CHAPTER 4 SOLUTIONS

2/17/10

4-1) Load:

2Vm Vo 2 2(120) / 
Vo = ; Io = = = 6.0 A.
 R 18
Vm 120 2 9.43
I o , peak = = = 9.43 A.; I o ,rms = = 6.67 A.
R 18 2

Each diode:

I o 6.0 I
I D ,avg = = = 3.0 A.; I D , peak = I o , peak = 9.43 A.; I D ,rms = o ,rms = 4.71 A.
2 2 2

4-2)

2Vm 2 (120 ) 2 Vo 108

bridge : Vo = = = 108 V .; I o = = = 4.32 A.
  R 25
PIV = Vm = 120 2 = 170 V .
Center tapped : Vm = 120 2, I o = 4.32 A.; PIV = 2Vm = 2(120) 2 = 340 V .

4-3)

V0 2Vm 200
=I0 = = = 4.24 A.
R  R  15
2V  1 1  Vn
Vn = m  −  ; Z n = R + (no L) ; I n =
2 2

  n −1 n +1  Zn
V2 = 42.4, V4 = 8.49, ; Z 2 = 47.7 , Z 4 = 91.7 
42.4 V
I2 = = 0.890 A., I 4 = 4 = 0.0925 A.
47.7 Z4
2
 0.890  (0.0925) 2
I rms =  I = 4.24 + 
2
n
2
 + = 4.29 A.
 2  2
1 I
I D = I 0 = 2.12 A.; I D ,rms = o ,rms = 3.03 A.
2 2
I s = 0; I s ,rms = I o ,rms = 4.29 A.
4-4)

V0 2Vm 340
I0 == = = 10.8 A.
R  R  10
2V  1 1  Vn
Vn = m  −  ; Z n = R + (no L ) ; I n =
2 2

  n −1 n +1  Zn
V2 = 72.2, V4 = 14.4, ; Z 2 = 21.3 , Z 4 = 39.0 
72.2 V
I2 = = 3.38 A., I 4 = 4 = 0.37 A.
21.3 Z4
2
 3.38  (0.37) 2
I rms =  I = 10.8 + 
2
n
2
 + = 11.1 A.
 2  2
1 I
I D = I 0 = 5.4 A.; I D ,rms = o ,rms = 7.84 A.
2 2
I s = 0; I s ,rms = I o ,rms = 11.1 A.

4-5)

a) Average load current : R = 15 L = 30 mH

Vo 2Vm /  2(120) 2 /  108
Io = = = = = 7.20 A.
R R 15 15

b) Power is determined from the Fourier series. Using Eq. 4-4 and 4-5.

n Vn, V. Zn. Ω In, A.

2 72.0 27.1 2.65
4 14.4 47.7 0.302

2 2
 2.65   0.32 
I rms  7.20 + 
2
 +  = 7.45 A.; P = I rms R = (7.45) 25 = 832 W .
2 2

 2   2 
P 832
I s ,rms = I o ,rms = 7.45 A.; pf = = = 0.93
S (120)(7.45)
4-6

a) Average load current : R = 12 L = 20 mH

Vo 2Vm /  2(120) 2 /  108
Io = = = = = 9.0 A.
R R 12 12

b) Power is determined from the Fourier series. Using Eq. 4-4 and 4-5.

n Vn, V. Zn. Ω In, A.

2 72.0 19.3 3.74
4 14.4 32.5 0.444

2 2
 3.74   0.444 
I rms  9.0 + 
2
 +  = 9.39 A.; P = I rms R = (9.39) 12 = 1, 058 W .
2 2

 2   2 
P 1058
I s ,rms = I o ,rms = 9.39 A.; pf = = = 0.94
S (120)(9.39)

4-7)

Vo 2Vm 2 2(40)
a) I o = = = = 9.0 A.
R R 4
I V 40
b) I rms = m = o ,rms = = 10 A.
2 R 4
c) I s ,avg = 0
N   40 
d ) I s ,rms = I o ,rms  2  = 10   = 1.67 A.
 N1   240 
4-8) Load:

