6.6 - Circuits with Capacitors and Inductors (LC Circuits)

Introduction to LC Circuits

An LC circuit has just a charged capacitor and an inductor, with no resistor and no battery. They’re also called oscillating circuits, because the energy sloshes back and forth between the capacitor and the inductor.

The defining feature: energy keeps moving between: - Electric field energy in the capacitor. - Magnetic field energy in the inductor.

An analogy: LC circuits behave exactly like a mass on a spring, with energy trading between kinetic and potential forms.

Conservation of Energy in LC Circuits

You can find the maximum current in an LC circuit with nothing more than conservation of energy.

The Energy Balance

Total energy stays fixed:

U_{total} = U_C + U_L = \text{constant}

Energy in the capacitor:

U_C = \frac{1}{2}\frac{Q^2}{C}

Energy in the inductor:

U_L = \frac{1}{2}LI^2

At different moments:

Moment	Capacitor	Inductor	Charge	Current
Maximum charge	$U_{C,max} = \frac{Q_0^2}{2C}$	$U_L = 0$	$Q = Q_0$	$I = 0$
Midpoint	$U_C = \frac{U_{total}}{2}$	$U_L = \frac{U_{total}}{2}$	$Q = \frac{Q_0}{\sqrt{2}}$	$I = \frac{I_0}{\sqrt{2}}$
Maximum current	$U_C = 0$	$U_{L,max} = \frac{LI_0^2}{2}$	$Q = 0$	$I = I_0$

The energy conservation equation:

\frac{Q_0^2}{2C} = \frac{1}{2}LI_0^2

That’s what lets you get the maximum current from the initial charge.

The Simple Harmonic Motion Model

The charge on the capacitor oscillates with simple harmonic motion.

The Differential Equation

Apply Kirchhoff’s Loop Rule to an LC circuit:

\frac{Q}{C} + L\frac{dI}{dt} = 0

Since

I = \frac{dQ}{dt}

\frac{dI}{dt} = \frac{d^2Q}{dt^2}

Substituting:

\frac{Q}{C} + L\frac{d^2Q}{dt^2} = 0

Rearranging:

\frac{d^2Q}{dt^2} = -\frac{1}{LC}Q

That’s exactly the shape of the simple harmonic oscillator equation:

\frac{d^2Q}{dt^2} = -\omega^2 Q

The thing to notice: the second derivative of

Q

is proportional to

-Q

. The only functions that pull that off are the trig functions, sine and cosine.

Angular Frequency and Period

Angular Frequency

Match the differential equation up with the SHM form:

\omega^2 = \frac{1}{LC}

\omega = \frac{1}{\sqrt{LC}}

Check it by plugging in: try

Q(t) = Q_0\cos(\omega t)

. Its second derivative is

\frac{d^2Q}{dt^2} = -\omega^2Q_0\cos(\omega t) = -\omega^2Q

. That matches

\frac{d^2Q}{dt^2} = -\frac{1}{LC}Q

only if

\omega^2 = \frac{1}{LC}

, so the circuit can only oscillate at

\omega = \frac{1}{\sqrt{LC}}

. This is the

\omega_{LC}

on the equation sheet.

where: -

\omega

= angular frequency (rad/s). -

L

= inductance (H). -

C

= capacitance (F).

What it means: how fast the charge and current oscillate.

Period of Oscillation

The period is how long one full cycle takes:

T = \frac{2\pi}{\omega}

Substituting

\omega = \frac{1}{\sqrt{LC}}

T = \frac{2\pi}{\frac{1}{\sqrt{LC}}}

T = 2\pi\sqrt{LC}

where: -

T

= period of oscillation (s). -

L

= inductance (H). -

C

= capacitance (F).

What it means: the time for the charge to go through one full cycle, from

+Q_0

-Q_0

and back to

+Q_0

. The energy moves from the capacitor to the inductor and back twice in that time, since the capacitor stores the same energy whether its charge is

+Q_0

-Q_0

The Oscillation Cycle

The energy trades back and forth between electric and magnetic forms:

6.6 - Circuits with Capacitors and Inductors (LC Circuits)

The key points: - When

Q

is at its max,

I = 0

, since all the energy is in the capacitor. - When

I

is at its max,

Q = 0

, since all the energy is in the inductor. - The total energy

U_{total} = U_C + U_L

never changes. - In an ideal circuit with no resistance, this goes on forever.

The Analogy: Mass on a Spring

LC circuits are mathematically identical to a mass bouncing on a spring.

Mass-Spring System	LC Circuit
Mass $m$	Inductance $L$
Spring constant $k$	$1/C$
Position $x$	Charge $Q$
Velocity $v = \frac{dx}{dt}$	Current $I = \frac{dQ}{dt}$
Kinetic energy $\frac{1}{2}mv^2$	Magnetic energy $\frac{1}{2}LI^2$
Potential energy $\frac{1}{2}kx^2$	Electric energy $\frac{Q^2}{2C}$
$\omega = \sqrt{\frac{k}{m}}$	$\omega = \frac{1}{\sqrt{LC}}$
$\frac{d^2x}{dt^2} = -\frac{k}{m}x$	$\frac{d^2Q}{dt^2} = -\frac{1}{LC}Q$

Same math, different physics. Everything you know about simple harmonic motion carries straight over to LC circuits.

