A capacitor C is charged to an initial potential of 70.0 V, with an initial char
ID: 1265254 • 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 = 4.93
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 = 4.9310^-2 H and resistance RL. 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 12 periods. Note that the curve passes through a grid intersection point. Calculate the time required for 91% of the initial energy to be dissipated.Explanation / Answer
8 periods in 0.7 ms
So period = 0.7 / 8 = 0.0875 ms
T = 0.0000875 s
w = 2pi / T = 71807 Hz
peak at t1 => v1 = 50 V
v0 = 70V , v1 = 50 v at t1 = 0.7 ms
V = V0 e^(-Rt/2L) coswt
at t = 8T coswt = 1
R = -2L ln(v1/v0) / t1
R = -2 x 4.93 x 10^-2 ln (50/70) / 0.7x10-3
R = 47.39 Ohm
w0 = 71807 = 1/sqrt(LC)
C = 3.93 x 10^-9 F
After 12 periods T12 = 0.00105 s
V(After 12 periods T12 = 0.00105 s ) = 70 e^(-47.39x0.00105/2x4.93x10-2)
V = 42.26 Volt
Energy = CV^2 /2 = 3.93 x 10^-9 x 42.26^2 /2 = 3.51 x 10^-6 J .........Ans
energy will be 0.91 when V = 0.954V
t1 = -2L ln(v1/v0) / R = -2 x 4.93 x 10-2 x ln (0.954) / 47.39 = 9.81 x 10^-5 sec ........Ans
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