AP Physics C: E&M · Topic 13.6

Topic 13.6: Circuits with Capacitors and Inductors (LC Circuits)

Unit 13: Electromagnetic Induction10-20% of the multiple-choice section

In an LC circuit a charged capacitor and an inductor trade energy back and forth. The charge on the capacitor obeys the same differential equation a mass on a spring does, so it oscillates, and the angular frequency is one over the square root of the inductance times the capacitance.

AP Physics: Unit 13 (topics 13.6 Circuits with Capacitors and Inductors (LC Circuits)). AP Physics C: Electricity and Magnetism Unit 13, Topic 13.6, the last topic of the course. One learning objective, 13.6.A, describe the physical and electrical properties of a circuit containing a combination of capacitors and a single inductor. Supported by three essential-knowledge statements, the smallest block in Unit 13: 13.6.A.1 (in circuits containing only a charged capacitor and an inductor, the maximum current in the inductor can be determined using conservation of energy within the circuit), 13.6.A.2 (in LC circuits the time dependence of the charge stored in the capacitor can be modeled as simple harmonic motion, DERIVED equation d2q/dt2 = -(1/LC) q), and 13.6.A.3 (the angular frequency of an oscillating LC circuit can be derived from the differential equation that describes an LC circuit, DERIVED equation omega = 1/sqrt(LC)). Topic 13.6 prints NO boundary statement; the whole of Unit 13 carries exactly one, under Topic 13.2, about not deriving the speed of light from Maxwell's equations. The differential equation is NOT on the equation sheet; the angular frequency IS, printed as omega_LC = 1/sqrt(LC) with no equivalent subscripts even though the learning objective says a combination of capacitors, unlike the two time constants which are printed with eq subscripts. No solved charge-versus-time function is printed anywhere in Unit 13. The words RLC, damping and resonance appear nowhere in the AP Physics C: Electricity and Magnetism course description, and the word oscillating appears exactly once, in 13.6.A.3. The correspondence with simple harmonic motion is exact: AP Physics C: Mechanics essential knowledge 7.3.A.2 prints d2x/dt2 = -omega^2 x as its own derived equation, so matching coefficients gives omega^2 = 1/LC. Charge corresponds to position, current to velocity, inductance to mass, and the reciprocal of capacitance to the spring constant. Suggested skills are 1.B, 2.B, 2.C, 3.A and 3.B. Only Topics 13.2 and 13.6 in Unit 13 list skill 3.A. No sample multiple-choice or free-response question in the CED aligns to 13.6.A. There is no AP Physics 2 counterpart, since the words inductance and inductor appear nowhere in the AP Physics 2 course description.

What Topic 13.6 requires

Topic 13.6 closes Unit 13 and the whole AP Physics C: Electricity and Magnetism course. One learning objective, three essential-knowledge statements, and no boundary statement.

13.6.A, describe the physical and electrical properties of a circuit containing a combination of capacitors and a single inductor.

  • 13.6.A.1 in circuits containing only a charged capacitor and an inductor (LC circuits), the maximum current in the inductor can be determined using conservation of energy within the circuit.
  • 13.6.A.2 in LC circuits, the time dependence of the charge stored in the capacitor can be modeled as simple harmonic motion. The derived equation is
d2qdt2=1LCq\frac{d^2 q}{dt^2} = -\frac{1}{LC}q
  • 13.6.A.3 the angular frequency of an oscillating LC circuit can be derived from the differential equation that describes an LC circuit. The derived equation is
ω=1LC\omega = \frac{1}{\sqrt{LC}}

The suggested skills are 1.B, create quantitative graphs with appropriate scales and units, including plotting data; 2.B, calculate or estimate an unknown quantity with units from known quantities by selecting and following a logical computational pathway; 2.C, compare physical quantities between two or more scenarios or at different times and locations in a single scenario; 3.A, create experimental procedures that are appropriate for a given scientific question; and 3.B, apply an appropriate law, definition, theoretical relationship, or model to make a claim.

Three statements is the smallest required-content block in Unit 13, and 13.6.A.2 does almost all of the work by naming the model: simple harmonic motion. That single phrase imports an entire chapter of AP Physics C: Mechanics into this topic, which is what makes it the least new material and the most leverage of anything in the unit.

Note the objective's wording, exactly as in Topic 13.5: a combination of capacitors and a single inductor. Plural capacitors, one inductor. That has consequences, and they come later on this page.

How far the CED takes the differential equation

This is the question worth settling first, because it decides how much of a university treatment belongs here.

Topic 13.6 prints no boundary statement. Neither does 13.1, 13.3, 13.4 or 13.5. The whole of Unit 13 carries exactly one, and it sits under Topic 13.2, about the speed of light. So there is no sentence anywhere fencing off the LC mathematics, and the level is set by the essential-knowledge statements themselves and by the labels on their equations.

Both equations in this topic carry the Derived Equation label. The CED's Required Equations page defines it: not all equations in the framework appear on the sheet, and many "are provided for reference and guidance, or to demonstrate the final results of derivations expected of students on the exam. These equations are denoted as 'Derived Equations.'"

Now check them against the sheet, one at a time, because the label and the sheet do not track each other in this course.

EquationCED labelOn the C: E&M sheet
d2qdt2=1LCq\dfrac{d^2q}{dt^2} = -\dfrac{1}{LC}qderived, 13.6.A.2no
ω=1LC\omega = \dfrac{1}{\sqrt{LC}}derived, 13.6.A.3yes, printed as ωLC=1LC\omega_{LC} = \dfrac{1}{\sqrt{LC}}

So the result of the derivation is handed to you and the derivation is not. Read together with 13.6.A.3, which says the angular frequency "can be derived from the differential equation that describes an LC circuit", the expectation is clear enough: produce the differential equation, then get ω\omega from it, even though you could have read ω\omega off the sheet.

