5 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 5
Two capacitors, of capacitance C₁ and C₂ with C₁ greater than C₂, are connected in parallel with each other across an ideal battery of emf ε. After a long time, which statement about the capacitors is correct?
Answer and reasoning
AEach has ΔV equal to ε/2, and C₁ carries more charge than C₂ does. A student who thinks capacitors in parallel share the battery's potential difference picks this. Each capacitor is connected directly across the battery, so each has the full ε.
BBoth carry the same charge, and C₂ has the larger ΔV across it. A student who thinks capacitors connected to the same battery carry equal charges, whatever the connection, picks this. Equal charges belong to capacitors in series; in parallel the potential differences are equal and the charges are C₁ε and C₂ε.
CEach carries charge (C₁ + C₂)ε, the charge found for Ceq. A student who gives each capacitor the charge of the equivalent capacitor picks this. (C₁ + C₂)ε is the total charge on the pair; each capacitor holds only its own part, C₁ε or C₂ε.
DEach has ΔV equal to ε, and C₁ carries more charge than C₂.Correct Capacitors in parallel are connected between the same two points, here the battery's terminals, so each has ΔV = ε. Their charges are C₁ε and C₂ε, so the larger capacitance carries more charge; together they hold (C₁ + C₂)ε, which is why Ceq,p = C₁ + C₂.
Two initially uncharged capacitors of different capacitance are connected in series with an ideal battery. After they are charged, each plate of both capacitors carries charge of the same magnitude Q. Which reasoning correctly explains why the magnitudes are equal?
Answer and reasoning
AThe same current crosses the gap in each capacitor, so equal charge passes through each of them. A student who thinks charge flows across the gap between a capacitor's plates, as it flows through a resistor, picks this. No charge crosses the gap; charge moves onto one plate and off the facing plate, through the wires.
BThe battery hands out the charge it stores in equal amounts to the capacitors connected to it. A student who thinks the battery stores charge and hands it out picks this. A battery moves charge that is already in the circuit; the equal magnitudes come from conservation of charge on the isolated conductor between the capacitors.
CThe first capacitor fills up and then passes the same amount of charge on to the second one. A student who pictures the capacitors charging one after another picks this. Both capacitors charge at the same time, with equal charges at every instant, because the conductor between them stays neutral throughout.
DThe inner plates and the wire joining them are isolated, so their net charge stays zero.Correct The plate of one capacitor that is wired to a plate of the other forms, with the connecting wire, a conductor with no connection to the battery. It starts neutral and charge is conserved, so if one of its plates has −Q the other must have +Q. Both capacitors therefore carry charge of magnitude Q, whatever their capacitances.
An uncharged capacitor of capacitance C is connected in series with a resistor of resistance R, a switch and an ideal battery of emf ε, and the switch is closed at t = 0. Let q be the charge on the capacitor. At the instant when q = Cε/4, what is dq/dt?
Answer and reasoning
A0.25 ε/R A student who gives the resistor the same potential difference as the capacitor in series with it picks this: (ε/4)/R = 0.25 ε/R. The loop rule makes the two potential differences add to ε, so the resistor has what is left, 3ε/4.
B0.75 ε/RCorrect Rearranging the loop rule, dq/dt = (ε − q/C)/R. When q = Cε/4 the capacitor has ΔVC = q/C = ε/4, so the resistor has ε − ε/4 = 3ε/4 across it and dq/dt = 0.75 ε/R.
C1.25 ε/R A student who treats the charging capacitor as a second battery helping the first picks this: (ε + ε/4)/R = 1.25 ε/R. Around the loop the capacitor's potential difference opposes the battery's emf, leaving ε − q/C across the resistor.
D1.00 ε/R A student who thinks the battery keeps the current at its initial value, ε/R, until the capacitor is full picks this. The loop rule gives dq/dt = (ε − q/C)/R, which has already fallen to (3/4)ε/R when q = Cε/4.
Working Loop rule: ε − (dq/dt)R − q/C = 0, so dq/dt = (ε − q/C)/R. With q = Cε/4, q/C = ε/4, so dq/dt = (ε − ε/4)/R = (3/4)ε/R.
In trial 1, an uncharged capacitor is charged through a resistor by an ideal battery, and its charge takes a time t₁ to reach 90 percent of its final value. In trial 2, with the same battery, the resistance is doubled and the capacitance is tripled, and the charge takes a time t₂ to reach 90 percent of its final value. What is t₂/t₁?
Answer and reasoning
A2.00 A student who thinks the resistance alone sets how long charging takes, and the capacitance only how much charge is stored, picks this. The time constant is the product RC, so tripling C also triples every charging time.
B0.67 A student who uses R/C as the time constant picks this: (2R)/(3C) is 2/3 of R/C. The time constant is RC; a larger capacitance makes charging slower, not faster.
C3.00 A student who thinks the capacitance alone sets how long charging takes, and the resistance only limits the current, picks this. A larger resistance lowers the current at every value of q, so doubling R also doubles every charging time: τ = RC.
D6.00Correct The fraction of the final charge depends only on t/τ: q/qfinal = 1 − e−t/τ. Reaching 90 percent therefore takes the same number of time constants, ln 10 ≈ 2.3, in both trials, so t₂/t₁ = τ₂/τ₁ = (2R)(3C)/(RC) = 6.00. The larger final charge in trial 2 does not change the fraction reached at a given t/τ.
Working q/qfinal = 1 − e−t/τ = 0.90 at t = τ ln 10 ≈ 2.3τ in both trials, so the time to reach a given fraction is proportional to τ = RC. τ₂/τ₁ = (2R)(3C)/(RC) = 6, so t₂/t₁ = 6.00.
The graph shows the potential difference ΔVC across a capacitor as a function of time t while it charges through a resistor from an ideal battery of emf ε. Which statement about the current in the capacitor's branch is supported by the graph?
Answer and reasoning
AIt stays constant, because the battery keeps driving charge at the same rate as at the start. A student who thinks the battery supplies a constant current picks this. A constant current would make ΔVC grow at a steady rate, giving a straight line; the graph bends over because the current falls.
BIt increases, because the resistor's ΔV equals ΔVC, which the graph shows rising over time. A student who thinks the resistor has the same potential difference as the capacitor in series with it picks this. By the loop rule the resistor has ε − ΔVC, which shrinks as ΔVC grows, so the current falls.
CIt decreases, because the slope of the graph, which is proportional to it, decreases.Correct The current is the rate at which charge arrives on the plates: I = dq/dt = C(dΔVC/dt), which is C times the slope of the graph. The slope is steepest at t = 0 and becomes smaller and smaller as ΔVC approaches ε, so the current decreases toward zero.
DIt is zero throughout, because charge cannot cross the gap between the plates of the capacitor. A student who thinks there is never a current in a capacitor's branch picks this. ΔVC = q/C is rising, so charge is arriving on the plates through the wires: there is a current in the branch until the capacitor is fully charged.
Working I = dq/dt = C(dΔVC/dt): proportional to the slope of the graph, which decreases toward zero as ΔVC approaches ε.
In preparation: 0 of 5 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
11.8.A.1 Equivalent capacitance, CeqFix
Equivalent capacitance, Ceq
The capacitance of a single capacitor that, connected in place of a group of capacitors, would store the same total charge for the same potential difference across the group. Unit: farad (F).
Capacitors in series
Capacitors connected one after another, with no junction between them, so that each pair of adjacent plates of neighboring capacitors, with the wire joining them, forms an isolated conductor. For capacitors in series, 1/Ceq,s = Σi 1/Ci; Ceq,s is the reciprocal of that sum.
Series Ceq is less than the smallest capacitance
Every capacitor in series adds a positive term to 1/Ceq,s, so 1/Ceq,s is greater than 1/Cmin and Ceq,s is less than the smallest capacitance Cmin in the set. Physically, the capacitors share one charge Q while the potential difference across the set exceeds that across any one of them.