2Vm
Vo = I o R = (10)(15) = 150 V =

Vo 150
Vm = = = 2.36 V .
2 2
V 236
Vo ,rms = m = = 166.6 V . on each sec. tap, 333.2 V . overall
2 2
N V 120
120 V . source : 1 = 1 = = 0.36:1 or 1:2.78
N 2 V2 333.2
N1 240
240 V . source : = = 0.72:1 or 1:1.39
N 2 333.2

4-9)

2Vm
V = I o R = (5)(10) = 50 V . =

Vo 50
Vm = = = 78.5 V .
2 2
V 78.5
Vo ,rms = m = = 55.5 V . on each sec. tap, 111 V . overall
2 2
N V 120
120 V . source : 1 = 1 = = 1.08:1
N 2 V2 111
N1 240
240 V . source : = = 2.16:1
N 2 111

4-10)

Vo 2Vm
a ) I o = 10 A. = =
R R
2Vm 2(120) 2
R= = = 10.8  total
 Io  10
Rx = 10.8 − 4 = 6.8 
b) V2 = 72 Z 2 = 151
V2 72
I2 = = = 0.4764
Z 2 151
I o  2 I 2 = 2(0.4764) = 0.953 A.
4-11)

2Vm 2(120) 2
− Vdc − 48
V − Vdc
a) I o = o =  =  = 20.0 A.
R R 3
Pdc = I oVdc = (20.0)(48) = 961 W .

b) Fourier Series

n Vn, V. Zn. Ω In, A.

2 72.2 11.7 6.16
4 14.4 22.8 0.631

2 2
 6.16   0.631 
I rms = 20.0 + 
2
 +  = 20.5 A.
 2   2 
PR = I rms
2
R = (20.5) 2 (3) = 1, 259 W .
 170  P 961 + 1259
c) S = Vrms I rms =   (20.5) = 2460 VA; pf = S = 2460 = .90
 2
d ) I o = 2 I 2 = 6.16 = 12.32 A.

_____________________________________________________________________________________

4-12

2Vm 2(340)
− Vdc − 96
Vo − Vdc
a) I o = =  =  = 24.1 A.
R R 5
Pdc = I oVdc = (24.1)(96) = 2,313 W .

b) Fourier Series

n Vn, V. Zn. Ω In, A.

2 144.3 30.6 4.72
4 28.9 60.5 0.477
 2 2
 4.72   0.477 
I rms = 24.1 + 
2
 +  = 24.3 A.
 2   2 
PR = I rms
2
R = (24.1) 2 (5) = 1,958 W .
 340  P 2313 + 1958
c) S = Vrms I rms =   (24.3) = 5,848 VA; pf = S = = .90
 2 5848
d ) I o = 2 I 2 = 4.72 = 9.44 A.

4-13) I 0 = 7.03 A. from PSpice

4-14) a) Continuous current; P=474 W.

b) Discontinuous current; P=805 W.

4-15

2Vm 2(120 2)
− Vdc − 24
Vo − Vdc
a) I o = =  =  = 21.0 A.
R R 4
Pdc = I oVdc = (21.0)(24) = 504 W .

b) Fourier Series

n Vn, V. Zn. Ω In, A.

2 72.0 30.4 2.37
4 14.4 60.5 0.238

2 2
 2.37   0.238 
I rms = 212 +   +  = 21.1 A.
 2   2 
PR = I rms
2
R = (21.1) 2 (4) = 1, 777 W .
P 504 + 1777
c) S = Vrms I rms = (120 ) (21.1) = 2,529 VA; pf = = = .90
S 2529
4-16

2Vm 2(120 2)
− Vdc − 36
Vo − Vdc
a) I o = =  =  = 14.4 A.
R R 5
Pdc = I oVdc = (14.4)(36) = 519 W .

b) Fourier Series

n Vn, V. Zn. Ω In, A.

2 72.0 45.5 1.58
4 14.4 90.6 0.159

2 2
 1.58   0.159 
I rms = 14.4 + 
2
 +  = 14.45 A.
 2   2 
PR = I rms
2
R = (14.45) 2 (5) = 1, 044 W .
P 519 + 1044
c) S = Vrms I rms = (120 ) (14.45) = 1, 734 VA; pf = = = .90
S 1734

_____________________________________________________________________________________

4-17)

26.5A

100uH

20.0A

40mH

10.0A

0A
150ms 152ms 154ms 156ms 158ms 160ms 162ms 164ms 166ms 168ms 170ms 172ms 174ms
I(L1)
Time
The current with the 100 μH inductor is discontinuous.

