LC oscillations

A swing with no moving part

Pull a swing back and let go. It speeds up as it comes down, passes the lowest point, and rises on the other side. All the time its energy changes from potential to kinetic and back. A circuit can do the same. A capacitor stores energy in its electric field. A coil stores energy in its magnetic field. If we join them, the energy can swing between the two.

One cycle of the swing

Charge a capacitor, and then join it to a coil. There is no battery and no resistance. Follow the energy through one period $T$.

1. $t = 0$. The capacitor holds the whole charge $q_0$. There is no current. All the energy $\dfrac{q_0^2}{2C}$ is in the electric field of the capacitor.
2. The capacitor discharges. The charge starts to flow through the coil. The coil opposes any change in the current with a back emf, so the current builds up gradually.
3. $t = T/4$. The capacitor is empty, and the current is largest. All the energy is in the magnetic field of the coil, $\tfrac{1}{2}LI_0^2$.
4. The coil keeps the current going. The current cannot stop at once, so it goes on and charges the capacitor the other way, until the charge is $-q_0$ at $t = T/2$. The current is zero again.
5. The second half. The same happens backwards. At $t = 3T/4$ the current is largest in the other direction, and at $t = T$ the circuit is back where it started.

The charge and the current are a quarter of a cycle apart. When one is largest the other is zero.

The equation of the oscillation

Take the charge $q$ on the capacitor and the current $i = \dfrac{dq}{dt}$. In the loop there is no source. Going round the loop, the voltage across the capacitor is $\dfrac{q}{C}$ and the voltage across the coil is $L\dfrac{di}{dt}$. The loop rule says they add to zero:

This is the equation of simple harmonic motion, which you know from oscillations, with $\omega_0^2 = \dfrac{1}{LC}$. For a capacitor that starts with the charge $q_0$ the answer is:

with

The largest current is $I_0 = \omega_0q_0$. A larger $L$ or a larger $C$ gives a slower swing. Make $L$ four times larger and the period is only twice as long, because $T$ is proportional to the square root of $LC$.

The energy at every instant

The energy in the capacitor is:

The energy in the coil is $U_B = \tfrac{1}{2}Li^2$. Put in $i = \omega_0q_0\sin\omega_0 t$ and use $L\omega_0^2 = \dfrac{1}{C}$:

Add them. Since $\cos^2 + \sin^2 = 1$, the sum is the constant $\dfrac{q_0^2}{2C}$. It is also $\tfrac{1}{2}LI_0^2$. The energy is not lost. It only moves from one place to the other, twice in each period of the charge.

The spring and the mass

An LC circuit and a mass on a spring obey the same equation. Each quantity of one has a partner in the other.

• Charge $q$ is the partner of the displacement $x$.
• Current $i$ is the partner of the velocity $v$.
• Inductance $L$ is the partner of the mass $m$. Both resist a change of the flow.
• The inverse $1/C$ is the partner of the spring constant $k$. A smaller capacitor is a stiffer spring.
• Electric energy $\dfrac{q^2}{2C}$ is the partner of $\tfrac{1}{2}kx^2$. Magnetic energy $\tfrac{1}{2}Li^2$ is the partner of $\tfrac{1}{2}mv^2$.

So $\omega_0 = \sqrt{\dfrac{k}{m}}$ for the spring becomes $\omega_0 = \sqrt{\dfrac{1}{LC}}$ for the circuit. The coil is the inertia of the circuit, and the capacitor is its spring.

A worked example

A coil of $40$ mH is joined to a capacitor of $25\ \mu\text{F}$ that has been charged to $16$ V. There is no resistance. Find the frequency, the starting charge, the largest current and the total energy.

