Problem — Inductor and capacitor (LC)

Problem

In an ideal $LC$ circuit (no resistance) the inductor has $L=0{,}10\ \text{H}$ and the capacitor $C=10\ \mu\text{F}$. At the initial instant the capacitor is charged to voltage $V_0=12\ \text{V}$ and no current flows. Calculate: (a) the oscillation frequency; (b) the maximum current in the circuit; (c) the total energy, writing it in two equivalent forms.

Ideal $LC$ circuit: capacitor $C$ (on the left) and inductor $L$ (on the right) closed in a loop.

Solution

(a) Oscillation frequency. The natural angular frequency is $\omega=1/\sqrt{LC}$, so $f=\frac{1}{2\pi\sqrt{LC}},\qquad LC=0{,}10\cdot 10\cdot 10^{-6}=10^{-6}\ \text{s}^2,\quad \sqrt{LC}=10^{-3}\ \text{s}$ $f=\frac{1}{2\pi\cdot 10^{-3}}=\frac{1}{6{,}28\cdot 10^{-3}}\approx 159\ \text{Hz}$ $\ev{f\approx 159\ \text{Hz}}$

(c) Total energy. At the start it is entirely electrical, stored in the capacitor: $E=\tfrac{1}{2}C V_0^2=\tfrac{1}{2}\cdot 10\cdot 10^{-6}\cdot (12)^2=7{,}2\cdot 10^{-4}\ \text{J}$ In the ideal circuit it is conserved and oscillates between electrical and magnetic form: $\ev{E=\tfrac{1}{2}C V_0^2=\tfrac{1}{2}L\,i_\text{max}^2=7{,}2\cdot 10^{-4}\ \text{J}}$

(b) Maximum current. When the capacitor is discharged all the energy is magnetic, $\tfrac{1}{2}L\,i_\text{max}^2=E$: $i_\text{max}=\sqrt{\frac{2E}{L}}=\sqrt{\frac{2\cdot 7{,}2\cdot 10^{-4}}{0{,}10}}=\sqrt{1{,}44\cdot 10^{-2}}=0{,}12\ \text{A}$ $\ev{i_\text{max}=0{,}12\ \text{A}}$ The same result follows from $i_\text{max}=\omega C V_0=V_0\sqrt{C/L}$.

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Topics: Induzione elettromagnetica Concepts: Oscillazioni LC e RLC · Densità di energia del campo · Risonanza