4-18)

Vm
V  ; Vo  Vm 120 2 = 169.7 V .; 0.01Vo  1.7 V .
2 fRC
Vm 169.7
C= = = 4160  F .
2 fRVo 2(60)(200)(1.7)
I o Vo 169.7
ID = =  = 0.43 A.
2 2 R 2(200)
I D , peak : from Eq. 4 − 11,
 Vo −1  1.7 
 = sin −1 1 −  = sin 1 −  = 81.9
 Vm  169.7 
 sin  
From Eq. 3 − 48, I D , peak = Vm  C cos  + 
 R 
 sin 81.9 
= 120 2  377(8.32)(10) −3 cos81.9 +  = 38.5 A.
 200 

4-19)

Vo 100
Req = = = 200 ; Vm = 100  Vo ; Vo = 1 V .
I o 0.5
Vm 100
C= = = 4167  F .
2 fRC Vo 2(60)(200)(1)
Io
ID = = 0.2 A.
2
 V 
From Eq. 4 − 11,  = sin −1 1 − o −1
 = sin (0.99) = 81.9
 Vo 
 sin  
From Eq. 3 − 48, I D , peak = Vm   C cos  + 
 R 
 sin 81.9 
= 100  377(4167)(10) −6 cos81.9 +  = 22.6 A.
 200 

4-20) C ≈ 3333/2 = 1667 µF. Peak diode currents are the same. Fullwave circuit has
advantages of zero average source current, smaller capacitor, and average diode current ½ that
for the halfwave. The halfwave circuit has fewer diodes, and has only one diode voltage drop
rather than two.

4-21)

3 L 3(377)(.01)
a) R = 7  : = = 1.62
R 7
> 1  continuous current
2Vm
Vo = = 108 V .


3 L 3(377)(.01)
b) R = 20  : = = 0.57
R 20
From Fig . 4 − 8, Vo  0.7Vm = 0.7(120) 2 = 119 V .
V   119 
(1) Eq. 4 − 18 :  = sin −1  o  = sin −1   = 0.777 rad .
 Vm   120 2 
1
(2) Eq. 4 − 20 : I L (t ) = [V (cos  − cos t ) − Vo (t −  )] < 1 
L m
iL (  ) = 0 = Vm (cos  − cos  ) − Vo (  −  )
= 120 2(cos(.777) − cos  ) − 119(  − .777) →  = 3.216 rad .

1
(3) I L =
  i (t )d (t ) = 6.14 A.

L

(4) Vo = I L R = (6.14)(20) = 122.9 V . 119 V .

Calculated Vo is slightly larger than initial estimate. Try Vo=120 V.:

(1) Vo = 120 V . From Eq. 4 − 18,  = 0.785

(2) From Eq. 4 − 20,
i (  ) = 0 = 120 2[cos(.785) − cos  ] − 120(  − .785) →  = 3.197 rad .

1
(3) I L =
  i(t )d (t ) = 5.895 A.
(4) Vo = I L R = (5.895)(20) = 117.9 V .  120

Therefore, 119 < Vo < 120 V. (Vo=119.6 with more iterations.)

c) PSpice results:

127

R=20

R=7
100

50

R=7

R=20
0
559ms 560ms 561ms 562ms 563ms 564ms 565ms 566ms 567ms 568ms 569ms 570ms
V(OUT+,OUT-) I(L1)
Time

R = 7 results in continuous current with Vo = 108 V. R = 20 results in discontinuous current with Vo =

120 V. The simulation was done with C = 10,000 μF.

4-22) PSpice results with a 0.5 Ω resistance in series with the inductance: For Rload = 5 Ω,
Vo=56.6 V. (compared to 63.7 volts with an ideal inductor); for Rload = 50 Ω, Vo=82.7 V.
(compared to 84.1 volts with an ideal inductor).