1. The angular frequency. $LC = 40 \times 10^{-3}\ \text{H} \times 25 \times 10^{-6}\ \text{F} = 1.0 \times 10^{-6}\ \text{s}^2$. So $\omega_0 = \dfrac{1}{\sqrt{LC}} = 1000$ rad/s.
2. The frequency. $f = \dfrac{\omega_0}{2\pi} = \dfrac{1000\ \text{rad/s}}{2\pi} = 159$ Hz. The period is $T = \dfrac{1}{f} = 6.28$ ms.
3. The starting charge. $q_0 = CV_0 = 25 \times 10^{-6}\ \text{F} \times 16\ \text{V} = 400\ \mu\text{C}$.
4. The largest current. $I_0 = \omega_0q_0 = 1000\ \text{rad/s} \times 400 \times 10^{-6}\ \text{C} = 0.40$ A.
5. The total energy. $U = \dfrac{q_0^2}{2C} = \tfrac{1}{2}CV_0^2 = \tfrac{1}{2} \times 25 \times 10^{-6}\ \text{F} \times (16\ \text{V})^2 = 3.2\ \text{mJ}$.

Check with the coil. $\tfrac{1}{2}LI_0^2 = \tfrac{1}{2} \times 0.040\ \text{H} \times (0.40\ \text{A})^2 = 3.2\ \text{mJ}$. It matches.

Start with the defaults. Which curve is highest at $t = 0$? Make the capacitance four times larger, $100\ \mu\text{F}$. By what factor does the period change? Then change the starting voltage. Which readouts change, and which do not?

Practise with new numbers

Resistance makes the swing die away

Every real coil has some resistance, and so does every wire. The resistor turns electrical energy into heat. Each swing is then a little smaller than the one before. This is a damped oscillation. For a small resistance $R$ the charge follows:

The cosine still swings, but its height falls steadily inside the envelope $q_0e^{-Rt/2L}$. The energy falls as $e^{-Rt/L}$. For a small $R$ the frequency is almost the same as $\dfrac{1}{2\pi\sqrt{LC}}$. A circuit also loses energy by sending out radio waves, which a radio transmitter uses on purpose.

Try $R = 1\ \Omega$, and then $R = 30\ \Omega$. How does the time to halve the amplitude change? Is the frequency changed much?

A series LCR circuit joined to an AC source is this circuit with a source added. When the source has the frequency $\dfrac{1}{2\pi\sqrt{LC}}$ it pushes at the right moment in every swing. It supplies exactly the energy that the resistor wastes. This is the idea of resonance.

Try it in Python

The program follows the charge and the current of an LC circuit in small steps of time, with no resistor. It finds the period from the times at which the charge passes through zero, and compares it with the formula.

Common mistakes

• Forgetting the $2\pi$. $\omega_0 = \dfrac{1}{\sqrt{LC}}$ is in rad/s. The frequency is $f = \dfrac{\omega_0}{2\pi}$ in hertz.
• Changing mH and $\mu\text{F}$ wrongly. $1\ \text{mH} = 10^{-3}\ \text{H}$ and $1\ \mu\text{F} = 10^{-6}\ \text{F}$.
• Saying the current is largest when the charge is largest. The two are a quarter of a cycle apart.
• Saying energy is lost. With no resistance the total energy stays constant. It moves between the capacitor and the coil.
• Taking half the charge for half the energy. The energy depends on $q^2$, so $q = \dfrac{q_0}{2}$ holds only a quarter of the energy.
• Mixing up the partners of the spring. The coil $L$ is the mass, and $1/C$ is the spring constant.

Remember these

• Charge and current in an LC circuit oscillate: $q = q_0\cos\omega_0 t$ and $i$ is largest when $q = 0$.
• $\omega_0 = \dfrac{1}{\sqrt{LC}}$, $f = \dfrac{1}{2\pi\sqrt{LC}}$ and $T = 2\pi\sqrt{LC}$.
• Total energy $\dfrac{q_0^2}{2C} = \tfrac{1}{2}LI_0^2$ stays constant. It moves between the capacitor and the coil.
• The partners are $q$ and $x$, $i$ and $v$, $L$ and $m$, $1/C$ and $k$.
• A resistance damps the oscillation: $q_0e^{-Rt/2L}$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. A coil of $9.0$ mH and a capacitor of $1.0\ \mu\text{F}$ form an LC circuit. What is the angular frequency of the oscillation?

2. At one moment the charge on the capacitor of an LC circuit is zero. What is the current, and where is the energy?

3. The capacitance of an LC circuit is made four times larger. What happens to the frequency?

4. In which way is an LC circuit like a mass on a spring? Name the partner of the inductance.