What the CED does not ask for, and it is worth being precise about the absences:

  • No solved function of time is printed anywhere in Unit 13. Not for the LR current and not for the LC charge. Statement 13.6.A.2 says the time dependence "can be modeled as simple harmonic motion", which points you at x=xmaxcos(ωt+ϕ)x = x_{\max}\cos(\omega t + \phi) in the reprinted mechanics table rather than printing an electrical version of it.
  • The words RLC, damping and resonance appear nowhere in the AP Physics C: Electricity and Magnetism course description. Statement 13.6.A.1's own wording is "circuits containing only a charged capacitor and an inductor". A resistor in the loop is Topic 13.5's business, and a circuit with all three is outside this course.
  • No driven or forced oscillation. The word "oscillating" appears exactly once in the whole course description, in 13.6.A.3.

Two supporting signals about register. The unit's Building the Science Practices page says students may benefit from "practice crafting clear, concise arguments, derivations, and calculations that follow a logical pathway", and names deriving a mathematical equation that shows the current in a solenoid as a function of time as a valuable skill. And the CED's own sample free-response Question 4, on an RC circuit, instructs students to "Derive, but do not solve, a differential equation" and then asks them to justify a prediction by referring to the equation they wrote. That is the shape to prepare for: the setup and the reasoning carry the points.

The oscillator you already own

Statement 13.6.A.2 says the charge can be modeled as simple harmonic motion. It is worth seeing exactly how literally that is meant, because the correspondence is not an analogy that mostly works. It is the same differential equation with different letters.

The AP Physics C: Mechanics course description, at essential knowledge 7.3.A.2, says the position as a function of time for an object exhibiting SHM is a solution of the second-order differential equation derived from the application of Newton's second law, and prints as its derived equation

d2xdt2=ω2x\frac{d^2 x}{dt^2} = -\omega^2 x

Set that beside 13.6.A.2's d2qdt2=1LCq\dfrac{d^2q}{dt^2} = -\dfrac{1}{LC}q and the identification is immediate: qq plays the part of xx, and ω2=1LC\omega^2 = \dfrac{1}{LC}. Statement 13.6.A.3 is that one line of algebra.

Here is the full dictionary. Everything in the right-hand column that is printed on the C: E&M sheet is marked, and because that sheet reprints the entire C: Mechanics table, everything in the middle column is on the same page as everything in the right.

IdeaMass on a springLC circuit
the oscillating variableposition xxcharge on the capacitor qq
its rate of changevelocity v=dx/dtv = dx/dtcurrent I=dq/dtI = dq/dt (on the sheet)
what resists changemass mm, inertiainductance LL, electrical inertia
what pushes backspring constant kk, stiffness1/C1/C, the reciprocal of capacitance
the restoring relationFs=kΔx\vec{F}_s = -k\Delta\vec{x} (on the sheet)ΔVC=q/C\Delta V_C = q/C from C=Q/ΔVC = Q/\Delta V (on the sheet)
the differential equationd2xdt2=kmx\dfrac{d^2x}{dt^2} = -\dfrac{k}{m}xd2qdt2=1LCq\dfrac{d^2q}{dt^2} = -\dfrac{1}{LC}q
angular frequencyω=k/m\omega = \sqrt{k/m}ω=1/LC\omega = 1/\sqrt{LC} (on the sheet)
periodTs=2πm/kT_s = 2\pi\sqrt{m/k} (on the sheet)T=2πLCT = 2\pi\sqrt{LC}
the solutionx=xmaxcos(ωt+ϕ)x = x_{\max}\cos(\omega t + \phi) (on the sheet)q=qmaxcos(ωt+ϕ)q = q_{\max}\cos(\omega t + \phi)
energy in the "spring"Us=12k(Δx)2U_s = \frac{1}{2}k(\Delta x)^2 (on the sheet)UC=12QΔV=q22CU_C = \frac{1}{2}Q\Delta V = \dfrac{q^2}{2C}
energy in the "mass"K=12mv2K = \frac{1}{2}mv^2 (on the sheet)UL=12LI2U_L = \frac{1}{2}LI^2 (on the sheet)
maximum of the ratevmax=ωxmaxv_{\max} = \omega x_{\max}Imax=ωqmaxI_{\max} = \omega q_{\max}

Three things to take from the table rather than from the individual rows.

The inductor is the mass. That is the row students find least obvious and it is the most useful one. Inductance opposes a change in current exactly as mass opposes a change in velocity, and both store energy in proportion to the square of the rate, 12LI2\frac{1}{2}LI^2 against 12mv2\frac{1}{2}mv^2. The identification is already made on the Topic 13.4 page, where inductance is called electrical inertia.

Capacitance is the reciprocal of stiffness. A large capacitor is a weak spring: it takes a lot of charge to build up a given potential difference, just as a floppy spring takes a lot of stretch to build up a given force. That reciprocal is the reason CC appears under the square root in the numerator's denominator rather than where kk sits, and it is why a bigger capacitor gives a slower oscillation, which catches people who pattern-match from ω=k/m\omega = \sqrt{k/m} without checking.

The period follows from two printed equations. T=2πLCT = 2\pi\sqrt{LC} is not printed anywhere, but ωLC=1/LC\omega_{LC} = 1/\sqrt{LC} is on the sheet and T=2π/ω=1/fT = 2\pi/\omega = 1/f is on the same sheet in the reprinted mechanics table. One substitution.

