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Safari File Edit View History Bookmarks Window Help 00% Sat Nov 7 7:22 PM E s4, lite msu.edu Search Q&A; I Chegg.com Pandora Radio Listen to Free Internet Radio, Find New Music 4) LON-CAPA Damped oscillations Evaluate Feedback L urse Contents HW o9 (11/09) Damped oscillations otes mel A capacitor C is charged to an initial potential of 70.0 with an initial charge of Qo. It is in a circuit with a switch and an inductor with inductance L 2.65x10 H and resistance RL. 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. 70 60 50 40 30 20 10 10 20 30 40 50 60 70 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 t, in milliseconds) Calculate the energy in the circuit after a time of 17 periods. Note that the curve passes through a grid intersection point Submit Answer Incorrect. Tries 2/12 Previous Tries Calculate the time required for 92% of the initial energy to be dissipated Submit Answer Tries 0/12

Explanation / Answer

here,

the figure shows case of light damped oscillations.The frequecy in damped oscillations is given as :

fo = 1/sqrt(LC) ------------------(1)

solving for fo : there are 10 complete(20pi radians) cycles in 0.8ms (0.0008s)
therefore,
fo = 20 * pi /t
fo = 20*3.14/0.0008
fo = 78500 rad/s

Solving for capacitance from eqn 1:
C = 1 / fo^2*L
C = 1/(2.65 * 10^-2 *78500)
C = 4.807*10^-4 F

Therefore, Initial Energy :
Eo = 0.5 * C * V^2
Eo = 0.5 * 4.807*10^-4 * 70^2
Eo = 1.177 J

also from graph V dissipate fro 70 to 40 exponentially in 0.8ms (0.0008s)
Vo/V(t) = exp(-kt) ---------------------(2)
taking log on both sides we get,
Kt = ln(Vo/V(t))
k = ln(70/40) / 0.0008
k = 699.520

time for 17 periods = 0.0008 * 17 / no of periods in 0.0008 s
t17 = 0.0008*17/10
t17 = 0.00136 s

Value of kt after 17 periods:
Kt = 0.00136*699.520 = 0.9513

Therefore Energy after 17 peroid will be :
E(t)/Eo = ( V(t)/Vo )^2 --------------(3)

Using equation 2 we get,
E(t)/Eo = exp(-2Kt)
E(t) = Eo*exp(-2Kt)
E(t) = 1.177 / exp(-2*0.9513)
E(t) = 0.1755 J

PARt B:
WHen 92 % of energy be dissipitated
from eqn 3

E(t)/Eo = (100-92) / 100
       = 8/100
       = 0.08

Therefore, from eqn 3
V(t)/Vo = sqrt(0.08)
       = 0.2828

also.

exp(-kt) = V(t)/Vo
-kt = ln(0.2828 )
kt = 1.26302

solving fot time t,

t = 1.26302 / k
t = 1.26302 / 699.520
t = 0.001805 s or 1.805*10^-3s

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