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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 / 2 , and the spring, which has potential energy k x 2 / 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 ) = d q ( t ) / d t , the total electromagnetic energy U is

U = 1 2 L i 2 + 1 2 q 2 C .

For the mass-spring system, v ( t ) = d x ( t ) / d t , the total mechanical energy E is

E = 1 2 m v 2 + 1 2 k x 2 .

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

x ( t ) = A cos ( ω t + ϕ )

where ω = k / m . Hence, the charge on the capacitor in an LC circuit is given by

q ( t ) = q 0 cos ( ω t + ϕ )

where the angular frequency of the oscillations in the circuit is

ω = 1 L C .

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

i ( t ) = d q ( t ) d t = ω q 0 sin ( ω t + ϕ ) .

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

An LC Circuit

In an LC circuit, the self-inductance is 2.0 × 10 −2 H and the capacitance is 8.0 × 10 −6 F. At t = 0 , all of the energy is stored in the capacitor, which has charge 1.2 × 10 −5 C. (a) What is the angular frequency of the oscillations in the circuit? (b) What is the maximum current flowing through circuit? (c) How long does it take the capacitor to become completely discharged? (d) Find an equation that represents q(t) .

Strategy

The angular frequency of the LC circuit is given by [link] . To find the maximum current, the maximum energy in the capacitor is set equal to the maximum energy in the inductor. The time for the capacitor to become discharged if it is initially charged is a quarter of the period of the cycle, so if we calculate the period of the oscillation, we can find out what a quarter of that is to find this time. Lastly, knowing the initial charge and angular frequency, we can set up a cosine equation to find q ( t ).

Solution

  1. From [link] , the angular frequency of the oscillations is
    ω = 1 L C = 1 ( 2.0 × 10 −2 H ) ( 8.0 × 10 −6 F ) = 2.5 × 10 3 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,
    1 2 L I 0 2 = 1 2 q 0 2 C ,

    so
    I 0 = 1 L C q 0 = ( 2.5 × 10 3 rad/s ) ( 1.2 × 10 −5 C ) = 3.0 × 10 −2 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 = 2 π ω = 2 π 2.5 × 10 3 rad/s = 2.5 × 10 −3 s ,

    the time taken for the capacitor to become fully discharged is ( 2.5 × 10 −3 s ) / 4 = 6.3 × 10 −4 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 cos ϕ .

    Thus, ϕ = 0 , and
    q ( t ) = ( 1.2 × 10 −5 C ) cos ( 2.5 × 10 3 t ) .

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Source:  OpenStax, University physics volume 2. OpenStax CNX. Oct 06, 2016 Download for free at http://cnx.org/content/col12074/1.3
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