Electric nets 20210721 tecnicas de analisis 15.21 16.35
- 1. Problema 1 (15.41)
Utilice Laplace para encontrar π£(π‘) para π‘ > 0.
SoluciΓ³n:
π(0β ) = 12
3
3 + 6
= 4 π
π (
1
6
+
1
3
+
π
104
) =
4
π
β
π
104
π =
4 β 10β4
π
104 +
1
2
=
4
π + 5 000
β΄ π£(π‘) = 4 β π-5 000βπ‘
β π’(π‘) [π] β
Problema 2 (15.42)
Encuentre π(π‘), π‘ > 0 mediante Laplace.
SoluciΓ³n:
π‘ β 0β
: π
= 2 + 3 β₯ (4 + 6) =
56
13
β¦
ππ =
12 π
56/13 β¦
=
39
14
π΄
π(0β ) = - (
39
14
π΄) β
3β¦
3β¦ + 4β¦ + 2β¦
= -
13
14
π΄
π = -
13/7
2π + 9
= -
13
14
1
π +
9
2
π(π‘) = -
13
14
β π
-
9
2
π‘
β π’(π‘) [π΄] β
104
π
4
π
3
6
+
-
π
13/7
2s
9
π
- 2. Problema 3 (15.43)
Use Laplace para encontrar π£0
(π‘), π‘ > 0.
SoluciΓ³n:
π£π
(0β ) = 12π β
6 πβ¦
6 πβ¦ + 3 πβ¦
= 8 π
π£(0β ) = 8 π β
2 πβ¦
2 πβ¦ + 4 πβ¦
=
8
3
π
π0 = (
8
π
) β
2
2 + 4 +
104
π
=
8
3π + 5 000
π£0
(π‘) =
8
3
β π
β
5000
3
π‘
β π’(π‘) [π] β
Problema 4 (16.1)
Determine el valor del voltaje de salida cuando π‘ β β.
SoluciΓ³n:
π0 = (
12
π
β
1
2
1
2
+
1
3
+
1
π
) β
2
π
2
π
+ 3
= (
6
π (
5
6
+
1
π
)
) β
2
π (
2
π
+ 3)
=
6
5π
6
+ 1
β
2
2 + 3π
=
72
(5π + 6)(3π + 2)
π£(β) = πππ
π β0
π β
72
(5π + 6)(3π + 2)
= 0 β
Problema 5 (16.2)
Determine el voltaje de salida cuando π‘ β β.
SoluciΓ³n:
π£(0β ) = 0
π0 = (
72
π (π + 6)
1
π + 6
+ π +
1
3
) β
1
1 + 2
=
24
π β (1 + (π +
1
3
) (π + 6))
π£(β) = πππ
π β0
24
1 + (π +
1
3
) (π + 6)
= 8 β
8
π
104
π
4
2
π0
12
π
2
π
π
2 3
π0
72
π
6 π 2
1
π0
1
π
- 3. Problema 6 (16.3)
Determine el voltaje de salida cuando π‘ β β.
SoluciΓ³n:
π0 =
(
2
π
1
2
+
1
1 +
2
π
+
1
π + 2
)
β
π
π + 2
π0 =
2
(π + 2) (
1
2
+
π
π + 2
+
1
π + 2
)
π£0
(β) = πππ
π β0
2 β π
(π + 2) (
1
2
+
π
π + 2
+
1
π + 2
)
= 0 β
Problema 7 (16.4)
Mediante Laplace, encuentre π1
(π‘), π‘ > 0 suponiendo condiciones iniciales nulas.
SoluciΓ³n:
π1 =
1
2
(
12
π
β
6
π
β
6
π
β
π
2
1
2
+
1
π
+
π
2
) =
1
2
(
12
π
β
6
π
β 3
1
2
+
1
π
+
π
2
)
1
2
(
12
π
β
6 β 3π
1
2
π + 1 +
π 2
2
) =
1
2
(
12
π
β
12 β 6π
π 2 + π + 2
) =
6
π
+
3(π +
1
2
) β 6 β
3
2
(π +
1
2
)
2
+
7
4
β΄ π1
(π‘) = 6 β π’(π‘) + π
β
1
2
π‘
β [3 πΆππ (
β7
2
π‘) β
15
β7
πππ (
β7
2
π‘)] β π’(π‘) [π΄] β
Problema 8 (16.5)
Encuentre π£0
(π‘), π‘ > 0.