If you want the mechanical side refreshed, the simple harmonic motion guide covers the routine, and the AP Physics C: Mechanics Unit 7 hub covers what that course requires.

Deriving the differential equation, twice

Statement 13.6.A.2 gives the equation and not the route. There are two routes, and both are worth having, because a question can hand you either starting point.

Route one: Kirchhoff's loop rule. This is the same move that produced the LR equation at 13.5.A.2.

Take a capacitor of capacitance CC carrying charge qq on one plate, in a closed loop with an inductor of inductance LL and nothing else. Declare the convention first and keep it: let qq be the charge on the plate the current flows into, so that I=dq/dtI = dq/dt, which is the definition the sheet prints.

Going once around the loop, the two potential changes must sum to zero:

  1. Across the capacitor, from C=Q/ΔVC = Q/\Delta V on the sheet, the magnitude of the potential difference is q/Cq/C.
  2. Across the inductor, from 13.4.A.3 and the sheet, the self-induced emf is LdI/dt-L\,dI/dt.
qC+LdIdt=0\frac{q}{C} + L\frac{dI}{dt} = 0

Now substitute I=dq/dtI = dq/dt, so dI/dt=d2q/dt2dI/dt = d^2q/dt^2:

qC+Ld2qdt2=0d2qdt2=1LCq\frac{q}{C} + L\frac{d^2q}{dt^2} = 0 \quad \Longrightarrow \quad \frac{d^2q}{dt^2} = -\frac{1}{LC}q

which is exactly the derived equation at 13.6.A.2.

Route two: conservation of energy. This one starts from 13.6.A.1, and it is the more physical of the two.

There is no resistor in the loop, so nothing dissipates energy. Statement 13.4.A.2.ii says the transfer of energy generated in an inductor to other forms obeys conservation laws, and 13.4.A.2.i names one of those forms explicitly: the stored energy can be "used to charge a capacitor". So the total is constant:

U=q22C+12LI2=constantU = \frac{q^2}{2C} + \frac{1}{2}LI^2 = \text{constant}

Differentiate both sides with respect to time. A constant differentiates to zero, and the chain rule handles both terms:

qCdqdt+LIdIdt=0\frac{q}{C}\frac{dq}{dt} + LI\frac{dI}{dt} = 0

Substitute I=dq/dtI = dq/dt throughout and factor out that common II:

I(qC+Ld2qdt2)=0I\left(\frac{q}{C} + L\frac{d^2q}{dt^2}\right) = 0

The current is not zero at all times, so the bracket must vanish, and the same differential equation drops out.

Why the minus sign is the whole point. The equation says the second derivative of the charge is proportional to the charge and points the other way. That is the definition of a restoring relationship, and AP Physics C: Mechanics statement 7.1.A.2 gives the general form: SHM results when the magnitude of the restoring force exerted on an object is proportional to that object's displacement from its equilibrium position. Here "displacement from equilibrium" is the charge on the capacitor, since q=0q = 0 is the state with no stored electric energy, and the further the charge is from zero the harder the circuit pushes it back. A plus sign would give runaway exponential growth rather than oscillation, so the sign is not a bookkeeping detail.

From the equation to the frequency

Statement 13.6.A.3 says the angular frequency "can be derived from the differential equation that describes an LC circuit". Two ways to run that derivation, and the second is worth knowing because it works even if you have forgotten the mechanical result.

By comparison. The C: Mechanics derived equation is d2x/dt2=ω2xd^2x/dt^2 = -\omega^2 x. Matching coefficients with d2q/dt2=(1/LC)qd^2q/dt^2 = -(1/LC)q gives ω2=1/LC\omega^2 = 1/LC, so

ω=1LC\omega = \frac{1}{\sqrt{LC}}

By substitution. Propose the solution q=qmaxcos(ωt+ϕ)q = q_{\max}\cos(\omega t + \phi), which is the electrical version of the x=xmaxcos(ωt+ϕ)x = x_{\max}\cos(\omega t + \phi) printed in the reprinted mechanics table. Differentiate twice, using the derivative rules from the booklet's CALCULUS box, which prints ddx[cos(ax)]=asin(ax)\frac{d}{dx}\left[\cos(ax)\right] = -a\sin(ax) and ddx[sin(ax)]=acos(ax)\frac{d}{dx}\left[\sin(ax)\right] = a\cos(ax):

dqdt=ωqmaxsin(ωt+ϕ),d2qdt2=ω2qmaxcos(ωt+ϕ)=ω2q\frac{dq}{dt} = -\omega q_{\max}\sin(\omega t + \phi), \qquad \frac{d^2q}{dt^2} = -\omega^2 q_{\max}\cos(\omega t + \phi) = -\omega^2 q

Substituting into the differential equation gives ω2q=(1/LC)q-\omega^2 q = -(1/LC)q, which holds at all times only if ω2=1/LC\omega^2 = 1/LC. The substitution route also hands you the current for free, since I=dq/dtI = dq/dt:

I=ωqmaxsin(ωt+ϕ),Imax=ωqmaxI = -\omega q_{\max}\sin(\omega t + \phi), \qquad I_{\max} = \omega q_{\max}

Check the units, because this is a place where a slip is easy to miss. LL is in henries and CC in farads. A henry is Vs/A\mathrm{V \cdot s/A} and a farad is C/V=As/V\mathrm{C/V} = \mathrm{A \cdot s / V}, so a henry-farad is s2\mathrm{s^2}, its square root is a second, and its reciprocal is s1\mathrm{s^{-1}}. Angular frequency in radians per second, as it should be.

