LC Oscillations
Share
LC Oscillations
Electromagnetic Induction of Class 12
LC OSCILLATIONS
The ability of an inductor and a capacitor to store energy leads to the important phenomenon of electrical oscillations. Figure (4.26 a) shows a capacitor with initial charge Qo connected to an ideal inductor having no resistance. All the energy in the system is stored in the electrical field:
UE = Q20/2C
At t = 0, the switch is closed and the capacitor begins to discharge (see Fig. 4.26 b). As the current increases, it sets up a magnetic field in the inductor, and so part of the energy is stored in the magnetic field, UB = 1/2 LI2. When the current reaches its maximum value Io, as in Fig. (4.26 c), all the energy is in the magnetic field : UB = 1/2LI02. The capacitor now has no energy, which means Q = 0. Thus, I =0 when Q = Q0 and Q = 0 when I = I0. The current now starts to charge the capacitor, as in figure (4.26 d). In figure (4.26 e), the capacitor is fully charged, but with polarity opposite to its initial state in figure (4.26 a). The process just described will now repeat itself till the system reverts to its original state. Therefore the energy in the system oscillates between the capacitor and the inductor. As the block−spring system shown in the diagram suggests, the current and the charge in fact undergo simple harmonic oscillations. We will pursue the analogy later.
|
U = 1/2 kA2, K = 0 (a) |
U ≠ 0 , K ≠ 0 (b) |
|
U = 0, K ≠ 0 (c) |
U ≠ 0, K ≠ 0 (d) |
|
U ≠ 0, K = 0 (e) |
|
|
Consider the situation depicted in Fig (4.26 b) and redrawn in figure (4.27). The current is increasing (dI/dt > 0), which means the induced emf in the inductor has the polarity shown, so Vb < Va. According to Kirchoff’s loop rule,
In order to relate the current in the wire to the charge on the capacitor, we note that the current causes the charge Q on the capacitor to decrease, so I = − dQ/dt With this, the loop rule becomes
|
Fig. (4.27)At the instant depicted the current is increasing, so the polarity of the induced emf in the inductor is an shown., |
This has the same form as equation for simple harmonic oscillation.
The charge therefore oscillates with a natural angular frequency
ωo = 1/√LC
Example : 4.12
In an LC circuit, as in fig. (4.27), L = 40 mH, C = 20 µF, and the maximum potential difference across the capacitor is 80 V. Find
- the maximum charge on C
- the angular frequency of the oscillation
- the maximum current
- the total energy
Solution
(a)Qo = CVo = (2 × 10−5F ) (80 V) = 1.6 × 10−3 C
(b)The angular frequency is
ωo = 
= 1120 rad/s
(c)The maximum current is
Io = ωoQo = (1120 rad/s) (1.6 × 10−3 C) = 1.79 A
(d)The total energy is simply the initial energy of the capacitor
U = 
J