4-23)

Vm 120 2
a) I o = (1 + cos  ) = (1 + cos 45) = 4.61 A.
R  20
V 1  sin  120 2 1 45( /180) sin 90
b) I rms = m − + = − + = 5.72 A.
R 2 2 4 20 2 2 4
c) I s ,rms = I o ,rms = 5.72 A.
d ) P = I rms
2
R = (5.72) 2 20 = 655 W .; S = Vrms I rms = (120)(5.72) = 686 VA.
P 655
pf = = = 0.954
S 686
4-24)

Vm 1  sin 2
I rms = − +
R 2 2 4
V 
S = Vrms I rms =  m  ; P = I rms
2
R
 2
Vm 1  sin 2
2 − +
2 2 4
2
P I R 2 I rms R R
pf = = rms
= =
S  Vm  Vm Vm
  I rms
 2
1  sin 2  sin 2
= 2 − + = 1− +
2 2 4  2

4-25) a) α = 15° : Check for continuous current. First period:

Vm
i (t ) = sin((t ) −  ) + Ae −t / = 10.84sin(t − 0.646) + 5.75e −t /.754
Z
i (  ) = 0 →  = 217;  − 180 = 37   → continuous current
Or
L377(0.050)
 = tan −1 = tan −1 = 37   → continuous current
R 25
2V V 208.7
Vo = m cos  = 208.7 V .; I o = o = = 8.35 A.
 R 25

b) α = 75° Check for continuous current. First period:

 = 37 from part a,  = 75 → discontinuous current

Vm
i (t ) = sin((t ) −  ) + Ae −t / = 10.84sin(t − 0.646) − 37.9e −t /0.754
Z
i (  ) = 0 →  = 216 ;  − 180 = 36   → discontinuous current

1
Io =
 
 i(t )d (t ) = 2.32 A.
4-26)a) α = 20°: Check for continuous current. First period:

Vm
i (t ) = sin((t ) −  ) + Ae −t / = 4.12sin(t − 0.756) + 2.36e −t /0.943
Z
i (  ) = 0 →  = 224,  − 180 = 42   → continuous current
L377(0.075)
 = tan −1 = tan −1 = 43   → continuous current
R 30
2V V 101.5
Vo = m cos  = 101.5 V .; I o = o = = 3.38 A.
 R 30

b) α = 80°: Check for continuous current. First period:

Vm
i (t ) = sin((t ) −  ) + Ae −t / = 4.12sin(t − 0.756) − 10.8e −t /0.943
Z
i (  ) = 0 →  = 221;  − 180 = 41   → discontinuous current
L
 = tan −1 = 37   → discontinuous current
R

1
Io =
  i(t )d (t ) = 0.838 A.


4-27) The source current is a square wave of ±Io.

Vo 2Vm
P = I rms
2
R = I o2 R; I o = =
pf R R
2 2
 2V   2V  1
P= m  R= m 
 R     R
Vm  2Vm  2Vm2
S = Vs ,rms I s ,rms = Vs ,rms I o =   =
2  R  R
2
 2Vm  1
P    R 2 2
pf = = =
S 2Vm2 
R
4-28)

I o = 4.5 A.  Vo = I o R = 4.5(20) = 90 V .

I o = 8 A.  Vo = I o R = 8(20) = 160 V .
Vm 160
Eq. 4 − 23 : Vo = (1 + cos  ): forVo = 160 V . and  = 0, Vm = = 251 V .
 2
V    90 
forVm = 251 and Vo = 90,  = cos −1  o − 1 = cos −1  − 1 = 82.7
 Vm   251 
Vm' 120 2
turns ratio = = = 0.68 :1 or 1:148
Vm 251

Note that the turns ratio could be lower (higher secondary voltage) and α adjusted accordingly.

4-29)

Vo = I o R = 10(5) = 50 V .; from Eq. 4 − 30,

 Vo   50 
 = cos −1  −1
 = cos   = 62.5
 m
2V 2 2 (120 ) 
 
 L  −1  377(.1) 
check for continuous current : tan −1   = tan   = 82.4
 R   5 
62.5  82.4  continuous
V2 = 132 V . Z 2 = 75.6  I 2 = 1.75 A.
I o  2(1.75) = 3.5 A.
4-30)

2Vm 2 2(240)
Vo = cos  = cos105 = −56 V .
 
