4 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 4
In a circuit with a steady current, positive charge passes through a resistor in the direction of the current. Which statement describes what happens?
Answer and reasoning
AThe electric potential energy decreases, and the resistor's internal energy increases.Correct Current in a resistor is directed from higher to lower potential, so charge moving with the current moves to a lower potential and ΔUE = qΔV is negative. That energy is transferred to the resistor, whose internal energy increases.
BSome of the charge is used up, and it becomes internal energy of the resistor. A student who thinks current is used up picks this. As much charge leaves the resistor each second as enters it; what the resistor takes is energy, the decrease in electric potential energy.
CThe electric potential energy is unchanged, because the current does not change. A student who treats current as if it were energy picks this. The current is the same on both sides, but the potential is lower on the downstream side, so each charge has less electric potential energy there.
DThe electric potential energy increases as the charge is pushed through the resistor. A student who thinks charge rises in potential through a resistor, as it does through a battery, picks this. In a resistor the current goes from higher to lower potential, so the electric potential energy decreases.
In a circuit with steady currents, a small positive charge starts at point P, travels once around a closed loop that contains one battery and several resistors, and returns to P. Which statement about the changes in the electric potential energy of the system as the charge travels around the loop is correct?
Answer and reasoning
AThe decreases in the resistors add up to more than the increase in the battery. A student who thinks resistors use up more energy than the battery supplies picks this. The charge would then return to P at a lower potential than it left, but P has a single potential. The gains and losses around a loop are equal.
BThe increase in the battery is greater than the decreases in the resistors. A student who thinks charges keep the energy the battery gives them picks this. The energy gained in the battery is transferred to the resistors on the way around, so the charge returns to P with the energy it started with.
CThere are no changes, because the current is the same all the way around. A student who treats an unchanging current as unchanging energy picks this. The current is the same all the way around, but the potential rises across the battery and falls across each resistor, so the energy changes on the way.
DThe increase in the battery equals the sum of the decreases in the resistors.Correct The potential at P has a single value, so the charge returns to P with the same electric potential energy it left with. By conservation of energy, the energy gained in the battery equals the energy transferred in the resistors; this is why the potential differences around the loop sum to zero.
An ideal battery and three resistors of different resistances are connected in series in a single loop. Which statement about the potential differences across the three resistors must be true?
Answer and reasoning
AEach one is equal to the emf of the battery. A student who thinks every element gets the battery's full potential difference picks this. Around the loop the drops must add up to ε, so three drops of ε each would add up to 3ε.
BThe potential differences are all equal. A student who thinks series elements share the potential difference equally picks this. They carry the same current, but their resistances differ, so ΔV = IR differs.
CTheir sum is equal to the emf of the battery.Correct By the loop rule, the drops across the three resistors add up to the rise across the battery, so their sum is ε. The share each receives, ΔV = IR, depends on its resistance.
DThe one nearest the positive terminal is greatest. A student who thinks the first resistor the current reaches takes the most picks this. Each ΔV = IR depends on the resistance, not the position; the largest resistance has the largest potential difference.
An ideal battery and three resistors, X, Y and Z, are connected in series by ideal wires in a single loop. The graph shows the electric potential V at points around the loop, starting at the battery's negative terminal and passing through the battery and then X, Y and Z in turn; the shaded bands mark the battery and the resistors. Which ranking of the resistances is correct?
Answer and reasoning
ARY > RX > RZCorrect The resistors are in series, so they carry the same current, and R = ΔV/I is proportional to the drop in potential across each. The graph falls 3.0 V across X, 5.0 V across Y and 1.0 V across Z, so RY > RX > RZ.
BRX > RY > RZ A student who reads the width of each band as the length of the resistor picks this, ranking X, Y and Z by how wide they are drawn. The widths only show how the graph is laid out; the drops, 3.0 V, 5.0 V and 1.0 V, show that Y has the greatest resistance.
CRZ > RX > RY A student who thinks a larger resistance has the smaller potential difference across it picks this. With the same current in each, ΔV = IR, so the largest drop, across Y, marks the largest resistance.
DRY > RZ > RX A student who ranks by how steeply the graph falls across each resistor picks this. The steepness depends on how wide each resistor is drawn along the position axis; the resistance depends only on the size of the drop.
