Cálculos em Circuitos Elétricos

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CHAPTER 3 SOLUTIONS 
2/20/10 
3-1) 
 
0
0
2 2
170 /
) 3.60 .
15
170
) 5.66 .
2 2(15)
) 5.66 (15) 480 .
170
) (5.66) 679 .
2
480
) 0.707 70.7%
679
m
rms m
rms
rms rms
V V
a I A
R R
V V
b I A
R R
c P I R W
d S V I VA
P W
e pf
S VA
 
   
   
  
 
   
 
    
3-2) 
 
0
0 0 0 0
1
2
2
1
) 12 .; (12)(20) 240 .
; 240 754 .
754
533 .
2
240
0.45
533
12
) 26.7 .
0.45
m
o m o
rms
o o
V
a I A I V I R V
R
V
V V V V
V V
N
N
N
b I I A
N
 

     
   
 
 
    
 
 
 
 
 
 
 
3-3) 
 
,
, ,
,
2
,
,
1
1
1 1 1
) ; ; ;
2 2
/
/ 2 12
2 2
/
22
) cos( ) cos(0) 1
1 1
0; cos( ) ;
2 2
s rms m m
rms R rms s rms
s rms rms
m
R rms
m ms rms rms
m
V V VP P
a pf I V V
S V I R
V
R
V R
pf
V VV I
R
b Displacement pf
VV
I pf DF DF
R R
 
 
    
 
 
    
  
  
  
   
      
 
3-4) Using Eq. 3-15, 
 
/
2 2 2 2
1 1
/0.377
) ( ) sin( ) (sin )
( ) 12 (377(0.012)) 12.8
377(0.012)
tan tan 0.361
12
377(0.012)
0.377
12
( ) 13.2sin( 0.361) 4.67 : 3.50 201
)
tm m
t
V V
a i t t e
Z Z
Z R L
L
rad
R
L
R
i t t e rad
b
 

   





  

 

  
     
   
     
   
  
     
2 2
4.36 . ( )
) 6.70 . ( ) (6.70) (12) 538 .
538
) 0.67
(120)(6.70)
avg
rms rms
I A numerical integration
c I A numerical integration P I R W
P
d pf
S

   
   
 
 
 
 
 
 
3-5) Using Eq. 3-15, 
 
/
2 2 2 2
1 1
/0.565
) ( ) sin( ) (sin )
( ) 10 (377(0.015)) 11.5
377(0.015)
tan tan 0.515
10
377(0.015)
0.565
10
( ) 14.8sin( 0.515) 7.27 : 3.657 209.5
tm m
t
V V
a i t t e
Z Z
Z R L
L
rad
R
L
R
i t t e rad
 

   





  

 

  
     
   
     
   
  
    
2 2
) 5.05 . ( )
) 7.65 . ( ) (7.65) (10) 584 .
584
) 0.637 63.7%
(120)(7.65)
avg
rms rms
b I A numerical integration
c I A numerical integration P I R W
P
d pf
S


   
   
 
3-6) Using Eq. 3-15, 
 
 
/
2 2 2 2
1 1
/2.01
) ( ) sin( ) (sin )
( ) 15 (377(0.08)) 33.7
377(0.08)
tan tan 1.11
15
377(0.08)
2.01
15
( ) 10.1sin( 1.11) 9.02 : 4.35 250
) 4.
tm m
t
avg
V V
a i t t e
Z Z
Z R L
L
rad
R
L
R
i t t e rad
b I
 

   





  

 

  
     
   
     
   
  
     

2 2
87 . ( )
) 6.84 . ( ) (6.84) (15) 701 .
701
) 0.427 42.7%
(240)(6.84)
rms rms
A numerical integration
c I A numerical integration P I R W
P
d pf
S
   
   
 
 
 
 
3-7) Using an ideal diode model, R = 48 Ω for an average current of 2 A. 
 
