3 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 3
In an ideal LC circuit, at the instant when the current is at its maximum, which statement about the energy stored in the circuit is correct?
Answer and reasoning
AHalf of it is in the inductor and half is in the capacitor then. A student who thinks the energy is shared equally at maximum current picks this. Equal sharing happens only when q = Q₀/√2; at maximum current the charge is zero.
BAll of it is in the inductor, since the capacitor is uncharged then.Correct The current is greatest at the instant the capacitor's charge passes through zero, so UC = q²/(2C) = 0 and all of the circuit's constant total energy is in the inductor, (1/2)LImax².
CMost of it is in the capacitor, since its charge is greatest then. A student who thinks the current is greatest when the charge is greatest picks this. At maximum current the charge on the capacitor is zero, so the capacitor stores no energy.
DNone of it is in the inductor, since its current is not changing then. A student who thinks an inductor stores energy only while its current changes picks this. At maximum current dI/dt = 0, so the inductor's emf is zero, but its stored energy (1/2)LI² is at its greatest.
A charged capacitor is connected across an ideal inductor at t = 0, with no resistance in the circuit. Which describes the charge on the capacitor after t = 0?
Answer and reasoning
AIt decreases exponentially to zero and then stays zero, as it does in an RC circuit. A student who pictures every capacitor discharge as an RC discharge picks this. The inductor keeps the current going when q reaches zero, so the capacitor recharges with the opposite polarity.
BIt oscillates, reversing its polarity each half cycle, with a constant amplitude.Correct The loop rule gives d²q/dt² = −q/(LC), the equation of simple harmonic motion, so q(t) = Q₀ cos(ωt): it passes through zero, reverses sign and returns, and with no resistance to dissipate energy the amplitude stays Q₀.
CIt oscillates with an amplitude that shrinks as the current uses up energy flowing around. A student who thinks current uses up energy as it flows picks this. With no resistance nothing dissipates energy, so the energy passes back and forth between capacitor and inductor and the amplitude stays constant.
DIt falls to zero and then returns to its first value, keeping the same polarity throughout. A student who pictures the charge flowing out of the capacitor and back in picks this. When q reaches zero the current is at its maximum and carries on, charging the capacitor with the opposite polarity.
In an ideal LC circuit, both the inductance and the capacitance are doubled. By what factor is the angular frequency of the oscillations multiplied?
Answer and reasoning
A×¼ A student who uses ω = 1/(LC) picks this: LC is four times larger. The angular frequency is 1/√(LC), so it changes by 1/√4 = ½.
B×1 A student who maps L to the spring constant and C to the mass uses ω = √(L/C), in which the two doublings cancel. In the analogy L plays the role of mass and 1/C of the spring constant, giving ω = 1/√(LC).
C×2 A student who thinks ω grows with √(LC), as the period does, picks this. The period doubles, so the angular frequency is halved.
D×½Correct ω = 1/√(LC). Doubling both L and C multiplies LC by 4 and √(LC) by 2, so ω is halved.
In preparation: 0 of 3 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
13.6.A.1 LC circuit Fix
LC circuit
A circuit containing only a charged capacitor (or a combination of capacitors) and an inductor, with negligible resistance. Charge flows back and forth between the capacitor's plates through the inductor, and energy passes back and forth between the capacitor's electric field and the inductor's magnetic field.
Energy in an LC circuit
With no resistance the total energy is constant: q²/(2C) + (1/2)LI² = Q₀²/(2C), where Q₀ is the maximum charge. When the capacitor's charge is maximum the current is zero and all the energy is in the capacitor; when the charge is zero the current is maximum and all the energy is in the inductor.
Maximum current, Imax
From conservation of energy, (1/2)LImax² = Q₀²/(2C), so Imax = Q₀/√(LC) = ΔV₀√(C/L), where ΔV₀ = Q₀/C is the capacitor's initial potential difference. SI unit: A.
