Question

Find the charge on the capacitor in an LRC-series circuit at t = 0.03 s when L = 0.05 h, R = 312, C = 0.008 f, E(t) = 0 V, 9(0) = 3 C, and i(0) = 0 A. (Round your answer to four decimal places.) 7.8289 X C Determine the first time at which the charge on the capacitor is equal to zero. (Round your answer to four decimal places.) 0.000404 XS

Answer 1

Given data: t=0.035 L = 0.05H, R=3.12.(=0.608F E(t) = OV / 900)=34, 1(O)=0 A For LRC Series (kt, R L mm C Ect) i da Apply dt Loop. KVL in the L + Rit <lidt = 664) dt Liga +R di dt 2 с ECA) dt? Subtitute the given values, 9 0.05 d²a dt2 + 3 de dt + 0.008 d'e dtz + 3 da 0.05 at + 9 0.00 8 X0-05 da وه +60 + 2500 q = 0 dt dt Auxilary equation is, D²+ GOD + 2500 =0

Roots, - Бt /ь час -D =) 2a - 60 I (601²4(1)(2500) 2(1) = -601 3600 - 10000 2 a 60 1 80i 2 - 30 + 40i = 2 = 8 Since the roots are complex, Then the General Solution is y = A ( A cos(pt) +B sin (pt) 900) = [A cos (40) + B Sin (406) - 30t e - 30t 30t de ar -30 e (A Cos ( 40t)+B Sinc40t)] + e 7-40 Sinczost + 40 B Song (40+)] 3000) 9co)= (ACOS(4000))+B Sin (4060)] 8 = 1 [A to] A=63, -3000) - 30e -3000 ico) (A COS(4000)) +B sin (4070)] + é (-40 Sin (401040B Cos(4000)] O = - 30 [A +0] + 40 ( B+o] - 30A + 40 B 40B = 30A = B = 0 30x3 40 ala

A = 3, ala B = -30t act) = e (3 cos (401) + V4 Sin (40+] So, the charge on the Capacitor in. LRc Series Ск at t = Orozs -3060-03), q (0.03) =ė [ 3 cos (40x9-03) + 14 Sin (qoxo-osj] = 0.4065 (360-9997)+ (0.0209)] = 0.4065 ( 2.9991+ 0.047025) = 0.4065 (3.0461] 460-03) = 1.238 coloumb's b) The change on the capacitor is Equal to Zero, When, CE) = 0 -Bot é ( 3 Cos(40+)+ V4 Sin(qot)] = 0 3 cos (40t) +2.25 Sin (404) = 0 2.25 Sin (40t) = -3 Cos (406) tan 40t) = -3/25 tan (40t) = -4/ 40t = -0.92729+ k7 The first time (ie k=1) t = -0-02319 + LT t = 0.055365
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