4 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 4
The diagram shows three identical capacitors connected between terminals X and Y. What is the equivalent capacitance of the combination between X and Y?
Answer and reasoning
A4.5 μF A student who combines capacitors by the resistor rules picks this: (3.0 × 3.0)/(3.0 + 3.0) = 1.5 μF for the parallel pair, plus 3.0 μF for C₁ in series, gives 4.5 μF. For capacitors the rules are the other way round: capacitances in parallel add, and capacitances in series combine by their reciprocals.
B2.0 μFCorrect C₂ and C₃ are joined between the same two points, so they are in parallel and their capacitances add: 3.0 μF + 3.0 μF = 6.0 μF. All the charge on C₁ must go to that pair, so C₁ is in series with it: 1/Ceq = 1/(3.0 μF) + 1/(6.0 μF) = 1/(2.0 μF), giving Ceq = 2.0 μF, less than the 3.0 μF of C₁.
C9.0 μF A student who adds all the capacitances, whatever the connection, picks this: 3.0 + 3.0 + 3.0 = 9.0 μF. Only C₂ and C₃ are in parallel; C₁ is in series with that pair, which makes the equivalent capacitance smaller than C₁ alone, not larger.
D3.0 μF A student who thinks the smaller part of a series connection limits it, so that Ceq equals the smaller of C₁ (3.0 μF) and the parallel pair (6.0 μF), picks this. The equivalent capacitance of capacitors in series is less than the smallest: 1/Ceq = 1/3.0 + 1/6.0 gives 2.0 μF.
Working C₂ and C₃ in parallel: 3.0 μF + 3.0 μF = 6.0 μF. C₁ in series with that pair: 1/Ceq = 1/(3.0 μF) + 1/(6.0 μF) = 0.50 μF⁻¹, so Ceq = 2.0 μF.
Two capacitors of different capacitance are connected in series with a battery and become charged. Which reasoning correctly explains why the two capacitors carry the same magnitude of charge on their plates?
Answer and reasoning
AThe plates joined by the wire between the capacitors form an isolated conductor, so they hold +Q and −Q.Correct The inner plates of the two capacitors and the wire joining them are isolated from the rest of the circuit and start with zero net charge. Charge is conserved, so when −Q collects on one inner plate, +Q is left on the other. Each capacitor therefore has +Q and −Q on its plates.
BThe same amount of charge flows across the gap in each capacitor, so each one ends up holding that amount. A student who thinks charge crosses the gap between the plates picks this. The gap is an insulator; no charge crosses it. Charge flows onto one plate and off the other, and the equal charges follow from conservation of charge on the isolated inner conductor.
CThe battery creates a fixed amount of charge and gives equal amounts of it to each of the two capacitors. A student who thinks a battery creates charge picks this. The battery only moves charge that is already in the wires and plates; the equal charges result from conservation of charge on the isolated conductor between the capacitors.
DCapacitors connected to one battery carry equal charges, whatever the connection, series or parallel. A student who thinks capacitors on one battery carry equal charges whatever the connection picks this. Capacitors in parallel have equal potential differences and carry charges in proportion to their capacitances; equal charge is special to the series connection.
In the circuit shown, the capacitors are initially uncharged, and switch S is then closed. What is the time constant of the circuit?
Answer and reasoning
A3.6 × 10⁻¹ s A student who adds capacitances in series picks this: 30 μF + 60 μF = 90 μF, and (4.0 × 10³ Ω)(90 × 10⁻⁶ F) = 0.36 s. Capacitors in series combine by their reciprocals, giving 20 μF.
B8.0 × 10⁻² sCorrect The two capacitors are in series: Ceq = (30 μF × 60 μF)/(30 μF + 60 μF) = 20 μF. Then τ = Req Ceq = (4.0 × 10³ Ω)(20 × 10⁻⁶ F) = 8.0 × 10⁻² s.
C1.2 × 10⁻¹ s A student who takes the equivalent capacitance of the series pair to be the smaller capacitance, 30 μF, picks this: (4.0 × 10³ Ω)(30 × 10⁻⁶ F) = 0.12 s. The equivalent capacitance of capacitors in series is less than the smaller one: 20 μF.
D2.0 × 10⁻⁴ s A student who uses 1/30 + 1/60 = 0.050 as the equivalent capacitance in μF picks this: (4.0 × 10³ Ω)(0.050 × 10⁻⁶ F). The sum 0.050 μF⁻¹ is 1/Ceq; its reciprocal gives Ceq = 20 μF.
An uncharged capacitor, a resistor, an open switch and an ideal battery are connected in series. The switch is closed at t = 0. The graphs, labeled (1) to (4), show four predictions of the current I in the resistor as a function of time t. Which graph best represents the current?
Answer and reasoning
AGraph (1) A student who thinks a capacitor charges at a steady rate until it is full picks this straight-line fall to zero at a definite time. The current falls fastest at the start and more and more slowly later, approaching zero without a sudden stop.
BGraph (2) A student who thinks an uncharged capacitor acts like a break at first, and lets current through later, picks this rising curve. The order is the reverse: uncharged, it acts like a wire (largest current); fully charged, it acts like a break (zero current).