Capacitors in parallel
Capacitors connected between the same two points, so each has the same potential difference across it. The charges add, so Ceq,p = Σi Ci, and each capacitor's charge is Ci ΔV.
Students often think Capacitances simply add whatever the connection, because more capacitors can store more charge. In fact No. Capacitors in parallel give a larger equivalent capacitance than any one of them, but capacitors in series give a smaller one, less than the smallest capacitance in the series.
Students often think The charge found from the equivalent capacitance, Q = Ceq ΔV, is the charge on each individual capacitor of the group. In fact Not in general. The equivalent capacitor carries the total charge the battery moves for the group. Capacitors in parallel share that total; only capacitors in series each carry the equivalent capacitor's charge.
11.8.A.2 Equal charge on capacitors in series Fix
Equal charge on capacitors in series
The adjacent plates of two neighboring capacitors in series (one plate of each), with the wire joining them, form a conductor that is isolated from the battery and starts neutral. By conservation of charge its net charge stays zero, so if one of its plates carries −Q the other carries +Q: every capacitor in series has charge of the same magnitude Q.
Students often think Capacitance plays the role that resistance plays in a series circuit, so the capacitor with the larger capacitance takes the larger share of the battery's potential difference. In fact No. Capacitors in series carry equal charges, so ΔV = Q/C is smallest across the largest capacitance.
Students often think Every capacitor or resistor connected to a battery has the battery's full potential difference across it, whatever else is connected. In fact No. Only elements connected directly across the battery's terminals have the full ε. Elements in series with others share ε among them.
11.8.B.1 Loop-rule equation for a charging RC circuit Fix
Loop-rule equation for a charging RC circuit
For a capacitor C charging through a resistor R from an ideal battery of emf ε in series, Kirchhoff's loop rule gives ε = (dq/dt)R + q/C, where q is the charge on the capacitor, dq/dt = I is the current, (dq/dt)R is the potential difference across the resistor and q/C is that across the capacitor.
Solutions for a charging capacitor
With q = 0 at t = 0, the solution of ε = (dq/dt)R + q/C is q(t) = Cε(1 − e−t/(RC)). Then ΔVC = ε(1 − e−t/(RC)) and I = dq/dt = (ε/R)e−t/(RC): the charge and ΔVC grow toward their final values while the current decays toward zero.
Students often think A charging capacitor acts like a second battery that helps the first, so the potential difference across the resistor is ε + ΔVC. In fact No. Around the loop, the charging capacitor's potential difference opposes the battery's emf, so the resistor has ε − q/C across it, not ε + q/C.
Students often think Elements in series have the same potential difference across them, so the resistor in series with a capacitor has the capacitor's potential difference across it. In fact No. Elements in series carry the same current, not the same potential difference. By the loop rule their potential differences add to ε: ΔVR = ε − q/C.
11.8.B.2 Time constant, τ Fix
Time constant, τ
The quantity τ = Req Ceq that sets the time scale of charging and discharging in an RC circuit. Unit: second (Ω·F = s). Every RC quantity changes as a function of t/τ, so the time to reach any given fraction of a change is proportional to τ.
Req and Ceq in the time constant
In τ = Req Ceq, Req is the equivalent resistance of the path through which the capacitor charges or discharges, and Ceq is the equivalent capacitance of the capacitors that charge or discharge together. Opening or closing a switch can change the path, and with it Req.
The 63 percent value
For a charging capacitor, q/qfinal = 1 − e−t/τ, so at t = τ the charge (and ΔVC) has reached 1 − e−1 ≈ 0.63 of its final asymptotic value.
The 37 percent value
For a discharging capacitor, q = Q₀e−t/τ, so at t = τ the charge (and ΔVC and the current) has fallen to e−1 ≈ 0.37 of its initial value. Each further time constant multiplies what remains by e−1 again.
Students often think The capacitance sets only how much charge the capacitor finally stores, and the resistance alone sets how long charging takes. In fact No. The time scale is the time constant τ = RC, so the capacitance matters as much as the resistance: doubling either doubles every charging time.
Students often think The capacitance alone sets how long charging takes, because a bigger capacitor takes longer to fill; the resistance only limits the size of the current. In fact No. The time constant is τ = RC: a larger resistance gives a smaller current at every stage of charging, so it lengthens every charging time as much as a larger capacitance does.
11.8.B.3 Steady state in an RC circuit Fix
Steady state in an RC circuit
The condition approached after a time much longer than τ, in which the charge, potential difference and current no longer change measurably. A charging capacitor is then modeled as fully charged, with zero current in its branch; a discharging one as uncharged.
Uncharged capacitor acts like a wire
At the instant an uncharged capacitor is placed in a circuit, q = 0, so ΔVC = q/C = 0. It has no potential difference across it, as a wire has none, so charge flows onto and off its plates at once and the current in its branch is set by the rest of the circuit.
Energy stored in a capacitor, UC
UC = (1/2)QΔV = q²/(2C) = (1/2)CΔVC². Unit: joule (J). Because UC is proportional to q², it changes by a different fraction than q does: a charging capacitor holds Ufinal(1 − e−t/τ)² at time t.
Rate of energy storage in a capacitor
The rate at which energy is stored in a capacitor is dUC/dt = (q/C)(dq/dt) = ΔVC I. While a capacitor charges from a battery through a resistor, the battery supplies εI, of which I²R is dissipated in the resistor and ΔVC I is stored.
Fully charged capacitor
A capacitor that has been charging for a time much longer than τ has zero current in its branch and its maximum potential difference, equal to the potential difference across whatever it is connected in parallel with (ε in a single series loop with an ideal battery).
Discharging capacitor
A charged capacitor connected across a resistor drives a current I = ΔVC/R. The charge leaves at a rate proportional to the charge remaining, so q = Q₀e−t/(RC), and ΔVC, I and UC all decrease from the first instant, more and more slowly, toward zero.
Students often think The battery delivers a constant current, so the current stays at its starting value until the capacitor is full. In fact No. An ideal battery keeps a fixed potential difference, not a fixed current. As the capacitor charges, its potential difference rises and the resistor's falls, so the current decreases continuously.
Students often think A capacitor blocks direct current at all times, so there is never any current in its branch, even while it is charging. In fact No. While a capacitor charges or discharges, charge flows onto or off its plates, so there is a current in its branch. The current is zero only in the steady state.
30 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 30
The diagram shows three capacitors connected between terminals X and Y. What is the equivalent capacitance of the combination between X and Y?
Answer and reasoning
A6.4 μF A student who combines capacitors by the resistor rules picks this: (0.50 × 2.5)/(0.50 + 2.5) ≈ 0.42 μF for the parallel pair, plus 6.0 μF for C₁ in series, gives 6.4 μF. For capacitors the rules are the other way round: capacitances in parallel add, and capacitances in series combine by their reciprocals.
B9.0 μF A student who adds all the capacitances, whatever the connection, picks this: 6.0 + 0.50 + 2.5 = 9.0 μF. Only C₂ and C₃ are in parallel; C₁ is in series with that pair, which makes the equivalent capacitance smaller than either part.
C2.0 μFCorrect C₂ and C₃ are joined between the same two points, so they are in parallel and their capacitances add: 0.50 μF + 2.5 μF = 3.0 μF. All the charge on C₁ must go to that pair, so C₁ is in series with it: 1/Ceq = 1/(6.0 μF) + 1/(3.0 μF) = 1/(2.0 μF). Ceq = 2.0 μF, less than either 6.0 μF or 3.0 μF.