Students often think At the instant of maximum current, the circuit's energy is shared equally between the capacitor and the inductor, so (1/2)LImax² is half of Q₀²/(2C). In fact No. The current is maximum when the capacitor's charge is zero, so at that instant all of the energy, Q₀²/(2C), is in the inductor. The energy is shared equally only at the instants when q = Q₀/√2.
Students often think The energy stored in a capacitor is QΔV (or Q²/C), the charge multiplied by the final potential difference, with no factor of 1/2. In fact No. UC = (1/2)QΔV = Q²/(2C) = (1/2)CΔV². The potential difference grows from zero as the capacitor charges, so the work done is half of QΔV.
13.6.A.2 Differential equation of an LC circuit Fix
Differential equation of an LC circuit
Kirchhoff's loop rule, q/C + L dI/dt = 0 with I = dq/dt, gives d²q/dt² = −(1/(LC))q: the second derivative of the charge is proportional to the charge and opposite in sign, the defining equation of simple harmonic motion (compare d²x/dt² = −(k/m)x).
Charge as a function of time
The solution of the LC equation is sinusoidal, q(t) = Q₀ cos(ωt + φ). If the capacitor holds Q₀ at t = 0 with zero current, q(t) = Q₀ cos(ωt): the charge reverses sign each half period, and in an ideal circuit the amplitude stays Q₀.
Current in an LC circuit
I = dq/dt = −ωQ₀ sin(ωt + φ). The current's magnitude is greatest, ωQ₀, when the charge is zero, and zero when the charge is at ±Q₀: current and charge are a quarter period out of step.
Students often think The current in an LC circuit is greatest when the capacitor has its greatest charge, because the charge drives the current. In fact No. At maximum charge the current is zero (the charge momentarily stops changing); the current is greatest when the charge is zero. Current and charge are a quarter period out of step.
Students often think With no resistance in the circuit, nothing changes the current once it has started, so it has the same value at all times. In fact No. The current oscillates sinusoidally. The capacitor's potential difference makes the inductor's current change: dI/dt = −q/(LC), which is zero only when q = 0.
13.6.A.3 Angular frequency of an LC circuit, ωLCFix
Angular frequency of an LC circuit, ωLC
Comparing d²q/dt² = −(1/(LC))q with the simple-harmonic equation d²x/dt² = −ω²x gives ωLC = 1/√(LC); the inductance plays the role of mass and 1/C the role of spring constant. SI unit: rad/s.
Period and frequency
T = 2π/ω = 2π√(LC) and f = 1/T = ω/(2π). The angular frequency ω (rad/s) and the frequency f (Hz) differ by the factor 2π.
Equivalent capacitance in an LC circuit
When several capacitors are combined with a single inductor, the circuit oscillates as if it had one capacitor of the equivalent capacitance: capacitors in series combine as 1/Ceq = Σ 1/Ci, in parallel as Ceq = Σ Ci, and ω = 1/√(LCeq).
Students often think The angular frequency and the frequency are the same quantity, so f = 1/T can be used where ω is needed, and 1/√(LC) is taken to be the frequency f. In fact No. ω = 2πf. For an LC circuit ω = 1/√(LC) is the angular frequency, in rad/s; the frequency is f = 1/(2π√(LC)) in Hz, and from a graph ω = 2π/T.
Students often think Capacitors combine by the same rules as resistors, so the capacitances of capacitors in series add. In fact No. For capacitors in series 1/Ceq = Σ 1/Ci, so Ceq is less than the smallest capacitance; capacitances add in parallel.
8 more questions. Every wrong answer here is a real mistake students make, and you see why it is wrong as soon as you answer.
Question 1 of 8
A 20 μF capacitor is charged to a potential difference of 12 V and then connected across an ideal 50 mH inductor, with no other elements in the circuit. What is the maximum current in the inductor?
Answer and reasoning
A1.7 × 10⁻¹ A A student who thinks the energy is shared equally at the moment of maximum current sets (1/2)LI² equal to half of 1.44 × 10⁻³ J and gets 0.17 A. At maximum current the capacitor's charge is zero, so all the energy is in the inductor.
