Question

••24. A single-loop circuit consists of a 7.20 ? resistor, a 12.0 H inductor, and a 3.20 ?F capacitor. Initially the capacitor has a charge of 6.20 ?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.

Answer 1

As mentioned in the problem, we have an undriven RLC circuit with the initial condition of a charged capacitor. Please 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. 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 given by –

A = A0 * EXP(-??0t), where ? is the damping coefficient.

Since we are interested in the decay over N cycles, so 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, 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.

Now,
?0 = 1 / sqrt(12x3.20x10^-6) = 1000 / 6.197 = 161.4 rad/s

? = 7.20 / (2x161.4x12) = 0.002

? = 2*pi*0.002 = 0.01256

(a) For N = 5 –

Q(5) = Q0 * EXP(-?N) = 6.20*EXP(-0.01256*5) = 5.823 ?C

(b) For N = 10 –

Q(5) = Q0 * EXP(-?N) = 6.20*EXP(-0.01256*10) = 5.468 ?C

(c) For N = 100 –

Q(5) = Q0 * EXP(-?N) = 6.20*EXP(-0.01256*100) = 1.766 ?C