D3.0 μF A student who thinks the smaller part of a series connection limits it, so that Ceq equals the smaller of C₁ (6.0 μF) and the parallel pair (3.0 μF), picks this. The equivalent capacitance of capacitors in series is less than the smallest: 1/Ceq = 1/6.0 + 1/3.0 gives 2.0 μF.
Working C₂ and C₃ are in parallel: 0.50 μF + 2.5 μF = 3.0 μF. C₁ is in series with that pair: 1/Ceq = 1/(6.0 μF) + 1/(3.0 μF) = 0.50 μF⁻¹, so Ceq = 2.0 μF.
A capacitor of capacitance C is connected in series with a parallel combination of two capacitors, of capacitance C and 2C, and the arrangement is connected to an ideal battery of emf ε. The capacitors are initially uncharged. After a long time, what is the magnitude of the charge on each plate of the 2C capacitor?
Answer and reasoning
A3Cε/4 A student who gives each capacitor the charge of the equivalent capacitor picks this: Q = Ceq ε = 3Cε/4. That is the total charge on the parallel pair, which the pair shares in proportion to capacitance; the 2C capacitor has Cε/2 of it.
BCε/2Correct The parallel pair acts as 3C, and it is in series with C, so Ceq = (C)(3C)/(C + 3C) = 3C/4 and the battery moves Q = 3Cε/4. That charge is on the single capacitor C and is the total on the pair, so the pair has ΔV = Q/(3C) = ε/4. The 2C capacitor therefore carries 2C × ε/4 = Cε/2.
CCε A student who splits ε equally between the two groups in series picks this: ε/2 across the pair gives 2C × ε/2 = Cε. The single capacitor and the pair carry equal charge, so the larger capacitance, 3C for the pair, has the smaller share of ε, ε/4.
D2Cε A student who puts the battery's full emf across every capacitor picks this: 2C × ε = 2Cε. The pair is in series with the single capacitor C, so only part of ε, here ε/4, is across it.
Working Parallel pair: C + 2C = 3C. In series with C: 1/Ceq = 1/C + 1/(3C) = 4/(3C), so Ceq = 3C/4. Charge moved by the battery: Q = Ceq ε = 3Cε/4, the charge on the single capacitor C and the total on the pair. ΔV across the pair = Q/(3C) = ε/4, so Q2C = 2C(ε/4) = Cε/2. Check: ΔV across the single C = 3ε/4, and 3ε/4 + ε/4 = ε.
Three capacitors, of capacitance C, 2C and 6C, are connected in series with an ideal battery of emf ε. After the capacitors have been charging for a long time, what is the potential difference ΔV across the 2C capacitor?
Answer and reasoning
AΔV = 3ε/10Correct The reciprocals add: 1/Ceq = (6 + 3 + 1)/(6C), so Ceq = 3C/5, and each capacitor in series carries Q = Ceq ε = 3Cε/5. Then ΔV = Q/(2C) = 3ε/10. The three potential differences, 3ε/5, 3ε/10 and ε/10, add to ε.
BΔV = 2ε/9 A student who gives each capacitor a share of ε in proportion to its capacitance, as resistors in series share in proportion to resistance, picks this: 2C/(C + 2C + 6C) × ε = 2ε/9. With equal charges, ΔV = Q/C is largest for the smallest capacitance, so the shares go inversely with C.
CΔV = ε/3 A student who shares ε equally among the three capacitors picks this. Capacitors in series carry equal charges, not equal potential differences: the 2C capacitor has Q/(2C) = 3ε/10.
DΔV = ε A student who puts the battery's full emf across each capacitor picks this. The potential differences across the three capacitors in series must add to ε, so none of them can be ε.
Two identical capacitors are connected in series with an ideal battery and allowed to charge fully; the magnitude of the charge on each plate is then Q₁. The capacitors are discharged, a third identical capacitor is added in series with them, and all three are charged fully by the same battery; the magnitude of the charge on each plate is then Q₂. What is Q₂/Q₁?
Answer and reasoning
A1.50 A student who adds capacitances in series, as resistances in series add, picks this: Ceq would go from 2C to 3C, and the charge would rise by a factor of 3/2. Capacitors in series combine by their reciprocals, so adding one in series lowers Ceq.
B1.00 A student who takes the equivalent capacitance of a series chain to be its smallest capacitance, C in both cases, picks this. Every added capacitor adds a positive term to 1/Ceq, so Ceq falls from C/2 to C/3.
C0.44 A student who thinks the charge the battery moves is shared equally among the capacitors picks this: (Cε/3) ÷ 3 compared with (Cε/2) ÷ 2 gives 4/9 ≈ 0.44. Capacitors in series do not share the charge: each carries the full charge of the equivalent capacitor.
D0.67Correct For N identical capacitors in series, 1/Ceq = N/C, so Ceq = C/N: C/2 with two and C/3 with three. Every capacitor in series carries the charge on the equivalent capacitor, Ceq ε, so Q₂/Q₁ = (C/3)/(C/2) = 2/3 ≈ 0.67.
Working Each capacitance C, battery emf ε. Two in series: 1/Ceq = 2/C, Ceq = C/2, Q₁ = Cε/2. Three in series: Ceq = C/3, Q₂ = Cε/3. Q₂/Q₁ = (1/3)/(1/2) = 2/3 ≈ 0.67.
A student claims that the equivalent capacitance Ceq of any set of capacitors connected in series is less than the smallest capacitance in the set, Cmin. Which reasoning correctly supports the claim?
Answer and reasoning
AThe reciprocal of Ceq is that of Cmin plus more positive terms, so Ceq is less than Cmin.Correct For capacitors in series, 1/Ceq = Σi 1/Ci. The sum contains 1/Cmin and at least one more positive term, so 1/Ceq > 1/Cmin, and taking reciprocals reverses the inequality: Ceq < Cmin. The argument holds for any number of capacitors and any values.
BThe capacitor next to the battery charges first and then stops any more charge from reaching the rest. A student who pictures capacitors in series charging one after another picks this. All the capacitors in series charge at the same time and carry the same charge at every instant; none fills first and blocks the others.
CEach capacitor further along the chain gets less charge, so the whole set stores less than its smallest one. A student who thinks charge is stored progressively along the chain picks this. Each pair of plates joined by a wire forms an isolated conductor whose net charge stays zero, so every capacitor in series carries the same charge.
DThe battery's ΔV is split equally among them, so each one stores less charge than it would alone. A student who thinks the battery's potential difference is shared equally picks this. The shares are not equal (ΔV = Q/C is largest across the smallest capacitance), and comparing each capacitor with itself connected alone says nothing about how Ceq compares with Cmin.
The diagram shows three arrangements, X, Y and Z, of identical capacitors, each of capacitance C. CX, CY and CZ are the equivalent capacitances between the two terminals of each arrangement. Which ranking of the equivalent capacitances is correct?
Answer and reasoning
ACX > CY > CZ A student who uses the resistor rules for capacitors picks this: X gives 3C, Y gives C + C/2 = 3C/2, and Z gives (2C)(C)/(3C) = 2C/3. For capacitors the rules are reversed, which reverses the ranking.
BCZ > CY > CXCorrect In X the three capacitors are in series, so CX = C/3. In Y, C is in series with a parallel pair of capacitance 2C, so CY = (C)(2C)/(3C) = 2C/3, less than C. In Z, a series pair (C/2) is in parallel with C: CZ = 3C/2. So CZ > CY > CX.
CCX = CY = CZ A student who adds the capacitances whatever the connection picks this, since each arrangement has three capacitors and so 3C. Only capacitances in parallel add; in series the reciprocals add, which makes CX, CY and CZ all different.
DCZ > CX = CY A student who sets the equivalent capacitance of a series connection equal to its smallest capacitance picks this: X gives C, Y gives the smaller of C and 2C, which is C, and Z gives C + C = 2C. A series connection has Ceq less than its smallest capacitance, so CX = C/3 and CY = 2C/3 are not equal.
