2 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 2
A small positive test charge is carried once around a circuit consisting of an ideal battery, ideal wires and one resistor, in the direction of the conventional current. Which statement correctly describes the electric potential energy of the test charge?
Answer and reasoning
AIt increases in the battery and decreases steadily along both the wires and the resistor. A student who thinks the potential falls along every part of the circuit picks this. Ideal wires have negligible resistance, so the potential is the same all along each wire; the whole decrease is in the resistor.
BIt decreases in the resistor, where part of the test charge itself is used up. A student who thinks charge is used up in a resistor picks this. The charge passes through the resistor unchanged; what it gives up is electric potential energy.
CIt stays the same all the way around, because the current is the same everywhere. A student who thinks a constant current means constant energy picks this. The current is the same in every part of a single loop, but the potential, and with it the charge's potential energy, rises in the battery and falls in the resistor.
DIt increases in the battery and decreases by the same amount in the resistor.Correct In the battery the charge moves from the − to the + terminal, to a higher potential, so its electric potential energy increases by qε. Along ideal wires the potential does not change. In the resistor it moves to a lower potential and loses qε, the same amount, so it returns with its starting energy.
The circuit shown contains two ideal batteries and two resistors in a single loop. In terms of ε and R, what is the magnitude of the current in the loop?
Answer and reasoning
A(4/5)ε/R A student who adds the two emfs whatever their orientation gets (3ε + ε)/(5R). The + terminals of the two batteries face the same way on the page, so around the loop they oppose each other: the net emf is 3ε − ε.
B(2/5)ε/RCorrect Going clockwise around the loop, up through the 3ε battery (− to +): +3ε; down through the ε battery (+ to −): −ε; and −I(2R) − I(3R) across the resistors. Loop rule: 3ε − ε − 5IR = 0, so I = 2ε/(5R) = (2/5)ε/R, clockwise.
C(5/3)ε/R A student who treats 2R and 3R as parallel, because they are drawn in separate vertical branches, uses Req = 6R/5 and gets 2ε/(6R/5). Each resistor is in series with a battery in its branch, and the loop is a single path, so the resistors are in series: Req = 5R.
D(3/2)ε/R A student who takes the current to be set by the 3ε battery and the resistor in its own branch alone gets 3ε/(2R). The current is the same all around the single loop and is set by the net emf and the total resistance.
Working Clockwise: +3ε − I(3R) − ε − I(2R) = 0 → I = 2ε/(5R). (Emfs added: 4ε/(5R); resistors treated as parallel: 2ε/(6R/5) = 5ε/(3R); local: 3ε/(2R).)
In preparation: 0 of 2 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
11.6.A.1 Electric potential energy change, ΔUE = qΔV Fix
Electric potential energy change, ΔUE = qΔV
When a charge q moves through a potential difference ΔV, its electric potential energy changes by qΔV. SI unit: joule (J). A positive charge moving to a higher potential gains electric potential energy; a negative charge moving to a higher potential loses it.
Energy transfer in circuit elements
In a battery, charge moving from the − to the + terminal gains electric potential energy (the battery converts its internal energy to electrical energy); in a resistor, charge moving in the direction of the current loses electric potential energy, which becomes thermal energy.
Potential difference as energy per unit charge
The potential difference across an element is the energy transferred per unit charge passing through it: 1 V = 1 J/C. The energy transferred when charge q passes is qΔV.
Students often think The potential difference across the external circuit always equals the battery's emf, even when the battery has internal resistance and a current is in it. In fact No. Part of the emf is across the battery's own internal resistance, Ir, so the potential difference across the external circuit is ε − Ir. Each coulomb that passes through the external resistor therefore transfers ε − Ir joules there, not ε.
Students often think The potential difference across a circuit element is the energy transferred in that element, so ΔV volts means ΔV joules whatever charge passes. In fact No. Potential difference is energy transferred per unit charge (1 V = 1 J/C). The energy transferred when a charge q passes through an element is ΔUE = qΔV, so it depends on how much charge passes as well as on ΔV.
11.6.A.2 Closed loop Fix
Closed loop
Any closed path through a circuit that starts and ends at the same point. A charge carried around a closed loop returns to its starting potential.
Kirchhoff's loop rule
The sum of the potential differences across all the elements of any closed loop is zero, ΣΔV = 0. It follows from conservation of energy: a charge carried around a loop returns with the same electric potential energy it started with.
Sign conventions for the loop rule
Going around a loop in a chosen direction: across a battery from − to +, ΔV = +ε; from + to −, ΔV = −ε. Across a resistor in the direction of the current, ΔV = −IR; against the current, ΔV = +IR.
