2

Idc :=

I p

Vm

Vm

Vs :=

Diodes

Vo = 364.57

R := 10 Vdc

:=

Vx = 13.61

Vdc − Vx

Idc

Id = 12

2


−3

:= 0.5⋅ 10

I p

IR := Transformer

Vm

Vs :=

Vs = 266.58

2 I p

Is :=

VI

IR = 18.85

2

Is = 26.66

2

:=

3 VI = 7.107 × 10

Vs⋅ I s

Using Eq. (3-1) Pdc :=

Using Eq. (3-2)

TUF :=

1 TUF Vdc := 750

Using Eq. (3-33) Vm

:=

Vs :=

Diodes

I p

:=

2 3 Pdc = 5.76 × 10

R

3 Pac = 7.11 × 10

Pac := Vs⋅ I s

Using Eq. (3-8)

Prob 3-8

(0.6366⋅ Vm)

Pdc

TUF = 0.8105

Pac

= 1.23 6000 Idc :=

Vdc

Vm = 453.39

1.6542 Vm

Vs = 320.6

2 Idc

Id :=

I p

Idc

IR :=

3

= 6 × 10

3 Id = 3 × 10

2 I p

3 IR = 4.24 × 10

2


Transformer

Vs :=

Vm

Vs = 320.6

2

Is := I p

6 VI = 1.924 × 10

Pdc := Vdc⋅ I dc

6 Pdc = 4.5 × 10

Pac := 3Vs⋅ I s

6 Pac = 5.77 × 10

Using Eq. (3-2) Using Eq. (3-8)

3

= 6 × 10

Vs⋅ I s

VI

Using Eq. (3-1)

Is

:=

Pdc

TUF :=

1 TUF

TUF = 0.78

Pac

= 1.28

Prob 3-9 Vm

:= 170

f := 60

R := 10

ω := 2⋅ π ⋅ f

ω = 376.99

0.04 RF :=

Guess L

−3

:= 5⋅ 10

Given

Using Eq. (3-67)

2

⋅ L ⎞ ⎛ 2⋅ ω − 0.481 RF⋅ 1 + ⎝ R ⎠ 3 Find( L ) ⋅ 10

= 158.93

f := 60

R := 10

=

0

mH

Prob 3-10 Vm

:= 170

ω := 2⋅ π ⋅ f

ω = 376.99

RF := 0.02 Guess L

−3

:= 5⋅ 10


Given

Using Eq. (3-67) 2

⎛ 6⋅ ω ⋅ L ⎞ − RF⋅ 2⋅ 1 + ⎝ R ⎠ 3 Find( L ) ⋅ 10

2 35

=

0

= 7.76

mH

Prob 3-11 E

:= 20

V p

V p

Vs := n ( a)

:= 120

Vs

α :=

asin

Cap

= 60

⎛ E ⎞ ⎝ Vm ⎠

β := π − α δ := β − α ( b)

Vm

2π ⋅ Idc

Idc := 10

:= 2⋅ V s

α⋅

180 = 13.63

β⋅

180 = 166.37

δ

1

R :=

:= 200

n

:= 2

f := 60

Vm = 84.85

π

π

180

π

= 152.73

⋅ ( 2⋅ V m⋅ cos ( α ) + 2⋅ E ⋅ α − π ⋅ E)

R = 1.776

( c)

Irms

:=

1 2 2⋅ π ⋅ R

⋅

⎡⎛ Vm2 ⎣⎝

2

⎞

2

+ E ⋅ ( π − 2⋅ α ) +

⎠

Vm

⎤

2

2

⋅ sin ( 2 ⋅ α ) − 4⋅ V m⋅ E⋅ cos ( α )

⎦

Irms = 16.98 2 PR := Irms ⋅ R ( d )

Pdc := E⋅ Idc ho

( e)

PR = 512.06

:=

η :=

Cap Pdc Pdc Pdc + PR

Pdc ho

=1

η⋅ 100 = 28.09 Chapter 3-Diodes Rectifiers Page # 3 -5

%

= 200

Ω

PIV := Vm

( f )

+E

PIV = 104.85

Prob 3-12 E

:= 12

:= 100

Cap

V p

Vs := n

Vs

α :=

( a)

asin

= 60

Vm

⎛ E ⎞ ⎝ Vm ⎠

β := π − α δ := β − α ( b)

2π ⋅ Idc

( c)

1

Irms :=

2 2⋅ π ⋅ R

⋅

:= 2⋅ V s

180 = 8.13

β⋅

180 = 171.87

n := 2

π

π

180

π

= 163.74

⎞

2

+ E ⋅ ( π − 2⋅ α ) +

⎠

2

f := 60

Vm = 84.85

⋅ ( 2⋅ V m⋅ cos ( α ) + 2⋅ E ⋅ α − π ⋅ E)