Working X: three in series, CX = C/3. Y: C in series with a parallel pair (2C): CY = (C)(2C)/(3C) = 2C/3. Z: a series pair (C/2) in parallel with C: CZ = 3C/2. So CZ > CY > CX.
In the circuit shown, the ideal battery has been connected to the initially uncharged capacitors for a long time. Q₁, Q₂ and Q₃ are the magnitudes of the charge on each plate of C₁, C₂ and C₃. Which ranking of the charges is correct?
Answer and reasoning
AQ₂ > Q₁ > Q₃ A student who shares the branch's 12 V between C₂ and C₃ in proportion to capacitance picks this: 8 V across C₂ gives 48 μC and 4 V across C₃ gives 12 μC, with Q₁ = 36 μC. Capacitors in series carry equal charge, so the larger capacitance has the smaller potential difference.
BQ₁ = Q₂ = Q₃ A student who thinks capacitors on the same battery share the charge equally, whatever the connection, picks this. C₂ and C₃ in series do carry one common charge, but C₁ has the full 12 V across it and carries a larger charge.
CQ₁ > Q₂ = Q₃Correct C₁ and the branch containing C₂ and C₃ are both connected across the battery, so each has 12 V across it. C₁ has Q₁ = (3.0 μF)(12 V) = 36 μC. C₂ and C₃ are in series in their branch, so they carry equal charge: their Ceq = 2.0 μF gives Q₂ = Q₃ = 24 μC. So Q₁ > Q₂ = Q₃.
DQ₂ > Q₁ = Q₃ A student who puts the battery's full 12 V across every capacitor picks this: Q₂ = 72 μC and Q₁ = Q₃ = 36 μC. C₂ and C₃ are in series, so the 12 V is divided between them.
Working C₁ is directly across the battery: Q₁ = (3.0 μF)(12 V) = 36 μC. C₂ and C₃ are in series with each other, and that branch is across the battery: C₂₃ = (6.0)(3.0)/(9.0) μF = 2.0 μF, so Q₂ = Q₃ = (2.0 μF)(12 V) = 24 μC. Q₁ > Q₂ = Q₃.
In the circuit shown, the ideal battery has been connected for a long time to three initially uncharged capacitors, X, Y and Z, with the capacitances labeled. Which ranking of the potential differences across the capacitors is correct?
Answer and reasoning
AΔVX > ΔVZ > ΔVY A student who gives the larger capacitance the larger share of the potential difference, as a larger resistance takes a larger share, picks this. With equal charges, ΔV = Q/C is smallest across the largest capacitance.
BΔVX = ΔVY = ΔVZ A student who thinks the battery's potential difference is shared equally among the capacitors picks this. Equal potential differences would need charges in proportion to capacitance, but capacitors in series carry equal charges.
CΔVY > ΔVZ > ΔVXCorrect The three capacitors are in a single loop with the battery, so they are in series. The wire and plates between two neighboring capacitors form an isolated conductor that starts neutral, so all three carry the same charge Q. Then ΔV = Q/C is largest for the smallest capacitance: ΔVY = Q/C, ΔVZ = Q/(2C), ΔVX = Q/(3C).
DΔVZ > ΔVX > ΔVY A student who thinks Y, which has no plate wired to the battery, receives little or no charge picks this, ranking Y last and ranking X and Z by Q/C. Y carries the same charge as X and Z: as they charge, charge separates in the isolated conductors on either side of Y.
Working Single loop: series. The conductors between capacitors are isolated and neutral, so each capacitor carries the same charge Q. ΔV = Q/C: ΔVY = Q/C > ΔVZ = Q/(2C) > ΔVX = Q/(3C).
An uncharged capacitor of capacitance C, a resistor of resistance R and an ideal battery of emf ε are connected in series, and the circuit is completed at t = 0. By solving the loop-rule equation ε = (dq/dt)R + q/C, find the time t at which the current in the resistor is half its initial value.
Answer and reasoning
ARC ln 2Correct With q(0) = 0 the solution is q = Cε(1 − e−t/(RC)), so I = dq/dt = (ε/R)e−t/(RC). The current is half its initial value when e−t/(RC) = 1/2, that is t = RC ln 2 ≈ 0.69RC, a little less than one time constant.
BRC/2 A student who thinks the current falls at a steady rate, equal to its initial rate, picks this: at the initial rate of decrease, (ε/R)/(RC), the current would halve at RC/2. The rate of decrease itself shrinks as the current falls, so halving takes longer: RC ln 2.
CRC A student who takes the time constant as the time for the current to fall to half picks this. At t = RC the current has fallen to e−1, about 37 percent of its initial value; it passed one half earlier, at RC ln 2.
D(R/C) ln 2 A student who uses R/C as the time constant picks this. The exponent in the solution must be dimensionless, and the loop rule gives the time constant RC, which has units of seconds (Ω·F = s); R/C does not.
Working With q(0) = 0: q = Cε(1 − e−t/(RC)), so I = dq/dt = (ε/R)e−t/(RC). I = I₀/2 when e−t/(RC) = 1/2, so t = RC ln 2 ≈ 0.69RC.
An uncharged 50 μF capacitor is connected in series with a 20 kΩ resistor, an open switch and an ideal 9.0 V battery. The switch is closed at t = 0. What is the current in the resistor at t = 2.0 s?
Answer and reasoning
A3.9 × 10⁻⁴ A A student who thinks the current starts at zero and builds up picks this, using the growing form (ε/R)(1 − e−t/τ). The uncharged capacitor has no potential difference at t = 0, so the current starts at its largest value, ε/R, and decays.
B6.1 × 10⁻⁵ ACorrect Differentiating the charging solution q = Cε(1 − e−t/τ) gives I = (ε/R)e−t/τ, with τ = RC = (2.0 × 10⁴ Ω)(5.0 × 10⁻⁵ F) = 1.0 s. At t = 2.0 s, I = (4.5 × 10⁻⁴ A)e−2.0 ≈ 6.1 × 10⁻⁵ A, about 14 percent of the initial current.
C4.5 × 10⁻⁴ A A student who thinks the battery drives a constant current picks ε/R = 4.5 × 10⁻⁴ A. As the capacitor charges, its potential difference rises and the resistor's, ε − q/C, falls, so the current decreases.
D1.1 × 10⁻⁴ A A student who takes the time constant as the time for the current to halve picks this: two halvings give (4.5 × 10⁻⁴ A)/4 ≈ 1.1 × 10⁻⁴ A. Each time constant multiplies the current by e−1 ≈ 0.37, not by 1/2.
Working τ = RC = (2.0 × 10⁴ Ω)(5.0 × 10⁻⁵ F) = 1.0 s. From q = Cε(1 − e−t/τ), I = dq/dt = (ε/R)e−t/τ = (9.0 V/2.0 × 10⁴ Ω)e−2.0 = (4.5 × 10⁻⁴ A)(0.135) = 6.1 × 10⁻⁵ A.
In the circuit shown, the capacitors are initially uncharged and the battery is ideal. Switch S is closed at t = 0. What is the time constant of the circuit?
Answer and reasoning
A4.0 × 10⁻² s A student who adds capacitances in series, as resistances in series add, picks this: Ceq = 8.0 μF and τ = 4.0 × 10⁻² s. Capacitors in series combine by their reciprocals, giving 1.5 μF.
B1.0 × 10⁻² s A student who takes the equivalent capacitance of the series pair to be the smaller capacitance, 2.0 μF, picks this. The equivalent capacitance of capacitors in series is less than the smallest: 1.5 μF.