B2.4 × 10⁻¹ ACorrect Energy is conserved. Initially all the energy is in the capacitor: (1/2)CΔV² = (1/2)(20 × 10⁻⁶ F)(12 V)² = 1.44 × 10⁻³ J. At maximum current the capacitor is discharged, so (1/2)LImax² = 1.44 × 10⁻³ J and Imax = √(2(1.44 × 10⁻³ J)/(0.050 H)) = 0.24 A.
C3.4 × 10⁻¹ A A student who takes the capacitor's energy as CΔV² (that is, QΔV), without the factor 1/2, gets 2.88 × 10⁻³ J and 0.34 A. The capacitor stores (1/2)CΔV².
D5.8 × 10⁻² A A student who writes the inductor's energy as (1/2)LI sets (1/2)LI = (1/2)CΔV² and gets I = CΔV²/L = 0.058 A. The inductor stores (1/2)LI².
Working U = ½CΔV² = ½(20e-6)(12²) = 1.44e-3 J = ½LI² → I = √(2U/L) = √(0.0576) = 0.24 A (= ΔV√(C/L)).
In an ideal LC circuit, the capacitor (capacitance C) initially has charge Q₀ and the current is zero; the inductor has inductance L. What is the magnitude of the current at an instant when the charge on the capacitor is Q₀/2?
Answer and reasoning
A0.87Q₀/√(LC)Correct Energy is conserved: Q₀²/(2C) = (Q₀/2)²/(2C) + (1/2)LI². So (1/2)LI² = (3/4)Q₀²/(2C), I² = 3Q₀²/(4LC), and I = (√3/2)Q₀/√(LC) ≈ 0.87Q₀/√(LC).
B0.71Q₀/√(LC) A student who thinks half the energy has left the capacitor when half its charge has gone sets (1/2)LI² = Q₀²/(4C) and gets I = Q₀/√(2LC) ≈ 0.71Q₀/√(LC). The capacitor's energy depends on q², so at Q₀/2 it keeps only one-quarter.
C0.50Q₀/√(LC) A student who thinks the current grows in proportion to the charge that has left the capacitor takes half of Imax = Q₀/√(LC). The current is set by energy conservation, not by a proportion; at Q₀/2 it is already 0.87 of its maximum.
D1.22Q₀/√(LC) A student who takes the capacitor's energy as q²/C, without the factor 1/2, gets a total of Q₀²/C, leaves Q₀²/(4C) in the capacitor and gives (1/2)LI² = 3Q₀²/(4C), so I = √(3/2)Q₀/√(LC) — more than the true maximum current. The capacitor's energy is q²/(2C).
Working Q₀²/(2C) = (Q₀/2)²/(2C) + ½LI² → ½LI² = 3Q₀²/(8C) → I = √(3/(4LC))·Q₀ = 0.866 Q₀/√(LC).
The graph shows the charge q on the capacitor of an ideal LC circuit as a function of time t. Which ranks the magnitudes of the current in the circuit at the instants P, Q and R?
Answer and reasoning
AP > R > Q A student who thinks the current is greatest when the charge is greatest ranks by the size of q: P, then R, then Q. The current depends on the slope, which is zero at P and greatest at Q.
BP = Q = R A student who thinks the current stays constant in a circuit with no resistance picks this. The current changes because the capacitor's potential difference acts on the inductor; it is zero at P and greatest at Q.
CQ > R > PCorrect The current is I = dq/dt, the slope of the q–t graph. At P, a maximum of the charge, the slope and the current are zero. At Q the graph crosses zero at its steepest, so the current is greatest. At R the graph is still steep but less so than at Q.
DR > Q > P A student who takes the current to be proportional to the charge that has flowed off the capacitor since P ranks R (1.5Q₀ has flowed) above Q (Q₀) above P (none). The current is the rate of flow, which is greatest at Q.
Working I = dq/dt ∝ |sin(2πt/T)|: P (t = 0) → 0; Q (t = T/4) → 1; R (t = T/3) → 0.87. Q > R > P.