SoluciΓ³n:
π0 =
(
4
π 2 +
2
π
1
π
+
1
2 +
2
π )
β
1
1 + 1 +
2
π
=
(
4 + 2π
π +
π 2
2 +
2
π )
β
π
2π + 2
= (
2 + π
π +
π 3
2π + 2
) β
π
π + 1
= (
π + 2
π (π + 1) +
1
2
π 3
) β π = 2
π + 2
2(π + 1) + π 2
= 2
π + 2
π 2 + 2π + 2
= 2
(π + 1) + 1
(π + 1)2 + 1
π£0
(π‘) = 2πβπ‘
β (πΆππ π‘ + πππ π‘) β π’(π‘) π β
4
π
2 2
s
1
2
π
ππ
- 4. Problema 9 (16.6)
Encuentre π£0
(π‘), π‘ > 0.
SoluciΓ³n:
ππ (1 +
1
π
) β ππ
(0) =
4
π
β
2
π
ππ
(0) + ππ
(1 + π ) =
2
π
+
1
π + 1
}
ππ =
2
π (π + 1)
+
1
(π + 1)2
ππ =
2
π
β
2
π + 1
+
1
(π + 1)2
π£π
(π‘) = π£0
(π‘) = [2 β πβπ‘
β (2 β π‘)] β π’(π‘) π β
Problema 10 (16.7)
Encuentre π£0
(π‘), π‘ > 0.
SoluciΓ³n:
π0
(1 + π + 1) =
12
π
β
4
π + 1
+
6
π
1
+
-
4
π + 1
+
6
π
1/π
π0
(π + 2) =
18
π
β
4
π + 1
β
4π
π + 1
+ 6 β
π0 =
18
π (π + 2)
β
4
π + 2
+
6
π + 2
=
18
π (π + 2)
+
2
π + 2
=
9
π
β
9
π + 2
+
2
π + 2
π£0
(π‘) = (9 β 7 β πβ2π‘) β π’(π‘) π β
4
π
1
s
2/s
1 1/s
1
π + 1
a b
0
6
π
4
π + 1
12
π 1
π
1
1
1
+
π0
β
0
- 5. Problema 11 (16.8)
Encuentre π£0
(π‘), π‘ > 0.
SoluciΓ³n:
π1 (1 +
1
2
+
1
π
) = -2π1 β
4
π
(
1
2
+
1
π
)
π1 (
7
2
+
1
π
) = -
2
π
β
4
π 2
π1
(7π 2
+ 2π ) = -4π β 8
π1 = -4
π + 2
π (7π + 2)
ππ =
4
π
+ π1 =
4
π
β 4
π + 2
π (7π + 2)
ππ =
4
π
β 4 β
2
π (2)
β
4
7
β
-
2
7
+ 2
π +
2
7
= -
48
49
β
1
π +
2
7
π£(π‘) = -
48
49
β π
2
7
π‘
β π’(π‘) π β
ππ
π1
2π1
π
4
π
- 6. Problema 12 (16.9)
Encuentre π£0
(π‘), π‘ > 0.
SoluciΓ³n:
ππ₯
(1 + π + 1 + 2) =
4
π
β
2
π
(1 + π ) β ππ₯
(π + 4) =
2
π
β 2 β ππ₯ =
2 β 2π
π (π + 4)
ππ = 2ππ₯ = 4 β
1 β π
π (π + 4)
= 4 β
1/4
π
+ 4 β
5
-4(π + 4)
=
1
π
β
5
π + 4
π£π(π‘) = (1 β 5 β πβ4π‘) β π’(π‘) π β
Problema 13 (16.10)
Encuentre π£0
(π‘), π‘ > 0.
