Unit 13 · Topic 13.6

13.6 Circuits with Capacitors and Inductors (LC Circuits)

Connect a charged capacitor to an inductor and the energy swings back and forth between the capacitor's electric field and the inductor's magnetic field. The charge oscillates exactly like a mass on a spring, with angular frequency $\omega = \dfrac{1}{\sqrt{LC}}$ .

Key terms

LC circuit
electromagnetic oscillation
simple harmonic motion
angular frequency
energy conservation

What happens in an LC circuit

Start with a capacitor holding charge $Q_{max}$ , then connect it to an inductor. The capacitor begins to discharge, but the inductor keeps the current from rising instantly. The current grows as the capacitor empties.

When the capacitor is empty, the current is at its maximum and all the energy is in the inductor. The inductor then keeps the current going, so charge piles up on the capacitor again with the opposite sign. Then the process reverses. With no resistance, this repeats forever.

The equation and its solution

Loop rule around the circuit: $\dfrac{q}{C} + L\dfrac{dI}{dt} = 0$ . With $I = \dfrac{dq}{dt}$ , this becomes $\dfrac{d^2q}{dt^2} = -\dfrac{1}{LC}q$ .

That's the same form as a mass on a spring, $\dfrac{d^2x}{dt^2} = -\dfrac{k}{m}x$ , so the solution is simple harmonic: $q(t) = Q_{max}\cos(\omega t + \phi)$ , with $\omega = \dfrac{1}{\sqrt{LC}}$ . The period is $T = 2\pi\sqrt{LC}$ and the frequency is $f = \dfrac{1}{2\pi\sqrt{LC}}$ .

The current is the derivative: $I(t) = -\omega Q_{max}\sin(\omega t + \phi)$ , so $I_{max} = \omega Q_{max} = \dfrac{Q_{max}}{\sqrt{LC}}$ . The current is a quarter cycle out of step with the charge: it's largest when the charge is zero, and zero when the charge is largest.

Mass on a spring	LC circuit
position x	charge q
velocity v	current I
mass m	inductance L
spring constant k	1/C
kinetic energy ½mv²	inductor energy ½LI²
spring energy ½kx²	capacitor energy q²/2C

Energy conservation

With no resistance, the total energy stays constant: $\dfrac{q^2}{2C} + \dfrac{1}{2}LI^2 = \dfrac{Q_{max}^2}{2C} = \dfrac{1}{2}LI_{max}^2$ .

This is often the fastest route to the maximum current, or to the current at a given charge. Energy moves from capacitor to inductor and back twice in each period, so the energy in each element oscillates at twice the frequency of the charge.

Real circuits have some resistance, which turns energy into thermal energy and makes the oscillations die out. On the exam, assume ideal LC circuits unless a resistor is shown.

One cycle, a quarter at a time

Starting with the capacitor fully charged and no current, here's where things stand at each quarter of a period. Notice that the capacitor is full twice per cycle, once with each sign.

Time	Capacitor charge	Current	Where the energy is
0	$+Q_{max}$	0	all in the capacitor
T/4	0	maximum, one direction	all in the inductor
T/2	$-Q_{max}$	0	all in the capacitor
3T/4	0	maximum, opposite direction	all in the inductor
T	$+Q_{max}$	0	all in the capacitor

Worked examples

Try each one yourself first, then open the solution.

Example 1Calculator allowed
Frequency and maximum current
A 10 μF capacitor is charged to 20 V, then connected to a 25 mH inductor at t = 0. Find the angular frequency, the period, the maximum current, and the first time the capacitor is completely discharged.
Show the solutionHide the solution
1. Step 1: $Q_{max} = C\Delta V = (10 \times 10^{-6})(20) = 2.0 \times 10^{-4}\text{ C}$ .
2. Step 2: $\omega = \dfrac{1}{\sqrt{LC}} = \dfrac{1}{\sqrt{(25 \times 10^{-3})(10 \times 10^{-6})}} = 2000\text{ rad/s}$ , so $T = \dfrac{2\pi}{\omega} \approx 3.1 \times 10^{-3}\text{ s}$ .
3. Step 3: $I_{max} = \omega Q_{max} = (2000)(2.0 \times 10^{-4}) = 0.40\text{ A}$ . Check with energy: $\tfrac{1}{2}CV^2 = \tfrac{1}{2}(10 \times 10^{-6})(20)^2 = 2.0\text{ mJ}$ and $\tfrac{1}{2}LI_{max}^2 = \tfrac{1}{2}(0.025)(0.40)^2 = 2.0\text{ mJ}$ .
4. Step 4: The capacitor first empties a quarter period after starting full: $\dfrac{T}{4} \approx 7.9 \times 10^{-4}\text{ s}$ .
Answer: ω = 2000 rad/s, T ≈ 3.1 ms, $I_{max}$ = 0.40 A, first empty at about 0.79 ms
Example 2Calculator allowed
Half the charge isn't half the energy (classic trap)
In the circuit from the previous example, what is the current at a moment when the capacitor's charge is half its maximum?
Show the solutionHide the solution
1. Step 1: Capacitor energy depends on $q^2$ : at $q = \dfrac{Q_{max}}{2}$ , it holds $\left(\tfrac{1}{2}\right)^2 = \tfrac{1}{4}$ of the total energy.
2. Step 2: The inductor has the other $\tfrac{3}{4}$ : $\tfrac{1}{2}LI^2 = \tfrac{3}{4}\cdot\tfrac{1}{2}LI_{max}^2$ , so $I = \dfrac{\sqrt{3}}{2}I_{max}$ .
3. Step 3: $I = (0.866)(0.40) \approx 0.35\text{ A}$ .
4. Step 4: The trap is assuming half the charge means half the energy, which gives about 0.28 A.
Answer: About 0.35 A

Common mistakes

Thinking the current is largest when the capacitor is fully charged. At that moment the current is zero; it peaks when the capacitor is empty.
Mixing up ω and f. $\omega = \dfrac{1}{\sqrt{LC}}$ is in rad/s; divide by 2π to get f in Hz.
Splitting energy in proportion to charge. Capacitor energy goes as q², so half the charge is a quarter of the energy.
Using a decaying exponential for an LC circuit. With no resistance, the charge oscillates; it doesn't decay.

On the exam

Expect to derive the differential equation from the loop rule and to identify ω by comparing it with the spring equation.
Energy-conservation questions are common: find the maximum current, or the charge when the current is a given fraction of its maximum.
Graph questions ask you to sketch q, I, $U_C$ or $U_L$ against time. Check that q and I are a quarter cycle apart and that the energies never go negative.

Connected topics

Videos

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