The graph shows the charge q on the capacitor of an ideal LC circuit as a function of time t. What is the maximum current in the circuit?
Answer and reasoning
A5.0 × 10⁻³ A A student who uses the frequency f = 1/T in place of the angular frequency gets Q₀/T = (20 × 10⁻⁶ C)/(0.0040 s) = 5.0 × 10⁻³ A. The maximum of dq/dt is ωQ₀ with ω = 2π/T.
B2.0 × 10⁻² A A student who divides the charge by the time the capacitor takes to discharge, a quarter period, gets (20 × 10⁻⁶ C)/(0.0010 s) = 2.0 × 10⁻² A. That is the average current during the discharge; the maximum is π/2 times larger.
C6.3 × 10⁻² A A student who reads the period as the time between successive zeros, 2.0 ms, gets ω = 2π/0.0020 s and twice the true current. Successive zeros are half a period apart; the period is 4.0 ms.
D3.1 × 10⁻² ACorrect From the graph, Q₀ = 20 μC and the period is T = 4.0 ms (maximum to maximum). With q = Q₀ cos(ωt), I = dq/dt has maximum ωQ₀ = (2π/T)Q₀ = (2π/0.0040 s)(20 × 10⁻⁶ C) = 3.1 × 10⁻² A.
Working Q₀ = 20 μC; T = 4.0 ms (peaks at 0, 4, 8 ms). Imax = ωQ₀ = (2π/0.0040)(20e-6) = 0.0314 A.
Capacitor C in the circuit shown is initially charged, capacitor 2C is initially uncharged, and the inductor is ideal. Switch S is then closed. What is the angular frequency of the resulting oscillations of charge?
Answer and reasoning
A1.22/√(LC)Correct With S closed, L, C and 2C form a single loop, so the capacitors are in series: 1/Ceq = 1/C + 1/(2C) = 3/(2C), Ceq = 2C/3. The loop rule gives d²q/dt² = −q/(LCeq), so ω = 1/√(L(2C/3)) = √(3/2)/√(LC) ≈ 1.22/√(LC).
B0.58/√(LC) A student who adds series capacitances, as for resistors in series, uses Ceq = 3C and gets ω = 1/√(3LC) ≈ 0.58/√(LC). Capacitors in series combine by reciprocals, giving Ceq = 2C/3.
C1.00/√(LC) A student who takes the smaller capacitance, C, as the equivalent capacitance of the series pair picks this. The equivalent capacitance of capacitors in series is less than the smallest, 2C/3.
D7.70/√(LC) A student who takes 1/√(LCeq) to be the frequency f in hertz multiplies it by 2π to get ω: 2π(1.22)/√(LC) ≈ 7.70/√(LC). The expression 1/√(LCeq) is already the angular frequency.
In an ideal LC circuit, the capacitor (capacitance C) initially holds charge Q₀ and the current is zero. A student claims that a larger inductance L gives a smaller maximum current. Which reasoning correctly supports the claim?
Answer and reasoning
AL limits the current as a resistance would, so Imax = (Q₀/C)/L, smaller for larger L. A student who treats the inductance as a resistance picks this. The conclusion happens to agree with the claim, but the reasoning is wrong: an inductor sets dI/dt, not I, and (Q₀/C)/L has units of A/s.
BAt maximum current the energy is shared, so (1/2)LImax² = Q₀²/(4C) and Imax ∝ 1/√L. A student who thinks the energy is shared equally at the current maximum picks this. At that instant the capacitor's charge is zero, so all of the energy, not half, is in the inductor; the reasoning gives the wrong Imax.
CAt maximum current all the energy is in L, so (1/2)LImax² = Q₀²/(2C) and Imax ∝ 1/√L.Correct When the current is greatest, the capacitor is discharged, so all the initial energy Q₀²/(2C) is in the inductor. Then Imax = Q₀/√(LC), which decreases as L increases, supporting the claim with a correct energy argument.