SoluciΓ³n:
ππ₯ (π + 1 +
2
π
+ 1) =
4
π
β
2
π
(1 +
2
π
+ 1) β ππ₯ (π + 2 +
2
π
) = -
4
π 2
β ππ₯
(π 2
+ 2π + 2) = β
4
π
β ππ₯ = -4
1
π ((π + 1)2 + 1)
ππ₯ = -4
1
π (π + 1 β π)(π + 1 + π)
= -4
1
π (2)
β 4
1
(-1 + π)(π + 1 β π)(π2)
β 4
1
(-1 β π)(π + 1 + π)(-π2)
= -
2
π
β
1 + π
π + 1 β π
β
1 β π
π + 1 + π
ππ₯
(π‘) = (-2 β πβπ‘
β [(1 + π)πβππ‘
+ (1 β π)πππ‘ ]) β π’(π‘) = (-2 β πβπ‘
β [2πΆππ π‘ β 2 ππππ‘]) β π’(π‘) π΄
ππ = (1) (ππ₯ +
2
π
) β π£π
(π‘) = (-2 β πβπ‘
β [2 πΆππ π‘ β 2 πππ π‘] + 2) β π’(π‘)
π£π
(π‘) = 2πβπ‘
β (πππ π‘ β πΆππ π‘) β π’(π‘) π β
- 7. Problema 14 (16.11)
Encuentre π£0
(π‘), π‘ > 0.
SoluciΓ³n:
[1 + 2 + 1 -1
-1 1 + π + 1
] β [
ππ₯
ππ¦
] = [
4
π
+
2
π
0
] β [ 4 -1
-1 π + 2
] β [
ππ₯
ππ¦
] = [
6
π
0
] β ππ¦ =
|
6
π
-1
0 π + 2
|
|4 -1
-1 π + 2
|
=
6 +
12
π
4π + 8 β 1
=
6π + 12
π β (4π + 7)
ππ = (1) β ππ¦ =
6
4
β
π + 2
π β (π +
7
4
)
=
3
2
β
π + 2
π β (π +
7
4
)
ππ =
3
2
β
2
π (
7
4
)
+
3
2
β
-
7
4
+ 2
(-
7
4
) (π +
7
4
)
=
12
7
β
1
π
β
3
14
β
1
π +
7
4
π£π
(π‘) = (
12
7
β
3
14
π
β
7
4
π‘
) β π’(π‘) π β
- 8. Problema 15 (16.12)
Encuentre ππ
(π‘),π‘ > 0.
SoluciΓ³n:
[
1 +
1
π
-
1
π
-
1
π
2 + 1 +
1
π
] β [
ππ₯
ππ¦
] = [
4
π
1
π + 1
]
[
ππ₯
ππ¦
] =
π
3π + 4
β [
3 +
1
π
1
π
1
π
1 +
1
π
] β [
4
π
1
π + 1
] =
1
3π + 4
β [3π + 1 1
1 π + 1
] β [
4
π
1
π + 1
]
ππ = ππ¦ =
1
3π + 4
β [
4
π
+ 1] =
4
π (3π + 4)
+
1
3π + 4
=
4 + π
π (3π + 4)
ππ =
4
π (4)
+
1
3
β
4 β
4
3
-
4
3
β (π +
4
3
)
=
1
π
β
2
3 (π +
4
3
)
ππ
(π‘) = (1 β
2
3
β π
β
4
3
π‘
) β π’(π‘) π΄ β
- 9. Problema 16 (16.13)
Encuentre ππ
(π‘),π‘ > 0.
