A capacitor C is charged to an initial potential of 60.0 V, with an initial char

Question

A capacitor C is charged to an initial potential of 60.0 V, with an initial charge of Qo. It is in a circuit with a switch and an inductor with inductance L = 3.85×10-2 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. 50A:- 40 30. -10 -30 -4? -5? t, (in milliseconds) Calculate the energy in the circuit after a time of 12 periods. Note that the curve passes through a grid intersection point 0.000001701J Submit Answer Incorrect. Tries 6/12 Previous Tries Calculate the time required for 86% of the initial energy to be dissipated. 7.16x10-4 s You are correct

Explanation / Answer

Period here T=0.9/11 =0.0818 ms this is From grapgh

angular frequency W=2pi/T =2pi/(0.0818*10-3)

W=76794.5 rad/s

from graph at to=0 ms -----------> Vo=60 Volts

at t1=0.8 ms ----------------> V1=20 Volts

Voltage in a damped oscillating circuit is given by

V1 = Vo e^(-Rt/2L) Cos Wt

here Cos Wt = 1

V1 = Vo e^-RT1/2L

R = 2L * ln(V1/Vo)/t1

R = -2 * 0.0385 * ln(20/60)/(0.9*10^-3)

R = 94 ohms

Angular resosnant Frequency Wo^2 = 1/LC

Capacitance C = 1/wo^2 L

C = 1/(76794.5^2*0.0385)

C = 4.4 nF

intial emergy Stored Uo = 0.5 CV^2

Uo = 0.5* 4.4 *10^-9* 60^2

Uo = 7.92 *10^-6 Joules

Charge across Capacitor decays according to

U = Uo e^-t/T

where T is time constant = L/R

T = 0.0385/94

T = 0.41 ms

so after 12 periods

t = 12* 0.00818 = 0.982 ms

so U = 7.92*10^-6 * e^(-0.98/0.41)

U = 7.25 *10^-7 Joules

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