Time Dependence Functions

The charge and current both wobble sinusoidally with time.

Charge on the Capacitor

Q(t) = Q_0\cos(\omega t)

Q(t) = Q_0\cos\left(\frac{t}{\sqrt{LC}}\right)

where

Q_0

is the maximum charge, the amount it starts with at

t = 0

Current Through the Circuit

Since

I = \frac{dQ}{dt}

I(t) = -Q_0\omega\sin(\omega t)

I(t) = -\frac{Q_0}{\sqrt{LC}}\sin\left(\frac{t}{\sqrt{LC}}\right)

And the maximum current:

I_0 = \frac{Q_0}{\sqrt{LC}} = Q_0\omega

Starting with Current Instead of Charge

Q(t) = Q_0\cos(\omega t)

fits a circuit that starts with a fully charged capacitor and no current. In general

Q(t) = Q_{max}\cos(\omega t + \phi)

, and the starting conditions set

Q_{max}

and

\phi

. A common case: the capacitor starts uncharged and the inductor starts with current

I_0

. Then:

Q(t) = \frac{I_0}{\omega}\sin(\omega t), \qquad I(t) = \frac{dQ}{dt} = I_0\cos(\omega t)

The charge on the capacitor peaks at

Q_{max} = \frac{I_0}{\omega} = I_0\sqrt{LC}

, a quarter period after the start.

Example Problem: Maximum Current from Energy Conservation

Problem: A capacitor with capacitance

C = 50

μF gets fully charged to

Q_0 = 40

μC. It’s then wired in series with an inductor of inductance

L = 20

mH through an open switch. After the switch closes, what’s the maximum current through the inductor?

Solution

Step 1: apply conservation of energy

When the capacitor is fully charged, all the energy is in it:

U_{C,initial} = \frac{Q_0^2}{2C}

When the current peaks, all the energy is in the inductor:

U_{L,max} = \frac{1}{2}LI_0^2

Energy conservation says those are equal:

\frac{Q_0^2}{2C} = \frac{1}{2}LI_0^2

Step 2: solve for the maximum current

Cancel the

\frac{1}{2}

on both sides:

\frac{Q_0^2}{C} = LI_0^2

I_0^2 = \frac{Q_0^2}{LC}

I_0 = \frac{Q_0}{\sqrt{LC}}

Step 3: plug in the numbers

Convert everything to SI first: -

Q_0 = 40

μC

= 40 \times 10^{-6}

C -

C = 50

μF

= 50 \times 10^{-6}

F -

L = 20

= 20 \times 10^{-3}

I_0 = \frac{40 \times 10^{-6}}{\sqrt{(20 \times 10^{-3})(50 \times 10^{-6})}}

I_0 = \frac{40 \times 10^{-6}}{\sqrt{1000 \times 10^{-9}}}

I_0 = \frac{40 \times 10^{-6}}{\sqrt{10^{-6}}}

I_0 = \frac{40 \times 10^{-6}}{10^{-3}}

I_0 = 40 \times 10^{-3} \text{ A}

I_0 = 0.04 \text{ A} = 40 \text{ mA}

Answer: the maximum current through the inductor is 40 mA, or 0.04 A.

Example Problem: An Inductor Charging a Capacitor

Problem: In the circuit below, the switch has been at

a

for a long time, so a steady current flows through the battery, the resistor, and the inductor. The capacitor is uncharged. At

t = 0

, the switch moves to

b

. That cuts out the battery and resistor and connects the inductor to the capacitor.

Find the current in the inductor just before the switch moves, and the energy it stores.

Find the maximum charge on the capacitor and the maximum voltage across it.

Find the angular frequency, the period, and the first time the capacitor is fully charged.

Find the charge and the current at $t = 1.0$ ms, and check that energy is conserved.

Solution

(a) Before the switch moves

After a long time, the inductor acts like a wire, so:

I_0 = \frac{\mathcal{E}}{R} = \frac{10 \text{ V}}{20\ \Omega} = 0.50 \text{ A}, \qquad U_L = \frac{1}{2}LI_0^2 = \frac{1}{2}(0.20 \text{ H})(0.50 \text{ A})^2 = 0.025 \text{ J}

(b) Maximum charge and voltage

The inductor’s current can’t jump, so right after the switch moves, the 0.50 A flows onto the capacitor. When the capacitor is fully charged,

I = 0

and all the energy is in the capacitor:

\frac{Q_{max}^2}{2C} = 0.025 \text{ J} \quad\Rightarrow\quad Q_{max} = \sqrt{2(20 \times 10^{-6} \text{ F})(0.025 \text{ J})} = 1.0 \times 10^{-3} \text{ C}

V_{max} = \frac{Q_{max}}{C} = \frac{1.0 \times 10^{-3} \text{ C}}{20 \times 10^{-6} \text{ F}} = 50 \text{ V}

The capacitor ends up at 50 V even though the battery was only 10 V. That energy came from the inductor’s magnetic field.