DThe energy in L is (1/2)LImax, so (1/2)LImax = Q₀²/(2C) and Imax ∝ 1/L. A student who writes the inductor's energy as proportional to the current picks this. The inductor's energy is (1/2)LI², so Imax ∝ 1/√L, not 1/L.
Working At Imax, q = 0, so UC = 0: (1/2)LImax² = Q₀²/(2C) → Imax = Q₀/√(LC) ∝ L−1/2. Distractor reasonings: I = ΔV/L (units A/s); equal sharing; UL = (1/2)LI.
An ideal LC circuit has an inductance of 80 mH and a capacitance of 5.0 μF. What is the angular frequency of its oscillations?
Answer and reasoning
A9.9 × 10³ rad/s A student who takes 1/√(LC) to be the frequency in hertz multiplies by 2π: 2π(1.58 × 10³) = 9.9 × 10³. The expression 1/√(LC) is already the angular frequency.
B2.5 × 10⁶ rad/s A student who uses ω = 1/(LC) picks this: 1/(4.0 × 10⁻⁷). That has units of s⁻², not rad/s; the angular frequency is 1/√(LC).
C1.3 × 10² rad/s A student who maps L to the spring constant and C to the mass uses ω = √(L/C) = √(0.080/5.0 × 10⁻⁶) = 1.3 × 10². In the analogy L corresponds to mass and 1/C to the spring constant.
In an ideal LC circuit, at t = 0 the capacitor (capacitance C) has charge Q₀ and the current is zero; the inductor has inductance L. At what time t does the energy stored in the inductor first equal the energy stored in the capacitor?
Answer and reasoning
A1.05√(LC) A student who takes the capacitor’s energy to be proportional to its charge thinks the energies are equal when q = Q₀/2, solves cos(ωt) = 1/2 to get ωt = π/3, and picks this. The capacitor’s energy is q²/(2C), so at Q₀/2 it still holds one-quarter of the total; the energies are equal when q = Q₀/√2.
B0.13√(LC) A student who takes 1/√(LC) to be the frequency f and writes q = Q₀cos(2πft) solves 2πt/√(LC) = π/4 and picks this. 1/√(LC) is the angular frequency ω, so the phase is ωt = t/√(LC).
C0.79√(LC)Correct With q = Q₀cos(ωt) and ω = 1/√(LC), the capacitor stores (Q₀²/(2C))cos²(ωt) and the inductor the rest, (Q₀²/(2C))sin²(ωt). They are first equal when ωt = π/4, at t = (π/4)√(LC) ≈ 0.79√(LC).
D1.57√(LC) A student who believes the energy is shared equally at the instant of maximum current picks the first time the current is greatest, a quarter period, (π/2)√(LC). At that instant the capacitor is uncharged and all the energy is in the inductor.
Working d²q/dt² = −(1/(LC))q with q(0) = Q₀, dq/dt(0) = 0 gives q = Q₀cos(ωt), ω = 1/√(LC). UC = q²/(2C) = (Q₀²/(2C))cos²(ωt); by conservation of energy UL = (Q₀²/(2C)) − UC = (Q₀²/(2C))sin²(ωt) (the same as (1/2)L(dq/dt)²). UL = UC ⇔ tan²(ωt) = 1; first at ωt = π/4, t = (π/4)√(LC) = 0.785√(LC) ≈ 0.79√(LC). Distractors (sympy): energies taken equal at q = Q₀/2, cos(ωt) = 1/2, t = (π/3)√(LC) = 1.05√(LC); 1/√(LC) taken as f with q = Q₀cos(2πft), 2πt/√(LC) = π/4, t = √(LC)/8 = 0.125√(LC) → 0.13; energies taken equal at maximum current, t = T/4 = (π/2)√(LC) = 1.57√(LC).
Compiled from the AP Physics C: Electricity and Magnetism Course and Exam Description (effective Fall 2024, 2026 reissue) and our question bank · Specialist review in progress. How these pages are made · Free, no account