100 − 56
Io = = 4.4 A.; Pdc = I oVdc = (4.4)(100) = 440 W .
10
Pac = Pbridge = I oVo = (4.4)(56) = 246 W .
PR = Pdc − Pac = 440 − 246 = 194 W .
V2
From Fig . 4 − 12,  0.83 for  = 105
Vm
V2 = 0.83 Vm = 0.83 2(240) = 281 V .
Z 2 = R + j 2 L = 10 + j 2(377)(.8) = 603 − R
V2 281
I2 = = = 0.47 A.; I o  2 I 2 = 0.94 A. p − p
Z 2 603
4-31)

Vo − Vdc
a) I o =
R
 V −V  1
Pbridge (absorbed )  I o (−Vo ) =  o dc  (−Vo ) = −   (Vo2 − VoVdc )
 R  R
Vo − VoVdc + Pbr R = 0
2

Vo2 + 100Vo + 2000(0.8) = 0

Vo = −20 V or − 80 V .
2000 2000
with Vo = −20, I o = = 100 A.; with Vo = −80, I o = = 25 A.
20 80
choose Vo = −80 V . to minimize losses
 Vo  −1  −80 
 =cos −1   = cos  137.8
 2Vm   2 2(120 
V2
b) at  = 137.8, from Fig . 4 − 12,  0.65  V2 = 0.65 2(120) = 110 V .
Vm
I o  (.1) I o = (.1)(25) = 2.5 A.
I o
I2 = = 1.25 A.
2
V 110
Z2 = 2 = = 88  = R + j 2o L  2o L
I 2 1.25
Z2 88
L= = = 0.117 H = 117 mH
2o 2(377)

Choose L somewhat larger, say 120 mH, to allow for approximations.

4-32) In Fig. 4-14, Pac = Pbridge = -VoIo = 1000 W. Using Vdc = -96 V gives this solution:

Kirchhoff ' s voltage law gives − Vo + (1) I o − 96 = 0

−1000
Vo =
Io

1000
+ I o − 96 = 0
Io

I o2 − 96 I o + 1000 = 0

I o = 84.11 or 11.89 A. Use11.89 A.

then Vo = −84.11V .

 Vo   −84.11 
 = cos −1  −1
 = cos   = 141.1
 2 (120 ) 2 
 2Vm  

From Fig .4 − 14,

V2
Vm
 0.64 gives ( )
V2 = 0.64 120 2 = 109 V .

1.189
I o  2 I 2 = 0.10 ( I o ) = 1.189 A. I2 = = 0.595 A
2
V2 109
Z2 = = = 183 = R + j L = 1 + j L
I 2 0.595

183 183
 L  183 L = = 0.48 H .
 377

_____________________________________________________________________________________
4-33)

a ) Pdc = 5000 W . absorbed → I oVdc = −5000

−5000
Io = = 33.3 A.
−150
Vo = −150 + 0.6 I o = −150 + 0.6(33.3) = −130 V .
2Vm V    −130 
Vo = cos  →  = cos −1  o  = cos −1   = 127
  2Vm   2 2(240) 
b) Pbridge = I o (−Vo ) = (33.3)(130) = 4329 W .
V2
c) From Fig . 4 − 12, at 127,  0.73 → V2 = 0.73(240) 2 = 248 V .
Vm
I o
I o = 0.1I o = 0.1(33.3) A.; I 2 = = 1.67 A.
2
V2 248
Z2 = = = 149   2o L
I 2 1.67
149
L= = 0.197 H  200 mH
2(377)

4-34)

3Vm 3 2(480)
a ) Vo = = = 648 V .
 
Vo 648
Io = = = 12.96 A.
R 50
V 480 2  2
b) io (t ) = m sin t = sin t = 13.6sin t for  t 
R 50 3 3
2 /3
1
I rms =
 /3 
 /3
(13.6sin t ) 2 d (t ) = 12.98 A.