Reading the dependence, which is skill 2.C. Everything sits under a square root, so factors of change are gentle.

ChangeEffect on ω\omegaEffect on T=2πLCT = 2\pi\sqrt{LC}
double LLdivided by 2\sqrt{2}multiplied by 2\sqrt{2}
double CCdivided by 2\sqrt{2}multiplied by 2\sqrt{2}
quadruple LLhalveddoubled
double both LL and CChalveddoubled
change the initial chargeno effectno effect

The last row is the one worth stating out loud. The frequency of an LC circuit does not depend on how much charge you started with. That is the electrical version of the fact that a mass on a spring has the same period whatever its amplitude, and it comes from the same place: the differential equation contains LL and CC and nothing about the initial conditions. Change the starting charge and every current and voltage scales, while the timing does not move at all.

The energy exchange, quarter period by quarter period

Statement 13.6.A.1 says the maximum current in the inductor can be determined using conservation of energy within the circuit, so this is the topic's headline calculation. It takes one line:

Q22C=12LImax2Imax=Q1LC=ωQ\frac{Q^2}{2C} = \frac{1}{2}LI_{\max}^2 \quad \Longrightarrow \quad I_{\max} = Q\sqrt{\frac{1}{LC}} = \omega Q

All of the energy that was in the capacitor's electric field when the charge was greatest is in the inductor's magnetic field when the current is greatest, because there is nowhere else for it to go. That the answer equals ωQ\omega Q is a useful cross-check, and it also confirms that the energy route and the SHM route are the same physics.

Follow one full cycle, starting from a fully charged capacitor at rest.

TimeChargeCurrentUCU_CULU_LMechanical twin
t=0t = 0+Q+Q, maximumzeroall of itzeromass at maximum displacement, at rest
t=T/4t = T/4zeromaximum one wayzeroall of itmass at equilibrium, moving fastest
t=T/2t = T/2Q-Q, maximum the other wayzeroall of itzeromass at the other extreme, at rest
t=3T/4t = 3T/4zeromaximum the other wayzeroall of itmass back at equilibrium, moving the other way
t=Tt = T+Q+Q againzeroall of itzeroback to the start

Four facts to carry out of that table:

  1. The two maxima are a quarter period apart. Charge is greatest exactly when current is zero, and current is greatest exactly when charge is zero. Nothing in an LC circuit peaks at the same moment as anything else except by that quarter-cycle rule.
  2. The capacitor's charge reverses. After half a period the plate that started positive is negative. This is not decay, it is oscillation, and it is the row that distinguishes an LC circuit from the RC and LR circuits in the rest of the course, which approach an asymptote and stay there.
  3. Energy oscillates at twice the frequency of charge. UCq2U_C \propto q^2, and squaring a cosine doubles the frequency. Over one period of the charge, the energy makes two complete round trips between the two stores. This is the single most common mistake in a sketch of UCU_C against time.
  4. Nothing is lost. With no resistor there is no dissipation, so the oscillation in this model continues indefinitely. Statement 13.6.A.1's phrasing, "circuits containing only a charged capacitor and an inductor", is what licenses that, and the sheet's exam conventions add that wires and batteries are ideal.

At an arbitrary instant, with q=Qcos(ωt)q = Q\cos(\omega t) and I=ωQsin(ωt)I = -\omega Q \sin(\omega t):

UC=Q22Ccos2(ωt),UL=Q22Csin2(ωt)U_C = \frac{Q^2}{2C}\cos^2(\omega t), \qquad U_L = \frac{Q^2}{2C}\sin^2(\omega t)

To see that the second of those is right, substitute Imax=ωQI_{\max} = \omega Q and ω2=1/LC\omega^2 = 1/LC into 12LI2\frac{1}{2}LI^2. Their sum is constant because sin2θ+cos2θ=1\sin^2\theta + \cos^2\theta = 1, which is one of the four identities printed in the booklet's IDENTITIES box. That is a genuinely satisfying place for a printed trigonometric identity to earn its keep.

The energy splits evenly when cos2=sin2=12\cos^2 = \sin^2 = \frac{1}{2}, which is at ωt=45\omega t = 45 degrees, one eighth of a period after the start, and then again every quarter period after that.

A combination of capacitors and a single inductor

Read the learning objective again: a combination of capacitors and a single inductor. Plural capacitors. That is deliberate, it matches the wording of 13.5.A exactly (a combination of resistors and a single inductor), and it creates the same trap.

The CC in ω=1/LC\omega = 1/\sqrt{LC} is the equivalent capacitance the inductor sees, not the value printed on any one component. The sheet gives you the two combination rules:

1Ceq,s=i1CiCeq,p=iCi\frac{1}{C_{\text{eq,s}}} = \sum_i \frac{1}{C_i} \qquad C_{\text{eq,p}} = \sum_i C_i

and the direction of each is the reverse of the resistor case, which is where the errors come from. Capacitors in parallel add; capacitors in series add reciprocally.

There is a notation detail on the sheet worth flagging, because it is the sort of thing that makes a student doubt a correct method. The two time constants are printed with explicit equivalent subscripts, τ=ReqCeq\tau = R_{\text{eq}}C_{\text{eq}} and τ=L/Req\tau = L/R_{\text{eq}}, while the LC angular frequency is printed as ωLC=1/LC\omega_{LC} = 1/\sqrt{LC} with no subscripts at all. Do not read that as permission to use a single capacitor's value when there are several. The learning objective says "a combination of capacitors", and the physics is unambiguous: what governs the oscillation is the total capacitance in the loop.

