Halliday, Fundamentals of Physics, 10e Help I PRINTER VERSION BACK NEXT Chapter

Question

Halliday, Fundamentals of Physics, 10e Help I PRINTER VERSION BACK NEXT Chapter 31, Problem 024 A single-loop circuit consists of a 7.1 2 resistor, 11.9 H inductor, and a 3.4 uF capacitor. Initially the capacitor has a charge of 6.1 uC 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. (a) Number (b) Number )Number Click if you would like to Show Work for this question: Units Units Units SHOW HINT GO TUTORIAL LINK TO TEXT LINK TO SAMPLE PROBLEM

Explanation / Answer

Given that -

R = 7.1 ohm, L = 11.9 H, C = 3.4 micro F = 3.4 x 10^-6 F

Here, we have an undriven RLC circuit with the initial condition of a charged capacitor. Note that this is analogous to an unforced spring-mass-damper system with an initially stretched spring, and the effect of damping can be treated similarly. We know natural frequency ?0 = 1/sqrt(LC), and the oscillation period T = 2?/?0.

so. w0 = 1 / sqrt(11.9 x 3.4 x 10^-6) = 157.2 rad/s

After an integer number of cycles N the system is in the same phase of the oscillation as it was at t=0. So the charge Q at such a time is just the initial charge Q0 * the term representing reduced amplitude (decay) due to damping.
Amplitude of a system subject to exponential decay is
A = A0 * EXP(-??0t), where ? is the damping coefficient.
Since we are interested in the decay over N cycles, we can rewrite this as
A = A0 * EXP(-?N) where ? is the per-cycle decay ("logarithmic decrement") = 2??, and N = t/T.
Damping in an RLC circuit is defined to be at the critical value when ?0 = R/2L (ref.), or R = 2?0L;

thus we can define ? (actual/critical damping) = R/2?0L.
So to solve the problem, we find ?, ?, and the values of Q0*EXP(-?N) for N = 5, 10 and 100.

We get -

?0 = 157.2 rad/s
? = R/2?0L = 7.1 / (2*157.2*11.9) = 0.002
? = 0.01256

(a) Q(5) = 6.1*e^-(0.01256*5) = 5.729 ?C

(b) Q(10) = 6.1*e^-(0.01256*10) = 5.380 ?C

(c) Q(100) = 6.1*e^-(0.01256*100) = 1.737 ?C

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