CourseNotes.50 - r L C i 2 cos 2 ( t )-sin 2 ( t ) Setting...

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Part a In order to match the initial conditions, we will choose to use a sine instead of a cosine to describe the charge on the capacitor. q ( t ) = Q sin( ωt ) Note that we could also have met the initial conditions by choosing the proper phase shift in the cosine term, but this will simplify our life. We do not know the initial charge on the capacitor, but we do know the initial current i 0 . If we take a derivative of the charge dq dt = ωQ cos( ωt ) we see that the maximum current is I = i 0 = ωQ Q = i 0 ω = i 0 LC Part b This problem asks about the rate of change of the energy on the capacitor, so we must take a derivative of the energy. dU dt = d dt q 2 2 C = q C dq dt = 1 C q ( t ) i ( t ) We have used the chain rule to simplify before plugging in the actual quantities. Hence, the rate of change of the energy is dU dt = 1 C Q 2 ω sin( ωt )cos( ωt ) = r L C i 2 0 sin( ωt )cos( ωt ) We want to maximize this with respect to time, so we need to take one more derivative. d dt ± dU dt ² =
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Unformatted text preview: r L C i 2 cos 2 ( t )-sin 2 ( t ) Setting this equal to zero and solving for t gives tan( t ) = 1 t = 1 4 = 4 LC Part c For the nal part of this question, we simply need to plug our time into our equation for dU dt . dU dt max = r L C i 2 cos 4 sin 4 = 1 2 r L C i 2 Problem 31.26 A single-loop circuit consists of a 7 . 2 resistor, a 12 H inductor, and a 3 . 2 F capacitor. Initially the capacitor has a charge of 6 . 2 C and the current is zero. Calculate the charge on the capacitor N complete cycles later for (a) N = 5, (b) N = 10, and (c) N = 100. We showed earlier in this chapter that the charge on the capacitor in an RLC circuit has a cosine oscillation which is exponentially damped. q ( t ) = q e-Rt 2 L cos( t ) 50...
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