Question

An L-R-C series circuit has L = 0.500 H. C = 5.00 pF, and R = 310 22. At t=0 the current is zero and the initial charge on the capacitor is 2.80 x 10c. Part A What is the values of the constant A in the followin equetion? q=Ae - (R/2L) cos Express your answer to three significant figures and include the appropriate units. ANSWER: A- Part B What is the value of the constant ? Express your answer in radians to three significant figures. ANSWER: rad Part How much time does it take for each complete current oscillation after the switch in this circuit is closed? Express your answer to three significant figures and include the appropriate units. ANSWER Part D What is the charge on the capacitor after the first complete current oscillation? Express your answer to three significant figures and include the appropriate units. ANSWER:

Answer 1

C = 5-4F a '1-0.511, A to R=3102 9;. = 2.8x10c In i = O A Part A 가 my С. L di also, + q + IR = 0 dt с Iu da dt da + dr + 9 =0 dt? Lo da let d (ydt dt (022 JD + 19 O 2 Lo D + RDF I t Tc I INTR-4 2V 12 -R 22 Solution is + R? D. given by 4L2 C q= A (PRL) cos I R? At t=0 q=2-8x10 (D) A cosa 1-922) = I = tal dt Al-Pe Le Cos R2 912 R? A 18/21/t sin to 412 1-R2 LC 412 t+o 7t-o, T=0 0 = A | PL AIR 21 coso - Aft - R2 ILC 4L² sind

= tand R2 41² A) / Ver tomo 310 2x0.5 0.5X 5xlo - (31032 4x10.52 Om 56 234 3103-29.4 below horizontalo - from equation ④ 2.3x16: ACOS (-294) 14 = 30212xloc) 7 ant B = 0.513 red below harmontal) Parte Since to = 27 = ) T=27 I > TE 2 2x II R2 to bo-3012 0.5x5x10 4x40-532 T: 0. 01l4 sec Part D 9 = Bu 212 x10") - 310 10.0114 2x0.5) Cos 127 -0,513 9- 0. 082x16 C