 
 
 
 
 
 Time
0s 5ms 10ms 15ms 20ms
I(R1) AVG(I(L1))
0A
4.0A
8.0A
Current
Average Current
Iavg = 2 A for R = 48 ohms
(16.700m,2.0030)
 
 
 
 
 
3-8) Using Eqs. 3-22 and 3-23, 
 
/
/
2 2 2 2
1 1
1
) ( ) sin( )
sin( )
( ) 10 (377(.075) 30.0
377(.075)
tan tan 1.23
10
377(0.075)
2.83
10
100
sin 0.299 17.
240 2
tm dc
m dc
dc
m
V V
a i t t Ae
Z R
V V
A e
Z R
Z R L
L
rad
R
L
R
V
rad
V
 
 
  
 







 

   
 
    
 
     
   
     
   
  
   
/2.83
2 2
1
( ) 11.3sin( 1.23) 10 21.2 ; 3.94 226
3.13 . ( (100)(3.13) 313 .
) 4.81 . ( (4.81) (10) 231 .
313 231
)
(240
t
avg dc dc avg
rms R rms
i t t e rad
I A numerical integration), P V I W
b I A numerical integration) P I R W
P
c pf
S
  

      
   
   

  0.472 47.2%
)(4.81)
 
 
3-9) Using Eqs. 3-22 and 3-23, 
 
/
/
2 2 2 2
1 1
1
) ( ) sin( )
sin( )
( ) 12 (377(0.12) 46.8
377(0.12)
tan tan 1.31
12
377(0.12)
3.77
12
48
sin 0.287 16.4
120 2
tm dc
m dc
dc
m
V V
a i t t Ae
Z R
V V
A e
Z R
Z R L
L
rad
R
L
R
V
rad
V
 
 
  
 







 

   
 
    
 
     
   
     
   
  
    
/3.77
2 2
( ) 3.63sin( 1.31) 4.0 7.66 ; 4.06 233
1.124 . ( (48)(1.124) 54.0 .
) 1.70 . ( (1.70) (12) 34.5 .
54.0 34.5
)
t
avg dc dc avg
rms R rms
i t t e rad
I A numerical integration), P V I W
b I A numerical integration) P I R W
P
c pf
S
        
   
   

  0.435 43.5%
(120)(1.70)
 
 
 
 
3-10) Using Eq. 3-33, 
 
1 1
( ) (cos cos ) ( )
48
sin sin 0.287 .
120 2
( ) 4.68 4.50cos( ) 1.23 .; 4.483 257
1
( ) ( ) 2.00 .; 2.00(48) 96 .
2
m dc
dc
m
o dc o dc
V V
i t t t
L L
V
rad
V
i t t t A rad
I i t d t A P I V W


    
 

   
 

 
   
   
     
  
     
    
 
 
3-11) 
 
 Time
0s 5ms 10ms 15ms 20ms
AVG(W(Vdc))
0W
100W
200W
300W
L = 0.25 H
 
 
 
 
 
3-12) L ≈ 0.14 H for 50 W (51 W). 
 
 
 
 Time
0s 5ms 10ms 15ms 20ms
AVG(W(Vdc))
0W
50W
100W
L = 0.14 H
(16.670m,51.156)
 
 
 
3-13) Using Eq. 3-34, 
 a) 
 
0
0 0
120 2 54
54.0 .; 4.50 .
12
mV VV V I A
R      
 
b) 
n Vn Zn In 
0 54.02 12.00 4.50 
1 84.85 25.6 3.31 
2 36.01 46.8 0.77 
4 7.20 91.3 0.08 
The terms beyond n = 1 are insignificant. 
 
3-14) 
 
Run a transient response long enough to achieve steady-state results (e.g., 1000ms). The peak-to-
peak load current is approximately 1.48 A, somewhat larger than the 1.35 A obtained using only 
the first harmonic. (The inductance should be slightly larger, about 0.7 H, to compensate for the 
approximation of the calculation.) 
 
 
 
 
 
 
 
3-15) 
a) 
 
0
1
1 0
2 2 2 2
1
2 2 2
50
3.98 .
4
/ 2 25
0.05 0.199 .
( ) ( )
25
( ) 9 ( ) 125
0.199
125
0.33
2 60
m
m
V
I A
R
VV
I I A
Z R L R L
R L L L
L H
 
 
  

  
    
 
      
 
 
b) A PSpice simulation using an ideal diode model gives 0.443 A p-p in the steady state. This 
compares with 2(I1)=2(0.199)=0.398 A p-p. 
 
