A capacitor C is charged to an initial potential of 70.0 V, with an initial char
ID: 1769766 • Letter: A
Question
A capacitor C is charged to an initial potential of 70.0 V, with an initial charge of Q0. It is in a circuit with a switch and an inductor with inductance L = 5.31×10-2 H and resistance RL.
P Pandora Radio Liste × D WeBWorK: mth-234-f LON CAPA Damped Osc ×e A: ACapacitor CIs Char G 621.7112 Google Searc 1 × × . C f secure https://s2lite.msu.edu/enc/67/6e7454ccb062e93ba5428cSdeba855a6352f2e0a3c33e771 56be0ed89693f230ddc7b80f66fa08ee49c09b7ab8afa40bee893cbc33ed71 a : Apps D Prereq Flowchart-CE D Mathematics Michi ACS Style Guide Ch y's The 7 Day Shredding 0Company Search FS17 MTH 234 A capacitor C is charged to an initial potential of 70.0 v, with an initial charge of Qo. It is in a circuit with a switch and an inductor with inductance L- 5.31×10.2 H and resistance R At t=0, the switch is closed, and the curve below shows the potential V across the capacitor as a function of time t. 60 H 3e 20 -10 10 20 30 50 -60 t, (in illiseconds) Calculate the energy in the circuit after a time of 16 periods. Note that the curve passes through a grid intersection point. 4.04 104J Subrrnit Answer Incorrect. Tries 1/12 Previous Tries Calculate the time required for 93% of the initial energy to be dissipated 7:45 PM lype here to search 11/5/2017Explanation / Answer
11 oscillations completed in 0.80 ms
Time period, T = 0.80/11 = 0.0727 ms
w = 2 pi / T = 86394 rad/s
and w = 1/sqrt(LC )
L = 5.31 10^-2
C = 2.523 x 10^-9 F
For voltage oscillations,
V = V0 e^(-b t)
at t = 11 T , V = 40 volt
40 = 70 e^(-11bT)
bT = 0.05087
at t = 16 T
V = 70 e^(-16 x 0.05087) = 31 Volt
Energy = C V^2 / 2 = (2.523 x 10^-9)(31^2) /2
= 1.21 x 10^-6 J or 1.21 uJ ..........Ans
C V^2 /2 = 0.93 C V0^2 /2
V = 0.964 V0
0.964 V0= V0 e^(-bt)
ln(0.964) = - n b T
0.0363 = n (0.05087)
n = 0.714
t = 0.052 ms = 5.2 x 10^-5 s
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