Graph of potential around a loop
A graph of electric potential against position around a loop, starting and ending at the same point. It rises across a battery (− to +), falls across each resistor in the direction of the current, is flat along ideal wires, and returns to its starting value. The position axis shows the order of elements, not their sizes.
Students often think The battery's potential difference is shared equally among the elements in a series loop, whatever their resistances. In fact No. By the loop rule the potential differences add up to the battery's emf, but each resistor's share is IR, so the resistors share it in proportion to their resistances: a larger resistance has a larger potential difference.
Students often think The electric potential decreases steadily along every part of a circuit, including ideal connecting wires, as the charges move away from the battery. In fact No. An ideal wire has negligible resistance, so ΔV = IR = 0 along it: the potential is the same at every point of a wire between two elements, and the graph of potential against position is flat there.
8 more questions. Every wrong answer here is a real mistake students make, and you see why it is wrong as soon as you answer.
Question 1 of 8
A battery with an emf of 12.0 V and an internal resistance of 1.0 Ω drives a current of 2.0 A through an external resistor. How much electric energy is transferred to the external resistor as 5.0 C of charge passes through it?
Answer and reasoning
A50 JCorrect The potential difference across the external resistor equals the terminal potential difference: ε − Ir = 12.0 V − (2.0 A)(1.0 Ω) = 10.0 V. Each coulomb passing through the resistor loses 10.0 J of electric potential energy, so ΔUE = qΔV = (5.0 C)(10.0 V) = 50 J.
B60 J A student who takes the full emf, 12.0 V, to be across the external resistor gets (5.0 C)(12.0 V) = 60 J. The battery's internal resistance takes Ir = 2.0 V of the emf, leaving 10.0 V for the external resistor.
C30 J A student who shares the emf equally between the internal and external resistances gives each 6.0 V and gets (5.0 C)(6.0 V) = 30 J. The shares are in proportion to the resistances: Ir = 2.0 V across the internal resistance, 10.0 V across the external resistor.
D10 J A student who takes the potential difference across the resistor, 10.0 V, as the energy transferred picks this. A volt is a joule per coulomb, so 5.0 C passing through 10.0 V transfers 50 J.
Working ΔVext = ε − Ir = 12.0 − 2.0 × 1.0 = 10.0 V. ΔUE = qΔV = 5.0 × 10.0 = 50 J. (Full emf: 60 J; equal split: 30 J; ΔV as energy: 10 J.)
A student claims that Kirchhoff's loop rule, ΣΔV = 0 around any closed loop, is a consequence of conservation of energy. Which reasoning correctly supports the student's claim?
Answer and reasoning
ACharge is conserved, so the charge flowing into each element per second equals the charge flowing out of it. A student who attaches the loop rule to conservation of charge picks this. The statement is true, but it supports the junction rule, not the loop rule; the loop rule is about potential differences, which describe energy per unit charge.
BThe battery supplies a fixed amount of charge, which the resistors in the loop use up before it returns. A student who thinks charge is used up in resistors picks this. Charge is conserved and is not used up; what the resistors take is energy, transferred from the charges' electric potential energy.
CA charge carried around a loop returns with its original potential energy, so the qΔV terms must add to zero.Correct The change in a charge's electric potential energy across each element is qΔV. After a complete loop the charge is back where it started, at the same potential, with the same electric potential energy, so the changes add to zero: qΣΔV = 0, and ΣΔV = 0.
DEnergy is shared equally around the loop, so every element in it has the same potential difference. A student who thinks the battery's potential difference is shared equally picks this. The potential differences across the elements are generally different; what the loop rule requires is that they add to zero.
A battery and two resistors, R₁ and R₂, form a single loop. The graph shows the electric potential V as a function of position around the loop, starting and ending at the battery's negative terminal. The resistance of R₁ is 2.0 Ω. What is the resistance of R₂?
Answer and reasoning
A6.0 Ω A student who reads the potential at the start of R₂, 9.0 V, as the potential difference across it gets 9.0 V/1.5 A = 6.0 Ω. The potential difference across R₂ is the change in V along its segment: 9 V − 3 V = 6.0 V.
B1.0 Ω A student who thinks the larger potential drop in a series loop is across the smaller resistance gets R₂ = R₁ × (3.0 V/6.0 V) = 1.0 Ω. With the same current in both, the larger drop is across the larger resistance: R₂/R₁ = 6.0 V/3.0 V.
C4.0 ΩCorrect From the graph, the potential falls by 3.0 V across R₁ (from 3 V to 0) and by 6.0 V across R₂ (from 9 V to 3 V). The current through R₁ is ΔV/R = 3.0 V/2.0 Ω = 1.5 A, and in a single loop the same current passes through R₂, so R₂ = 6.0 V/1.5 A = 4.0 Ω.
