krit.club logo

Alternating Current - LC Oscillations

Grade 12CBSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

•

An LC circuit consists of a capacitor CC and an inductor LL. When a charged capacitor is connected to an inductor, the charge on the capacitor and the current in the circuit exhibit electrical oscillations, known as LC oscillations.

•

The energy in the circuit oscillates between the electric field of the capacitor (UE=q22CU_E = \frac{q^2}{2C}) and the magnetic field of the inductor (UB=12Li2U_B = \frac{1}{2}Li^2).

•

In an ideal LC circuit (zero resistance), the total energy U=UE+UBU = U_E + U_B remains constant over time.

•

The oscillations are analogous to mechanical Simple Harmonic Motion (SHM). In this analogy, charge qq corresponds to displacement xx, inductance LL corresponds to mass mm, and capacitance 1/C1/C corresponds to the spring constant kk.

•

The differential equation governing the circuit is Ld2qdt2+qC=0L\frac{d^2q}{dt^2} + \frac{q}{C} = 0, which is similar to the SHM equation md2xdt2+kx=0m\frac{d^2x}{dt^2} + kx = 0.

📐Formulae

ω0=1LC\omega_0 = \frac{1}{\sqrt{LC}}

f=12πLCf = \frac{1}{2\pi\sqrt{LC}}

q(t)=Q0cos⁡(ω0t)q(t) = Q_0 \cos(\omega_0 t) (assuming q=Q0q=Q_0 at t=0t=0)

i(t)=−dqdt=I0sin⁡(ω0t)i(t) = -\frac{dq}{dt} = I_0 \sin(\omega_0 t) where I0=ω0Q0I_0 = \omega_0 Q_0

Utotal=Q022C=12LI02U_{total} = \frac{Q_0^2}{2C} = \frac{1}{2}LI_0^2

UE=Q022Ccos⁡2(ω0t)U_E = \frac{Q_0^2}{2C} \cos^2(\omega_0 t) and UB=Q022Csin⁡2(ω0t)U_B = \frac{Q_0^2}{2C} \sin^2(\omega_0 t)

💡Examples

Problem 1:

A 20 mH20\text{ mH} inductor is connected to a fully charged 50 \muF50\text{ \mu F} capacitor. If the initial charge on the capacitor is 10 mC10\text{ mC}, what is the natural frequency of the circuit and the maximum current?

Solution:

Given: L=20×10−3 HL = 20 \times 10^{-3}\text{ H}, C=50×10−6 FC = 50 \times 10^{-6}\text{ F}, Q0=10×10−3 CQ_0 = 10 \times 10^{-3}\text{ C}.

  1. Frequency f=12πLC=12π20×10−3×50×10−6=12π10−6=1032π≈159.15 Hzf = \frac{1}{2\pi\sqrt{LC}} = \frac{1}{2\pi\sqrt{20 \times 10^{-3} \times 50 \times 10^{-6}}} = \frac{1}{2\pi\sqrt{10^{-6}}} = \frac{10^3}{2\pi} \approx 159.15\text{ Hz}.
  2. Angular frequency ω0=1LC=1000 rad/s\omega_0 = \frac{1}{\sqrt{LC}} = 1000\text{ rad/s}.
  3. Max Current I0=ω0Q0=1000×10×10−3=10 AI_0 = \omega_0 Q_0 = 1000 \times 10 \times 10^{-3} = 10\text{ A}.

Explanation:

The natural frequency is determined by the resonance condition of the LL and CC components. The maximum current is reached when all the electrical energy from the capacitor is converted into magnetic energy in the inductor.

Problem 2:

At what time tt is the energy stored in the capacitor and inductor shared equally for the first time, if the capacitor was fully charged at t=0t=0?

Solution:

For equal energy, UE=UB=12UtotalU_E = U_B = \frac{1}{2} U_{total}. q22C=12(Q022C)  ⟹  q2=Q022  ⟹  q=Q02\frac{q^2}{2C} = \frac{1}{2} \left( \frac{Q_0^2}{2C} \right) \implies q^2 = \frac{Q_0^2}{2} \implies q = \frac{Q_0}{\sqrt{2}}. Since q(t)=Q0cos⁡(ω0t)q(t) = Q_0 \cos(\omega_0 t), we have Q0cos⁡(ω0t)=Q02Q_0 \cos(\omega_0 t) = \frac{Q_0}{\sqrt{2}}. cos⁡(ω0t)=12  ⟹  ω0t=π4\cos(\omega_0 t) = \frac{1}{\sqrt{2}} \implies \omega_0 t = \frac{\pi}{4}. t=π4ω0=πLC4t = \frac{\pi}{4\omega_0} = \frac{\pi \sqrt{LC}}{4} or t=T8t = \frac{T}{8} where TT is the time period.

Explanation:

Energy is shared equally when the charge drops to 1/21/\sqrt{2} of its maximum value. This occurs at one-eighth of the total time period TT of the oscillation.