A capacitor C is charged to an initial potential of 70.0 V, with an initial char
ID: 1265454 • 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 = 5.63
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 = 5.63x10^-2 H and resistance RL. At t=0 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 89% of the initial energy to be dissipated.Explanation / Answer
9 periods in 0.7 mS
T = 0.7 /9 =0.078 ms = 0.000078 s
V = VO ( e^(-Rt / 2L) coswt
w = 2pi / w
for 9 period, coswt=1
and V = 40 volt at t = 0.7 ms and V0 = 70 volt
R = - 2L ln(V/V0) / t
R = - 2x 5.63 x 10^-2 ln(40/70) / 0.000078
R = 807.86 Ohm
w = 1 / sqrt(LC)
2pi / 0.000078 = 1 / sqrt(5.63 x 10^-2 x C)
C = 2.74 x 10^-9 F
V at t = 15T
V = 70 ( e^(-15/2))
V = 0.0387 Volt
Energy = CV^2 /2 = 2.74 x 10^-9 x 0.0387^2 /2
= 2.05 x 10^-12 J
.................Ans
CV^2/2 = 0.89CV0^2 /2 ( Given)
V = 0.943 V0
V / V0 = 0.943
t = - 2L ln(V/V0) / R
t = -2 x 5.63 x10-2 x ln0.943 / 807.86
t = 9.12 x 10^-6 sec
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