Which way the combination pushes the frequency:

  • Capacitors in parallel give a larger CeqC_{\text{eq}}, so a smaller ω\omega and a longer period. More total capacitance is a floppier spring.
  • Capacitors in series give a CeqC_{\text{eq}} smaller than the smallest of them, so a larger ω\omega and a shorter period.

The second worked example below runs the same two capacitors both ways and gets angular frequencies differing by a factor of 4.5\sqrt{4.5}, with neither equal to what either capacitor alone would give.

One inductor, though. The objective says "a single inductor" and the framework never combines them, so there is no equivalent-inductance rule anywhere in the course description or on the sheet. If a question shows two inductors, look again at the circuit before assuming a combination rule you have not been given.

Where the initial charge comes from. Statement 13.4.A.2.i says the energy stored in an inductor can be dissipated through a resistor or used to charge a capacitor, which is the handoff into this topic. In practice an LC problem starts one of two ways: a capacitor charged by a battery which is then switched out of the loop, or an inductor already carrying a current which is then switched into a loop with an uncharged capacitor. The second is the same oscillation started a quarter period later, so q=qmaxsin(ωt)q = q_{\max}\sin(\omega t) rather than a cosine, and the phase constant ϕ\phi in q=qmaxcos(ωt+ϕ)q = q_{\max}\cos(\omega t + \phi) is what encodes that choice.

Graphs, and how Topic 13.6 is tested

Skill 1.B, create quantitative graphs with appropriate scales and units including plotting data, is listed first for this topic. Note that it is 1.B and not 1.C: this is the quantitative-graph skill, so axes get labels, units and a numerical scale, not just a shape.

Five curves for one LC circuit started from a fully charged capacitor at t=0t = 0:

QuantityStarts atShapePeriod
charge qq+Q+QcosineTT
current IIzeronegative sine, a quarter period behind the chargeTT
potential difference across the capacitor, q/Cq/CQ/CQ/Csame shape as the charge, scaledTT
energy in the capacitor, q2/2Cq^2/2Cmaximumcosine squared, never negativeT/2T/2
energy in the inductor, 12LI2\frac{1}{2}LI^2zerosine squared, never negativeT/2T/2

Four things a scoring guideline can identify:

  1. The two energy curves are never negative and both oscillate between zero and the same maximum. Drawing an energy curve that dips below the axis is an immediate error.
  2. The two energy curves are at half the period of the charge, so two full humps per cycle of qq.
  3. They sum to a horizontal line at the total energy. If you draw both on one set of axes, that is a free self-check, and it is the same trick that works for the two voltages in an LR circuit.
  4. Charge and current are a quarter period out of step, with the current at its extreme where the charge crosses zero and vice versa. Marking one such pair of points is often what earns the alignment mark.

Skill 3.A, create experimental procedures. Only two topics in Unit 13 list 3.A, this one and 13.2. The natural question here is how to measure an unknown inductance or capacitance from an oscillation. Build the LC loop with the known component, charge the capacitor, switch it into the loop, record the potential difference across the capacitor against time, measure the period from the trace, and invert T=2πLCT = 2\pi\sqrt{LC} to get the unknown. Reading several cycles and dividing gives a better period than reading one. That mirrors the CED's own Unit 13 sample activity for Topic 13.5, which has students measure a solenoid's inductance from an LR time constant, and the unit's exam-preparation page says laboratory investigations about the behavior of circuits when a solenoid is in the circuit are valuable exercises that concretely demonstrate principles and ideas that are often abstract.

Skill 2.B is a straight calculation with units, and skill 2.C is a comparison between two circuits or two instants. Skill 3.B is a claim backed by the model, and the strongest one available is that an LC circuit oscillates because its governing equation is the simple harmonic motion equation.

None of the fifteen sample multiple-choice questions or four sample free-response questions in the CED aligns to 13.6.A. Unit 13 as a whole is weighted at 10 to 20 percent of the multiple-choice section over about 10 to 20 class periods.

Four traps.

  • Thinking a bigger capacitor means a faster oscillation. It does not. CC is under a square root in the denominator, and capacitance behaves like the reciprocal of stiffness.
  • Drawing the energy curves at the same period as the charge. They run at twice the frequency.
  • Using one capacitor's value when the circuit holds several. Use the equivalent capacitance, and remember that capacitors combine the opposite way to resistors.
  • Expecting the oscillation to decay. Not in this model. Statement 13.6.A.1 specifies a circuit containing only a charged capacitor and an inductor, and the words RLC, damping and resonance do not appear anywhere in the AP Physics C: Electricity and Magnetism course description.

How this topic sits against the rest of the unit. Topic 13.5 and the RC circuits of Unit 11 are both first-order: one derivative, an exponential approach to an asymptote, and a time constant. Topic 13.6 is the only second-order circuit in the course, and second order is what makes oscillation possible rather than mere approach. There is no AP Physics 2 counterpart, because that course contains no inductors at all: the words inductance and inductor appear nowhere in its course description. The nearest algebra-based relative is AP Physics 2 Topic 11.8 on RC circuits, which is for AP Physics 2 students and covers a first-order circuit; this page is for AP Physics C: Electricity and Magnetism students.

A full LC circuit from a charged capacitor

A 40 μF40\ \mu\mathrm{F} capacitor is charged to 600 μC600\ \mu\mathrm{C} and then connected across a 2525 mH inductor with negligible resistance. Find (a) the angular frequency, frequency and period, (b) the initial potential difference across the capacitor, (c) the total energy in the circuit, (d) the maximum current, computed two ways, and (e) the time at which the current first reaches that maximum.

