A series circuit contains an inductor, a resistor, and a capacitor for which L = 0.5h, R = 102, and C = 0.01f , respectively. s10, 05 The voltage E (t) = {0" instantaneous charge q (t)on the capacitor for t > 0, if q (0) = 0 and d (0) = 0 is applied to the circuit. Determine the O q(t) = 10- 10 cos(10t) – e-10t sin(10t) – -e-10(t-5) cos(10 (t – 5)) – me-10(t-6) sin(10 (t – 5))] U (2 – 5) -10t е O g (t) 3D3-글e과 cos(26)-글e-2 sin(2t)- [높 - 놀e-2(-5) cos(2 (t -5)) - 글어-3) sin(2 (t-5)] U(t-5) -2(t-5) O q(t) = 1- e-10t cos(10t) e-10t sin(10t) – [1-e-10(t-5) cos(10 (t – 5)) – e-10(t-5) sin(10 (t – 5))] U (t – 5) Og (t) %3D -공e-와 cos(5t)-1e-st sin(5t)- [끊-좋e-st-5) cos(5 (t - 5)) - 금e- St-5) sin(5 (t-5)] U(t -5)
A series circuit contains an inductor, a resistor, and a capacitor for which L = 0.5h, R = 102, and C = 0.01f , respectively. s10, 05 The voltage E (t) = {0" instantaneous charge q (t)on the capacitor for t > 0, if q (0) = 0 and d (0) = 0 is applied to the circuit. Determine the O q(t) = 10- 10 cos(10t) – e-10t sin(10t) – -e-10(t-5) cos(10 (t – 5)) – me-10(t-6) sin(10 (t – 5))] U (2 – 5) -10t е O g (t) 3D3-글e과 cos(26)-글e-2 sin(2t)- [높 - 놀e-2(-5) cos(2 (t -5)) - 글어-3) sin(2 (t-5)] U(t-5) -2(t-5) O q(t) = 1- e-10t cos(10t) e-10t sin(10t) – [1-e-10(t-5) cos(10 (t – 5)) – e-10(t-5) sin(10 (t – 5))] U (t – 5) Og (t) %3D -공e-와 cos(5t)-1e-st sin(5t)- [끊-좋e-st-5) cos(5 (t - 5)) - 금e- St-5) sin(5 (t-5)] U(t -5)
Linear Algebra: A Modern Introduction
4th Edition
ISBN:9781285463247
Author:David Poole
Publisher:David Poole
Chapter6: Vector Spaces
Section6.7: Applications
Problem 18EQ
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16) Please help with following multiple choice ASAP!
![A series circuit contains an inductor, a resistor, and a capacitor for which
= 0.01f, respectively.
L = 0.5h, R = 10N, and C
The voltage E (t) = {o"
{o. t25
s10,
0<t<5
is applied to the circuit. Determine the
instantaneous charge q (t)on the capacitor for t > 0, if q (0) = 0 and
q (0) = 0
O q (t) = 10
- be-10t cos(10t) – e-10t sin(10€) – [ - e-10t-5) cos(10 (t – 5)) – e-10(t-5) şin(10 (t – 5))] U (t – 5)
te-10(t-5) cos(10 (t – 5)) - e-10(t-5) sin(10 (t – 5))]U (t – 5)
COS
Og(t) =D-e-과 cos(2t)-글e-2 sin(2t)-[-글e-2(-8) cos (2 (t-5))-글1e-26-5) sin(2 (t-5))] U (t -5)
O q(t)
-2(t-5) sin(2 (t – 5))] U (t – 5)
O q (t) = 1- e-10t cos(10t) – e-10t sin(10t) – [1- e-10(t-5) cos(10 (t - 5)) – e-10(t-5) sin(10 (t – 5))] U (t - 5)
O q(t) = -te-5t cos(5t) – e-5t sin(5t) – - te-5(-5) cos(5 (t – 5)) – e-5(t-5) sin(5 (t – 5)] U (t – 5)
1/2](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F74a43f44-8ce4-41e9-89f6-5a77f02dff59%2F60c0d90b-93b9-4a39-90ba-1c995414e55a%2Fp68izkq_processed.jpeg&w=3840&q=75)
Transcribed Image Text:A series circuit contains an inductor, a resistor, and a capacitor for which
= 0.01f, respectively.
L = 0.5h, R = 10N, and C
The voltage E (t) = {o"
{o. t25
s10,
0<t<5
is applied to the circuit. Determine the
instantaneous charge q (t)on the capacitor for t > 0, if q (0) = 0 and
q (0) = 0
O q (t) = 10
- be-10t cos(10t) – e-10t sin(10€) – [ - e-10t-5) cos(10 (t – 5)) – e-10(t-5) şin(10 (t – 5))] U (t – 5)
te-10(t-5) cos(10 (t – 5)) - e-10(t-5) sin(10 (t – 5))]U (t – 5)
COS
Og(t) =D-e-과 cos(2t)-글e-2 sin(2t)-[-글e-2(-8) cos (2 (t-5))-글1e-26-5) sin(2 (t-5))] U (t -5)
O q(t)
-2(t-5) sin(2 (t – 5))] U (t – 5)
O q (t) = 1- e-10t cos(10t) – e-10t sin(10t) – [1- e-10(t-5) cos(10 (t - 5)) – e-10(t-5) sin(10 (t – 5))] U (t - 5)
O q(t) = -te-5t cos(5t) – e-5t sin(5t) – - te-5(-5) cos(5 (t – 5)) – e-5(t-5) sin(5 (t – 5)] U (t – 5)
1/2
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