CGraph (3)Correct At t = 0 the uncharged capacitor acts like a wire, so the current starts at its largest value, ε/R. As the capacitor's potential difference grows, less is left across the resistor, so the current falls, quickly at first and then more and more slowly, approaching zero: graph (3).
DGraph (4) A student who thinks the battery supplies a constant current picks this. The battery keeps a fixed potential difference; as the capacitor charges, the potential difference across the resistor, and so the current, falls.
Working At t = 0, ΔVC = 0, so I = ε/R (largest). Then I = (ε − ΔVC)/R falls as ΔVC rises, more and more slowly, approaching zero: graph (3).
In preparation: 0 of 4 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 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 the plates joined by each connecting wire are isolated from the rest of the circuit. For capacitors in series, 1/Ceq,s = Σi (1/Ci): the reciprocals add, and Ceq,s is the reciprocal of that sum.
Series Ceq is less than the smallest capacitance
Capacitors in series all carry the same charge Q, but the potential difference across the whole group is larger than that across any one of them, so Ceq,s = Q/ΔVtotal is less than the smallest individual capacitance.
Capacitors in parallel
Capacitors connected between the same two points, so each has the same potential difference across it. For capacitors in parallel, Ceq,p = Σi Ci, and the total charge is the sum of the charges on the individual capacitors.
Students often think Capacitances simply add whatever the connection, because more capacitors can always store more charge. In fact No. Capacitors connected in parallel give a larger equivalent capacitance than any one of them, but capacitors connected 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 No. The equivalent capacitor stores the TOTAL charge that the battery moves for the group. In parallel, that total is shared among the capacitors; only in series does each capacitor carry the same charge as the equivalent capacitor.
11.8.A.2 Charge on capacitors in series Fix
Charge on capacitors in series
The two plates joined by the wire between two capacitors in series, together with that wire, form an isolated conductor that starts with zero net charge. Conservation of charge then requires +Q on one of those plates and −Q on the other, so every capacitor in the series carries the same magnitude of charge Q on each plate.
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 the same charge Q, so ΔVi = Q/Ci: the capacitor with the SMALLER capacitance has the larger potential difference across it.
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. Elements in series share the battery's potential difference: by the loop rule, their potential differences add up to the battery's emf. An element has the full emf of an ideal battery across it only if it is connected directly across the battery or if every element in series with it has zero potential difference, as for a resistor in series with an uncharged capacitor, or a fully charged capacitor in series with a resistor that carries no current.
11.8.B.1 RC circuit Fix
RC circuit
A circuit that contains resistors and capacitors. When it is switched on or off, the charge on each capacitor, the potential differences and the currents change over a time interval set by the time constant, rather than instantly.
Time constant, τ
A measure of how quickly a capacitor in an RC circuit charges or discharges, defined as τ = Req Ceq, where Req and Ceq are the equivalent resistance and capacitance of the charging or discharging path. Unit: second (s); 1 Ω·F = 1 s.
Time constant for charging
For a capacitor charging from zero, τ is the time for the charge to rise to approximately 63 percent of its final value. The charge keeps rising after t = τ, more and more slowly, toward its final value.
Time constant for discharging
For a fully charged capacitor discharging, τ is the time for the charge to fall to approximately 37 percent of its initial value, that is, to lose approximately 63 percent of it.
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 Yes. The time constant is τ = Req Ceq, so it is proportional to both the equivalent resistance and the equivalent capacitance: a larger capacitance takes longer to charge through the same resistance.
Students often think Adding any resistor to a circuit increases the equivalent resistance, because there is more resistance in the circuit. In fact No. A resistor added in series increases the equivalent resistance, but a resistor added in parallel gives charge another path and decreases it.
11.8.B.2 Steady state Fix
Steady state
The condition an RC circuit approaches a long time after a change: the charge on each capacitor, the potential differences and the currents no longer change with time.
Uncharged capacitor at switch-on
An uncharged capacitor has zero potential difference across it, so immediately after it is placed in a circuit it acts like a wire: charge flows onto one plate and off the other freely, and the current in its branch is set by the rest of the circuit.
Charging: how the quantities change together
As a capacitor charges, the potential difference across it grows (ΔVC = Q/C), so the charge on its plates and its stored energy, UC = (1/2)QΔVC, grow. By the loop rule, less of the battery's potential difference is left for the resistance in series with it, so the current in its branch falls.
Asymptotic approach to steady state
During charging or discharging, the charge, the potential difference across the capacitor and the current in its branch change quickly at first and then more and more slowly; these quantities and the stored energy all approach their steady-state values gradually, without a sudden stop.
Fully charged capacitor
A long time after charging begins, the potential difference across the capacitor reaches its maximum and there is zero current in the capacitor's branch; the branch then acts as a break in the circuit. The maximum ΔVC equals the potential difference across whatever the capacitor's branch is connected between.
Start of discharging
When a charged capacitor is connected across a resistor, charge begins to flow off its plates at once, so the charge on the plates and the stored energy begin to decrease immediately; the initial current is set by the capacitor's potential difference and the resistance of the path.