⎡⎛ Vm2 ⎣⎝

Idc := 5

α⋅

δ

1

R :=

:= 120

V p

Vm

2

R = 4.26

Ω

⎤

2

⋅ sin( 2 ⋅ α ) − 4⋅ Vm⋅ E⋅ cos (α )

⎦

Irms = 8.21 2 PR := Irms ⋅ R

PR = 287.03

Pdc := E⋅ Idc

( d )

ho ( e)

Cap

:=

= 60

ho = 1.67

Pdc

η :=

( f )

Pdc

Pdc Pdc + PR

PIV := Vm

+E

η⋅ 100 = 17.29

%

PIV = 96.85

Prob 3-13 L

−3

:= 4.5⋅ 10

R := 4

f := 60

ω := 2⋅ π ⋅ 60


ω = 376.99

Vs := 120

E

:= 20

T

:=

1

3 T ⋅ 10

f

= 16.67

1 2 + ( ω ⋅ L)

:=

2 R

θ :=

atan

Z

2

⎛ ω⋅ L ⎞ ⎝ R ⎠

Z = 4.34

θ = 0.4

R π ⎡ − ⎛ ⎞ ⋅ ⎛ ⎞ ⎤ L ω Vs 1 + e ⎝ ⎠ ⎝ ⎠ ⎦ ⎣ ⋅ sin(θ )⋅ − Y := 2⋅ R ⎞ ⎛ π ⎞ ⎤ Z ⎛ ⎡ − ⋅ L ⎠ ⎝ ω ⎠ ⎝ ⎣1 − e ⎦

( a)

iL ( t)

:= 2⋅

Vs Z

E R

+

E R

− 2⋅

⎛ R ⎞ ⋅ ⎛ π ⎞ − ⎛ R ⎞ ⋅ t L ⋅ sin( ω⋅ t − θ ) + ( Y) ⋅ e⎝ ⎠ ⎝ ω ⎠ ⋅ e ⎝ L ⎠ −

Vs Z

⋅ sin( θ ) Y = 0.02

E R iL ( 0) = 10.27

( b)

The average diode current can be found from Equation 3-16 as : 1

⌠ 120 1 ⎮ Id := ⋅ i ( t) ⎮ T L ⌡0 ( c)

Id = 11

d t

The rms diode current can be found using Equation 3 -16 1

⎛ 1 ⎞ ⎜⌠ 120 ⎟ 1 2 ⎮ Ir := ⎜ ⋅ iL ( t) d t⎟ ⎮ T ⎜⌡ ⎟ 0 ⎝ ⎠ ( d )

2

Ir = 17.04

The rms output current can then be found by


Irms := Ir ⋅ 2

Irms = 24.1

The max min values of output current

⎡

⎡ − ⎛ R ⎞ ⋅ ⎛ π ⎞ ⎤⎤ ⎛ R ⎞ ⋅ ⎛ π − ⎞ t ⎝ L ⎠ ⎝ ω ⎠ Vs⋅ 2 L ⎠ ⎝ ω ⎠ ⎢ 1 + e ⎝ ⎢ i1 ( t) := ⋅ sin( ω⋅ t − θ ) + sin( θ )⋅ e ⋅ − 1⎥⎥ − Z ⎢ ⎢ − ⎛ R ⎞ ⋅ ⎛ π ⎞ ⎥⎥ L ω ⎣ ⎣1 − e ⎝ ⎠ ⎝ ⎠ ⎦⎦

E R

Guess t

−3

:= 5⋅ 10

Given

⎡

⎡

⎡ − ⎛ R ⎞ ⋅ ⎛ π ⎞ ⎤⎤ ⎛ R ⎞ ⋅ ⎛ π − t ⎞ ⎝ L ⎠ ⎝ ω ⎠ V⋅ 2 L ⎠ ⎝ ω ⎠ ⎢ 1 + e d ⎢ s ⎝ ⎢ ⋅ sin ( ω⋅ t − θ) + sin( θ ) ⋅ e ⋅ − 1⎥ ⎥ − ⎢ ⎢ − ⎛ R ⎞ ⋅ ⎛ π ⎞ ⎥⎥ d t ⎢ Z L ω ⎣ ⎣ ⎣1 − e ⎝ ⎠ ⎝ ⎠ ⎦⎦

( )

Imin

:=

=

0

⎦

Imax = 34.36

t1

at

1.15

( )

