The solution are attached. Can you please explain if the circuit is a natural or step responce. I;m having difficultu understanding. I think it should be natural responce because once the switch is closed, all the electrical components will discharge through the resistor? Thank you
The solution are attached. Can you please explain if the circuit is a natural or step responce. I;m having difficultu understanding. I think it should be natural responce because once the switch is closed, all the electrical components will discharge through the resistor? Thank you
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
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The solution are attached. Can you please explain if the circuit is a natural or step responce. I;m having difficultu understanding. I think it should be natural responce because once the switch is closed, all the electrical components will discharge through the resistor?
Thank you
Transcribed Image Text:8.44 In the circuit in Fig. P8.44, the resistor is adjusted
for critical damping. The initial capacitor voltage is
15 V, and the initial inductor current is 6 mA.
a) Find the numerical value of R.
(Ans: 1250 (2)
b) Find the numerical values of i and di/dt immedi-
ately after the switch is closed.
(Ans: 60 A/s)
c) Find vc(t) for t≥ 0. (Ans: 56,250 t e-5,000t + 15×e-5,000t)
Figure P8.44
+
VC
t = 0)
320 nF
R
125 mH
![P 8.44 [a] w²
=
α =
R
2L
... R = (5000) (2) L = 1250 N
di
1
LC
[c] vc
dt
(0)
[b] i(0) — iz(0) = 6 mA
=
10⁹
(125)(0.32)
wwwww
Wo
VL
vL (0) — 15 — (0.006) (1250) = 7.5 V
60 A/s
+ D₂e-5000t
7.5
0.125
Dite-5000t
vc (0) = D₂ = 15 V
-
-
W
=
A
5000 rad/s
duc (0)
dt
.. D₁ = 56,250 V/s
VC 56,250te
²(0) = D₁ – 5000D₂
=
-5000t
25 x 106
=
ic(0)
C
V₁
+ 15e-5000t V.
-iL (0)
с
= -18,750
t≥0.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F861b1720-99dc-46a2-8dbc-12fe90922958%2F4aa92cc6-8747-4170-961d-775315249b89%2F53sh12q_processed.jpeg&w=3840&q=75)
Transcribed Image Text:P 8.44 [a] w²
=
α =
R
2L
... R = (5000) (2) L = 1250 N
di
1
LC
[c] vc
dt
(0)
[b] i(0) — iz(0) = 6 mA
=
10⁹
(125)(0.32)
wwwww
Wo
VL
vL (0) — 15 — (0.006) (1250) = 7.5 V
60 A/s
+ D₂e-5000t
7.5
0.125
Dite-5000t
vc (0) = D₂ = 15 V
-
-
W
=
A
5000 rad/s
duc (0)
dt
.. D₁ = 56,250 V/s
VC 56,250te
²(0) = D₁ – 5000D₂
=
-5000t
25 x 106
=
ic(0)
C
V₁
+ 15e-5000t V.
-iL (0)
с
= -18,750
t≥0.
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