SoluciΓ³n:
ππ₯ (2 + 1 +
1
π
+ π ) =
2
π
β
1
π + 1
β (π +
1
π
)
ππ₯ (π + 3 +
1
π
) =
2
π
β
1
π + 1
β
π 2
+ 1
π
β ππ₯
(π 2
+ 3π + 1) = 2 β
π 2
+ 1
π + 1
ππ₯ ([π +
3
2
]
2
β
5
4
) =
2π + 2 β π 2
β 1
π + 1
ππ₯ =
1 + 2π β π 2
(π + 1) (π +
3 β β5
2
) (π +
3 + β5
2
)
ππ = ππ₯ β
2
π
=
2
(π + 1)
+
1 β 3 + β5 β (
-3 + β5
2
)
2
-3 + β5
2
β (π +
3 β β5
2
) (β5)
+
1 β 3 β β5 β (
-3 β β5
2
)
2
-3 β β5
2
β (π +
3 + β5
2
) (-β5)
β
2
π
ππ =
2
π + 1
+
7β5 β 15
10
β
1
π +
3 β β5
2
+
-7β5 β 15
10
β
1
π +
3 + β5
2
ππ
(π‘) = (2πβπ‘
+ π
β
3
2
π‘
β [
7β5 β 15
10
β π
β
β5
2
π‘
+
-7β5 β 15
10
β π
β5
2
π‘
]) β π’(π‘)
π0
(π‘) = (2πβπ‘
+ π
β
3
2
π‘
β [
7
β5
β πππβ
β5
2
π‘ β 3 β πΆππ β
β5
2
π‘]) β π’(π‘) π΄ β
- 10. Problema 17 (16.14)
Encuentre π£π
(π‘), π‘ > 0.
SoluciΓ³n:
π1 (1 +
1
π
+ 2) =
2
π
+ 2π1 + 4π1
π1 (-3 +
1
π
) =
2
π
π1
(-3π + 1) = 2 β π1 =
2
-3π + 1
π0 = 2(π1 β 2ππ΄
) = 2(
2
π
β
4
-3π + 1
) =
4
π
+
8
3
β
1
π β
1
3
π£π
(π‘) = (4 +
8
3
β π
π‘
3) β π’(π‘) π β
- 11. Problema 18 (16.15)
Encuentre π£0
(π‘), π‘ > 0 mediante superposiciΓ³n.
SoluciΓ³n:
Actuando la fuente de voltaje:
ππ
β²
= (
4
π
) β
1
1 + π + 1 +
2
π
=
4
π 2 + 2π + 2
Actuando la fuente de corriente:
ππ
β²β²
= (1) β {(
2
π
) β
π
π + 1 +
2
π
+ 1
} =
2π
π 2 + 2π + 2
Superponiendo:
ππ = ππ
β²
+ ππ
β²β²
=
4
π 2 + 2π + 2
+
2π
π 2 + 2π + 2
= 2 β
π + 2
π 2 + 2π + 2
= 2 β
(π + 1) + 1
(π + 1)2 + 1
π£0
(π‘) = 2πβπ‘
β (πΆππ π‘ + πππ π‘) β π’(π‘) π β
- 12. Problema 19 (16.16)
Encuentre π£0
(π‘), π‘ > 0 mediante transformaciΓ³n de fuentes.
SoluciΓ³n:
ππ β
4
π 2
+
2
π
; β€π β π ; ππ β ππ β€π =
4
π
+ 2
β€π β π + 1 +
2
π
ππ = ππ β
1
1 + β€π
=
4 + 2π
π
β
1
π + 2 +
2
π
= 2 β
π + 2
π 2 + 2π + 2
= 2
(π + 1) + 1
(π + 1)2 + 1
β π£0
(π‘) = 2πβπ‘
β (πΆππ π‘ + πππ π‘) β π’(π‘) π β
Problema 20 (16.17)
Use Thevenin para encontrar π£0
(π‘), π‘ > 0.
SoluciΓ³n:
ππ‘β = (π ) β (
4
π 2
+
2
π
) =
4
π
+ 2
β€π‘β = π + 1 +
2
π
ππ = ππ‘β β
β€π
β€π + β€π‘β
= (
4 + 2π
π
) β
1
1 + π + 1 +
2
π
=
4 + 2π
π 2 + 2π + 2
ππ = 2 β
(π + 1) + 1
(π + 1)2 + 1
π£0
(π‘) = 2πβπ‘
β (πΆππ π‘ + πππ π‘) β π’(π‘) π β
- 13. Problema 21 (16.18)
Encuentre π£0
(π‘), π‘ > 0 mediante Thevenin.
