Ejemplo cálculo de Impedancia equivalente

Ejemplo cálculo de Impedancia equivalente
Problem 9.51
(a)
20 mH

→
12.5 µF 
→
jωL = j (103 )( 20 × 10 -3 ) = j20
1
1
=
= - j80
3
j ωC j (10 )(12.5 × 10 - 6 )
Z in = 60 + j20 || ( 60 − j80)
( j20)( 60 − j80)
Z in = 60 +
60 − j60
Zin = 63.33 + j23.33 = 67.494∠ 20.22°
Yin =
(b)
1
=
∠
°
Zin 0.0148 -20.22 S
10 mH

→
20 µF 
→
30 || 60 = 20
j ωL = j (10 3 )(10 × 10 -3 ) = j10
1
1
=
= - j50
3
j ωC j (10 )( 20 × 10 - 6 )
Z in = - j50 + 20 || ( 40 + j10)
(20)( 40 + j10)
Z in = - j50 +
60 + j10
Z in = 13.5 − j48.92 = 50.75∠ - 74.56°
Yin =
1
=
Zin 0.0197∠ 74.56° S = 5.24 + j18.99 mS
Ejemplo cálculo de Impedancia equivalente
Problem 9.48

→ j ωL = j (10 × 10 3 )( 2 × 10 -3 ) = j20
1
1
1 µF 
→
=
= - j100
j ωC j (10 ×10 3 )(1× 10-6 )
2 mH
50 Ω
+
1∠ 0° A
+
V
j20 Ω
−
+
−
Vin
−
-j100 Ω
V = (1∠0°)( 50) = 50
Vin = (1∠0°)(50 + j20 − j100 ) + ( 2)(50)
Vin = 50 − j80 + 100 = 150 − j80
Z in =
Vin
= 150 – j80 Ω
1∠0°
2V
Ejemplo división de voltajes
ω = 200
1
1
50 µF 
→
=
= - j100
jωC j (200)(50 × 10 -6 )
Problem 9.35

→ jωL = j (200 )( 0.1) = j20
(50)(-j100 ) - j100
50 || -j100 =
=
= 40 − j20
50 − j100
1 - j2
0.1 H
j20
j20
Vo =
( 60 ∠0°) =
( 60 ∠0°) = 17 .14 ∠90°
j20 + 30 + 40 − j20
70
Thus, v o ( t ) = 17.14 sin(200t + 90° ) V
v o ( t ) = 17.14 cos(200t) V
or
Ejemplo división de corrientes
Problem 9.41
0.1 F 
→
1
1
=
= - j5
jωC j ( 2)( 0. 1)
0.5 H 
→
jωL = j ( 2)(0.5) = j
The current I through the 2-Ω
Ω resistor is
I=
I
1
Is = s ,
1 − j5 + j + 2
3 − j4
I s = (10)( 3 − j4) = 50∠ - 53.13°
Therefore,
i s (t ) = 50 cos(2t – 53.13°) A
where I = 10∠0°