Connect a charged capacitor to an inductor, with no resistor and no battery. The capacitor starts to discharge, but the inductor will not let the current stop: when the capacitor is empty, the current keeps going and charges it the other way. Energy sloshes back and forth between the capacitor's electric field and the inductor's magnetic field, exactly the way it moves between a spring and a mass.
Why it works
Around the loop, the capacitor's voltage equals the inductor's emf: $\dfrac{q}{C} = -L\dfrac{dI}{dt}$, with $I = dq/dt$. So
$$L\frac{d^2q}{dt^2} + \frac{q}{C} = 0,$$
the equation of simple harmonic motion with $q$ in place of $x$. Its solution is $q = Q_0\cos\omega t$ with $\omega^2 = \dfrac{1}{LC}$.
Key idea
$$\omega = \frac{1}{\sqrt{LC}}, \qquad q = Q_0\cos\omega t, \qquad I = -Q_0\omega\sin\omega t.$$
The total energy $\dfrac{q^2}{2C} + \tfrac{1}{2}LI^2 = \dfrac{Q_0^2}{2C}$ stays constant. The current is largest, $Q_0\omega$, when the capacitor is empty.
| Mass on a spring | LC circuit |
|---|---|
| position $x$ | charge $q$ |
| velocity $v$ | current $I$ |
| mass $m$ | inductance $L$ |
| stiffness $k$ | $1/C$ |
| $\tfrac{1}{2}kx^2$ | $q^2/(2C)$ |
| $\tfrac{1}{2}mv^2$ | $\tfrac{1}{2}LI^2$ |
| $\omega = \sqrt{k/m}$ | $\omega = 1/\sqrt{LC}$ |
Worked examples
Worked example: Frequency and peak current
A $5.0\ \mu\text{F}$ capacitor charged to $10\ \text{V}$ is connected to a $20\ \text{mH}$ inductor. Find the frequency and the largest current.
- $\omega = \dfrac{1}{\sqrt{(0.020)(5.0 \times 10^{-6})}} = 3160\ \text{rad/s}$, so $f = \omega/2\pi = 503\ \text{Hz}$.
- $Q_0 = CV_0 = 50\ \mu\text{C}$, so $I_{\max} = Q_0\omega = 0.158\ \text{A}$.
- Check with energy: $\tfrac{1}{2}CV_0^2 = 2.5 \times 10^{-4}\ \text{J}$ and $\tfrac{1}{2}LI_{\max}^2 = 2.50 \times 10^{-4}\ \text{J}$. They match.
Answer: $f \approx 503\ \text{Hz}$, $I_{\max} \approx 0.158\ \text{A}$
Worked example: Tuning a radio
A tuner uses a $200\ \text{pF}$ capacitor. What inductance makes it resonate at $1.0\ \text{MHz}$, in the AM band?
- $f = \dfrac{1}{2\pi\sqrt{LC}}$, so $L = \dfrac{1}{(2\pi f)^2C} = \dfrac{1}{(2\pi \times 10^{6})^2(200 \times 10^{-12})} = 127\ \text{μH}$.
Answer: $L \approx 127\ \text{μH}$
Common mistake
"The current is greatest when the capacitor holds the most charge." It is the other way around. When the capacitor is fully charged, all the energy is in its electric field and the current is momentarily zero, like a pendulum at the end of its swing. The current peaks a quarter cycle later, when the capacitor is empty and all the energy is in the inductor. Check: $q = Q_0\cos\omega t$ and $I = -Q_0\omega\sin\omega t$ are never largest at the same time.