2
I s ,rms = (12.98) = 10.6 A.
3
c) P = I rms
2
R = (12.98) 2 50 = 8419 W .
S = 3VI = 3(480)(10.6) = 8808 VA
P 8419
pf = = = 0.956
S 8808
4-35)

3Vm 3 2(240) Vo 324

a ) Vo = = = 324 V .; I o = = = 4.05 A.
  R 80
6Vm
b) V6 = = 0.055Vm = 0.055 2(240) = 18.5 V .
 (62 − 1)
Z 6 = R = 80
V6 18.5
I6 = = = 0.23 A.
Z6 80
2
 0.23 
I rms  I + I 6 rms = 4.05 + 
2
o
2
 = 4.06 A.
 2 
I 4.04
c) I D = o = = 2.02 A.
2 2
I 4.05
d ) I D ,rms = o ,rms = = 2.87 A.
2 2
I o ,rms 2 4.06 2
e) I s ,rms = = = 3.31 A.
3 3
f) P=I 2
o , rms R = (4.06) 2 80 = 1315 W .; S = 3VI = 3(240)(3.31) = 1376 VA
P 1315
pf = = = 0.956
S 1376
4-36)

3Vm 3 2(480) Vo 649

a ) Vo = = = 649 V .; I o = = = 6.49 A.
  R 100
6Vm
b) V6 = = 0.055Vm = 0.055 2(480) = 37.1 V .
 (62 − 1)
Z 6 = R + j 6o L = 100 + j 6(377)(.015) = 100 + j 37.9 = 106 
V6 37.1
I6 = = = 0.35 A.
Z 6 106
2
 0.35 
I rms  I + I 6 rms = 6.49 + 
2
o
2
 = 6.49 A.
 2 
I 6.49
c) I D = o = = 3.25 A.
2 2
I 6.49
d ) I D ,rms = o ,rms = = 4.59 A.
2 2
I o ,rms 2 6.49 2
e) I s ,rms = = = 5.3 A.
3 3
f) P=I 2
o , rms R = (6.49) 2100 = 4212 W .; S = 3VI = 3(480)(5.3) = 4406 VA
P 4212
pf = = = 0.956
S 4406

4-37)

There are no differences between the calculations in Problem 4.36 and the PSpice results. The
power absorbed by each diode ia approximately 1.9 W.

4-38)Equation (4-46) gives values of of I1 = 28.6 A, I5 = 5.71 A, I7 = 4.08 A, I11 = 2.60 A, and I13
= 2.20 A. All compare well with the PSpice results. The total harmonic distortion (THD) is
27.2% when including harmonics through n = 13.

_____________________________________________________________________________________
4-39)

a) Vo = I o R = (25)(120) = 3000 V .
  Vo  −1   3000 
 = cos −1   = cos   = 57.7
 3Vm   3 2(4160) 
V6
b) From Fig . 4 − 21,  0.28  V6 = 0.28 2(4160) = 1640 V .
Vm
V12
 0.135  V12 = 794 V .
Vm
V18
 0.09  V18 = 525 V .
Vm

c)

50A

0A Load

-50A
I(R)

40A
S1

0A
I(S1)
80A

S4
SEL>>
0A
I(S4)
50A
Ia
0A

-50A
65ms 70ms 75ms 80ms 85ms 90ms 95ms 100ms
-I(VAN)
Time
4-40)

a ) Vo = I o R = (10)(50) = 500 V .
  Vo
 −1   500 
 = cos −1   = cos   = 39.5
 3Vm
  3 2(480) 
V
b) From Fig. 4 − 21, 6  0.21  V6 = 0.21 2(480) = 143 V .
Vm
V12
 0.1  V12 = 68 V .
Vm
V18
 0.07  V18 = 48 V .
Vm

c)

20A

SEL>> Load
-20A
I(R)

S1
10A

0A
I(S1)

10A S4

0A
I(S4)

10A
0A Ia

-10A

65ms 70ms 75ms 80ms 85ms 90ms 95ms 100ms

-I(VAN)
Time

_____________________________________________________________________________________
4-41)

3Vm 3 2(480)
a ) Vo = cos  = cos 35 = 531 V .
 
Vo 531
Io = = = 10.6 A.
R 50
V
b) 6  0.19  V6 = 0.19 2(480) = 130 V .
Vm
Z 6 = R + j 60 L = 50 + j 6(377)(0.05) = 124 
V6 130
I6 = = = 1.05 A.
Z 6 124
2 2
 I   1.05 
I o ,rms  i +  6  = 10.62 + 
2
o  = 10.65 A.
 2  2 
 2  2
I s ,rms =   I o ,rms =  10.65 = 8.6 A.
 3  3

4-42)

3Vm 3 2(480)
a ) Vo = cos  = cos 50 = 417 V .
 