Working Series loop: the same current I in X, Y and Z, so R = ΔV/I ∝ ΔV. From the graph: X, 9.0 V → 6.0 V (3.0 V); Y, 6.0 V → 1.0 V (5.0 V); Z, 1.0 V → 0 (1.0 V). RY > RX > RZ.
In preparation: 0 of 4 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
11.6.A.1 Change in electric potential energy, ΔUE = qΔV Fix
Change in electric potential energy, ΔUE = qΔV
When a charge q moves through a potential difference ΔV, the electric potential energy of the system changes by ΔUE = qΔV. A positive charge that moves to a higher potential gains electric potential energy; one that moves to a lower potential loses it. SI unit: joule (J).
Emf of a battery, ε
The energy per unit charge that a battery supplies to charge passing through it from its negative to its positive terminal. For an ideal battery, the potential difference across the terminals equals ε. SI unit: volt (V = J/C).
Energy conversion in a resistor
Charge moving through a resistor in the direction of the current goes from higher to lower potential, so the electric potential energy decreases by qΔV and the resistor's internal energy increases (a bulb also emits light). The current is the same on both sides of the resistor: charge is not used up.
Students often think Current (charge) is used up as it passes through a resistor or bulb, so less current leaves it than enters it. In fact No. The current leaving a resistor equals the current entering it: charge is not used up. What changes is the electric potential energy, which is transferred to the resistor as internal energy.
Students often think Because the current is the same on both sides of a resistor (and all around a series loop), the electric potential energy of the charges does not change: current and energy are treated as the same thing. In fact No. The current says how much charge passes each second, not how much energy each charge has. Charges passing through a resistor in the direction of the current move to a lower potential and lose electric potential energy, ΔUE = qΔV, even though the current is unchanged.
11.6.A.2 Loop rule and conservation of energy Fix
Loop rule and conservation of energy
A charge that travels once around a closed loop returns to its starting point, where the potential has not changed, so the electric potential energy gained in batteries equals the energy transferred in the other elements of the loop.
Students often think Charges are 'used up' as they travel around a loop, so they return to their starting point with less electric potential energy than they left with, and some of the drop in potential need not appear across any element. In fact No. The potential at a point in a circuit has a single value, so a charge that goes around a loop and returns has the same electric potential energy as when it left. Every drop in potential is across some element, and the drops add up to the rises across batteries.
Students often think The battery gives charges energy that they keep, so the energy gained in the battery is greater than the energy transferred in the resistors on a trip around the loop. In fact No. The energy a charge gains in the battery is transferred to the resistors as it goes around the loop; back at its starting point, its electric potential energy is the same as before.
11.6.A.3 Kirchhoff's loop rule, ΣΔV = 0 Fix
Kirchhoff's loop rule, ΣΔV = 0
The potential differences across all the elements of any closed loop sum to zero. Going through a battery from − to + counts +ε (from + to −, −ε); going through a resistor in the direction of the current counts −IR, and against it +IR.
Closed loop
Any path through a circuit that returns to its starting point without passing through any point twice. A loop through a battery that is in series with a set of parallel branches includes only one of those branches; a loop made of two parallel branches goes through one in each direction, so their potential differences cancel, which is why parallel branches have the same potential difference.
Students often think Elements in series share the battery's potential difference equally, whatever their resistances, because the same current passes through each. In fact Only if their resistances are equal. Series elements carry the same current, so ΔV = IR is larger across the larger resistance; the potential differences add up to the emf.
Students often think Every element in a circuit has the full potential difference of the battery across it. In fact No. Around any loop the potential differences add up to the emf, so elements in series share it. Only an element connected directly across an ideal battery has the full emf across it.
11.6.A.4 Graph of electric potential against position around a loop Fix
Graph of electric potential against position around a loop
A graph of the electric potential at points around a loop: flat along ideal wires, rising across a battery from − to +, falling across each resistor in the direction of the current, and returning to its starting value. In a series loop the fall across each resistor is proportional to its resistance.
Ideal wire
A connecting wire with negligible resistance, so the electric potential is the same at every point along it.
Students often think A larger resistance lets less 'through', so it has the smaller potential difference across it. In fact No. Series resistors carry the same current, so ΔV = IR is larger across the larger resistance.
Students often think On a graph of potential against position, the steeper the drop across an element, the greater its resistance. In fact No. In a series loop the resistance is proportional to the size of the drop across an element, not to how steep the graph is; the steepness also depends on how wide the element is drawn along the position axis.