Discharging
As a capacitor discharges, the charge on it, the potential difference across it and the current in its branch all decrease, more and more slowly, toward zero. The current in the capacitor's branch is in the opposite direction to the current while it was charging.
Steady-state model after t ≫ τ
After charging or discharging for times much greater than τ, the capacitor's branch may be modeled as carrying zero current (the capacitor acting as a break), with the capacitor either fully charged or fully discharged, and the rest of the circuit analyzed as a resistor circuit.
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. The battery keeps a fixed potential difference, not a fixed current. As the capacitor's potential difference grows, the potential difference across the resistance falls, so the current decreases toward zero.
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 Yes, while the capacitor's charge is changing. Immediately after switch-on, an uncharged capacitor acts like a wire, so its branch can carry a large current; the current falls as the capacitor charges and is zero only once it is fully charged.
21 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 21
Two capacitors, of capacitance C and 4C, are connected in series with an ideal battery of emf ε. After the capacitors have been charging for a long time, what is the magnitude of the charge on each plate of the 4C capacitor?
Answer and reasoning
AQ = 16Cε/5 A student who gives the larger capacitance the larger share of ε, as a larger resistance would get, picks this: ΔV = 4ε/5 across the 4C capacitor and Q = 4C × 4ε/5. With equal charges, ΔV = Q/C is largest across the SMALLER capacitance, so the 4C capacitor has only ε/5.
BQ = 5Cε A student who adds capacitances in series, as resistances in series add, picks this: Ceq = 5C and Q = 5Cε. Capacitors in series combine by their reciprocals: Ceq = 4C/5, less than either capacitance.
CQ = Cε A student who thinks the smaller capacitor limits the chain, so that Ceq equals C, picks this. The equivalent capacitance of capacitors in series is less than the smallest one: Ceq = 4C/5, giving Q = 4Cε/5.
DQ = 4Cε/5Correct The capacitors are in series, so 1/Ceq = 1/C + 1/(4C) = 5/(4C) and Ceq = 4C/5. Capacitors in series all carry the charge on the equivalent capacitor: Q = Ceq ε = 4Cε/5. As a check, ΔV across C is Q/C = 4ε/5 and across 4C is Q/(4C) = ε/5, which add up to ε.
Working Series: 1/Ceq = 1/C + 1/(4C) = 5/(4C), so Ceq = 4C/5. Each capacitor in series carries the equivalent capacitor's charge: Q = Ceq ε = 4Cε/5. (ΔV across C = 4ε/5, across 4C = ε/5; sum ε.)
A student claims that the equivalent capacitance of any set of capacitors connected in series is less than the smallest capacitance in the set. Which reasoning correctly supports the student's claim?
Answer and reasoning
AThe capacitor next to the battery fills up first and then stops any more charge from reaching the others. A student who pictures the capacitors charging one after another, the first filling before charge moves on, 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.
BThe battery's ΔV is shared equally among them, so each has less ΔV than it would if connected alone. A student who thinks the battery's potential difference is shared equally picks this. The shares are not equal: the smallest capacitor has the largest share, since ΔV = Q/C. The argument also compares each capacitor with itself alone, not the set with its smallest capacitor.
CEach carries the same charge Q, but the ΔV across the whole set exceeds that across the smallest, so Q/ΔV is smaller.Correct Ceq is the charge stored per volt across the whole set. Every capacitor in series carries the same charge Q, and the potential difference across the set is the sum of theirs, which is larger than Q/Csmallest alone. So Ceq = Q/ΔVtotal is less than Csmallest = Q/ΔVsmallest.
DLess charge reaches each capacitor further along the chain, so the set stores less than one capacitor. A student who thinks charge is stored progressively along the chain picks this. The plates between capacitors form isolated conductors, so every capacitor carries the same charge Q; nothing is lost along the chain.
Two capacitors, of capacitance C and 2C, are connected in parallel with each other across an ideal battery of emf ε. After they have been charging for a long time, what is the magnitude of the charge on each plate of the capacitor C?
Answer and reasoning
AQ = CεCorrect Both capacitors are connected directly across the battery, so each has the full potential difference ε across it. Capacitor C therefore carries Q = Cε. The 2C capacitor carries 2Cε, and together they hold 3Cε = Ceq ε, with Ceq = 3C.
BQ = Cε/2 A student who thinks capacitors in parallel share the battery's potential difference picks this, giving each ε/2 and so Q = Cε/2. Capacitors in parallel are joined between the same two points, so each has the full ε.
CQ = 3Cε/2 A student who thinks capacitors in parallel carry equal charges picks this: the total, Ceq ε = 3Cε, shared equally gives 3Cε/2 each. Each capacitor has the same ΔV, so the charges are in proportion to the capacitances: Cε and 2Cε.
DQ = 3Cε A student who gives each capacitor the charge of the equivalent capacitor picks this: Ceq = 3C, so Q = 3Cε. That is the total charge on the pair; capacitor C holds only its own part, Cε.
