L.C oscillations

L.C Oscillations :
A capacitor is charged to a p.d. of Vo by connecting it across a battery and then it is allowed to discharge through a pure inductor of inductance L.

Initial charge on the plates of the capacitor qo = CVo

At any instant, let the charge flown in the circuit be q and current in the circuit be i.

Applying Kirchhoff ‘s law

$ \displaystyle \frac{q_0 -q}{C} -L\frac{di}{dt} = 0 $

Differentiating w.r.t. time we get

$ \displaystyle -\frac{dq}{dt} -LC \frac{d^2 i}{dt^2} = 0 $

$ \displaystyle \frac{d^2 i}{dt^2} = -\frac{1}{LC}i $

$ \displaystyle \frac{d^2 i}{dt^2} = -\omega^2 i $

$ \displaystyle f = \frac{1}{2\pi \sqrt{LC}} $

∴ The charge q on the plates of the capacitor and current I in the circuit vary sinusoidally as

q = q0 sin (ωt + φ) and

i = q0 ω cos (ωt + φ).

where φ is the initial phase and it depends on initial situation of the circuit.

The angular frequency of the circuit $ \displaystyle \omega = \frac{1}{\sqrt{LC}} $

The total energy of the system remains conserved

$ \displaystyle \frac{1}{2}CV^2 + \frac{1}{2}L i^2 = constant =\frac{1}{2}CV_0^2 = \frac{1}{2}L i_0^2 $

In mechanical oscillation, there is mutual exchange between K.E. and P.E. whereas in electrical oscillation (L.C. circuit), there is exchange between the electrical energy (associated with the capacitor) and magnetic energy (associated with the inductor).

Illustration : A capacitor of capacitance 2µF is charged to a potential difference of 12V. It is then connected across an inductor of inductance 0.6 mH. What is the current in the circuit at a time when the potential difference across the capacitor is 6.0 V  ?

Solution : As the capacitor is charged to a p.d. of 12V, the initial charge on the capacitor is

q0 = CV0 = 2 × 10-6 × 12 C     . . . (i)

At any instant as the capacitor discharges through the inductor (LC circuit), the instantaneous charge on the capacitor is given by

q = qo cos ωt    . . . (ii) [because at t = 0 , q = qo]

But q = CV       …… (iii)

where V is the p.d. at the instant ‘ t ‘

From (i) and (iii) we obtain

$ \displaystyle \frac{q}{q_0} = \frac{V}{V_0} $

Putting the value of V and Vo we obtain,

$ \displaystyle \frac{q}{q_0} = \frac{1}{2}$

$ \displaystyle cos\omega t = \frac{1}{2}$

⇒ ωt = π/3 rad.          …………. (iv)

$ \displaystyle \omega = \frac{1}{\sqrt{LC}} $

On putting the values ,

$ \displaystyle \omega = \frac{(10)^5}{2\sqrt3} $ rad/s

The current through the circuit at that instant is given by,

i = dq/dt

⇒ i = − q0 ω sin ωt

⇒ | i | = q0 ω sin ωt

Putting the value of qo , ω & ωt ,

|i| = 0.6 A

Exercise : A capacitor discharges through an inductor of 0.1 henry. If the frequency of discharge is 1000 Hz, calculate the capacitance.

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