Vo 417
Io = = = 41.7 A.
R 10
V
b) 6  0.25  V6 = 0.25 2(480) = 170 V .
Vm
Z 6 = R + j 60 L = 10 + j 6(377)(0.01) = 24.7 
V6 170
I6 = = = 6.9 A.
Z 6 24.7
2 2
 I   6.9 
I o ,rms  i +  6  = 41.7 2 + 
2
o  = 42.3 A.
 2  2
 2  2
I s ,rms =   I o ,rms = 
  41.7 = 34 A.
 3  3
4-43)

a ) Vo = I o R = (20)(20) = 400 V .

 V
 −1   400 
a = cos −1  o = cos   = 52
 3Vm
  3 2(480) 
V
b) From Fig . 4 − 21, 6  0.25  V6 = 0.25( 2)(480) = 170 V .
Vm
2 2 2
 I 6   I12   I18 
  +  +   0.02 I o or I 62 + I122 + I182  0.02 2 I o
 2  2  2
Z 6 = R + j 6 L
V6
= I 6  0.02 I o = 0.02(20) = 0.4 A.
Z6
V6 170
Z6 = = = 425  = R + j 6 L = 20 + j 6(377) L
I 6 0.4
6(377) L  425
425
L= = 0.188 H
6(377)
L  190 mH

4-44)

 Vo   −280 
 = cos −1  −1
 = cos   = 149.8
 3 2 ( 280 ) 
 3Vm  

300V − 280V
Io = = 40 A
0.5
Pdc = Vdc I o = ( 300 )( 40 ) = 12,000 W supplied

PR = I o2 R = 402 ( 0.5 ) = 800 W absorbed

PBridge = Pac = ( 280 )( 40 ) = 11, 200 W absorbed

_____________________________________________________________________________________
4-45)

 1.5(10)6 W . 
Pbridge = 1.5 MW .; Vo =   = −1500 V .
 1000 A. 
  Vo  −1   ( −1500) 
 = cos −1   = cos   = 105.5
 3Vm   3 2(4160) 
 2
I s ,rms =  1000 = 816 A.
 3 

4-46)

−100,000
With Pac = Pbridge = 100 kW absorbed, - Vo I o = 100,000 or Vo =
Io

Kirchhoff's voltage law gives - Vo + ( I o ) ( 0.1 ) − 1000V = 0

−100,000
Substituting for Vo , − + 0.1I o − 1000 = 0
Io

0.1I o2 − 1000 I o + 100,000 = 0 yields the two roots Io = 9,890 A or 101 A

Choose Io = 101 A because this solution results in lower I o2 losses.

Vo = −1000V + I o ( 0.1 ) = −1000 + ( 0.1)101 = −989.9 V

3Vm
Vo = cos  , where Vm = 2 12,500 ( N 2 / N1 ) 


 Vo   −989.9 
 = cos −1   = cos −1
 
 3Vm   3 2 12,500 ( N 2 / N1 )  
  
N 2 / N1 = 1 will theoretically work, but  = 93.36, but the harmonic content will be large.

A better solution would be to choose N 2 / N1 to be perhaps 1/10 (step-down). Then  = 125.9

 V6
From Fig. 4-21,  0.3
Vm

Vm = 2 (12,500 ( N 2 / N1 ) ) = 2 (12,500 / 10 ) = 1768 V

V6  0.3Vm = 0.3 (1768 ) = 530V

I o  2 I 6  0.5I o = 0.5 (101) = 5.05 A → I 6 = 2.525 A

V6 530
Z6 = = = 210  = R + j L = 0.1 + j 377 L  377 L
I 6 2.525

210
L= = 0.56 H
377

_____________________________________________________________________________________

4-47)

a ) Vo1 =
3Vm, L − L
cos(1 ) =
(
3 230 2 ) cos(45) = 329.5 kV
 

Vo 2 =
3Vm, L − L
cos( 2 ) =
(
3 230 2 ) cos(134.4) = −326 kV
 
Vo1 + Vo 2 329.5kV − 326kV
Io = = = 231 A
R 15
P1 = Vo1 I o = 76.17 MW

P2 = Vo 2 I o = −75.37 MW

b) Pline = I o2 R = 800 kW

_____________________________________________________________________________________
4-48)

3Vm
a ) Vm = 230 2 kV ; Vo = cos( )


