A capacitor C is charged to an initial potential of 70.0 V, with an initial char

Question

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 = 5.61×10-2 H and resistance RL. Ri. At t=0, the switch is closed, and the curve below shows the potential V across the capacitor as a function of time t. 6 4 30 2 10 -10 -30 t, (in milliseconds) Calculate the energy in the circuit after a time of 12 periods. Note that the curve passes through a arid intersection point. Submit Answer Tries 0/12 Calculate the time required for 86% of the initial energy to be dissipated. Submit Answer Tries 0/12

Explanation / Answer

at 0.8 ms -> 9 osciilations.

time period. T = 0.8/9 ms

f = 1/T = 1 / (2 pi sqrt(LC))

9000 / 0.8 = 1 / (2 pi sqrt(5.61 x 10^-2 C))

C = 3.57 x 10^-9 F

V = V0 e^(-bt)

where V0 = 70 Volt

at t = 0.8 ms, V = 50 Volt

50 = 70 e^(-b x 0.8)

b = 0.42 ms^-1

at t = 12 x 0.8 / 9

V = 70 [ e^(-0.42 x 12 x 0.8 / 9)]

V = 44.7 Volt

Energy = C V^2 /2 = (3.57 x 10^-9)(44.7^2)/2

= 3.56 x 10^-6 J .........Ans

Emax = C (70)^2 / 2

E = (1 - 0.86)Emax

C V^2 /2 = 0.14 C (70)^2 / 2

V = 26.2 Volt

26.2 = 70 e^(-0.42t)

t = 2.34 ms ............Ans

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