D1.3 Ω A student who takes the current to be set by the battery and R₁ alone, 9.0 V/2.0 Ω = 4.5 A, and reads the 6.0 V drop across R₂ from the graph gets 6.0 V/4.5 A = 1.3 Ω. R₁ has only 3.0 V across it, so the current in the loop is 3.0 V/2.0 Ω = 1.5 A.
Working ΔV₁ = 3.0 V, ΔV₂ = 9.0 − 3.0 = 6.0 V (from graph). I = 3.0/2.0 = 1.5 A. R₂ = 6.0/1.5 = 4.0 Ω. (Potential at R₂'s start: 9.0/1.5 = 6.0 Ω; inverse share: 1.0 Ω; current from full emf across R₁: 6.0/4.5 = 1.3 Ω.)
An ideal battery is connected in series with two resistors, R₁ and R₂, that initially have equal resistances. R₂ is then replaced by a resistor with three times its resistance; R₁ is unchanged. How does the potential difference across R₂ change?
Answer and reasoning
AIt becomes 3/2 times its original valueCorrect Initially the resistances are R and R, so each has ε/2. Afterward they are R and 3R: the current is ε/(4R), and the potential difference across R₂ is (ε/(4R))(3R) = 3ε/4. The ratio is (3ε/4)/(ε/2) = 3/2.
BIt becomes three times its original value A student who thinks the battery keeps supplying the same current gets ΔV = IR₂ tripled. The larger resistance reduces the current from ε/(2R) to ε/(4R), so the potential difference increases by less.
CIt becomes 1/2 of its original value A student who thinks the larger resistance gets the smaller share of the potential difference shares ε in the ratio 1 : 1/3, giving R₂ only ε/4. In series the larger resistance has the larger potential difference: 3ε/4.
DIt remains at its original value A student who thinks the battery's potential difference is always shared equally keeps ε/2 across R₂. The shares are in proportion to the resistances, so after the change R₂ has three quarters of ε.
A battery and three resistors, R₁, R₂ and R₃, form a single loop. The graph shows the electric potential V as a function of position around the loop, starting at the battery's negative terminal; the position axis shows only the order of the elements. Which ranking of the resistances is correct?
Answer and reasoning
AR₃ > R₁ > R₂ A student who ranks by the steepness of each segment picks this: R₃'s narrow segment falls most steeply. The widths of the segments are arbitrary; the resistance is shown by the size of the drop, which is largest for R₁.
BR₁ > R₃ > R₂Correct The resistors are in series, so they carry the same current, and each resistance is proportional to the potential drop across it, R = ΔV/I. From the graph the drops are 12 V − 6 V = 6 V for R₁, 6 V − 4 V = 2 V for R₂ and 4 V − 0 = 4 V for R₃, so R₁ > R₃ > R₂.
CR₁ > R₂ > R₃ A student who ranks by the potential at each resistor's position (12 V, 6 V and 4 V at their starts) picks this. The potential difference across a resistor is the change in V along its segment, not the value of V there.
DR₂ > R₁ > R₃ A student who ranks by the length of each segment along the position axis picks this. The position axis only shows the order of the elements around the loop; the resistances are shown by the drops: 6 V, 2 V and 4 V.
Working Same current I. ΔV: R₁ 6 V, R₂ 2 V, R₃ 4 V → R₁ : R₂ : R₃ = 3 : 1 : 2 → R₁ > R₃ > R₂. (Slopes: R₃ steepest, then R₁, then R₂; starting potentials 12, 6, 4; segment widths R₂ > R₁ > R₃.)
An ideal battery of emf ε, a resistor R and a switch S are connected in series, and S is open. An ideal voltmeter V₁ is connected across R, and an ideal voltmeter V₂ is connected across the open switch. Which statement correctly describes the readings?
Answer and reasoning
AV₁ reads ε and V₂ reads zero, since R is connected directly to the battery A student who thinks a resistor connected to a battery always has the battery's full potential difference across it picks this. With no current, R has no potential difference; the emf appears across the gap.
BBoth read zero, since with no current there is no ΔV anywhere around the loop A student who thinks a potential difference needs a current picks this. The battery's emf is there whether or not there is a current; by the loop rule it is all across the open switch.
CEach reads ε/2, since the battery's ΔV is shared equally by R and S A student who thinks the emf is shared equally among the elements of a loop picks this. The share across R is IR, which is zero with no current, so the gap at S has all of ε.
DV₁ reads zero and V₂ reads ε, since R carries no current and so has no ΔVCorrect With S open there is no current, so the potential difference across R is IR = 0. The loop rule still holds around the loop through the battery, R and the gap at S: ε − 0 − ΔVS = 0, so the whole emf is across the open switch. V₁ reads zero and V₂ reads ε.