3-16) 
 
0
0
0
1 1
1
2 21
1
1
2 2 2 2
,
170
) 54.1
54.1 24
3.01 .
10
1 . 2 0.5 .
170
85
2 2
85
170 ( )
0.5
170
450 .
377
) (3.01)(24) 72.2 .
) ; (3.01) (0.5 / 2)
m
dc
o
m
dc avg dc
R rms rms n rms
V
a V V
V V
I A
R
i A I I A
V
V V
V
Z R L L
I
L mH
b P I V W
c P I R I I
 
 
  
 
  
    
  
      
 
  
    
2
3.12 .
(3.12) (10) 97.4 .R
A
P W 
 
3-17) a) τ = RC = 10310-3=1 s; τ/T = 60. With τ >> T, the exponential decay is very small 
and the output voltage has little variation.b) Exact equations: 
 
1 1
(2 )/
tan ( ) tan (377) 1.5573 90.15
sin 200sin(90.15 ) 199.9993
sin sin 0 1.391 79.72
(1 sin ) 3.21 .
m
RC
o m
RC rad
V
e rad
V V V
   
   

  

 
  
        
  
     
   
 
 c) Approximation of Eq. 3-51: 
 
3 3
200
3.33 .
(60)(10 )(10 )
m
o
V
V V
fRC 
   
 
3-18) a) R = 100 Ω: τ = RC (100)10-3 = 0.1 s; τ/T = 6. 
 
1 1
(2 )/ )
3
tan ( ) tan (37.7) 1.5973 91.52
sin 200sin(91.52 ) 199.93
sin sin 0 1.0338 59.23
(1 sin ) 28.16 . ( )
200
33.3 . ( )
(60)(100)(10 )
m
RC
o m
m
o
RC rad
V
e rad
V V V exact
V
V V approximation
fRC
   
   

  

 
  

        
  
     
   
    
 b) R = 10 Ω: τ = RC (10)10-3 = 0.01 s; τ/T = .6. 
 
1 1
(2 )/ )
3
tan ( ) tan (3.77) 1.830 104.9
sin 200sin(104.9 ) 193.3
sin sin 0 0.2883 16.5
(1 sin ) 143.2 . ( )
200
333 . ( )
(60)(10)(10 )
m
RC
o m
m
o
RC rad
V
e rad
V V V exact
V
V V approximation
fRC
   
   

  

 
  

        
  
     
   
   
 
 In (a) with τ/T=6, the approximation is much more reasonable than (b) where τ/T=0.6. 
 
 
 
3-19) a) With C = 4000 µF, RC = 4 s., and the approximation of Eq. 3-51 should be reasonable. 
 
120 2
0.707 .
(60)(4)
m
o
V
V V
fRC
   
 
 b) With C = 20 µF, RC = 0.02, which is on the order of one source period. Therefore, 
the approximation will not be reasonable and exact equations must be used. 
 
1 1 6tan ( ) tan ((377)(1000)(20(10) ) 1.703 97.6 )
0.5324 30.5 ( . 3 43)
sin 83.6 .o m m
RC rad
rad numerically from Eq
V V V V
   


          
   
   
 
3-20) a) With C = 4000 µF, RC = 2 s., and the approximation of Eq. 3-51 should be reasonable. 
 
120 2
1.41 .
(60)(2.0)
m
o
V
V V
fRC
   
 
 b) With C = 20 µF, RC = 0.01, which is on the order of one source period. Therefore, 
the approximation will not be reasonable and exact equations must be used. 
 
1 1 6tan ( ) tan ((377)(500)(20(10) ) 1.83 104.9 )
0.2883 16.5 ( . 3 43)
sin 121 .o m m
RC rad
rad numerically from Eq
V V V V
   


          
   
   
 
3-21) From Eq. 3-51 
 
1 1
,
,
120 2
1,886
60(750)(2)
2
sin 1 sin 1 1.417 81.2
120 2
sin
cos 18.7 .
0.226 .
m
o
o
m
D peak m
m
D avg
V
C F
fR V
V
rad
V
I V C A
R
V
I A
R



 
 
  

   
         
  
 
   
 
 
 
 
3-22) Assuming Vo is constant and equal to Vm, 
 2 2 2 2(120 2)
576
50
o m mV V VP R
R R P
      
 
 From Eq. 3-51 
 
1 1
,
,
120 2
3,270
60(576)(1.5)
1.5
sin 1 sin 1 1.438 82.4
120 2
sin
cos 28.1 .
0.295 .
m
o
o
m
D peak m
m
D avg
V
C F
fR V
V
rad
V
I V C A
R
V
I A
R



 
 
  

   
         
  
 
   
 
 
 
3-23) Using the definition of power factor and Vrms from Eq. 3-53, 
 
2 2
, , , ,
/ /
( )( ) ( )( / )
sin 2
1
1 sin 2 1 sin 22 2
1
2 2 2 4/ 2 2
rms rms rms
s rms s rms s rms rms s rms
m
m
V R V R VP
pf
S V I V V R V
V
V
 
