29, 8. Find the charge on the capaitor of an RLC series circuit at time t, if L = 1 henry, R C = 0.25 farad, and E(t) = 50 cos(t) volts, given the initial conditions q(0) = 150/13 C and q'(0)=200/13 C. Solution: q(t) 100/3 39 e sin (√3t) + 100 13 sin(t) + 150 13 -cos(t)
29, 8. Find the charge on the capaitor of an RLC series circuit at time t, if L = 1 henry, R C = 0.25 farad, and E(t) = 50 cos(t) volts, given the initial conditions q(0) = 150/13 C and q'(0)=200/13 C. Solution: q(t) 100/3 39 e sin (√3t) + 100 13 sin(t) + 150 13 -cos(t)
Introductory Circuit Analysis (13th Edition)
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ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
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![Find the charge on the capaitor of an RLC series circuit at time t, if L = 1 henry, R = 29,
C = 0.25 farad, and E(t) = 50 cos(t) volts, given the initial conditions q(0) 150/13 C and
q'(0) = 200/13 C.
Solution:
Solution:
q(t)=
9. Find the Laplace transform of the function
is
100√3
39
10. The inverse Laplace transform of
Solution:
150
-esin(√3t) + sin(t) + cos(t)
100
13
13
f(t) = (et-e-t)³
L{f(t)}
F(s)
84 1082 +9
6s +3
s4 + 58² +4
L-¹{F(s)} = sin(t) + 2 cos(t) +
85(5²7)
- sin(2t) - 4 cos(2t)
2](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F374fb018-a137-42da-b757-b1fa97ba03f1%2Feaeb4981-0534-4b49-ba6b-5c12d69a8e1e%2F7hyjing_processed.png&w=3840&q=75)
Transcribed Image Text:Find the charge on the capaitor of an RLC series circuit at time t, if L = 1 henry, R = 29,
C = 0.25 farad, and E(t) = 50 cos(t) volts, given the initial conditions q(0) 150/13 C and
q'(0) = 200/13 C.
Solution:
Solution:
q(t)=
9. Find the Laplace transform of the function
is
100√3
39
10. The inverse Laplace transform of
Solution:
150
-esin(√3t) + sin(t) + cos(t)
100
13
13
f(t) = (et-e-t)³
L{f(t)}
F(s)
84 1082 +9
6s +3
s4 + 58² +4
L-¹{F(s)} = sin(t) + 2 cos(t) +
85(5²7)
- sin(2t) - 4 cos(2t)
2
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