SoluciΓ³n:
ππ‘β =
4
π
β
2
π
β (1 + π ) =
2
π
β 2 = 2 β
1 β π
π
β€π‘β = 1 + π + 1 = π + 2
ππ = ππ‘β β
2
2 + β€π‘β
= 2 β
1 β π
π
β
2
π + 4
= 4 β
1 β π
π β (π + 4)
= 4 β
1/4
π
+ 4 β
5
-4(π + 4)
=
1
π
β
5
π + 4
π£π(π‘) = (1 β 5 β πβ4π‘) β π’(π‘) π β
Problema 22 (16.19)
Encuentre π£0
(π‘), π‘ > 0. Por medio de ecuaciones de malla.
SoluciΓ³n:
ππ₯ (π + 1 +
2
π
+ 1) =
4
π
β
2
π
(1 +
2
π
+ 1) β ππ₯ (π + 2 +
2
π
) = -
4
π 2
β ππ₯
(π 2
+ 2π + 2) = β
4
π
β ππ₯ = -4
1
π ((π + 1)2 + 1)
ππ₯ = -4
1
π (π + 1 β π)(π + 1 + π)
= -4
1
π (2)
β 4
1
(-1 + π)(π + 1 β π)(π2)
β 4
1
(-1 β π)(π + 1 + π)(-π2)
= -
2
π
β
1 + π
π + 1 β π
β
1 β π
π + 1 + π
ππ₯
(π‘) = (-2 β πβπ‘
β [(1 + π)πβππ‘
+ (1 β π)πππ‘ ]) β π’(π‘) = (-2 β πβπ‘
β [2πΆππ π‘ β 2 ππππ‘]) β π’(π‘) π΄
ππ = (1) (ππ₯ +
2
π
) β π£π
(π‘) = (-2 β πβπ‘
β [2 πΆππ π‘ β 2 πππ π‘] + 2) β π’(π‘)
π£π
(π‘) = 2πβπ‘
β (πππ π‘ β πΆππ π‘) β π’(π‘) π β
- 14. Problema 23 (16.20)
Encuentre ππ
(π‘),π‘ > 0 mediante Thevenin.
SoluciΓ³n:
ππ‘β = (
4
π
) β
1
π
1
π
+ 1
β (1) β (-
1
π + 1
) =
4
π β (π + 1)
+
1
π + 1
=
4 + π
π β (π + 1)
β€π‘β = 1 β₯
1
π
+ 1 =
1
π
1+
1
π
+ 1 =
1
π + 1
+ 1 =
π + 2
π + 1
ππ =
ππ‘β
β€π‘β + 2
=
4 + π
π β (π + 1)
β
1
π + 2
π + 1
+ 2
=
4 + π
π (π + 2 + 2π + 2)
=
π + 4
π β (3π + 4)
ππ =
4
π (4)
+
1
3
β
4 β
4
3
-
4
3
β (π +
4
3
)
=
1
π
β
2
3 (π +
4
3
)
ππ
(π‘) = (1 β
2
3
β π
β
4
3
π‘
) β π’(π‘) π΄ β
- 15. Problema 24 (16.21)
Encuentre π£π
(π‘), π‘ > 0 mediante Thevenin.
SoluciΓ³n:
ππ‘β =
4
π
+ (2π ) (
1
π + 2
) + (2) (
2
π
) =
8
π
+
2π
π + 2
=
8π + 16 + 2π 2
π β (π + 2)
β€π‘β = 2π + 2
ππ = ππ‘β β
1
1 + β€π‘β
= 2 β
π 2
+ 4π + 8
π β (π + 2)
β
1
2π + 3
ππ = 2 β
8
π β (6)
+ 2 β
4 β 8 + 8
(-2)(π + 2)(-1)
+
2
2
β
9
4
β 6 + 8
(-
3
2
) (-
3
2
+ 2) (π +
3
2
)
=
8
3π
+
4
π + 2
β
17
3(π +
3
2
)
π£π
(π‘) = (
8
3
+ 4πβ2π‘
β
17
3
π
β
3
2
π‘
) β π’(π‘) π β
- 16. Problema 25 (16.22)
Encuentre π£π
(π‘), π‘ > 0 mediante Thevenin.