    
   
   
 
      
 
3-24) 
 
2
2
,
120 2
) (1 cos ) (1 cos 45 ) 46.1 .
2 2
sin 2
) ; 1
2 2
120 2 0.785 sin(2(0.785))
1 80.9 .
2 2
80.9
65.5 .
100
80.9 65.5
) (120) 97.1 ; 0.674 67.4%
100 97.1
m
o
rms m
rms
s rms rms
V
a V V
V V
b P V
R
V
P W
P
c S V I VA pf
S

 
 
 
 
     
   
   
 
 
       
 
 
 
 
3-25) 
 
1 1
2
,
,
2
,
) (2.5)(30) 75 (1 cos )
2
2 2 (75)
cos 1 cos 1 65.5 1.143
240 2
)
sin 2 240 2 1.143 sin(2(1.143))
1 1 147.6 .
2 2 2 2
147.6
726 .
30
) (
m
o o
o
m
o rms
m
o rms
s rms rms
V
a v I R V
V
or rad
V
V
b P
R
V
V V
P W
c S V I


 

 
   
 
    
   
        
  

      
 
 
147.6 726
240) 1181 ; 0.615 61.5%
30 1181
P
VA pf
S
 
     
 
 
3-26) 
 
/0.754
2 2 2
) ( ) 5.42sin( 0.646) 1.33 .
25 0.524 , 3.79 217 ( )
1
) ( ) ( ) 1.80 .
2
1
) ( ) ( ) 2.80 .; (2.80) 25 193 .
2
t
o
rms o R rms
a i t t e A
rad rad numerically
b I i t d t A
c I i t d t A P P I R W





 
 
 

 

  
     
 
     


 
3-27) 
 
/0.707
2 2 2
) ( ) 3.46sin( 0.615) 6.38 .
60 1.047 , 3.748 215 ( )
1
) ( ) ( ) 0.893 .
2
1
) ( ) ( ) 1.50 .; (1.50) 40 90.3 .
2
t
o
rms o R rms
a i t t e A
rad rad numerically
b I i t d t A
c I i t d t A P P I R W





 
 
 

 

  
     
 
     


 
 
3-28) α ≈ 46°. Do a parametric sweep for alpha. Use the default (Dbreak) diode, and use Ron = 
0.01 for the switch. Alpha of 46 degrees results in approximately 2 A in the load. 
 
 
 
 
 
 
 
 
3-29) α ≈ 60.5°. Do a parametric sweep for alpha. Use the default (Dbreak) diode, and use Ron 
= 0.01 for the switch. Alpha of 60.5 degrees results in approximately 1.8 A in the load. 
 
 
 
 
 
 
 
 
3-30) From Eq. 3-61, 
 
/3.142
2 2 2
) ( ) 4.29sin( 1.263) 4.0 7.43 ., 0.873 3.95
1
( ) ( ) 1.04 ., (1.04)(48) 50.1 .
2
1
) ( ) ( ) 1.67 .; (1.67) 12 33.5 .
2
50.1 33.5
)
(120)(1.
t
o dc o dc
rms R rms
a i t t e A t rad
I i t d t A P I V W
b I i t d t A P I R W
P
c pf
S





  
 

 

     
    
    

 


0.417 41.7%
67)
 
 
3-31) From Eq. 3-61, 
 
/0.565
2 2 2
) ( ) 2.95sin( 0.515) 0.96 3.44 ., 1.047 3.32
1
( ) ( ) 0.454 ., (0.454)(96) 43.6 .
2
1
) ( ) ( ) 0.830 .; (0.830) 100 69.0 .
2
43.6 69.0
)
(2
t
o dc o dc
rms R rms
a i t t e A t rad
I i t d t A P I V W
b I i t d t A P I R W
P
c pf
S





  
 

 

     
    
    

 


0.565 56.5%
40)(0.830)
 
 
 
 
 
3-32) α ≈ 75°. Alpha = 75 degrees gives 35 W in the dc voltage source. An Ron = 0.01 for the 
switch and n = 0.001 for the diode (ideal model). 
 
