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CAPÍTULO 10 ANÁLISIS DE ESTADO ESTACIONARIO SINUSOIDAL. SECCIÓN 10.2 ANÁLISIS NODAL 10.1 Determine i en el circuito de la figura 10.50

2 cos(10 ) ! " = 2#0° $ = 10

1% ! &' =

1 1 = = ,*0.1[-] *$+ *10

1/ ! &3 = *$/ = *10[-] &4 = 1[-] &56' = &56' =

100 10 1 7 *10 =9 8* : [-] 101 101 1 8 *10 &563 =

&563 =

&4 7 &3 &4 7 &3

&56' 7 &' &56' 7 &'

&56' 7 ,*0.1 100 ;;0 =9 ,* : [-] &56' , *0.1 ;;01 ;;01 &56 = &563 8 1 &56 =

100 ;;0 ,* 81 ;;01 ;;01

10001 ;;0 &56 = 9 ,* : [-] = 1.01? = = = "#[B] 4 #;@ # A 4 A 1 4 1C : >D = #;C = # A 4 A 1 = #4[B]

*16& " 10°- " #4

*16& " 10°- " #4

F

EF + 0.5EF = E? F

+

+

*16& " 10°- " #8

*16& " 10°- " #8

F

F

"

=

F

1"# F

1"#

=0

*16& " 10°*16& " 10°F F F " + " " =0 #8 #4 #4 #8 1 " # F

+

F

+

#8 #8

F

+

F

+

1"# 1"# F

F

=

F

=

#4 #4

*16& " 10°- *16& " 10°+ #4 #8 *16& " 10°- *16& " 10°+ #4 #8

1 1 1 $ + + % = "1.04 " #5.91 #8 1 " # #4

1 1 ! + " # = $1.04 $ "5.91 2 8 % = $4.74 $ "10.63 &' = &' =

% 1$"

$4.74 $ "10.63 1$"

&' = 2.94 $ "7.69 %( = 1 ) &'

%( = 1 ) *2.94 $ "7.69, %( = 2.94 $ "7.69 %( = 8.23- $ 69.04°

10.5 Encontrar /: en el circuito de la Fig. 10.54.

%(

25 cos*4 ) 10; % = 25-0° ? = 4 ) 10;

2@A > B' =

1 1 = = $"125[F] DE "?C " ) 2 ) 10 ) 4 ) 10;

1 1 G > BH = "?G = " ) 4 ) 10; ) = "1000[F] 4 4 &' + &H = &(

&( =

25-0° $ %( 2I

%( %( $ 10&( 25-0° $ %( + = "1000 $"125 2I

$*0.5 + "7,%( + "0.04*25 $ %( , = $12.5 $*0.5 + "7.04,%( = $12.5 $ " %( = 0.2668 $ "1.7566

&( =

25-0° $ %( 25 $ 0.2668 + "1.7566 = 2I 2I &( = 0.012 + "8.78 ) 10DJ &( = 0.012-4.1°

= 12"#$(4000% + 4.1°)[&']

!

10.6 Determine *, en la figura 10.55.

"4 20

!

"

!

20 !

20

"

-

#

+

+

/

!

20 + $10

= 3%0°

4 # ! + = 3%0° 20 20 + $10 #

=

20

20 + $10

1

4 ! ! + = 3%0° 20 + $10 20 + $10

("2 + $0.5)

# =

= 350°

1

= 60 + $30

!