C7.5 × 10⁻³ sCorrect With S closed, every element is in one loop, so the resistors are in series (Req = 5.0 kΩ) and so are the capacitors: 1/Ceq = 1/(2.0 μF) + 1/(6.0 μF), giving Ceq = 1.5 μF. Then τ = Req Ceq = (5.0 × 10³ Ω)(1.5 × 10⁻⁶ F) = 7.5 × 10⁻³ s.
D3.3 × 10⁻³ s A student who stops after adding the reciprocals, 1/2.0 + 1/6.0 ≈ 0.67, and uses 0.67 μF as Ceq picks this. That sum is 1/Ceq, in μF⁻¹; its reciprocal, 1.5 μF, is the equivalent capacitance.
Working Every element is in one loop. Req = 2.0 kΩ + 3.0 kΩ = 5.0 kΩ. 1/Ceq = 1/(2.0 μF) + 1/(6.0 μF) = (2/3) μF⁻¹, so Ceq = 1.5 μF. τ = Req Ceq = (5.0 × 10³ Ω)(1.5 × 10⁻⁶ F) = 7.5 × 10⁻³ s.
In the circuit shown, the battery is ideal. Switch S has been closed for a long time and is then opened at t = 0. The time constant for the discharge of the capacitor after S is opened can be written as τ = kRC. What is the value of k?
Answer and reasoning
A4.00Correct Once S is open, the battery's branch is broken, so there is no current in R₁. The capacitor can discharge only around the loop through R₃, the top junction, R₂ and the bottom junction. R₂ and R₃ carry the same current one after the other, so they are in series: Req = 3R + R = 4R and τ = Req C = 4RC.
B5.00 A student who includes every resistor in the circuit picks this: R + 3R + R = 5R. R₁ is in the branch broken by the open switch, so it carries no current after t = 0 and is not part of the discharge path.
C1.00 A student who keeps the resistance that was in series with the capacitor while it charged, R₃, picks this. After S opens, the discharge current must also pass through R₂, so Req = 4R.
D0.75 A student who treats R₂ and R₃ as parallel because they are drawn in separate branches picks this: (3R)(R)/(4R) = 0.75R. With S open they form a single loop with the capacitor and carry the same current, so they are in series.
Working With S open, the battery's branch is broken, so R₁ carries no current. The capacitor's only loop: C → R₃ → top junction → R₂ → bottom junction → C. R₂ and R₃ carry the same current: series. Req = 3R + R = 4R, τ = 4RC, k = 4.00.
An initially uncharged capacitor is charged in a circuit containing resistors and an ideal battery of emf 12 V. The graph shows the potential difference ΔVC across the capacitor as a function of time t after the circuit is completed. What is the time constant for the charging?
Answer and reasoning
A4.2 × 10⁻³ s A student who takes τ as the time to reach half the final value picks this: the curve reaches 4.0 V at about 4.2 ms. At t = τ the capacitor has about 63 percent of its final potential difference, 5.1 V, which the curve reaches later.
B2.8 × 10⁻³ s A student who uses 37 percent for a charging capacitor picks this: 0.37 × 8.0 V ≈ 3.0 V, reached at about 2.8 ms. For charging, τ is the time to reach about 63 percent of the final value; 37 percent is what remains of a discharging capacitor's charge after one time constant.
C6.0 × 10⁻³ sCorrect The dashed line shows that ΔVC approaches 8.0 V, not the battery's 12 V, because of the other resistors. At t = τ a charging capacitor has reached about 63 percent of its final value: 0.63 × 8.0 V ≈ 5.1 V. The curve reaches 5.1 V at t = 6.0 ms, so τ = 6.0 × 10⁻³ s.
D1.8 × 10⁻² s A student who takes the final value to be the battery's emf picks this: 0.63 × 12 V ≈ 7.6 V, which the curve reaches only at about 18 ms. The graph shows that this capacitor approaches 8.0 V, and the 63 percent is of that final value.
Working The dashed line gives the final ΔVC = 8.0 V (less than 12 V because of the other resistors). At t = τ, ΔVC = (1 − e−1)(8.0 V) ≈ 0.63 × 8.0 V ≈ 5.1 V. The curve reaches 5.1 V at t = 6.0 ms, so τ = 6.0 × 10⁻³ s. (Curve drawn as ΔVC = (8.0 V)(1 − e−t/6.0 ms).)
An uncharged 470 μF capacitor is connected in series with a 10 kΩ resistor, an open switch and an ideal 9.0 V battery. The switch is closed at t = 0. What is the charge on the capacitor at t = 4.7 s?
Answer and reasoning
A2.7 × 10⁻³ CCorrect τ = RC = (1.0 × 10⁴ Ω)(4.7 × 10⁻⁴ F) = 4.7 s, so t = 4.7 s is exactly one time constant. The final charge is Cε = (4.7 × 10⁻⁴ F)(9.0 V) ≈ 4.2 × 10⁻³ C, and after one time constant a charging capacitor has about 63 percent of it: 0.63 × 4.23 × 10⁻³ C ≈ 2.7 × 10⁻³ C.
B1.6 × 10⁻³ C A student who uses the decaying exponential for every RC quantity picks this: Cεe−1 ≈ 0.37 × 4.23 × 10⁻³ C ≈ 1.6 × 10⁻³ C. A charging capacitor's charge grows from zero as Cε(1 − e−t/τ), so after one time constant it has gained about 63 percent of its final charge; the decaying form describes the current, or a discharging capacitor.
C2.1 × 10⁻³ C A student who takes the time constant as the time to reach half the final charge picks this: 0.50 × 4.23 × 10⁻³ C ≈ 2.1 × 10⁻³ C. At t = τ the charge is 1 − e−1 ≈ 63 percent of its final value.
D4.2 × 10⁻³ C A student who takes the time constant as the time to become fully charged picks the final charge, Cε ≈ 4.2 × 10⁻³ C. At t = τ the capacitor has only about 63 percent of that charge; it approaches Cε asymptotically.
Working τ = RC = (1.0 × 10⁴ Ω)(4.7 × 10⁻⁴ F) = 4.7 s, so t = τ. Final charge Cε = (4.7 × 10⁻⁴ F)(9.0 V) = 4.23 × 10⁻³ C. q(τ) = Cε(1 − e−1) ≈ 0.63 × 4.23 × 10⁻³ C = 2.7 × 10⁻³ C.
A 20 μF capacitor is charged to a potential difference of 9.0 V and disconnected from the battery. At t = 0 it is connected across a 5.0 kΩ resistor. What is the current in the resistor at t = 0.10 s?
Answer and reasoning
A1.1 × 10⁻³ A A student who thinks a discharging capacitor keeps 63 percent after one time constant picks this: 0.63 × 1.8 × 10⁻³ A ≈ 1.1 × 10⁻³ A. After one time constant the charge, and with it ΔVC and the current, has fallen to about 37 percent of its initial value.
B9.0 × 10⁻⁴ A A student who takes the time constant as the time for the current to halve picks this. At t = τ the current is e−1 ≈ 0.37 of its initial value, not 0.50.
C1.8 × 10⁻³ A A student who thinks a discharging capacitor acts like a battery, keeping its potential difference and current steady until it is nearly empty, picks the initial current. ΔVC = q/C falls from the moment charge starts to leave, so the current falls from the start.
D6.6 × 10⁻⁴ ACorrect τ = RC = (5.0 × 10³ Ω)(2.0 × 10⁻⁵ F) = 0.10 s, so t = τ. The current starts at ΔV₀/R = 1.8 × 10⁻³ A and decays as I₀e−t/τ, because I = ΔVC/R and ΔVC = q/C falls with the charge. After one time constant it is about 37 percent of I₀: 6.6 × 10⁻⁴ A.
