A capacitor C is charged to an initial potential of 60.0 V, with an initial char
ID: 1629690 • Letter: A
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 = 3.01×10-2 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.
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 89% of the initial energy to be dissipated.
Explanation / Answer
The system is lightly damped, ie the energy dissipates slowly, over many cycles. So we can approximate the frequency of the damped oscillations by the natural frequency without damping :
w0 = 1/sqrt(LC).
We can find w0 from the graph, we are given L, we can then calculate C. This helps us find the initial energy in the circuit E0, which is stored entirely in the capacitor and is (1/2)CV^2.
So we first find w0 then C. There are 11 complete cycles (=22 radians) in 0.9 ms, so
w0 = 22 rad /0.9ms
= 7.679x10^4 rad/s.
w0^2 = 5.897x10^9 (rad/s)^2.
Lw0^2 = 3.01x10^-2 x5.897x10^9
= 1.775x10^8.
C = 1/Lw0^2
= 1/1.775x10^8
= 5.63 nC.
E0 = (1/2)CV^2
= 0.5*5.63*60^2 nJ
= 10.134 uJ (micro-joules).
The voltage on the capacitor dissipates exponentially with time :
V(t) = V0exp(-Kt)
V0/V(t) = exp(Kt)
Kt = ln(V0/V(t)).
V decays from 60V to 40V in time t=0.9ms so
0.9K = ln(60/40) = 0.4055
K = 0.4055/0.9 = 0.4505
when t is in ms.
We need to find the energy in the circuit after 12 periods. 11 periods is 0.9ms so 12 periods is 12*0.9/11 = 0.9818ms.
Kt = 0.4505 *= 0.9818= 0.4423.
Now
E(t)/E0 = [V(t)/V0]^2
= [exp(-Kt)]^2
= exp(-2Kt)
= exp(-0.8846)
= 0.41287
E(t) = 0.41287 x 10.134 uJ
= 4.184 uJ.
B.
When 89% of the energy has been dissipated, then
E(t)/E0 = (100-89)/100
V(t)/V0 = sqrt(11/100)
= 0.33166.
exp(-Kt) = V(t)/V0
-Kt = ln( 0.33166)
= -1.1036
t = 1.060132/0.4505
= 2.45ms.
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