 
 
 
 
 
 
 
 
 
 
 
3-33) From Eq. 3-61, 
 
/14.1
2 2 2
) ( ) 5.99sin( 1.50) 24.0 29.3 ., 0.873 4.24
1
( ) ( ) 1.91 ., (1.91)(48) 91.6 .
2
1
) ( ) ( ) 2.93 .; (2.93) 2 17.1 .
2
t
o dc o dc
rms R rms
a i t t e A t rad
I i t d t A P I V W
b I i t d t A P I R W





  
 

 

     
    
    


 
3-34) α ≈ 81° 
3-35) 
 
( )
sin
( ) 1
[ sin ]
( ) 1
[ sin ]
( )
1
( ) ( sin ) ( )
(cos cos ) ( )
( ) 4.34 7.58cos 1.82 ., 1.309 4.249
1
( ) ( ) 1.9
2
m dc
m dc
m dc
t
m dc
m dc
o
di t
L V t V
dt
di t
V t V or
dt L
di t
V t V
d t L
i t V t V d t
L
V Vt t
L L
i t t t A t
I i t d t








 
  

   
 
   
 

 
 
 
 
   
    
 

 1 .A
 
3-36) 
 v0 = vs when S1 on, v0=0 when D2 on 
 
1
, sin( ) ( ) (1 cos )
2 2
(1 cos )
2
o m
o o m
m
o
V V
I V V t d t
R
V
I
R


  
 


   
  

 
 
 
 
3-37) 
 
1 3
1
cos 1 ; 377(1.5)(10) 0.566
5(0.452)
cos 1 10.47
120 2
120 2 5(.566)
1 1 53.57 .
2 2 2(120)
( 54.0 .)
L s
s s
m
m L s
o
m
m
I X
u X L
V
u
V X X
V V
V
V
compared to V

 

 

 
      
 
 
    
 
  
      
   

 
PSpice: 
Use a current source for the constant load current: 
 
 
 
 
 
 
 
 
 
D1 to D2 
 
 
 
 
 
 
D2 to D1 
 
 
 
 
 
 
3-38) u = 20°. Run the simulation long enough for steady-state results. From the Probe output, 
the commutation angle from D1 to D2 is about 20 degrees, and from D2 to D1 is about 
18 degrees. Note that the time axis is changed to angle in degrees here. 
 
 
 
 
 
 
 
 
 
 
 
3-39) Run the simulation long enough for steady-state results. From the Probe output, the 
commutation angle from D1 to D2 is about 16.5 degrees, and from D2 to D1 is about 
14.7 degrees. Note that the time axis is changed to angle in degrees here. 
 
 
 
 
 
 
 
 
 
3-40) At ωt = π, D2 turns on, D1 is on because of the current in LS (see Fig. 3-17). 
 
 
11
1
1
1
1
; sin
( )
sin( ) ( ) ( )
, 0 [ 1 cos( )]
cos( ) cos 0 ( 1 cos ) cos
cos 1
dD
LS m s S
t
mD
D
s
m
D L
s
m m
L L
s s
L s
m
didi
KVL v V t L L
dt d t
Vdi
t d t i
d t L
V
at t u i u I
L
V V
u u u I u I
L L
I X
u
V


 

  
 
  


 

  
 
       
          
 
   
 

 
 
 
 
 
3-41) At ωt = α, 
 
2
2
1
1
( ) sin( ) ( ) 0 [cos cos ]
( ) [cos cos ]
( ) 0 [cos cos( )]
cos cos( )
cos cos
t
m
s m
s s
m
D L s L
s
m
D L
s
L s
m
L s
m
V
i t V t d t t
L L
V
i wt I i I t
L
V
i u I u
L
I L
u
V
I X
u
V


    
 
 

  


 
 
   
    
     
    
 
   
 

 
3-42) A good solution is to use a controlled half-wave rectifier with an inductor in series with the 
48-V source and resistance (Fig. 3-15). The switch will change the delay angle of the 
SCR to produce the two required power levels. The values of the delay angle depend on 
the value selected for the inductor. This solution avoids adding resistance, thereby 
avoiding introducing power losses. 
3-43) Several circuit can accomplish this objective, including the half-wave rectifier of Fig. 3-2a 
and half-wave rectifier with a freewheeling diode of Fig. 3-7, each with resistance added. 
Another solution is to use the controlled half-wave rectifier of Fig. 3-14a but with no 
resistance. The analysis of that circuit is like that of Fig. 3-6 but without Vdc. The 
resulting value of α is 75°, obtain from a PSpice simulation. That solution is good 
because no resistance is needed, and losses are not introduced. 
3-44 and 3-45) The controlled half-wave rectifier of Fig. 3-15 (without the resistance) can be 
used to satisfy the design specification. The value of the delay angle depends on the value 
selected for the inductor.