Working τ = RC = (5.0 × 10³ Ω)(2.0 × 10⁻⁵ F) = 0.10 s, so t = τ. I₀ = ΔV₀/R = 9.0 V/5.0 × 10³ Ω = 1.8 × 10⁻³ A. I(τ) = I₀e−1 ≈ 0.37 × 1.8 × 10⁻³ A = 6.6 × 10⁻⁴ A.
A charged capacitor begins discharging through a resistor at t = 0. What fraction of its initial charge remains on the capacitor at a time equal to two time constants after t = 0?
Answer and reasoning
A0.25 A student who takes the time constant as the time for the charge to halve picks this: (1/2)² = 0.25. Each time constant leaves e−1 ≈ 0.37 of the charge, not one half.
B0.14Correct A discharging capacitor's charge is q = Q₀e−t/τ, so each time constant multiplies the charge by e−1 ≈ 0.37. After two time constants, q/Q₀ = e−2 ≈ 0.37 × 0.37 ≈ 0.14.
C0.40 A student who thinks a discharging capacitor keeps 63 percent of its charge in each time constant picks this: 0.63² ≈ 0.40. After one time constant about 37 percent remains; 63 percent is the fraction a charging capacitor has gained.
D0.00 A student who takes the time constant as the time to discharge fully picks this, since 2τ is longer than τ. The charge decreases asymptotically: after two time constants about 14 percent remains.
Working q = Q₀e−t/τ; at t = 2τ, q/Q₀ = e−2 ≈ 0.14 (0.37 of 0.37).
In the circuit shown, the battery is ideal and capacitor C is initially uncharged. Switch S is closed at t = 0. I₁, I₂ and I₃ are the magnitudes of the currents in R₁, R₂ and R₃ immediately after S is closed. Which ranking of the currents is correct?
Answer and reasoning
AI₁ = I₂ > I₃ = 0 A student who thinks an uncharged capacitor acts like a break when the switch is first closed picks this, with no current in R₃ and R₁ and R₂ in series. The uncharged capacitor has no potential difference across it, so charge flows onto its plates at once: it acts like a wire.
BI₁ > I₂ > I₃ > 0Correct Just after S closes, the capacitor is uncharged, so ΔVC = q/C = 0 and it acts like a wire: R₃ is then simply in parallel with R₂. The parallel branches have the same potential difference, so the smaller resistance carries more current: I₂ = 2.0 mA and I₃ = 1.0 mA. Both currents pass through R₁, so I₁ = 3.0 mA. Hence I₁ > I₂ > I₃ > 0.
CI₁ = I₂ = I₃ = 0 A student who thinks nothing changes at the instant the switch closes, so every current starts at zero and builds up, picks this. The battery sets up potential differences across the resistors as soon as the circuit is completed, so the currents are nonzero at once; what changes gradually is the charge on the capacitor.
DI₁ > I₂ = I₃ > 0 A student who thinks the current divides equally at a junction picks this, giving R₂ and R₃ 1.5 mA each. Branches in parallel have the same potential difference, so the branch with less resistance, R₂, carries more current.
Working At t = 0, q = 0 so ΔVC = 0: the capacitor acts like a wire and R₃ is in parallel with R₂. R₂ ∥ R₃ = (3.0)(6.0)/(9.0) kΩ = 2.0 kΩ; Req = 4.0 kΩ; I₁ = 12 V/4.0 kΩ = 3.0 mA. The parallel pair has (3.0 mA)(2.0 kΩ) = 6.0 V: I₂ = 2.0 mA, I₃ = 1.0 mA. I₁ > I₂ > I₃ > 0.
An uncharged capacitor of capacitance C, a resistor of resistance R, an open switch and an ideal battery of emf ε are connected in series. The switch is closed at t = 0. Immediately after the switch is closed, at what rate is the potential difference across the capacitor increasing?
Answer and reasoning
A0 A student who thinks a quantity that is zero cannot be changing picks this. ΔVC is zero at t = 0, but charge is arriving at the rate ε/R, so ΔVC is increasing at its fastest rate then.
Bε/(2RC) A student who splits ε equally between the resistor and the capacitor picks this, giving I = ε/(2R) and a rate of ε/(2RC). At t = 0 the capacitor has no potential difference, so the resistor has all of ε.
CεC/R A student who knows that ΔVC starts to rise at the rate ε/τ, but uses τ = R/C, picks this. The time constant is RC, so the initial rate is ε/(RC); εC/R does not even have units of volts per second.
Dε/(RC)Correct At t = 0 the capacitor is uncharged, so ΔVC = q/C = 0 and the resistor has the full ε: I₀ = ε/R. Since ΔVC = q/C, its rate of change is (1/C)(dq/dt) = I₀/C = ε/(RC). This is the steepest the ΔVC(t) graph ever is.
Working At t = 0, q = 0 so ΔVC = 0 and ΔVR = ε: I₀ = ε/R. ΔVC = q/C, so dΔVC/dt = (1/C)(dq/dt) = I₀/C = ε/(RC).
An uncharged 5.0 μF capacitor is charged through a 4.0 kΩ resistor by an ideal 12 V battery, all connected in series. At what rate is energy being stored in the capacitor at the instant when the potential difference across the capacitor is 8.0 V?
Answer and reasoning
A8.0 × 10⁻³ WCorrect By the loop rule the resistor has 12 V − 8.0 V = 4.0 V, so I = 4.0 V/4.0 × 10³ Ω = 1.0 × 10⁻³ A. The stored energy is UC = q²/(2C), so dUC/dt = (q/C)(dq/dt) = ΔVC I = (8.0 V)(1.0 × 10⁻³ A) = 8.0 × 10⁻³ W.
B1.6 × 10⁻² W A student who gives the resistor the capacitor's 8.0 V picks this: I = 2.0 × 10⁻³ A and (8.0 V)(2.0 × 10⁻³ A) = 1.6 × 10⁻² W. The resistor has what is left of the emf, 4.0 V, so the current is 1.0 × 10⁻³ A.
C2.4 × 10⁻² W A student who thinks the current stays at its initial value, ε/R = 3.0 × 10⁻³ A, picks this: (8.0 V)(3.0 × 10⁻³ A) = 2.4 × 10⁻² W. As ΔVC rises the current falls; at ΔVC = 8.0 V it is 1.0 × 10⁻³ A.
D1.2 × 10⁻² W A student who thinks all the energy the battery supplies is stored in the capacitor picks εI = (12 V)(1.0 × 10⁻³ A) = 1.2 × 10⁻² W. Part of it, I²R = 4.0 × 10⁻³ W, is dissipated in the resistor, leaving 8.0 × 10⁻³ W stored.
Working ΔVR = 12 V − 8.0 V = 4.0 V; I = 4.0 V/4.0 × 10³ Ω = 1.0 × 10⁻³ A. dUC/dt = d(q²/2C)/dt = (q/C)(dq/dt) = ΔVC I = (8.0 V)(1.0 × 10⁻³ A) = 8.0 × 10⁻³ W.
An uncharged capacitor is charged through a resistor by an ideal battery. A student claims that the rate at which energy is stored in the capacitor is zero at the instant charging begins, is greater than zero later, and approaches zero after a long time. Which reasoning correctly supports the claim?
Answer and reasoning
AThe current stays at ε/R while ΔVC rises, so the rate grows from zero until charging stops. A student who thinks the battery drives a constant current picks this. The current is largest at the start and decreases continuously; the rate approaches zero at the end because the current approaches zero, not because charging stops abruptly.
BΔVC is zero at first, so it is not changing then; it rises later and levels off at ε. A student who thinks a quantity that is zero cannot be changing picks this. At t = 0, ΔVC is zero but rising at its fastest rate, ε/(RC); the rate of energy storage is zero then because ΔVC itself is zero, not because ΔVC is not changing.
