160 0.5 H 수312.5 μF R=16 A. L-0,5 H. C-312,5 pF Vs(t) = 4 V Vc(t) = V,(t) Sabit gerilim kaynağı Kondansatör üzerindeki gerilim Kondansatörden geçen akım Endüktanstan gerilimi Endoktans akımı Direnç üzerindecki gerilim Direnç üzerinden akan akım 1- 0 anındaki kondansatör gerilimi V(C) Vc(0) – 3 V The switch is closed at t=0 after being open for a long time. Laplace transform of voltage Vc(t) 3 s Vc(s) = V. s2 + 200 s + 6400 Which of the following is the mathematical expression for IR(t) current? a) ir(t) = -40t + 4 e-160t %3D -e b) ig(t) = - + -40t 20 Ie-160t A 20 %3D 4. c) iR(t) = Le-40 V 2091– 160t %3D 80 d) iR (t) = 4 + e-40t 4 e-160t A + = (1)"1 () -401 -160t A

Introductory Circuit Analysis (13th Edition)
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ISBN:9780133923605
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160
4 V
0.5 H
312.5 µF
R= 16 N
L-0,5H.
C= 312,5 µF
Vs(t) = 4 V
Vct) = V,(t)
Sabit gerilim kaynağı
Kondansatör üzerindoki gerilim
Kondansatörden geçen akım
Endüktanstan gerilimi
Endüktans akımı
Direnç üzerindeki gerilim
Direnç üzerinden akan akım
-0 anındaki kondansatör gerilimi
V(G)
Vc(0) – 3 V
The switch is closed at t=0 after being open for a
long time. Laplace transform of voltage Vc(t)
3 s
Vc(s) =
s2 + 200 s + 6400
Which of the following is the mathematical
expression for IR(t) current?
a) ir(t) = -e'
40t
+ 4 e-160t
b) ig (t) =
1.
-40t
+
e-160t A
20
%3D
e
20
|
4
c) ip(t) = e-10t - e-160t A
d) iR (t) = 4 + e-40t
4 e-160t
e) ig(t) =
40t
-e-160t
16
Transcribed Image Text:160 4 V 0.5 H 312.5 µF R= 16 N L-0,5H. C= 312,5 µF Vs(t) = 4 V Vct) = V,(t) Sabit gerilim kaynağı Kondansatör üzerindoki gerilim Kondansatörden geçen akım Endüktanstan gerilimi Endüktans akımı Direnç üzerindeki gerilim Direnç üzerinden akan akım -0 anındaki kondansatör gerilimi V(G) Vc(0) – 3 V The switch is closed at t=0 after being open for a long time. Laplace transform of voltage Vc(t) 3 s Vc(s) = s2 + 200 s + 6400 Which of the following is the mathematical expression for IR(t) current? a) ir(t) = -e' 40t + 4 e-160t b) ig (t) = 1. -40t + e-160t A 20 %3D e 20 | 4 c) ip(t) = e-10t - e-160t A d) iR (t) = 4 + e-40t 4 e-160t e) ig(t) = 40t -e-160t 16
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