Question

Find the charge on the capacitor in an LRC-series circuit at

t = 0.04 s

when

L = 0.05 h,

R = 3 Ω,

C = 0.02 f,

E(t) = 0 V,

q(0) = 7 C,

and

i(0) = 0 A.

(Round your answer to four decimal places.)
  C

Determine the first time at which the charge on the capacitor is equal to zero. (Round your answer to four decimal places.)
  s

Answer 1

Givens Information: = 0·05 H La R = 32 c=0.02 Elt) = OV =7c 910) ((o)=0A To Find: = 9(+) 't=0.04sec ao 7 solution: consider RLC as shown in circuit Series below figure : R=32 L=0:05H mm C=0.02F Ht filt) +- Elt) = OV

circuit: in above KUL Applying L dict) + 3S Siltat Elt) R i(t) + Jt 't : Worot. Differentiating 1 dilt) ilt) R dilt) + dt 0 = + dt² с substituting values of Rolf c: (+) 0 = 3 diit) dt + + 0.05 dilt d4² 0.02 on simplifying dilt) + 60 dict) + 1000 îlt) = 0 dt² de This differential is second-order Equation. General solution Homogenous by: CF. (complerrentary Paret) Ps It's given ilt) calculating of of a let d dt Em.

as: It's Auxiliary Equation is given m2 + bom +1000 =0 → -60+ m = 60)2- 41000) -6077-400 2 211) → -30 troi ma Thus, -3ot i(t)= e (a cos(lot) + Cz sin (107) IC: Applying ilo) = 0 A la & qct) = Sict) de -30t sinclot) 914 = 628 € q(t) = ca le-sot sincrot) #Note :~ -at Jee é a sin(bt) dt = -at e a²+6² [asin (bt) + b cosihty]

Here, a= 30, b=10 -3ot (Bosin (10+) +10 Ls (10+)] 2 오 q t)=-62 e (30) +10) + cos(iot)) -30t ☺ 3 sin clot) 914) に - C2 e 100 910) = = 70 – 7 > -C2 910) 100 -700 C2 = ㅋ

becomes ² Hence, eq" (1) - Bot ilt) ==0 700 sin(lot) -30t ºlt) 700 e sin (lot) - can be across capacitor (c) voltage expressed as & i(t) dt U= ect) dt Thus, a(zoss capacitor is given by { get) charge c vit) } filtro = > 9ct) - 30t 1 (+) 700 e sinlot) dt # Note -at sin (bt) dt (asin bt + bcoubt] a²t6²

Here, we have: a=30, b=10 -30t 700 e OH) = ( 30 sin (lot) + 10 Cos(lot)] 30'+102 -3ot 7 e J1922) 81) = [3sin (lot) + cos(lot) At t=0.04 sec → 810.04) -30(0.04) 7 e (3 sin (0.4)+ cos(0.4) - 03(04)] ANSWER .. 910.04) 4.4050 C Past (2) which acoose C Charge Time at becomes Zero. charge From eq" (2) oy - 30t 8(t) = 7e ( 3 sin clot) + cos(lot) As per question git) = 0 -30t → 7e = 0 tamclot) = -3 TU → lot tas (-2) = 1.89254 t = 0.189254 20-1892 sec. | Answer