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 Q0. It is in a circuit with a switch and an inductor with inductance L = 5.33

A capacitor C is charged to an initial potential of 60.0 V, with an initial charge of Q0. It is in a circuit with a switch and an inductor with inductance L = 5.33?10-2 H and resistance RL. Dynamically generated plot Calculate the energy in the circuit after a time of 16 periods. Note that the curve passes through a grid intersection point? Calculate the time required for 93% of the initial energy to be dissipated? At t=0, the switch is closed, and the curve below shows the potential V across the capacitor as a function of time t.

Explanation / Answer

eenrgy in RL Ckt is given by U = 0.5 Li^2

here we need current i

so

from graph

V at 0.8 secs = 30 V

initial Volatge V = 60 V

so

use the equation V = Vo(1-e^-Rt/L)

1-e^-Rt/L = 30/60 = 0.5

e^-Rt/L   = 1-0.5 = 0.5

-Rt/L   = ln (0.5) = -0.693

so

R    = 0.693*L/t

R = 0.693 * 0.0533/16

R = 0.0023 ohms

so

intiial current i = V/R = 60/0.0023 = 25990 A

current after 16 periods

i = io e^-RT/L

i = 25990 * e^-(0.0023 * 16/0.053)

i = 12979 Amps

so energy U = 0.5 LI^2

U = 0.5 * 0.0533 * 12979*12979

U = 4.489 *10^6 Joules

-----------------

apply part B :

e^-Rt/L = 0.93

-Rt/L = ln(0.93)

t = 0.0725 * L/R

t = 0.0725 * 0.0533/0.0023

t = 1.68 secs

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