14.5 Oscillations in an lc circuit  (Page 2/4)

We have followed the circuit through one complete cycle. Its electromagnetic oscillations are analogous to the mechanical oscillations of a mass at the end of a spring. In this latter case, energy is transferred back and forth between the mass, which has kinetic energy $m{v}^2\text{/}2$ , and the spring, which has potential energy $k{x}^2\text{/}2$ . With the absence of friction in the mass-spring system, the oscillations would continue indefinitely. Similarly, the oscillations of an LC circuit with no resistance would continue forever if undisturbed; however, this ideal zero-resistance LC circuit is not practical, and any LC circuit will have at least a small resistance, which will radiate and lose energy over time.

The frequency of the oscillations in a resistance-free LC circuit may be found by analogy with the mass-spring system. For the circuit, $i(t)=dq(t)\text{/}dt$ , the total electromagnetic energy U is

For the mass-spring system, $v(t)=dx(t)\text{/}dt$ , the total mechanical energy E is

The equivalence of the two systems is clear. To go from the mechanical to the electromagnetic system, we simply replace m by L , v by i , k by 1/ C , and x by q . Now x(t) is given by

where $\omega =\sqrt{k\text{/}m}.$ Hence, the charge on the capacitor in an LC circuit is given by

where the angular frequency of the oscillations in the circuit is

Finally, the current in the LC circuit is found by taking the time derivative of q(t) :

The time variations of q and I are shown in [link] (e) for $\varphi =0$ .

An LC Circuit

Strategy

Solution

1. From [link] , the angular frequency of the oscillations is
   
   $\omega =\sqrt{\frac{1}{LC}}=\sqrt{\frac{1}{(2.0\times {10}^{\mathrm{-2}}\text{H})(8.0\times {10}^{\mathrm{-6}}\text{F})}}=2.5\times {10}^3\text{rad/s}.$
2. The current is at its maximum $I_0$ when all the energy is stored in the inductor. From the law of energy conservation,
   
   $\frac{1}{2}L{I}_0^2=\frac{1}{2}\frac{q_0^2}{C},$
   
   so
   
   $I_0=\sqrt{\frac{1}{LC}}{q}_0=(2.5\times {10}^3\text{rad/s})(1.2\times {10}^{\mathrm{-5}}\text{C})=3.0\times {10}^{\mathrm{-2}}\text{A}.$
   
   This result can also be found by an analogy to simple harmonic motion, where current and charge are the velocity and position of an oscillator.
3. The capacitor becomes completely discharged in one-fourth of a cycle, or during a time T /4, where T is the period of the oscillations. Since
   
   $T=\frac{2\pi }{\omega }=\frac{2\pi }{2.5\times {10}^3\text{rad/s}}=2.5\times {10}^{\mathrm{-3}}\text{s},$
   
   the time taken for the capacitor to become fully discharged is $(2.5\times {10}^{\mathrm{-3}}\text{s})\text{/}4=6.3\times {10}^{\mathrm{-4}}\text{s}.$
4. The capacitor is completely charged at $t=0,$ so $q(0)=q_0.$ Using [link] , we obtain
   
   $q(0)=q_0=q_0\text{cos}\varphi .$
   
   Thus, $\varphi =0,$ and
   
   $q(t)=(1.2\times {10}^{\mathrm{-5}}\text{C})\text{cos}(2.5\times {10}^{3}t).$