Vo,max =
3Vm
=
(
3 230 2 ) = 325.3 kV
 
V  
Let Vo 2 = −300 kV (arbitrarily ); Then  2 = cos −1  o 2  = 164.98
 3Vm 
P2 −80MW
Io = = = 267 A (linecurrent )
Vo 2 −300kW

Vo1 = I o R − Vo 2 = 267(12) − (−300kV ) = 303.2 kV

Pline = I o2 R = 853 kW

_____________________________________________________________________________________

4-49)

3Vm
a ) Vm = 345 2 kV ; Vo = cos( )


Vo ,max =
3Vm
=
(
3 345 2 ) = 465.9 kV
 
V  
Let Vo 2 = −425kV (arbitrarily ); Then  2 = cos −1  o 2  = 155.8
 3Vm 
P2 −300MW
Io = = = 706 A (linecurrent )
Vo 2 −425kW

Vo1 = I o R − Vo 2 = 706(20) − (−425kV ) = 439.1kV

Pline = I o2 R = 9.97 MW

_____________________________________________________________________________________
4-50)

8 A.  I o  12 A.
 2V 
Vo =  m  cos  = I o R; Vo1 = 8(8) = 64 V .; Vo 2 = 12(8) = 96 V .
  
 64 
120 − volt source : 1 = cos −1   = 53.7
 2 2(120) 
V
From Fig . 4 − 12, 2  0.73  V2 = 124 V .
Vm
 96 
 2 = cos −1   = 27.3
 2 2(120) 
V2
From Fig . 4 − 12,  0.54  V2 = 92 V .
Vm
V2 124
using V2 = 124 V . for 1 and I o  2.5 A., Z 2 = = = 99 
I 2 2.5 / 2
Z2 99
Z 2 = R + j 20 L  L = = = 0.13 H
20 2(377)
For the 240 − volt source,
 64 
1 = cos −1   = 72.8
 2 2(240) 
V
From Fig . 4 − 12, 2  0.83  V2 = 280 V .
Vm
 96 
 2 = cos −1   = 63.6
 2 2 ( 240 ) 
 
V
From Fig . 4 − 12, 2  0.78  V2 = 265 V .
Vm
V2 280
using V2 = 280 V . for 1 and I o  2.5 A., Z 2 = = = 224 
I 2 2.5 / 2
Z2 244
Z 2 = R + j 20 L  20 L  L = = = 0.3 H
20 2(377)
The 120-volt source requires a smaller filter inductor.

_____________________________________________________________________________________
4-51)

Io =15A in a 20- resistor. I o = 0.1(Io ) = 1.5 A.

First solution using the single-phase 480-V source with a controlled bridge rectifier:

 Vo 
 = cos −1   = 46
 2Vm 
I o  2 I 2 → I 2 = 1.5 / 2 = 0.75 A

From Fig. 4-12, V2 /Vm  0.71 → V2 = 0.71 2 ( 480 ) = 482 V

V2 482
Z2 = = = 643  = R + jn L = 20 + j 2 ( 377 ) L
I 2 0.75

642
2 ( 377 ) L  643 → L = = 851 mH
2 ( 377 )

Second solution using the three-phase 480-V source with a controlled 6-pulse bridge rectifier:

 Vo   300 
 = cos −1  −1
 = cos   = 62.4
 3V  3 2 ( 480 ) 
 m, L − L   
I o  2 I 6 → I 6 = 1.5 / 2 = 0.75 A

From Fig. 4-12, V6 /Vm  0.28 → V6 = 0.28 2 ( 480 ) = 190V

V6 190
Z6 = = = 253  = R + jn L = 20 + j ( 6 ) 377 L
I 6 0.75

253
6 ( 377 ) L  253 → L = = 112 mH
6 ( 377 )

Uncontrolled rectifiers with additional resistances added can also satisfy the specifications.
However, adding resistance would increase power loss and decrease efficiency.

_____________________________________________________________________________________