Working Parallel: each capacitor has ΔV = ε. QC = Cε (and Q2C = 2Cε; total 3Cε = Ceq ε with Ceq = 3C).
Four identical capacitors, each of capacitance C, are first connected all in parallel, giving an equivalent capacitance CX. They are then reconnected all in series, giving an equivalent capacitance CY. How does CX compare with CY?
Answer and reasoning
ACX is 1/16 times CY. A student who uses the resistor rules for capacitors picks this: CX = C/4 in parallel and CY = 4C in series. For capacitors it is the reverse: capacitances in parallel add, so CX is the larger.
BCX is 4 times CY. A student who takes the equivalent capacitance of the series set to be the smallest capacitance, C, picks this: 4C ÷ C = 4. The equivalent capacitance of capacitors in series is less than the smallest one: CY = C/4.
CCX is the same as CY. A student who adds the capacitances whatever the connection picks this, getting 4C both times. Only the parallel connection adds capacitances; the series connection gives C/4.
DCX is 16 times CY.Correct In parallel the capacitances add: CX = 4C. In series the reciprocals add: 1/CY = 4/C, so CY = C/4. The ratio is CX/CY = 4C ÷ (C/4) = 16.
Working CX = C + C + C + C = 4C. 1/CY = 4 × (1/C), so CY = C/4. CX/CY = 4C ÷ (C/4) = 16.
The circuit shown contains an ideal battery and three capacitors, initially uncharged. After the capacitors have been charging 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
AQ₁ > Q₃ > Q₂ A student who thinks elements in series have the same potential difference picks this: with equal ΔV, Q = CΔV makes the charges follow the capacitances, 3.0 > 2.0 > 1.0 μF. Capacitors in series share the charge, not the potential difference; C₂, the smallest, has the largest ΔV.
BQ₁ = Q₂ = Q₃Correct The three capacitors form a single loop with the battery, so they are in series. The right plate of C₁ and the top plate of C₂, with the wire joining them, are isolated from everything else and start neutral, so if one has −Q the other has +Q; the same holds between C₂ and C₃. All three carry the same charge, whatever their capacitances.
CQ₁ > Q₂ > Q₃ A student who thinks charge is stored progressively along the chain, starting at the positive terminal, picks this. The conductors between the capacitors are isolated and neutral, so the charge cannot shrink from one capacitor to the next.
DQ₁ = Q₃ > Q₂ A student who thinks C₂, which is not wired directly to the battery, receives less charge picks this. When C₁ and C₃ charge, charge separates in the isolated conductors on either side of C₂, so C₂ carries the same charge as the others.
Working All three are in one loop: series. Each inner conductor (C₁–C₂ and C₂–C₃ plates with their wire) is isolated and neutral, so Q₁ = Q₂ = Q₃ = Ceq ε, with 1/Ceq = 1/3.0 + 1/1.0 + 1/2.0 μF⁻¹. (Equal-ΔV belief gives Q ∝ C: Q₁ > Q₃ > Q₂.)
Two capacitors, of capacitance C and 3C, are connected in series with an ideal battery of emf ε. After they have been charging for a long time, what is the potential difference across the capacitor C?
Answer and reasoning
AΔV = ε/4 A student who gives the larger capacitance the larger share of ε, as resistors in series share it, picks this: C/(C + 3C) = 1/4 of ε for capacitor C. With equal charges, ΔV = Q/C is larger across the SMALLER capacitance.
BΔV = ε A student who thinks every element connected to the battery has its full emf across it picks this. The two capacitors are in series, so their potential differences must add up to ε; each has only part of it.
CΔV = 3ε/4Correct In series both capacitors carry the same charge, Q = Ceq ε with Ceq = (C × 3C)/(C + 3C) = 3C/4, so Q = 3Cε/4. Across capacitor C, ΔV = Q/C = 3ε/4; across 3C it is Q/(3C) = ε/4. The smaller capacitance has the larger share, and the two add up to ε.
DΔV = 4ε A student who adds capacitances in series picks this: Ceq = 4C, Q = 4Cε and ΔV = Q/C = 4ε. That is larger than the battery's emf, which the loop rule forbids; capacitors in series combine by their reciprocals.
Working Ceq = (C)(3C)/(C + 3C) = 3C/4. Q = Ceq ε = 3Cε/4 on each capacitor. ΔVC = Q/C = 3ε/4 (ΔV3C = Q/(3C) = ε/4; sum = ε).
The diagram shows three circuits, each with an identical ideal battery. Every resistor has resistance R and every capacitor has capacitance C. τ₁, τ₂ and τ₃ are the time constants of Circuits 1, 2 and 3. Which ranking of the time constants is correct?
Answer and reasoning
Aτ₃ > τ₁ > τ₂Correct τ = Req Ceq. Circuit 1: τ₁ = RC. Circuit 2: two capacitors in series give C/2, so τ₂ = RC/2. Circuit 3: two resistors in series give 2R and two capacitors in parallel give 2C, so τ₃ = 4RC.