CThe rate is IΔVC: ΔVC is zero at the start, I nears zero at the end, and both are nonzero between.Correct The stored energy is UC = q²/(2C), so its rate of change is (q/C)(dq/dt) = ΔVC I. At the start the capacitor is uncharged, so ΔVC = 0 and the rate is zero even though the current is at its largest. After a long time the current approaches zero, so the rate does too. In between, both factors are nonzero.
DThe capacitor acts as a break at first and admits charge later, until it is fully charged. A student who thinks an uncharged capacitor first acts like a break picks this. Uncharged, it acts like a wire and charge flows onto it at once; the rate of energy storage starts at zero only because ΔVC starts at zero.
An uncharged capacitor is charged through a resistor by an ideal battery, starting at t = 0. The graphs, labeled (1) to (4), show four predictions of the energy UC stored in the capacitor as a function of time t. Which graph best represents UC?
Answer and reasoning
AGraph (1) A student who thinks the stored energy changes in proportion to the charge picks this curve, which has the same shape as q(t) and is steepest at t = 0. UC is proportional to q², so it starts with zero slope and rises slowly at first.
BGraph (2)Correct UC = q²/(2C) with q = Cε(1 − e−t/τ), so UC = Ufinal(1 − e−t/τ)². Its slope, ΔVC I, is zero at t = 0 because ΔVC is zero, so the curve starts flat, then rises steeply and levels off toward its final value, approaching it without a sudden stop.
CGraph (3) A student who thinks a capacitor charges at a steady rate until it is full picks this straight-line rise that stops abruptly. The rate of energy storage, ΔVC I, changes continuously, and UC approaches its final value asymptotically.
DGraph (4) A student who thinks the stored energy keeps increasing as long as the battery is connected picks this ever-steepening curve. When the capacitor is fully charged the current in its branch is zero, so no more energy is transferred and UC levels off at (1/2)Cε².
Working UC = q²/(2C) = Ufinal(1 − e−t/τ)². dUC/dt = ΔVC I = 0 at t = 0 (ΔVC = 0), so the curve starts flat, rises steeply, then levels off toward Ufinal.
An uncharged capacitor is charged through a resistor by an ideal battery. At a time equal to two time constants after charging begins, what fraction of its final stored energy does the capacitor hold?
Answer and reasoning
A0.86 A student who thinks the stored energy is proportional to the charge picks this, the fraction the charge has reached. UC = q²/(2C), so the energy fraction is the square of the charge fraction: 0.86² ≈ 0.75.
B0.56 A student who takes the time constant as the time to reach half the final charge picks this: the charge would be 3/4 of its final value after two time constants, and (3/4)² ≈ 0.56. Each time constant multiplies the remaining shortfall in charge by e−1 ≈ 0.37, so at 2τ the charge is about 86 percent of its final value.
C1.00 A student who takes the time constant as the time to become fully charged picks this, since 2τ is longer than τ. The capacitor approaches its final charge and energy asymptotically; at 2τ it holds about 75 percent of its final energy.
D0.75Correct After two time constants q = qfinal(1 − e−2) ≈ 0.86 qfinal. The stored energy is UC = q²/(2C), proportional to q², so UC/Ufinal = (1 − e−2)² ≈ 0.75.
Working q/qfinal = 1 − e−2 ≈ 0.865. UC ∝ q², so UC/Ufinal = (1 − e−2)² ≈ 0.75.
In the circuit shown, the battery is ideal and the capacitor is initially uncharged. The circuit has been connected for a long time. What is the magnitude of the charge Q on each plate of the capacitor?
Answer and reasoning
AQ = 2Cε/3Correct After a long time the capacitor is fully charged and its branch carries no current, so the battery drives I = ε/(3R) through 2R and R in series. The capacitor is connected across the 2R resistor, so ΔVC = I(2R) = 2ε/3 and Q = CΔVC = 2Cε/3.
BQ = Cε A student who thinks a capacitor always charges until its potential difference equals the battery's emf picks this. This capacitor is connected across the 2R resistor, so its final potential difference is that resistor's, 2ε/3, while R takes the other ε/3.
CQ = Cε/2 A student who shares ε equally between the two resistors picks this: ΔVC = ε/2. The resistors carry the same current, so the larger resistance, 2R, takes the larger share, 2ε/3.
DQ = 0 A student who thinks a capacitor always acts like a wire picks this, since a wire across 2R would leave no potential difference across it. That is true only at the instant charging begins; after a long time the capacitor carries no current and has the 2R resistor's potential difference across it.
Working After a long time the capacitor is fully charged: no current in its branch. I = ε/(2R + R) = ε/(3R) through both resistors. The capacitor is in parallel with the 2R resistor: ΔVC = I(2R) = 2ε/3. Q = CΔVC = 2Cε/3.
An ideal battery, a switch S and a resistor R₁ are connected in series with a parallel combination of two branches: one branch contains a resistor R₂, and the other contains a resistor R₃ in series with an initially uncharged capacitor. S is closed and left closed for a long time. Which statement then describes the currents in the resistors?
Answer and reasoning
ANone of the resistors carries a current once C is charged. A student who thinks all current stops once the capacitor is fully charged picks this. Only the capacitor's branch stops carrying current; R₁ and R₂ still form a complete loop with the battery.
BAll three carry currents, with C's branch acting like a wire. A student who thinks a capacitor always acts like a wire picks this. It acts like a wire only while uncharged, just after S closes; fully charged, it has zero current in its branch.
CEach keeps the current it had just after S closed, as ε is steady. A student who thinks the battery drives constant currents picks this. The currents change as the capacitor charges: the current in R₃'s branch falls to zero, and the currents in R₁ and R₂ change until they are equal.
DR₁ and R₂ carry the same nonzero current, and R₃ carries no current.Correct After a long time the capacitor is fully charged and there is zero current in its branch, so R₃ carries none. R₁ and R₂ then form a single series path across the battery and carry the same current, ε/(R₁ + R₂), which is not zero.
A capacitor is charged, disconnected from the battery and then, at t = 0, connected across a resistor. Which statement correctly describes the charge q on the capacitor and the energy UC stored in it immediately after t = 0?
Answer and reasoning
ABoth begin to decrease at once, and UC falls by a larger fraction than q.Correct At t = 0 the capacitor's potential difference drives a current, ΔVC/R, through the resistor at once, so charge leaves the plates and q begins to decrease immediately; UC = q²/(2C) decreases with it. Because UC is proportional to q², a small fractional decrease in q gives about twice that fractional decrease in UC.
BBoth begin to decrease at once, and UC falls by the same fraction as q does. A student who thinks the stored energy is proportional to the charge picks this. UC = q²/(2C), so when q falls by a small fraction, UC falls by about twice that fraction.
Cq stays the same, since charge is conserved; only UC begins to decrease. A student who thinks the capacitor's charge cannot change because charge is conserved picks this. Charge is conserved, but it moves: charge flows from one plate through the resistor to the other, so the magnitude of the charge on each plate decreases, and the energy with it.
DNeither changes at first; both begin to decrease after a short delay. A student who thinks nothing changes at the instant a circuit is completed picks this. The charged capacitor has its full potential difference at t = 0, so the current is largest at that instant and q starts to fall at once.
A charged capacitor discharges through a resistor. Which statement describes how the charge on the capacitor, the potential difference across it and the current in the resistor change during the discharge?
Answer and reasoning
AΔVC and the current stay nearly steady until the charge is almost gone, then drop. A student who thinks a discharging capacitor acts like a battery picks this. A capacitor's potential difference is q/C, so it falls as soon as charge leaves, and the current, ΔVC/R, falls with it.
BAll three decrease at steady rates and all reach zero together at a definite time. A student who thinks a capacitor discharges at a steady rate until it is empty picks this. The rate of discharge is proportional to the charge remaining, so it slows as the charge falls, and the quantities approach zero asymptotically.