= 32.53% " 139.40°

#

= 29.10% " 165.96°

20

20 + $10

(32.53% " 139.40°)

10.7 Utilice análisis nodal para encontrar V en el circuito de la Fig. 10.56

&! + &' + &* + 6%30° = 0

" 120% " 15° + 6%30° + + =0 "$30 50 40 + $20

40 + $20

+

"$30

+

50

=

120% " 15° " 6%30° 40 + $20

1 7 , +$ - = "3.19 " $4.78 25 300 / = "111.49 " $54.47

= 124.08! " 153.96°

10.8. Use el análisis nodal para encontrar la corriente #$ en el circuito de la fig. 10.57. Si #% = & '(%)*$$+ , -/7 [:]

;< = 6 cos)200> , 157 = 6!15° ? = 200

50@A B CD =

1 1 = = "E100[J] E?F E G 50 G 10HI G 200

100KL B CM = E?L = E G 100 G 10HN G 200 = E20[J] ;O , ;D , ;M = 6!15° , 0.1

O

5

O

" 40

,E

O

)"2.5 , E7

D

O

,

"E100

, 2.5 O

O

" 2.5

O

,

D

20E "2E

,

D

O

O

,4

, )1 " 2E7

"2.5 , E U 3

= 600!15° , 10

O

= ;O

O

D D

O

=

,3

" 40 O O

D

"

D

=0

= 0 B PQRSQ;óT2

"2.5 600!15° X V U OV = W 1 " 2E 0 D

O

= 145.52! " 89.03°

D

= 195.23!154.39° ;O =

;O =

D

=

10

)1 " 2E7

3

O

10 O

= 600!15° B PQRSQ;óT1

;N , 0.1 D

= 6!15° ,

20

" 2.5

O

O

" 40

D

145.52! " 89.03° " 195.23!154.39° 40 ;O = Y.2Y! " 52.16°

10.9. Use análisis nodal para encontrar

!

en el circuito de la figura.

10 "#$(10% &) ' * = 10+0° , = 10%

50-. ' /2 =

1 1 = = 93:0[;] 3,4 3 6 50 6 1078 6 10%

10 ' /? = 3,> = 3 6 10 6 107% 6 10% = 310[;]

Nodo a: *@ = 10+0A

Nodo b: B2 = BC D B?

*@ 9 *E *E 9 0 *E 9 *F = 9 A :0 :0 :3

1 1 1 1 1 +0 = *E G 9 D H D *F G H : :0 :3 :0 :3 5+0 = (1 D 53)*E 9 53*F AAAAA(1)

Nodo c: B? = IBC D B%

9

*E 9 *F *F 9 *J *E = DIG H :3 1003 :0

503*E 9 503*F = 3*J 9 3*F D :0*E

0 = (:0 9 503)*E D IK3*F 9 3*J AAAAAA(:)

Nodo d: B% = BL *F 9 *J *J = 1003 M0

10 0 = 3*F D G 9 3H *J AAAAAA(M) M

El sistema nos queda:

Donde:

5!0 = (1 + 5")#$ % 5"#& 0 = (25 % 50")#$ + 49"#& % "#' 10 0 = "#& + * % ", #' 3 #' = 6.15!70.15 #' = #-

#- = 6.15!70.15

S

#- = 6.15 /8:(10; + 70.15)

10.10. Aplique el análisis nodal para hallar " = #$%&'(/).

!

en el circuito dela figura 10.59. Sea

36 cos*2000+, - 1 = 3640° 5 = 2000 278 =

9 9 = = B:2C0 :;< :2000 > 2?90@A

C0DE = :;F = :2000 > C0?90@G = :900 HIJI$9$ KL M KN M KG = 3640° 1L 1L B 1N 1L M M = 3640° 2O :900 B:2C0 0.92C1L B :2.C1L M :1L B :1N = P00040° *0.92C B :2.C,1L B :1N = P00040° - Q;°

10.11. Usando análisis nodal encuentre @A (t), en el circuito de la figura

3BC#DE = #2C 2 !"#$ = "4! !0.5% =

1 = '" "#&

0.25% =

1 = '"2 "#&

Nodo 1

!!!!!!!!!!!

(80) ' 60* ' +, +, +, ' +/ = '" "2 2 (1 - "*+, - "+/ = 8) ' 60

Nodo 2

1-

+, ' +/ (8) ' 60* ' +/ =0 "2 "4 ' "2

!