Bτ₂ > τ₁ = τ₃ A student who combines capacitors by the resistor rules picks this: 2C in series gives τ₂ = 2RC, and C/2 in parallel gives τ₃ = (2R)(C/2) = RC. Capacitors in series give C/2 and capacitors in parallel give 2C.
Cτ₃ > τ₁ = τ₂ A student who thinks only the resistance sets the charging time picks this, comparing R, R and 2R. The capacitance matters too: τ = Req Ceq, so Circuit 2's smaller Ceq gives it the shortest time constant.
Dτ₃ > τ₂ > τ₁ A student who adds capacitances whatever the connection picks this: 2C in Circuit 2 gives τ₂ = 2RC, and (2R)(2C) gives τ₃ = 4RC. Capacitors in series give C/2, so τ₂ = RC/2 is the smallest.
In trial 1, an uncharged capacitor of capacitance C is charged through a resistor of resistance R by an ideal battery, and its charge takes a time t₁ to reach 63 percent of its final value. In trial 2, the resistor is replaced by two resistors, each of resistance R, connected in parallel, and the capacitor by two uncharged capacitors, each of capacitance C, connected in parallel. How long does the charge take to reach 63 percent of its final value in trial 2?
Answer and reasoning
At₁/4, one quarter of the time A student who combines capacitors by the resistor rules picks this: C/2 for the parallel pair, so τ = (R/2)(C/2) = RC/4. Capacitances in parallel add, giving 2C and τ = RC.
Bt₁/2, half the time taken A student who thinks only the resistance sets the charging time picks this: Req halves, so the time halves. The capacitance doubles as well, and τ = Req Ceq returns to RC.
C4t₁, four times the time before A student who thinks adding a resistor always increases the resistance picks this: 2R and 2C give 4RC. A second resistor in parallel gives charge another path, so Req = R/2.
Dt₁, the same as in trial 1Correct The time to reach 63 percent of the final charge is the time constant. Two resistors R in parallel give Req = R/2, and two capacitors C in parallel give Ceq = 2C, so τ = (R/2)(2C) = RC: the same as before.
The graph shows the charge Q on a capacitor as a function of time t while it charges through a resistor from an ideal battery, starting from zero at t = 0. The dashed line shows the final value of the charge. Which of the marked points W, X, Y and Z is at a time equal to one time constant after charging began?
Answer and reasoning
APoint X A student who takes τ as the time to reach half the final charge picks this point, at 25 μC. After one time constant the charge is approximately 63 percent of its final value, about 32 μC, which the curve reaches later, at point Y.
BPoint W A student who uses 37 percent for a charging capacitor picks this point, at about 19 μC (0.37 × 50 μC). For charging, τ is the time to reach approximately 63 percent of the final value; 37 percent is what remains of a discharging capacitor's charge.
CPoint YCorrect For a charging capacitor, τ is the time for the charge to reach approximately 63 percent of its final value: 0.63 × 50 μC ≈ 32 μC. Point Y is the point at about 32 μC, at t = 3.0 ms, so τ = 3.0 ms.
DPoint Z A student who takes τ as the time to become fully charged picks this point, where the curve has almost reached the dashed line. The charge approaches 50 μC more and more slowly; τ is the much shorter time to reach approximately 63 percent of it.
Working Final charge 50 μC; 63 percent of it is 0.63 × 50 μC ≈ 32 μC. The marked point at about 32 μC is Y (t = 3.0 ms), so t = τ at Y. (W ≈ 19 μC = 37 percent; X = 25 μC = 50 percent; Z at 15 ms, almost 50 μC.)
A 2.0 μF capacitor is charged to a potential difference of 6.0 V. At t = 0 it is connected across a 1.0 MΩ resistor and begins to discharge. What is the magnitude of the charge on each plate of the capacitor at t = 2.0 s?
Answer and reasoning
A4.4 μCCorrect The initial charge is Q₀ = CΔV = (2.0 μF)(6.0 V) = 12 μC, and τ = RC = (1.0 × 10⁶ Ω)(2.0 × 10⁻⁶ F) = 2.0 s. At t = τ a discharging capacitor keeps approximately 37 percent of its initial charge: 0.37 × 12 μC ≈ 4.4 μC.
B7.6 μC A student who thinks a discharging capacitor still holds 63 percent after one time constant picks this: 0.63 × 12 μC ≈ 7.6 μC. After one time constant about 63 percent has been LOST, leaving about 37 percent, 4.4 μC.
C6.0 μC A student who takes τ as the time for the charge to halve picks this: 12 μC/2 = 6.0 μC. After one time constant approximately 37 percent remains, not 50 percent.
D0.0 μC A student who thinks τ is the time to discharge completely picks this. At t = τ the capacitor still holds approximately 37 percent of its initial charge; it approaches zero only after many time constants.
Working Q₀ = CΔV = (2.0 × 10⁻⁶ F)(6.0 V) = 1.2 × 10⁻⁵ C = 12 μC. τ = RC = (1.0 × 10⁶ Ω)(2.0 × 10⁻⁶ F) = 2.0 s, so t = 2.0 s = τ. Q ≈ 0.37 × 12 μC ≈ 4.4 μC.