An ideal 12.0 V battery is connected in series with a 2.0 Ω resistor and a 4.0 Ω resistor. What is the change in the electric potential energy of an electron as it passes through the 4.0 Ω resistor? Use e = 1.60 × 10⁻¹⁹ C.
Answer and reasoning
A−1.9 × 10⁻¹⁸ J A student who gives the 4.0 Ω resistor the battery's full 12.0 V gets (1.60 × 10⁻¹⁹ C)(12.0 V) = 1.9 × 10⁻¹⁸ J lost. The 12.0 V is divided between the two resistors; the 4.0 Ω resistor has 8.0 V.
B−1.3 × 10⁻¹⁸ JCorrect The current is 12.0 V/6.0 Ω = 2.0 A, so the potential difference across the 4.0 Ω resistor is 8.0 V. The electron moves opposite to the conventional current, from the lower-potential end to the higher-potential end, so ΔV = +8.0 V along its path and ΔUE = qΔV = (−1.60 × 10⁻¹⁹ C)(+8.0 V) = −1.3 × 10⁻¹⁸ J. Like any carrier, it loses electric potential energy in the resistor.
C−9.6 × 10⁻¹⁹ J A student who shares the 12.0 V equally between the resistors uses 6.0 V and gets 9.6 × 10⁻¹⁹ J lost. The shares are in proportion to the resistances: 4.0 V and 8.0 V.
D−6.4 × 10⁻¹⁹ J A student who gives the larger resistor the smaller share uses 4.0 V and gets 6.4 × 10⁻¹⁹ J lost. With the same current in both, the 4.0 Ω resistor has the larger potential difference, 8.0 V.
Working I = 12.0/(2.0 + 4.0) = 2.0 A; ΔV₄ = 2.0 × 4.0 = 8.0 V. Electron moves from low to high potential: ΔV = +8.0 V; ΔU = (−1.60e-19)(8.0) = −1.28 × 10⁻¹⁸ J ≈ −1.3 × 10⁻¹⁸ J. (Full 12 V: −1.9 × 10⁻¹⁸; 6.0 V: −9.6 × 10⁻¹⁹; 4.0 V: −6.4 × 10⁻¹⁹.)
A rechargeable battery of emf ε is charged by a battery of emf 3ε. The positive terminal of the 3ε battery is connected through a resistor of resistance R to the positive terminal of the rechargeable battery, and the two negative terminals are connected directly by a wire, forming a single loop. Both batteries have negligible internal resistance. At what rate is electrical energy transferred into the rechargeable battery?
Answer and reasoning
A2ε²/RCorrect Around the loop the two emfs oppose each other: 3ε − IR − ε = 0, so I = 2ε/R. This current passes through the rechargeable battery from its positive to its negative terminal, so each charge q loses electric potential energy qε there, and energy is transferred into the battery at the rate εI = 2ε²/R. The other 4ε²/R of the 6ε²/R supplied by the 3ε battery is transferred in the resistor.
B4ε²/R A student who adds the two emfs, whichever way round they are connected, takes I = (3ε + ε)/R = 4ε/R and so gets ε(4ε/R). With the positive terminals joined through the resistor, the emfs oppose each other around the loop: the net emf is 3ε − ε.
C3ε²/R A student who takes the current to be set by the 3ε battery and the resistor alone, I = 3ε/R, picks this: ε(3ε/R). The rechargeable battery is in the same loop, and its emf opposes the other battery's, so only 3ε − ε is across the resistor.
D6ε²/R A student who finds the current correctly, I = 2ε/R, but takes all the energy the 3ε battery supplies, at the rate 3εI, to go into the rechargeable battery picks this. Charges also lose potential energy IR = 2ε per unit charge in the resistor, so only ε per unit charge, the rate εI, goes into the rechargeable battery.
Working Loop rule, starting at the joined negative terminals and going up through the 3ε battery, through R and down through the rechargeable battery (+ to −): 3ε − IR − ε = 0, so I = 2ε/R, entering the rechargeable battery at its positive terminal. Each charge q passing through it from + to − loses electric potential energy qε, so energy is transferred into it at the rate εI = ε(2ε/R) = 2ε²/R. Check: the 3ε battery supplies 3εI = 6ε²/R = I²R (4ε²/R) + εI (2ε²/R). Distractors (sympy-checked): emfs added, I = 4ε/R → 4ε²/R; current from the 3ε battery and R alone, I = 3ε/R → 3ε²/R; all supplied energy taken as delivered, 3εI → 6ε²/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
