A capacitor C is charged to an initial potential of 50.0 V, with an initial char
ID: 1630358 • Letter: A
Question
A capacitor C is charged to an initial potential of 50.0 V, with an initial charge of Q_0. It is in a circuit with a switch and an inductor with inductance L = 4.53 times 10^-2 H and resistance R_L. At t=0, the switch is closed, and the curve below shows the potential V across the capacitor as a function of time t. Calculate the energy in the circuit after a time of 15 periods. Note that the curve passes through a grid intersection point. Calculate the time required for 86% of the initial energy to be dissipated.Explanation / Answer
V = Vo e^(-bt)
from Graph:
after 8 periods or at t = 0.7ms
V = 20 volt
20 = 50 e^(-b x 0.7 x 10^-3)
b = 1309
and E = Eo e^(-2bt)
1) t = 15T
T = 0.7ms/8 = 0.0875 ms
t = 1.3125 ms
E = Eo e^(-2 x 1309 x 1.3125 x 10^-3)
E = 0.032Eo
Eo = C V0^2 /2
T = 2pi sqrt(LC)
0.0875 x 10^-3 = 2pi sqrt(4.53 x 10^-2 C)
C = 4.28 x 10^-9 F
E = (0.032)(4.28 x 10^-9 x 50^2 / 2)
E = 1.712 x 10^-7 J
(2) E = (1- 0.86)Eo = 0.14Eo
0.14 Eo = Eo e^(-2bt)
ln(0.14) = - 2 x 1309 t
t = 0.75 x 10^-3 s or 0.75ms
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