SoluciΓ³n:
ππ‘β = (1) (
2
π
) + 2(1) (
2
π
) =
6
π
ππ‘β =
2
π
β (1) (
π
3
β ππ‘β )
ππ‘β (1 +
π
3
) =
2
π
ππ‘β
(3 + π ) =
6
π
ππ‘β =
6
π β (π + 3)
β€π‘β =
ππβ
ππ‘β
=
6
π
β
π β (π + 3)
6
= π + 3
ππ = ππ‘β β
1
1 + β€π‘β
=
6
π
β
1
1 + π + 3
ππ =
6
π β (π + 4)
ππ =
6
4π
+
6
-4(π + 4)
=
3
2π
β
3
2(π + 4)
π£π
(π‘) = (
3
2
β
3
2
β πβ4π‘ ) β π’(π‘) π β
- 17. Problema 26 (16.23)
Encuentre los parΓ‘metros de transmisiΓ³n para la red mostrada.
SoluciΓ³n:
[π«] = [
1 + 1 1
1 1 +
1
2π
] = [
2 1
1 1 +
1
2π
] β Ξπ§ = 2 +
1
π
β 1 = 1 +
1
π
=
π + 1
π
[π₯] =
1
π«21
β [
π«11 Ξπ§
1 π«22
] =
1
1
β [
2 1
1 1 +
1
2π
]
π₯11 = 2; π₯12 = 1; π₯21 = 1; π₯22 = 1 +
1
2π
β
Problema 27 (16.24)
Encuentre los parΓ‘metros β€ de la red (a). Mediante esos parΓ‘metros, determine πΌ2(π‘) en la red (b).
SoluciΓ³n:
[π«] = [
1 +
1
π
1
π
1
π
π +
1
π
] = [
π + 1
π
1
π
1
π
π 2
+ 1
π
]
[
π1
π2
] = [
4
π
β (1)π1
-(1)π2
] = [
4
π
0
] + [-1 0
0 -1
] β [
π1
π2
]
[π«][π] = [π] β [
π + 1
π
1
π
1
π
π 2
+ 1
π
] β [
π1
π2
] = [
4
π
0
] + [-1 0
0 -1
] β [
π1
π2
] β [
2π + 1
π
1
π
1
π
π 2
+ π + 1
π
] β [
π1
π2
] = [
4
π
0
]
π2 =
|
2π + 1
π
4
π
1
π
0
|
|
2π + 1
π
1
π
1
π
π 2 + π + 1
π
|
=
-4/π 2
2π 3 + 2π 2 + 2π + π 2 + π + 1
π 2 β
1
π 2
=
-4
π β (2π 2 + 3π + 3)
=
-2
π β ([π +
3
4
]
2
+
15
16
)
π2 =
-2
π β (π +
3 β πβ15
4
) β (π +
3 + πβ15
4
)
= -
4
3π
+
-2
(
-3 + πβ15
4
)(π +
3 β πβ15
4
)(
πβ15
2
)
+
-2
(
-3 + πβ15
4
)(π +
3 β πβ15
4
)(
πβ15
2
)
π2 = -
4
3π
+
10 β π2β15
15 β (π +
3 β πβ15
4
)
+
10 + π2β15
15 β (π +
3 + πβ15
4
)
π2(π‘) = (-
4
3
+ π
3
4
π‘
β [
10 β π2β15
15
πβ
πβ15
4
π‘
+
10 + π2β15
15
π
πβ15
4
π‘
]) β π’(π‘) = (-
4
3
+ πβ
3
4
π‘
β [
4
3
πΆππ
β15
4
π‘ β
4
β15
β πππ
β15
4
π‘]) β π’(π‘) β
- 18. Problema 28 (16.25)
Encuentre los parΓ‘metros de transmisiΓ³n de la red mostrada.