  1. Convert once and keep the values: C=4.0×105C = 4.0 \times 10^{-5} F, Q=6.0×104Q = 6.0 \times 10^{-4} C, L=2.5×102L = 2.5 \times 10^{-2} H. Take qq as the charge on the plate the current flows into, so I=dq/dtI = dq/dt.

  2. (a) LC=(2.5×102)(4.0×105)=1.0×106 s2LC = (2.5 \times 10^{-2})(4.0 \times 10^{-5}) = 1.0 \times 10^{-6}\ \mathrm{s^2}, so LC=1.0×103\sqrt{LC} = 1.0 \times 10^{-3} s and, from the sheet, ω=1LC=1.0×103\omega = \dfrac{1}{\sqrt{LC}} = 1.0 \times 10^{3} rad/s.

  3. f=ω2π=10006.2832=159.2f = \dfrac{\omega}{2\pi} = \dfrac{1000}{6.2832} = 159.2 Hz, and T=2πω=6.28321000=6.28×103T = \dfrac{2\pi}{\omega} = \dfrac{6.2832}{1000} = 6.28 \times 10^{-3} s, about 6.3 ms. Both come from T=2π/ω=1/fT = 2\pi/\omega = 1/f, which is printed in the mechanics table the C: E&M sheet reprints.

  4. (b) From C=Q/ΔVC = Q/\Delta V on the sheet, ΔV=QC=6.0×1044.0×105=15\Delta V = \dfrac{Q}{C} = \dfrac{6.0 \times 10^{-4}}{4.0 \times 10^{-5}} = 15 V.

  5. (c) At t=0t = 0 the current is zero, so all the energy is in the capacitor. Using UC=12QΔVU_C = \frac{1}{2}Q\Delta V from the sheet: U=12(6.0×104)(15)=4.5×103U = \frac{1}{2}(6.0 \times 10^{-4})(15) = 4.5 \times 10^{-3} J, that is 4.5 mJ. The same figure from Q2/2CQ^2/2C: (6.0×104)22(4.0×105)=3.6×1078.0×105=4.5×103\dfrac{(6.0 \times 10^{-4})^2}{2(4.0 \times 10^{-5})} = \dfrac{3.6 \times 10^{-7}}{8.0 \times 10^{-5}} = 4.5 \times 10^{-3} J.

  6. (d) First way, conservation of energy, which is what 13.6.A.1 names. There is no resistor, so when the capacitor is empty all 4.5 mJ is in the inductor: 12LImax2=4.5×103\frac{1}{2}LI_{\max}^2 = 4.5 \times 10^{-3}.

  7. Imax2=2(4.5×103)2.5×102=9.0×1032.5×102=0.36I_{\max}^2 = \dfrac{2(4.5 \times 10^{-3})}{2.5 \times 10^{-2}} = \dfrac{9.0 \times 10^{-3}}{2.5 \times 10^{-2}} = 0.36, so Imax=0.60I_{\max} = 0.60 A.

  8. Second way, from the simple harmonic motion model: Imax=ωQ=(1.0×103)(6.0×104)=0.60I_{\max} = \omega Q = (1.0 \times 10^{3})(6.0 \times 10^{-4}) = 0.60 A. The two agree, which they must, since the energy statement and the SHM statement describe the same circuit.

  9. (e) The charge starts at its maximum, so q=Qcos(ωt)q = Q\cos(\omega t) and the current first reaches its extreme a quarter period later, when the capacitor is fully discharged.

  10. t=T4=6.283×1034=1.57×103t = \dfrac{T}{4} = \dfrac{6.283 \times 10^{-3}}{4} = 1.57 \times 10^{-3} s, about 1.6 ms.

  11. Units check on the frequency, because it is the step most worth verifying: a henry is Vs/A\mathrm{V \cdot s/A} and a farad is As/V\mathrm{A \cdot s/V}, so a henry-farad is s2\mathrm{s^2}, its square root is a second, and one over it is s1\mathrm{s^{-1}}.

(a) ω=1.0×103\omega = 1.0 \times 10^{3} rad/s, f=159f = 159 Hz, T=6.3T = 6.3 ms. (b) 15 V. (c) 4.5 mJ. (d) 0.60 A, from conservation of energy and from Imax=ωQI_{\max} = \omega Q alike. (e) At t=T/4=1.6t = T/4 = 1.6 ms.

Two capacitors and one inductor, run both ways

A 0.0800.080 H inductor is connected to a 12 μF12\ \mu\mathrm{F} capacitor and a 6.0 μF6.0\ \mu\mathrm{F} capacitor. Find the angular frequency and period when the two capacitors are (a) in parallel and (b) in series, (c) the ratio of the two angular frequencies, and (d) what you would have got by wrongly using the 12 μF12\ \mu\mathrm{F} value alone.

  1. The learning objective says a combination of capacitors and a single inductor, so the CC in ω=1/LC\omega = 1/\sqrt{LC} is the equivalent capacitance of the combination. The two rules are on the sheet.

  2. (a) In parallel, Ceq,p=iCi=12+6.0=18 μF=1.8×105C_{\text{eq,p}} = \sum_i C_i = 12 + 6.0 = 18\ \mu\mathrm{F} = 1.8 \times 10^{-5} F.

  3. LC=(0.080)(1.8×105)=1.44×106 s2LC = (0.080)(1.8 \times 10^{-5}) = 1.44 \times 10^{-6}\ \mathrm{s^2}, so LC=1.20×103\sqrt{LC} = 1.20 \times 10^{-3} s and ω=11.20×103=833\omega = \dfrac{1}{1.20 \times 10^{-3}} = 833 rad/s.

