Applications of the Higher Ordered Linear Equations9. A'sinusoidal e.m.f. 110 sin 400t volts is connected inseries in a circuit with an inductance of 0.1 henry, a re-sistance of 10 ohms, and a capacitance of 250(10)6farad. Find the steady-state solutions of Q and I interms of the time t. Also find the maximum values ofthe steady-state charge and current.ANSWERS:Q = - 0.00275(3 sin 400t + cos 400t)I = 1,10(sin 400t- 3 cos 400t)Qmax = 0.0087 coul.9.%3DImax = 3:48 amp....2 amn.

Answered: Image /qna-images/answer/cb62dfd5-0f83-4833-8d02-0ae4f168eb78.jpg

circuit with an inductance of 0.1 henry, a f 10 ohms, and a capacitance of 250(10 d the steady-state solutions of Q and I he time t. Also find the maximum values -state charge and current. - 0.00275(3 sin 400t + cos 400t) 1.10(sin 400t - 3 cos 400t) 0 0087 coul.

circuit with an inductance of 0.1 henry, a f 10 ohms, and a capacitance of 250(10 d the steady-state solutions of Q and I he time t. Also find the maximum values -state charge and current. - 0.00275(3 sin 400t + cos 400t) 1.10(sin 400t - 3 cos 400t) 0 0087 coul.

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

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Question

Applications of the Higher Ordered Linear Equations
9. A'sinusoidal e.m.f. 110 sin 400t volts is connected in
series in a circuit with an inductance of 0.1 henry, a re-
sistance of 10 ohms, and a capacitance of 250(10)6
farad. Find the steady-state solutions of Q and I in
terms of the time t. Also find the maximum values of
the steady-state charge and current.
ANSWERS:
Q = - 0.00275(3 sin 400t + cos 400t)
I = 1,10(sin 400t- 3 cos 400t)
Qmax = 0.0087 coul.
9.
%3D
Imax = 3:48 amp.
...2 amn.

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