EJERCICIOS (CAPITULO 8)
Enviado por Ronald Jesus Lima Mataqque • 19 de Diciembre de 2018 • Examen • 1.430 Palabras (6 Páginas) • 618 Visitas
EJERCICIOS (CAPITULO 8)
EJERCICIO 8.1
[a]
α =1/(2RC)=1/(2(1000)(2 × 10−6))= 250
ωo2 =1/LC=1/((12.5)(2 × 10−6))= 40,000
s1,2 = −250 ±= −250 ± 150[pic 1]
s1 = −100 rad/s s2 = −400 rad/s
[b]
SOBREAMORTIGUADO
[c] Queremos, ωd = 120 rad/s:
ωd =[pic 2]
α2 = ω2o− ω2 = d = 40,000 − (120)2 = 25,600
α = 160
1/(2RC)= 160; · R =1/(2(160)(2 × 10−6))= 1562.5Ω
[d]
s1, s2 = −160 ±= −160 ± j120 rad/s[pic 3]
[e]
α ==1/(2RC)[pic 4]
R =1/(2(200)(2 × 10−6))= 1250Ω
EJERCICIO 8.2
[a]
−α +o = −250[pic 5]
−α -o = −1000[pic 6]
Sumando las ecuaciones anteriores. , − 2α = −1250
α = 625 rad/s
1/2RC=1/2R(0.1 × 10−6)= 625
R = 8kΩ
2 = 750[pic 7]
4(α2 − ω2o) = 562,500
ωo = 500 rad/s
ω2o = 25 × 104 =1/LC
L =1/((25 × 104)(0.1 × 10−6))= 40H
[b]
iR = v(t)/R= −1e−250t + 4e−1000t mA, t≥ 0+
iC = C(dv(t)/dt)= 0.2e−250t − 3.2e−1000t mA, t≥ 0+
iL = −(iR + iC) = 0.8e−250t − 0.8e−1000t mA, t≥ 0
EJERCICIO 8.3
[a]
iR(0) =15/200= 75mA
iL(0) = −45mA
iC(0) = −iL(0) − iR(0) = 45 − 75 = −30mA
[b]
α =1/2RC=1/(2(200)(0.2 × 10−6))= 12,500
ω2o =1/LC=1/((50 × 10−3)(0.2 × 10−6))= 108
s1,2 = −12,500 ±√(1.5625 × 108 – 108) = −12,500 ± 7500
s1 = −5000 rad/s; s2 = −20,000 rad/s
v = A1e−5000t + A2e−20,000t
v(0) = A1 + A2 = 15
dv/dt(0) = −5000A1 − 20,000A2 =(−30 × 10−3) /(0.2 × 10−6 )= −15 × 104V/s
Resolviendo, A1 = 10; A2 = 5
v = 10e−5000t + 5e−20,000t V, t≥ 0
[c]
iC = C(dv/dt) = 0.2 × 10−6[−50,000e−5000t − 100,000e−20,000t]
= −10e−5000t − 20e−20,000t mA
iR = 50e−5000t + 25e−20,000t mA
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