In the circuit shown, bulbs A and B are identical, the battery is ideal and capacitor C is initially uncharged. Which statement describes the bulbs immediately after switch S is closed?
Answer and reasoning
AA and B are lit and equally bright. A student who thinks a capacitor never lets current into its branch picks this, treating A and B as a series pair. That describes the circuit a long time later. At first, charge flows freely onto and off the uncharged capacitor's plates, and that path bypasses B.
BA and B are lit, B more dimly. A student who thinks charge passes through a capacitor as through a resistor picks this, sharing the current between B and C. At first the uncharged capacitor has zero potential difference, like a wire, so B has none either and is not lit.
CNeither bulb is lit at first. A student who thinks the current needs time to build up picks this. The current starts as soon as the switch closes and is largest at that instant, because the uncharged capacitor acts like a wire.
DA is lit, and B is not lit at all.Correct The uncharged capacitor has zero potential difference across it, so at that instant it acts like a wire connected across B. B then has no potential difference across it and no current, so it is not lit, while A has the battery's full potential difference across it and is lit.
Working At t = 0, Q = 0 so ΔVC = 0: C acts like a wire across B, so ΔVB = 0 and B is unlit; A has the full ε. (Long after: C acts as a break; A and B are in series and equally bright.)
An uncharged capacitor, a resistor, a switch and an ideal battery of emf ε are connected in series. Which statement gives the potential differences across the resistor, ΔVR, and across the capacitor, ΔVC, immediately after the switch is closed?
Answer and reasoning
AΔVR = 0, ΔVC = ε A student who thinks an uncharged capacitor acts like a break at first picks this. That describes the capacitor a long time later, when it is fully charged. With no charge on its plates, ΔVC = Q/C = 0.
BΔVR = ε, ΔVC = 0Correct The capacitor has no charge yet, so ΔVC = Q/C = 0: it acts like a wire. By the loop rule the resistor then has the battery's full emf, ΔVR = ε, and the current starts at its largest value, ε/R.
CΔVR = ε, ΔVC = ε A student who thinks every element connected to the battery gets its full emf picks this. The resistor and capacitor are in series, so by the loop rule their potential differences add up to ε; with ΔVC = 0, all of ε is across the resistor.
DΔVR = 0, ΔVC = 0 A student who thinks nothing happens until the current has built up picks this. The loop rule holds at every instant, ΔVR + ΔVC = ε, so both cannot be zero once the switch is closed; the current is largest at that moment.
Working At t = 0, Q = 0, so ΔVC = Q/C = 0. Loop rule: ε − ΔVR − ΔVC = 0, so ΔVR = ε and I = ε/R.
A capacitor is charged through a resistor by an ideal battery. As the capacitor charges, the current in the resistor decreases. Which statement correctly explains the decrease?
Answer and reasoning
AThe capacitor's resistance to charge crossing its gap grows larger as the capacitor fills. A student who thinks charge crosses the gap between the plates, as it passes through a resistor, picks this. No charge crosses the gap at any stage; the current falls because the capacitor's growing potential difference opposes the battery's.
BThe battery runs down as it gives up its stored charge to fill the capacitor's plates. A student who pictures the battery as a store of charge that empties picks this. An ideal battery keeps the same potential difference throughout and supplies no charge of its own; the current falls because ΔVC rises.
CThe growing ΔV across the capacitor leaves less of the battery's ΔV across the resistor.Correct By the loop rule, ε = ΔVR + ΔVC at every moment. As charge builds up on the plates, ΔVC = Q/C grows, so ΔVR = ε − ΔVC falls, and with it the current I = ΔVR/R.
DSome charge is used up in the resistor, so less charge gets through to the capacitor. A student who thinks charge is used up in the resistor picks this. Charge is conserved: all the charge that passes through the resistor reaches the capacitor's plate. The current is smaller because ΔVR = ε − ΔVC is smaller.
An uncharged capacitor is charged through a 3.0 kΩ resistor by an ideal battery of emf 9.0 V, all connected in series. What is the current in the resistor at the instant when the potential difference across the capacitor is 6.0 V?
Answer and reasoning
A3.0 mA A student who thinks the current stays at its starting value picks this: 9.0 V/3.0 kΩ = 3.0 mA, the current at t = 0. Once the capacitor has 6.0 V across it, only 3.0 V is left for the resistor.
B2.0 mA A student who thinks elements in series have the same potential difference picks this, putting 6.0 V across the resistor: 6.0 V/3.0 kΩ = 2.0 mA. In series the potential differences add up to the battery's: 3.0 V + 6.0 V = 9.0 V.
C5.0 mA A student who adds the capacitor's potential difference to the battery's picks this: (9.0 V + 6.0 V)/3.0 kΩ = 5.0 mA. While the capacitor charges, its potential difference opposes the battery's.
D1.0 mACorrect By the loop rule, the resistor's potential difference is what is left of the battery's: ΔVR = 9.0 V − 6.0 V = 3.0 V. Then I = ΔVR/R = 3.0 V/3.0 kΩ = 1.0 mA.