180

π

⋅ ω⋅ t1 = 111.93

degrees

Imin = 6.04

i1 tx

tx⋅ 1000

R ⎥

t1 = 5.18 × 10

Imax := i1 t1

−

E⎥

−3

t1 := Find ( t)

tx := t1

⎤

= 0.68

ms

at

180

π

⋅ ω⋅ tx = 14.6

degrees

⎡ − ⎛ R ⎞ ⋅ ⎛ π ⎞ ⎤⎤ ⎛ R ⎞ ⋅ ⎛ π − t ⎞ ⎝ L ⎠ ⎝ ω ⎠ Vs⋅ 2 L ⎠ ⎝ ω ⎠ ⎢ 1 + e ⎝ ⎢ i1 ( t) := ⋅ sin ( ω⋅ t − θ) + sin( θ ) ⋅ e ⋅ − 1⎥ ⎥ − Z ⎢ ⎢ − ⎛ R ⎞ ⋅ ⎛ π ⎞ ⎥⎥ L ω ⎣ ⎣1 − e ⎝ ⎠ ⎝ ⎠ ⎦⎦ ⎡


E R

ta

( )

i1 ta

40 38 36 34 32 30 28 26 24 22 20 18 16 14 12 10 8 6 4 2 0

:= 0 , .0001 .. 2π

0

9 18 27 36 45 54 63 72 81 90 99 108117126135144153162171 180 180

π Imin = 6.04 at 180

π

⋅ ω ⋅ ta

is the minimum current

⋅ ω⋅ tx = 14.6

Imax = 34.36

degrees is the maximum current

at

180

π

⋅ ω⋅ t1 = 111.93

degrees

Prob 3-14: Three-phase Diode Rectifie r

:= 120

Input Phase Voltage

VS

Load Resistance

R := 5

Load Inductance

L

f s := 60 E

:= 20

−3

:= 2.5⋅ 10

The ripple factor of the output voltage is to be limited to less than 5% of the average output voltage by an L-C filter.


1

VS

ω := 2⋅ π ⋅ f s

Z

ω = 376.99

Z = 5.09

= 120

VAB :=

T

:=

1

:=

2 R

+ ( ω ⋅ L)

2

2

⎛ ω⋅ L ⎞ ⎝ R ⎠

θ :=

atan

180.

θ = 10.67 π

T = 0.02

f s

Vm

3⋅ V S

:= 2⋅ V S

Vm = 169.71

VAB = 207.85 Line-to-Line RMS Input Voltage R π ⎡ − ⎛ ⎞ ⋅ ⎛ ⎞ ⎤ ⎛ 2⋅ π − θ ⎞ − sin⎛ π − θ ⎞ ⋅ e ⎝ L ⎠ ⎝ 3⋅ ω ⎠ sin ⎢ ⎥ 2⋅ VAB 3 3 ⎝ ⎠ ⎝ ⎠ ⎥− ⋅⎢ I1 := R Z π ⎡− ⎛ ⎞ ⋅ ⎛ ⎞⎤ ⎢ ⎥ L 3⋅ ω 1 − e⎣ ⎝ ⎠ ⎝ ⎠⎦ ⎣ ⎦

E R I1 = 50.56

⎡ ⎡ sin⎛ 2.⋅ π − θ ⎞ − sin⎛ π − θ ⎞ ⎤ ⎛ R ⎞ ⎛ π ⎞⎤ ⋅ −t 2⋅ VAB 3 3 L ⎠ ⎝ 3. ⋅ ω ⎠ ⎥ ⎝ ⎠ ⎝ ⎠ ⎝ ⎢ ⎢ ⎥ iL ( t) := ⋅ sin ( ωt − θ ) + ⋅e − Z R ⎞ ⎛ π ⎞ ⎢ ⎢ ⎥ ⎥ ⎛ − ⋅ L ⎠ ⎝ 3. ⋅ ω ⎠ ⎝ ⎣ ⎣ 1−e ⎦ ⎦ 1.

R

Calculations for I1 using Equation on bottom of page 74 R π ⎡ − ⎛ ⎞ ⋅ ⎛ ⎞ ⎤ ⎛ 2⋅ π − θ ⎞ − sin⎛ π − θ ⎞ ⋅ e ⎝ L ⎠ ⎝ 3⋅ ω ⎠ sin ⎢ ⎥ 2⋅ VAB 3 3 ⎝ ⎠ ⎝ ⎠ ⎥− I1 := ⋅⎢ Z R π ⎡− ⎛ ⎞ ⋅ ⎛ ⎞⎤ ⎢ ⎥ L 3⋅ ω 1 − e⎣ ⎝ ⎠ ⎝ ⎠⎦ ⎣ ⎦

2.

E

Calculations for Average Diode Current

⎛ T ⌠ 3 2 ⎮ ⎜ ⋅ Id := ⎜ T ⎮⎮T ⎝

⌡

6

⎞ iL ( t) d t⎟ ⎟ ⎠

Id = 17.38


E R I1 = 50.56

351115115-Solutions-Manual-Power-Electronics-Circuits-Devices-Applications-4th-Edition-Muhammad-H-Rashid.pdf

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