(c) Frequency and timing

\omega = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{(0.20 \text{ H})(20 \times 10^{-6} \text{ F})}} = 500 \text{ rad/s}

T = \frac{2\pi}{\omega} = \frac{2\pi}{500 \text{ rad/s}} = 0.013 \text{ s}

The capacitor starts uncharged, so it’s first fully charged a quarter period later:

\frac{T}{4} = 3.1 \times 10^{-3}

s (3.1 ms).

(d) At 1.0 ms

Use

Q = \frac{I_0}{\omega}\sin(\omega t)

and

I = I_0\cos(\omega t)

with

\omega t = (500 \text{ rad/s})(1.0 \times 10^{-3} \text{ s}) = 0.50

rad:

Q = (1.0 \times 10^{-3} \text{ C})\sin(0.50) = 4.8 \times 10^{-4} \text{ C}, \qquad I = (0.50 \text{ A})\cos(0.50) = 0.44 \text{ A}

U_C + U_L = \frac{(4.79 \times 10^{-4} \text{ C})^2}{2(20 \times 10^{-6} \text{ F})} + \frac{1}{2}(0.20 \text{ H})(0.439 \text{ A})^2 = 0.0057 \text{ J} + 0.0193 \text{ J} = 0.025 \text{ J}

The total is still 0.025 J, the same as at the start. ✓

The graphs show both functions with real scales. The red dots mark

t = 1.0

ms, and the dashed line marks

t = \frac{T}{4}

, where the charge peaks and the current is zero.

Ideal vs. Real LC Circuits

The Ideal LC Circuit

What we assume: - No resistance (

R = 0

). - No energy loss. - The oscillations run forever at a constant amplitude.

The result: energy bounces between the capacitor and inductor indefinitely.

The Real LC Circuit

What actually happens: - There’s always some resistance in the wires and the inductor coil. - Energy slowly leaks away as heat. - The oscillations get damped, so the amplitude shrinks over time. - Eventually all the energy is gone.

Summary of Key Formulas

Angular Frequency

\omega = \frac{1}{\sqrt{LC}}

Period

T = 2\pi\sqrt{LC}

Maximum Current from Energy

I_0 = \frac{Q_0}{\sqrt{LC}}

Energy Conservation

\frac{Q^2}{2C} + \frac{1}{2}LI^2 = \frac{Q_0^2}{2C} = \text{constant}

Charge and Current Functions

Q(t) = Q_0\cos(\omega t)

I(t) = -Q_0\omega\sin(\omega t) = -I_0\sin(\omega t)

Comparison: Three Circuit Types

Circuit	Components	Time Constant or Period	Energy Behavior
RC	Resistor + Capacitor	$\tau = RC$	Exponential climb (charging) or decay to zero (discharging)
LR	Inductor + Resistor	$\tau = \frac{L}{R}$	Exponential climb to steady state, or decay to zero when the source is removed
LC	Inductor + Capacitor	$T = 2\pi\sqrt{LC}$	Oscillates forever (in the ideal case)

Applications of LC Circuits

1. Radio Tuning Circuits

Twist the variable capacitor to change $C$ and shift the resonant frequency.

That’s how you pick a radio station.

$f = \frac{1}{2\pi\sqrt{LC}}$

2. Oscillators

They generate AC signals at specific frequencies.

Used in signal generators and transmitters.

Also in the clock circuits inside electronics.

3. Filters

LC circuits block or pass specific frequencies.

Used in audio gear and communication systems.

Resonance happens at $\omega = \frac{1}{\sqrt{LC}}$ .

4. Tank Circuits

They park energy temporarily.

Shuttle it between the capacitor and the inductor.

Used in switching power supplies.

Common Mistakes

Forgetting to convert units. LC circuits love microfarads (μF), millihenrys (mH), and microcoulombs (μC). Always get everything into SI (F, H, C) before you calculate. Skipping this is the number one source of numerical errors.

Using the period formula as if it were a time constant. LC circuits have a period, $T = 2\pi\sqrt{LC}$ , for their oscillations, not a time constant $\tau$ . Don’t confuse it with RC circuits ( $\tau = RC$ ) or LR circuits ( $\tau = L/R$ ), where you get exponential decay instead.

Thinking energy leaks away in an ideal LC circuit. In an ideal LC circuit with no resistance, the total energy is conserved and oscillates forever. It doesn’t decay, it just changes form. Only real circuits with resistance actually lose energy.

Using the wrong capacitor energy formula in terms of Q. It’s $U_C = \frac{Q^2}{2C}$ , NOT $\frac{1}{2}CQ^2$ . The $Q^2$ goes in the numerator. That’s different from $U_C = \frac{1}{2}CV^2$ , where the $V^2$ sits next to the $C$ up top.

Forgetting that current and charge are 90° out of phase. When the charge is at its max, the current is zero. When the current is at its max, the charge is zero. They never peak together. It’s just like position and velocity in SHM, a quarter cycle apart.

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