Working Loop rule: ΔVR = ε − ΔVC = 9.0 V − 6.0 V = 3.0 V. I = ΔVR/R = 3.0 V/(3.0 × 10³ Ω) = 1.0 × 10⁻³ A = 1.0 mA.
An uncharged capacitor is charged through a resistor by an ideal battery. At a time equal to one time constant after charging begins, the energy stored in the capacitor is approximately what percentage of its final stored energy?
Answer and reasoning
A63% A student who thinks the stored energy is proportional to the charge picks this. The potential difference grows along with the charge, so UC = Q²/(2C): 63 percent of the charge gives (0.63)² ≈ 40 percent of the energy.
B40%Correct At t = τ the charge is approximately 63 percent of its final value. Since ΔVC = Q/C, the stored energy is UC = (1/2)QΔVC = Q²/(2C), proportional to Q². So the energy is (0.63)² ≈ 0.40, about 40 percent, of its final value.
C25% A student who takes τ as the time for the charge to reach half its final value picks this: (0.5)² = 25 percent. At t = τ the charge is approximately 63 percent of its final value, which gives about 40 percent of the energy.
D14% A student who uses 37 percent for a charging capacitor picks this: (0.37)² ≈ 14 percent. For charging, the charge at t = τ is approximately 63 percent of its final value; 37 percent is what remains of a discharging capacitor's charge.
Working At t = τ, Q ≈ 0.63 Qfinal. UC = (1/2)QΔVC with ΔVC = Q/C gives UC = Q²/(2C) ∝ Q², so U/Ufinal ≈ (0.63)² ≈ 0.40 = 40%.
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. Points J, K and L are marked on the curve, and IJ, IK and IL are the currents in the resistor at those times. Which ranking of the currents is correct?
Answer and reasoning
AIJ > IK > IL > 0Correct By the loop rule, ΔVR = ε − ΔVC, so I = (ε − ΔVC)/R. The gap between the curve and the dashed ε line shrinks from J to K to L, so the current decreases; at L the curve is still below ε and still rising, so the current there is small but not zero.
BIL > IK > IJ > 0 A student who thinks the resistor has the same potential difference as the capacitor in series with it picks this, making I = ΔVC/R grow with the curve. The resistor has what is left of ε, ε − ΔVC, which shrinks as ΔVC grows.
CIJ = IK = IL > 0 A student who thinks the battery supplies a constant current picks this. A constant current would build up charge at a steady rate and make the graph a straight line; the curve flattens because the current decreases.
DIJ = IK = IL = 0 A student who thinks there is never any current in a capacitor's branch picks this. ΔVC = Q/C is rising, so charge is flowing onto the plates: there is a current in the branch until the capacitor is fully charged.
Working I = (ε − ΔVC)/R. ΔVC increases from J to K to L, so ε − ΔVC decreases: IJ > IK > IL. At L the curve is still rising and below ε, so IL > 0.
In the circuit shown, the battery is ideal and capacitor C is initially uncharged. Switch S is closed and left closed for a long time. What is then the potential difference across the capacitor?
Answer and reasoning
A9.0 V A student who thinks a capacitor always charges to the battery's full emf picks this. That holds when the capacitor is in a single loop with the battery; here it is connected across R₂ only, and R₁ takes 3.0 V of the 9.0 V.
B4.5 V A student who shares the 9.0 V equally between R₁ and the R₂–C group picks this. The shares follow the resistances: 1.5 mA through 2.0 kΩ gives 3.0 V, and through 4.0 kΩ gives 6.0 V.
C6.0 VCorrect After a long time the capacitor is fully charged and its branch carries no current, so R₁ and R₂ form a series loop with the battery: I = 9.0 V/6.0 kΩ = 1.5 mA. The capacitor is connected across R₂, so ΔVC = ΔVR₂ = (1.5 mA)(4.0 kΩ) = 6.0 V.
D0.0 V A student who thinks the capacitor still acts like a wire after a long time picks this, as if it shorted out R₂. It acts like a wire only while uncharged; fully charged, its branch carries no current and it has R₂'s 6.0 V across it.
Working Steady state: zero current in C's branch. R₁ and R₂ in series: I = 9.0 V/(2.0 kΩ + 4.0 kΩ) = 1.5 mA. C is in parallel with R₂: ΔVC = ΔVR₂ = (1.5 mA)(4.0 kΩ) = 6.0 V.
A fully charged capacitor is connected across a resistor at t = 0 and begins to discharge. Which statement correctly describes the capacitor immediately after t = 0?
Answer and reasoning
AIts charge stays the same, since charge is conserved; its energy falls. A student who reads conservation of charge as 'the charge on the plates cannot change' picks this. Conservation applies to the whole circuit: charge leaving the positive plate arrives at the negative plate, so each plate's charge falls while the total stays zero.
BIts charge and its stored energy both begin to decrease at once.Correct As soon as the path through the resistor is complete, the capacitor's potential difference drives charge off the positive plate, through the resistor, to the negative plate. The charge on the plates falls at once, and so does the stored energy, UC = (1/2)QΔVC.