7 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 7
Resistors R₁ and R₂ are connected in series with an ideal battery, and R₂ has the greater resistance. Charge leaving the battery's positive terminal passes through R₁ first and then R₂. How does the electric potential energy converted when one coulomb of charge passes through R₁ compare with that converted when it passes through R₂?
Answer and reasoning
AIt is greater in R₁ than in R₂. A student who thinks the first resistor takes the largest share picks this. Order does not matter in a series loop: the energy per coulomb in each resistor is ΔV = IR, which depends on its resistance.
BIt is less in R₁ than in R₂.Correct The energy converted per coulomb is ΔUE/q = ΔV. In series both resistors carry the same current, so ΔV = IR is greater across R₂, the larger resistance: each coulomb loses more electric potential energy in R₂.
CIt is the same in R₁ as in R₂. A student who thinks series resistors share the potential difference equally picks this. The same current passes through both, but ΔV = IR is larger across the larger resistance.
DIt is zero in both resistors. A student who thinks the charges lose no energy because the current is not used up picks this. The current is unchanged, but each coulomb moves to a lower potential in each resistor and loses electric potential energy there.
In the circuit shown, the battery and the voltmeter are ideal. The battery's emf and the voltmeter's reading are marked on the diagram. What is the potential difference across resistor R₁?
Answer and reasoning
A8.0 V A student who adds all the potential differences as positive numbers picks this: 6.0 V + 2.0 V. Around the loop the battery is a rise and the resistors are drops, so the drops must add up to the rise.
B6.0 V A student who thinks every element has the battery's full potential difference across it picks this. R₁ is in series with the parallel pair, so it shares the 6.0 V with them; if R₁ had 6.0 V across it, R₂ and R₃ would have none.
C4.0 VCorrect Apply the loop rule to the loop through the battery, R₁ and R₃: the rise of 6.0 V across the battery equals the drops across R₁ and R₃. The voltmeter shows 2.0 V across R₃, so ΔV₁ = 6.0 V − 2.0 V = 4.0 V.
D2.0 V A student who counts both parallel branches in one loop picks this: 6.0 V − 2.0 V − 2.0 V. A loop through the battery passes through only one of R₂ and R₃; they share the same 2.0 V potential difference, which is counted once.
Working Loop through the battery, R₁ and R₃: ε − ΔV₁ − ΔV₃ = 0. The voltmeter is across R₃ (and, since R₂ and R₃ are in parallel, across R₂): ΔV₃ = 2.0 V. ΔV₁ = 6.0 V − 2.0 V = 4.0 V.
The circuit shown contains two ideal batteries, of emfs ε₁ and ε₂, and two resistors, of resistances R₁ and R₂. The current I is taken as positive in the clockwise direction shown by the arrow. Which expression gives I?
Answer and reasoning
A(ε₁ + ε₂)/(R₁ + R₂) A student who adds the two emfs whatever their orientation picks this. Both positive terminals face the top wire, so going around the loop one battery is a rise and the other a drop; their emfs subtract.
B(ε₁ − ε₂)/(R₁ − R₂) A student who gives the R₂ term a positive sign because R₂ is crossed from right to left on the page picks this. Going clockwise, the loop passes through both resistors in the direction of the current, so both terms are −IR.
C(ε₂ − ε₁)/(R₁ + R₂) A student who treats going through a battery from − to + as a drop picks this, which reverses both emf terms. The positive terminal is at the higher potential, so going up through the left battery is a rise of ε₁.
D(ε₁ − ε₂)/(R₁ + R₂)Correct Going clockwise from the lower left: +ε₁ up through the left battery (− to +), −IR₁ through R₁, −ε₂ down through the right battery (+ to −) and −IR₂ through R₂. Setting the sum to zero gives I = (ε₁ − ε₂)/(R₁ + R₂): the batteries oppose each other around the loop.
Working Clockwise from the lower-left corner: +ε₁ (left battery, − to +), −IR₁ (with the current), −ε₂ (right battery, + to −), −IR₂ (with the current). ε₁ − IR₁ − ε₂ − IR₂ = 0, so I = (ε₁ − ε₂)/(R₁ + R₂). (All emfs added: (ε₁ + ε₂)/(R₁ + R₂). +IR₂ for the bottom resistor: (ε₁ − ε₂)/(R₁ − R₂). Battery signs reversed: (ε₂ − ε₁)/(R₁ + R₂).)