CAll three decrease, quickly at first and then more slowly, toward zero.Correct The current is ΔVC/R and ΔVC = q/C, so the rate at which charge leaves, |dq/dt| = q/(RC), is proportional to the charge that remains. All three quantities therefore decrease as e−t/(RC): fastest at the start, then more and more slowly, approaching zero, the steady state.
DEach stays at its initial value briefly, and then all three begin to decrease. A student who thinks nothing changes at the instant a circuit is completed picks this. The current is largest at the first instant, ΔVC/R, so the charge starts to fall immediately.
In the circuit shown, the battery is ideal. Switch S has been closed for a time much longer than the time constant and is then opened at t = 0. What is the magnitude of the current in R₂ immediately after S is opened?
Answer and reasoning
A3.0 mA A student who thinks the capacitor charged to the battery's full 12 V picks this: 12 V/4.0 kΩ = 3.0 mA. In the steady state the capacitor had R₂'s potential difference, 8.0 V, because R₁ had the other 4.0 V.
B2.0 mACorrect While S was closed for a long time the capacitor was fully charged, so its branch carried no current and I = 12 V/6.0 kΩ = 2.0 mA passed through R₁ and R₂. The capacitor, in parallel with R₂, had ΔVC = (2.0 mA)(4.0 kΩ) = 8.0 V. Once S opens, R₁ is cut off, and the capacitor discharges through R₂ alone: I = 8.0 V/4.0 kΩ = 2.0 mA.
C1.3 mA A student who includes both resistors in the discharge path picks this: 8.0 V/6.0 kΩ ≈ 1.3 mA. With S open, R₁ leads only to the open switch and carries no current; the capacitor discharges through R₂ alone.
D0.0 mA A student who thinks there can be no current once the battery is switched out picks this. The charged capacitor has 8.0 V across it and drives a current through R₂ until it has discharged.
Working Before t = 0 (steady state): no current in the capacitor's branch; I = 12 V/(2.0 kΩ + 4.0 kΩ) = 2.0 mA through R₁ and R₂; ΔVC = ΔVR₂ = (2.0 mA)(4.0 kΩ) = 8.0 V. After S opens: R₁ leads only to the open switch; C discharges through R₂ alone: I = 8.0 V/4.0 kΩ = 2.0 mA.
An uncharged capacitor is charged through a resistor by an ideal battery, and the time constant of the circuit is τ. How long after charging begins is the charge on the capacitor within 1 percent of its final value, so that the capacitor can be modeled as fully charged to that precision?
Answer and reasoning
A0.99τ A student who thinks the capacitor charges at a steady rate, its initial rate, and so is full at t = τ, picks this: 99 percent at 0.99τ. The charging rate falls as the capacitor charges, so the last few percent take several more time constants.
B6.64τ A student who takes the time constant as the time for the shortfall to halve picks this: going from 100 percent to 1 percent takes log₂100 ≈ 6.64 halvings. Each time constant multiplies the shortfall by e−1 ≈ 0.37, which is faster than halving.
C4.61τCorrect The charge is q = qfinal(1 − e−t/τ), so the shortfall from the final value is qfinal e−t/τ. It falls to 1 percent when e−t/τ = 0.01, that is at t = τ ln 100 ≈ 4.61τ. This is why, after several time constants, the capacitor can be modeled as fully charged, with zero current in its branch.
D1.00τ A student who takes the time constant as the time to become fully charged picks this. After one time constant the charge is only about 63 percent of its final value.
Working q = qfinal(1 − e−t/τ); shortfall qfinal e−t/τ = 0.01 qfinal ⇒ t = τ ln 100 ≈ 4.61τ.
A capacitor of capacitance C is charged, disconnected from the battery and then, at t = 0, connected across a resistor of resistance R. At what time t is the energy stored in the capacitor half of its value at t = 0?
Answer and reasoning
ARC ln 2 A student who takes the stored energy to fall by the same fraction as the charge picks this, the time at which the charge is half of Q₀. UC is proportional to q², so when the charge is half of Q₀ the energy is already a quarter of U₀; the energy halves earlier, at (RC ln 2)/2.
B(RC ln 2)/2Correct During the discharge q = Q₀e−t/(RC). The stored energy is UC = q²/(2C), so UC = U₀e−2t/(RC): the energy falls twice as fast, in the exponent, as the charge. UC = U₀/2 when 2t/(RC) = ln 2, that is t = (RC ln 2)/2 ≈ 0.35RC, when the charge is still about 71 percent of Q₀.
CRC/2 A student who takes the time constant RC as the time for the charge to fall to half picks this: with the charge halving every RC, the energy, proportional to q², halves every RC/2. In a time RC the charge falls to e−1, about 37 percent of Q₀, not to half.
D(1 − 1/√2)RC A student who thinks the charge falls at a steady rate, equal to its initial rate Q₀/(RC), picks this: q = Q₀(1 − t/(RC)), and UC = U₀/2 when q = Q₀/√2. The current, and so the rate at which the charge falls, decreases as the capacitor discharges, so the charge decays exponentially and the energy takes longer to halve.
Working Loop rule for the discharge: q/C + R(dq/dt) = 0, so q = Q₀e−t/(RC). UC = q²/(2C) = U₀e−2t/(RC). UC = U₀/2 when e−2t/(RC) = 1/2, so t = (RC ln 2)/2 ≈ 0.35RC. Distractors (sympy-checked): energy taken to fall like the charge, e−t/(RC) = 1/2 → RC ln 2 ≈ 0.69RC; τ taken as the charge half-life, q = Q₀2−t/(RC), U ∝ 4−t/(RC) = 1/2 → RC/2; charge falling at its initial rate, q = Q₀(1 − t/(RC)), (1 − t/(RC))² = 1/2 → (1 − 1/√2)RC ≈ 0.29RC.
An uncharged capacitor of capacitance C is charged through a resistor of resistance R by an ideal battery of emf ε, all connected in series. At the instant when the energy stored in the capacitor is 64 percent of its final value, what is the current in the resistor?
Answer and reasoning
A0.36 ε/R A student who takes the stored energy to be the same fraction of its final value as the charge picks this: q = 0.64Cε, ΔVC = 0.64ε and I = (ε − 0.64ε)/R. UC is proportional to ΔVC², so the energy is 64 percent of its final value when ΔVC is 80 percent of ε.
B0.80 ε/R A student who finds ΔVC = 0.80ε correctly but gives the resistor the same potential difference as the capacitor in series with it picks this: 0.80ε/R. The loop rule makes the two potential differences add to ε, so the resistor has 0.20ε.
C0.20 ε/RCorrect The stored energy is UC = CΔVC²/2 and its final value is Cε²/2, so 64 percent of the final energy means ΔVC² = 0.64ε², that is ΔVC = 0.80ε. By the loop rule the resistor has the rest of the emf, ε − 0.80ε = 0.20ε, so I = 0.20 ε/R.
D1.00 ε/R A student who thinks the battery keeps the current at its initial value, ε/R, until the capacitor is full picks this. As ΔVC rises, the potential difference left for the resistor, ε − ΔVC, falls, and the current falls with it.
Working Final energy Uf = Cε²/2. UC = CΔVC²/2 = 0.64(Cε²/2) → ΔVC = √0.64 ε = 0.80ε. Loop rule: ε − IR − ΔVC = 0 → I = (ε − 0.80ε)/R = 0.20 ε/R. (Consistent with the time solution: (1 − e−t/(RC))² = 0.64 at t = RC ln 5, where I = (ε/R)e−t/(RC) = ε/(5R).) Distractors (sympy-checked): energy fraction taken as the charge fraction, ΔVC = 0.64ε → 0.36 ε/R; resistor given the capacitor's ΔV → 0.80 ε/R; current constant at its initial value → 1.00 ε/R.
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