CIts ΔV stays steady, as a battery's does, until it is nearly empty. A student who treats the capacitor like a battery picks this. Its potential difference is ΔVC = Q/C, so it falls as soon as charge leaves the plates, and the charge and energy fall from the first instant.
DNothing changes at first, because the current needs time to build up. A student who thinks current takes time to start picks this. The current is largest at the first instant, ΔVC/R, so the charge and energy begin to fall immediately.
In the circuit shown, switch S has been in position 1 for a long time, so capacitor C is fully charged. At t = 0 the switch is moved to position 2. Which statement describes the magnitude of the current in resistor R after t = 0?
Answer and reasoning
AIt is largest at t = 0, then decreases more and more slowly toward zero.Correct In position 2, R, C and the wire from contact 2 form a loop without the battery. At t = 0 the capacitor has its full potential difference, so the current, ΔVC/R, is largest then. As charge leaves the plates, ΔVC = Q/C falls, so the current falls, more and more slowly, toward zero.
BIt stays nearly steady, like a battery's current, until C is almost empty. A student who treats the discharging capacitor like a battery picks this. As charge leaves the plates, ΔVC = Q/C falls, so the current falls from the start; it does not stay steady.
CIt is zero, because the battery is no longer in a loop with resistor R. A student who thinks there can be no current without a battery picks this. The charged capacitor has a potential difference across it, so it drives a current through R until it has discharged.
DIt starts at zero and builds up gradually as charge begins to leave C. A student who thinks current takes time to build up after a switch is moved picks this. The fully charged capacitor has its full potential difference at t = 0, so the current is largest at that instant.
Working Position 2: loop C → R → switch → contact 2 → bottom wire → C, without the battery. I = ΔVC/R, largest at t = 0; ΔVC = Q/C falls as charge leaves, so I decreases toward zero, more and more slowly.
In the circuit shown, bulbs X and Y are identical, the battery is ideal and capacitor C is initially uncharged. Switch S is closed and left closed for a long time. How do the bulbs then compare?
Answer and reasoning
AX and Y are both lit equally. A student who thinks the capacitor still acts like a wire picks this, giving Y the same potential difference as X. That describes the first instant after the switch closes. Once C is fully charged, its branch carries no current.
BX and Y are lit, Y more dimly. A student who thinks charge flows through a capacitor as through a resistor picks this, expecting a smaller steady current through Y. No charge crosses the gap; once C is fully charged there is no current in its branch at all, so Y is not lit.
CNeither X nor Y is lit any more. A student who thinks all current stops once the capacitor is fully charged picks this. Only the capacitor's own branch stops carrying current; X's branch is a complete loop with the battery and keeps its current.
DX is lit, and Y is not lit.Correct After a long time C is fully charged and its branch carries no current, so Y, in series with C, is not lit. X is connected directly across the ideal battery in its own branch, so it keeps the full potential difference and stays lit.
Working t ≫ τ: C is fully charged and the Y–C branch carries zero current, so Y is unlit. X is directly across the ideal battery, so it has ε across it and stays lit.
An ideal battery of emf ε, a switch S and a resistor of resistance R₁ are connected in series with a parallel combination of a resistor of resistance R₂ and an initially uncharged capacitor. Switch S is closed and left closed for a long time. S is then opened, which disconnects the battery and R₁, so that the capacitor is connected only across R₂. What is the magnitude of the current in R₂ immediately after S is opened?
Answer and reasoning
Aε/(R₁ + R₂)Correct In the steady state the capacitor branch carries no current, so the current ε/(R₁ + R₂) passes through R₁ and R₂, and the capacitor has R₂'s potential difference, εR₂/(R₁ + R₂). Just after S is opened the capacitor keeps that potential difference across R₂, so the current in R₂ is ε/(R₁ + R₂), the same magnitude as just before.
Bε/R₂ A student who thinks the capacitor charges until its potential difference equals the emf picks this: ε/R₂. The capacitor is in parallel with R₂, so in the steady state it has only R₂'s potential difference, εR₂/(R₁ + R₂).
Cε/(2R₂) A student who shares the emf equally between R₁ and the parallel group gives the capacitor ε/2 and picks this. The shares depend on the resistances: the parallel group has R₂/(R₁ + R₂) of the emf in the steady state.
D0 A student who thinks there can be no current once the battery is disconnected picks this. The charged capacitor has a potential difference across R₂, so it drives a current through R₂ as it begins to discharge.
Working After a long time with S closed, the capacitor branch carries no current, so the current ε/(R₁ + R₂) passes through R₁ and R₂ in series. The capacitor is in parallel with R₂, so ΔVC = ΔVR₂ = εR₂/(R₁ + R₂). Immediately after S is opened, the capacitor still has this potential difference and it is the only element connected across R₂, so the current in R₂ is ΔVC/R₂ = ε/(R₁ + R₂). (Errors: ΔVC = ε gives ε/R₂; ΔVC = ε/2 gives ε/(2R₂); 'no current without a battery' gives 0.)
Compiled from the AP Physics 2 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