= 4" # 60 + $ + 0.5

%

=

%

1 + 8" # 60 # 4"30 1.5 # $

&'()* = 5.02,'-(2) # 46.55*[/]

10.12. Mediante análisis nodal determinar io al circuito de la figura 10.62.

Pasamos al dominio fasorial 7 = 1000 209-&:910009)9 = 9209 < 9 #;00 209> = 209> 109>9 = 9109> 2&099 = 2?0 50[@A] 9 = 9 #$20 10[BC] 9 = 9$10

Nodo 1) #20$ =

1 1# 2 + + 29? 20 10 ?=

#20$ =

2 $910

1 1# 2 2 + +2 9 20 10 $910

#D400 = 1 + 2 1 # 2 2 # D4 29 #D400 = 2 1 + 2(#2 # D4* E'F'92*9 1# 2 2 2 2 +2 9= + 10 $910 #$20 $10

1! 2 2 2 + = " 10 !"10 #!20 2 1 # 2 2 # !2 2 = ! 2" 0 = 2 1 + 2(#2 # !3)

Resolviendo el Sistema 2" = "

400 " 1" + "!0.5

"$" = "

2" 10

"$ = 34.74 < " #166.6

10.13. Determine %& en el circuito de la figura aplicando el método de su elección.

Realizamos una transformación de fuente en la malla 3 '

-

= $' * , = (5)(10) = 50[ ]

# 40/30° - # 50 + + =0 3 18 + !6 #!2 9

-

= 2:.26/62.88°[;]

10.14. Calcule el voltaje de los nodos 1 y 2 en el circuito. Usando análisis nodal

En el nodo 1 0

!1 0 !1 !2 !1 + + = 20#$° "2 10 "4

(1 + "2.5)!1

"2.5!2 = %173.2% + %"100

En el nodo 2 !2 !2 !2 !1 + + = 20#$° "2 "5 "4 "5.5!2 + "2.5!1 = %173.2% + %"100

1 en 2 &1 = %28.93'135.38°

10.15. Resolver por la corriente I en el circuito de Fig.10.64 utilizando análisis nodal.

Nodo 1:

!

"

=

+

#

+

$

%&20 % '! '! '! % '" =5+ + 2 %&2 &1 %&20 % '! &'! =5+ % &('! % '" ) 2 2 %&20 % '! = 10 + &'! % &2'! + &2'" %10 % &20 = (1 % &)'! + &2'" (1)

Nodo 2:

= "

5+2 5+

'! %&2

+2 +

$

=

*

'! '! % '" '" + = %&2 &1 4 '! '! % '" '" + = %& &1 4

5 + &'! % &('! % '" ) =

'" 4

20 + &4'! % &4'! + 4&'" = '" '" =

20 ,(2) 1 % &4

Sustituyo (2) en (1) %10, % ,&20, = , (1, % ,&)'1, + ,&2'2 ,%10, % ,&20, = , (1, % ,&)'1, + ,&2,(, %10, % ,&20, % , ,%10, % ,&20, +

20 ,), 1, % ,&4

&40 , = , (1, % ,&)'1 1, % ,&4

160 &40 ,, % ,, = , (1, % ,&)'1, 17 17

%0.59, % ,&22.35, = , (-22 , % ,45°)'1 '1, =

22.36/358.49° (,-22 , % ,45°)

,

,'1, = ,15:81/313:5°

Entonces:

= ! =

!1 "#2

15,81"313,5° #$2

! = (0,5"90°)(15,81"313,5°) ! = 7,91"43,49°

10.16. Aplique el análisis nodal para hallar Vx en el siguiente circuito

Nodo 1 %1 = %2 + %3 2 =

&1 # &2 &1 + 5 4$

2 = #'

&1 # &2 &1 *+ 4$ 5

40 = (#5)(&1 # &2)$ + 4&1 40 = #5&1$ + 5&2$ + 4&1$ (40)/20 = ((4 # 5$)&1 + 5$&2)/20 (1)