  4. T=2πLC=(6.2832)(1.20×103)=7.54×103T = 2\pi\sqrt{LC} = (6.2832)(1.20 \times 10^{-3}) = 7.54 \times 10^{-3} s, about 7.5 ms.

  5. (b) In series, 1Ceq,s=112+16.0=112+212=312\dfrac{1}{C_{\text{eq,s}}} = \dfrac{1}{12} + \dfrac{1}{6.0} = \dfrac{1}{12} + \dfrac{2}{12} = \dfrac{3}{12}, so Ceq,s=4.0 μF=4.0×106C_{\text{eq,s}} = 4.0\ \mu\mathrm{F} = 4.0 \times 10^{-6} F. Note it is smaller than either capacitor, which is the check that separates the two rules.

  6. LC=(0.080)(4.0×106)=3.2×107 s2LC = (0.080)(4.0 \times 10^{-6}) = 3.2 \times 10^{-7}\ \mathrm{s^2}, so LC=5.657×104\sqrt{LC} = 5.657 \times 10^{-4} s and ω=1768\omega = 1768 rad/s.

  7. T=(6.2832)(5.657×104)=3.55×103T = (6.2832)(5.657 \times 10^{-4}) = 3.55 \times 10^{-3} s, about 3.6 ms.

  8. (c) ωseriesωparallel=1768833=2.12\dfrac{\omega_{\text{series}}}{\omega_{\text{parallel}}} = \dfrac{1768}{833} = 2.12. Symbolically it is Cparallel/Cseries=18/4.0=4.5=2.121\sqrt{C_{\text{parallel}}/C_{\text{series}}} = \sqrt{18/4.0} = \sqrt{4.5} = 2.121, and the two agree. The series arrangement, with the smaller equivalent capacitance, oscillates faster.

  9. (d) Using 12 μF12\ \mu\mathrm{F} alone: LC=(0.080)(1.2×105)=9.6×107LC = (0.080)(1.2 \times 10^{-5}) = 9.6 \times 10^{-7}, LC=9.798×104\sqrt{LC} = 9.798 \times 10^{-4}, ω=1021\omega = 1021 rad/s.

  10. That matches neither answer. It is 22 percent above the parallel value and 42 percent below the series value, so it is not even a usable estimate. The sheet prints ωLC=1/LC\omega_{LC} = 1/\sqrt{LC} without an equivalent subscript, unlike the two time constants which carry one, and this is why that omission must not be read as permission.

(a) In parallel, Ceq=18 μFC_{\text{eq}} = 18\ \mu\mathrm{F}, ω=833\omega = 833 rad/s and T=7.5T = 7.5 ms. (b) In series, Ceq=4.0 μFC_{\text{eq}} = 4.0\ \mu\mathrm{F}, ω=1768\omega = 1768 rad/s and T=3.6T = 3.6 ms. (c) The ratio is 4.5=2.12\sqrt{4.5} = 2.12, with the series circuit faster. (d) 1021 rad/s, which matches neither, so the equivalent capacitance is not optional.

The state of the circuit at one sixth of a period

For the circuit of the first example, with L=25L = 25 mH, C=40 μFC = 40\ \mu\mathrm{F}, Q=600 μCQ = 600\ \mu\mathrm{C}, ω=1.0×103\omega = 1.0 \times 10^{3} rad/s and total energy 4.5 mJ, find at t=T/6t = T/6 (a) the charge on the capacitor, (b) the current, (c) the energy in each of the two stores, and (d) the first time at which the energy is shared equally.

  1. The charge starts at its maximum with zero current, so the phase constant is zero and q=Qcos(ωt)q = Q\cos(\omega t), with I=dq/dt=ωQsin(ωt)I = dq/dt = -\omega Q\sin(\omega t).

  2. At t=T/6t = T/6 the phase is ωt=2πTT6=π3\omega t = \dfrac{2\pi}{T}\cdot\dfrac{T}{6} = \dfrac{\pi}{3}, which is 60 degrees. That is one of the seven angles in the booklet's table of trigonometric values, where cos60=1/2\cos 60^\circ = 1/2 and sin60=3/2=0.8660\sin 60^\circ = \sqrt{3}/2 = 0.8660.

  3. (a) q=(6.0×104)(0.500)=3.0×104q = (6.0 \times 10^{-4})(0.500) = 3.0 \times 10^{-4} C, so 300 μC300\ \mu\mathrm{C}, half the starting charge.

  4. (b) I=ωQsin60=(1.0×103)(6.0×104)(0.8660)=0.520I = -\omega Q \sin 60^\circ = -(1.0 \times 10^{3})(6.0 \times 10^{-4})(0.8660) = -0.520 A. The magnitude is 0.52 A, and the minus sign says the capacitor is discharging, which is consistent with the charge having fallen from its maximum.

  5. (c) UC=q22C=(3.0×104)22(4.0×105)=9.0×1088.0×105=1.125×103U_C = \dfrac{q^2}{2C} = \dfrac{(3.0 \times 10^{-4})^2}{2(4.0 \times 10^{-5})} = \dfrac{9.0 \times 10^{-8}}{8.0 \times 10^{-5}} = 1.125 \times 10^{-3} J, about 1.1 mJ.

  6. UL=12LI2=12(2.5×102)(0.5196)2=(1.25×102)(0.2700)=3.375×103U_L = \frac{1}{2}LI^2 = \frac{1}{2}(2.5 \times 10^{-2})(0.5196)^2 = (1.25 \times 10^{-2})(0.2700) = 3.375 \times 10^{-3} J, about 3.4 mJ.