SoluciΓ³n:
[π«] = [π + 1 π
π π + 1
] β Ξz = π 2
+ 2π + 1 β π 2
= 2π + 1
[π₯] =
1
π«21
β [
π«11 Ξπ§
1 π«22
] =
1
π
β [π + 1 2π + 1
1 π + 1
]
π₯11 =
π + 1
π
; π₯12 =
2π + 1
π
; π₯21 = 1; π₯22 = π + 1 β
Problema 29 (16.26)
Encuentre π£π
(π‘), π‘ > 0, mediante Laplace. Suponga que el circuito ha alcanzado el estado estable en t = 0.
SoluciΓ³n:
π£π
(0β ) = 15 β
3
2
3
2
+ 6
= 3 π; ππΏ
(0β ) =
3
3
= 1 π΄
6 β₯ 3 =
18
9
= 2
ππ = 2 β
3
3 + 2π + 2
=
6
2π + 5
ππ =
3
π +
5
2
π£0
(π‘) = (3 β π
β
5
2
π‘
) β π’(π‘) π β
- 19. Problema 30 (16.27)
Encuentre ππ
(π‘),π‘ > 0, en la red mostrada.
SoluciΓ³n:
π0
(0β ) =
14π
3 β¦ + 0.5 β¦
= 4 π΄
ππΏ
(0β ) =
12
2
+
4
2
= 8 π΄
π0 =
1
1 + 1
β
6
π
β
8
π
1
2π
+
1
2
+
1
2
= -
2
2π + 1
ππ
(π‘) = -π
-
1
2
π‘
β π’(π‘) π΄ β
Problema 31 (16.28)
Encuentre ππ
(π‘),π‘ > 0, en la red mostrada.
SoluciΓ³n:
π£π
(0β ) = 12 β
4
4 + 2
= 8 π
β€π β (3 + 1) β₯ 2π =
8π
2π + 4
ππ β (
-8
π
) β (
8π
2π + 4
) β
1
1
π
+ 4
= -
32
(π + 2)(1 + 4π )
ππ =
ππ
3 + 1
= -
8
(π + 2)(4π + 1)
= -
8
(π + 2)(-7)
-
8
(-
1
4
+ 2) (4) (π +
1
4
)
=
8
7(π + 2)
β
8
7 (π +
1
4
)
ππ
(π‘) = (
8
7
πβ2π‘
β
8
7
π
β
1
4
π‘
) β π’(π‘) π΄ β
- 20. Problema 32 (16.29)
Encuentre ππ
(π‘),π‘ > 0, en la red mostrada.
SoluciΓ³n:
ππ‘β = 12
4
4 + 2
= 8 π
β€π‘β = 4 β₯ 2 +
1
π
=
4
3
+
1
π
=
4π + 3
3π
β€π β 2π β₯ (1 + 3) =
8π
2π + 4
=
4π
π + 2
ππ = ππ‘β β
β€π
β€π + β€π‘β
= 8 β
4π
π + 2
β
1
4π
π + 2
+
4π + 3
3π
=
96π 2
12π 2 + 4π 2 + 11π + 6
=
96π 2
16π 2 + 11π + 6
ππ =
ππ
4
=
24π 2
16π 2 + 11π + 6
=
3
2
β
33
2
(π +
11
2
) + 9 β
363
4
16 [(π +
11
32
)
2
+
263
1024
]
ππ
(π‘) = {
3
2
πΏ(π‘) + π
β
11
32
π‘
β (-
33
32
πΆππ β
β262
32
π‘ β
327β263
526
πππβ
β262
32
π‘)} β π’(π‘) π΄ β
Problema 33
Los demΓ‘s ejercicios se refieren a unos diagramas que no fueron adjuntados, por tanto, no pueden resolverse.