2 = (0. 2 # 0. 25$)&1 + 0. 25$&2

Nodo 2 3|450 = 2.121 + 2.121$ %2 + %5 = %4

!1 " !2 !2 + 2.121 + 2.121# = 4# "3# !2 !1 " !2 " = "2.121 " 2.121# "3# 4# " 12(

!2 !1 " !2 # + # = "2.121 " 2.121# "3 4

!2 " !1 !2 # + #) = ("2.121 " 2.121#)12 4 "3

"3!1 # + (3 " 4)!2# = "25.452 " 25.452# ("3!1 # " !2#)/"12 = ("25.452 " 25.452#)/"12 (2) 0. 25#!1 + 0. 08333!2 = 2. 121 + 2. 121#

Sistema de ecuaciones dos variables dos incógnitas: !1 = 5.2793 " 5.4190# !2 = 9.6145 " 9.1955# !$ = !1 " !2 !$ = (5.2793 " 5.4190#) " (9.6145 " 9.1955#) !$ = "4.33 + 3.776# [! ] !$ = 5.748%138.950°

10.17. Mediante análisis nodal obtener I 0

Nodo 1. 100 20° ! "# "# "# ! "% = + $4 3 2 100 20° =

"# (3 + $10) ! $2"% &&&(1) 3

Nodo 2. 100 20° ! "% "# ! "% "% + = + 1 2 !$2 100 20° = !

"# 3 1 + ' + $ * "% &&&(2) 2 2 2

Resolviendo las ecuaciones:

= 64,74" # 13,08°

! $

= 81,17" # 6,35°

Para I0: I% =

V! # V$ = 9,25" # 162,12°&[A] 2

10.18. Aplique análisis nodal para determinar

!

en el circuito de la figura 10.67.

"# = "$

Para el nodo 1

4%45° =

"$ "$& "' + 2 8 + (6

200%45° = )29 * (3,"$ * )4 * (3,"'

Para el nodo 2 "$& "' "' "' + 2"# = + 8 + (6 *( 4 + (5 * (2 "$ =

12 + (41 " 104 * (3 '

Reemplazando 200%45° = )29 * (3,

12 + (41 " * )4 * (3,"' 104 * (3 '

"' =

200%45° 14.21%89.17°

"- =

*(2 " 4 + (5 * (2 '

"- = 5.63%189°["]

10.19. Obtener V0 en el circuito de la figura usando análisis por nodos.

Súper-nodo: ! "# = 12$$$$$$$(1) "# "% ! "# " ! "% " + + = 2 !&4 4 &2 !2(" ! "% )& + 2" + "# & = "% ! "# (!2& + 2)" + (& + 1)"# + (2& ! 1)"% = 0$$$$$$$$$$$$$$$$$$(2)

Nodo V2 "% ! "# " ! "% + 0.2" = 4 '2 (!2' + 0.8)" + "# + (2' ! 1)"% = 0$$$$$$$$$$$$$$$$$$$$$$$$(3)

Sustituyendo 3 en 2 0 = 1.2" + &"#

Reemplazamos 1 en 3 0 = 1.2" + &(" ! 12) 0 = 1.2" + &" ! 12& 12& =" (1.2 + &) " = 4,9180 + 5,9016& " = 7.6822*50,19°

10.20. Remítase a la figura 10.69. Si vs (t)=Vm senωt y v0 (t)=A sen (ωt+φ), derive las expresiones de A y φ

Primero convertimos los elementos al dominio fasorial. = !" =

1 !"

#$ (%) = #& '*+("%)

#, (%) = -'*+("% . /%)

Observamos que dos elementos están es paralelo 023

1 7 !45 6 5 !45 !4 = = = 8 1 94 5 . 1 !45 . !4 ! 8485 . 1 !4

Realizamos un divisor de tensión #, =

0 ;# = : . 023 &

#, =

!45

; #& =

94 8 5

.1 !45 :. 94 8 5 . 1 45#&

?@° 9 %A+BC

45 :(1 9 4 8 5 )

#, = ->/ -=

45#&