  7. Check the total: 1.125×103+3.375×103=4.50×1031.125 \times 10^{-3} + 3.375 \times 10^{-3} = 4.50 \times 10^{-3} J, which is the whole energy, as it must be with no resistor in the loop.

  8. Check the fractions the fast way: UC/U=cos260=0.250U_C/U = \cos^2 60^\circ = 0.250 and UL/U=sin260=0.750U_L/U = \sin^2 60^\circ = 0.750, and 0.250+0.750=10.250 + 0.750 = 1 by the identity sin2θ+cos2θ=1\sin^2\theta + \cos^2\theta = 1 that the booklet prints. Notice that the charge is at half its maximum while the capacitor holds only a quarter of the energy, because the energy goes as the square.

  9. (d) Equal sharing needs cos2(ωt)=sin2(ωt)=0.500\cos^2(\omega t) = \sin^2(\omega t) = 0.500, so ωt=45\omega t = 45 degrees, which is T/8T/8.

  10. t=6.283×1038=7.85×104t = \dfrac{6.283 \times 10^{-3}}{8} = 7.85 \times 10^{-4} s, about 0.79 ms. It happens again every quarter period after that, four times per cycle, which is the energy curves running at twice the frequency of the charge.

(a) 3.0×1043.0 \times 10^{-4} C, half the maximum. (b) 0.52 A in magnitude, with the capacitor discharging. (c) 1.1 mJ in the capacitor and 3.4 mJ in the inductor, summing to the full 4.5 mJ. (d) At t=T/8=0.79t = T/8 = 0.79 ms, and every quarter period thereafter.

Frequently asked questions

What is an LC circuit in AP Physics C?

A circuit containing only a charged capacitor and an inductor, with no resistance. Essential knowledge 13.6.A.1 uses exactly that wording. The capacitor discharges through the inductor, the inductor's magnetic field then recharges the capacitor with the opposite polarity, and the process repeats, so the energy moves back and forth between the electric field of the capacitor and the magnetic field of the inductor. Statement 13.6.A.2 says the time dependence of the charge stored in the capacitor can be modeled as simple harmonic motion, which makes it the only oscillating circuit in the course and the only second-order one.

What is the angular frequency of an LC circuit?

One divided by the square root of the inductance times the capacitance. It is printed on the AP Physics C: Electricity and Magnetism equation sheet as omega sub LC, and the course description gives it at essential knowledge 13.6.A.3, which says it can be derived from the differential equation that describes an LC circuit. The period follows in one substitution from the printed relation that the period is two pi over the angular frequency, giving two pi times the square root of the inductance times the capacitance. Neither depends on the initial charge, so an LC circuit oscillates at the same rate whatever amplitude you start it at.

Why is an LC circuit the same as simple harmonic motion?

Because it obeys the same differential equation. The AP Physics C: Mechanics course description prints the SHM equation as the second derivative of position equalling minus omega squared times position. The AP Physics C: Electricity and Magnetism course description prints the LC equation at essential knowledge 13.6.A.2 as the second derivative of charge equalling minus one over the inductance times capacitance, times the charge. Matching the two gives omega squared equal to one over LC. Term by term, charge plays the part of position, current plays the part of velocity, inductance plays the part of mass, and the reciprocal of capacitance plays the part of the spring constant. The energies match too: one half L I squared corresponds to one half m v squared, and the capacitor's stored energy corresponds to the spring's.

How do you find the maximum current in an LC circuit?

By conservation of energy, which is what essential knowledge 13.6.A.1 specifies. With no resistor there is nothing to dissipate energy, so all of the energy stored in the capacitor when its charge is greatest ends up in the inductor when the current is greatest. Set the capacitor's stored energy, the square of the maximum charge divided by twice the capacitance, equal to one half the inductance times the square of the maximum current, and solve. The answer is the maximum charge times the angular frequency, which is a useful cross-check and is the electrical version of the mechanical result that maximum speed equals angular frequency times amplitude.

Does a bigger capacitor make an LC circuit oscillate faster?

No, slower. The angular frequency is one over the square root of the inductance times the capacitance, so increasing either the inductance or the capacitance lowers the frequency and lengthens the period. Doubling the capacitance divides the angular frequency by the square root of two. The mechanical picture explains why: capacitance corresponds to the reciprocal of a spring constant, not to the spring constant itself, so a large capacitor behaves like a floppy spring. The confusion usually comes from pattern-matching the mechanical formula, where the angular frequency is the square root of k over m and a larger k means a faster oscillation.

Does AP Physics C cover RLC circuits or damped oscillations?

No. The words RLC, damping and resonance appear nowhere in the AP Physics C: Electricity and Magnetism course description, and essential knowledge 13.6.A.1 specifies circuits containing only a charged capacitor and an inductor. A resistor with an inductor is Topic 13.5, a resistor with a capacitor is the RC circuits topic in Unit 11, and a circuit containing all three is outside the course. Topic 13.6 prints no boundary statement, so nothing else fences the topic off; the limit comes from the wording of the essential knowledge statements themselves and from what the framework does and does not print.

How far does AP Physics C take the LC differential equation?

Topic 13.6 prints no boundary statement, so the level is set by the labels on its equations. Both are marked Derived Equations, a label the course description's Required Equations page defines as demonstrating the final results of derivations expected of students on the exam. The differential equation itself is not on the equation sheet, so producing it, either from Kirchhoff's loop rule or by differentiating the conserved total energy, is expected work. The angular frequency it leads to is on the sheet, and statement 13.6.A.3 still says it can be derived from the differential equation. No solved charge-versus-time function is printed anywhere in Unit 13; what the framework points you to instead is the simple harmonic motion solution printed in the mechanics table that the E and M sheet reprints.