A student builds a single loop from a battery, two resistors R₁ and R₂, a switch and connecting wires. Using an ideal voltmeter, the student measures 6.0 V across the battery's terminals, 2.5 V across R₁ and 2.5 V across R₂. Which conclusion is best supported by these measurements?
Answer and reasoning
AThe other 1.0 V was used up by the charges as they traveled all the way around the loop. A student who thinks charges can lose potential without it appearing across any element picks this. Every drop in potential is across some part of the loop, and the drops add up to the 6.0 V rise, so 1.0 V must be across an element that was not measured.
BSome other part of the loop, such as the switch or a connection, has 1.0 V across it.Correct The loop rule requires the drops around the loop to add up to the rise across the battery's terminals, 6.0 V. The resistors account for 5.0 V, so the remaining 1.0 V must be across some other element of the loop, such as the switch, a poor connection or the wires.
CThe battery has internal resistance, and the missing 1.0 V is dropped across that resistance. A student who thinks a voltmeter across the terminals reads the emf picks this. The reading across the terminals, 6.0 V, already allows for any drop inside the battery, so the external loop must account for all 6.0 V.
DThe meter misread both resistors, which should each have 3.0 V across them. A student who thinks series elements share the battery's potential difference equally picks this. Nothing requires equal shares, and with the switch and connections also in the loop, the resistors' share can be less than 6.0 V.
Working Loop rule around the external circuit: 6.0 V = 2.5 V + 2.5 V + ΔVother, so ΔVother = 1.0 V across some unmeasured element (switch, connection or wires).
An ideal battery of emf ε and two resistors, of resistances R and 4R, are connected in series in a single loop, and the current is steady. A positive charge Q passes through the resistor of resistance 4R in the direction of the current. What is the magnitude |ΔUE| of the change in the electric potential energy of this charge?
Answer and reasoning
A0.50Qε A student who shares the emf equally between the two resistors takes ΔV = ε/2 and picks this. The same current passes through both, so the larger resistance has the larger potential difference: 4ε/5 across 4R.
B1.00Qε A student who puts the battery's full emf across the 4R resistor picks this. Around the loop the potential differences across the two resistors add up to ε, so each has only part of it.
C0.20Qε A student who thinks the larger resistance lets less 'through' and so has the smaller potential difference shares the emf in inverse proportion to resistance, gives the 4R resistor ε/5, and picks this. Both resistors carry the same current, so ΔV = IR is larger across the larger resistance: 4ε/5.
D0.80QεCorrect By the loop rule the current is ε/(5R), so the potential difference across the 4R resistor is 4ε/5. From ΔUE = qΔV, charge Q moving through that potential difference to the lower potential loses electric potential energy 0.80Qε, which becomes thermal energy in the resistor.
Working Loop rule: ε − IR − I(4R) = 0, so I = ε/(5R). The potential difference across the 4R resistor is ΔV = I(4R) = 4ε/5 = 0.80ε. From ΔUE = qΔV, the charge Q moving through it to the lower potential changes its electric potential energy by −0.80Qε, so |ΔUE| = 0.80Qε; this energy is converted to thermal energy in the resistor. (Errors: equal shares of the emf give 0.50Qε; shares in inverse proportion to resistance give (1/5)Qε = 0.20Qε; the full emf across the resistor gives 1.00Qε.)
An ideal battery of emf ε and two resistors, X and Y, each of resistance R, are connected in series in a single loop. Resistor Y is then replaced by a resistor of resistance 3R, and X is unchanged. How does the potential difference across Y change?
Answer and reasoning
AIt rises to 3/2 of its first value.Correct Before, the current is ε/(2R) and Y has ε/2. After, the current is ε/(4R) and Y has (ε/(4R))(3R) = 3ε/4. The potential difference across Y rises by a factor of 3/2, while that across X falls from ε/2 to ε/4, keeping the sum equal to ε.
BIt increases to 3 times its first value. A student who treats the battery as a source of fixed current keeps I = ε/(2R) and picks this: ΔVY = (ε/(2R))(3R) = 3ε/2, which is more than the emf. The emf is fixed, so the larger total resistance lowers the current to ε/(4R).
CIt stays equal to its first value. A student who shares the emf equally between the two series resistors gives Y the value ε/2 both times and picks this. The same current passes through both, so the larger resistance, 3R, takes the larger share, 3ε/4.
DIt decreases to 1/2 of its first value. A student who notices that the current halves, from ε/(2R) to ε/(4R), and scales ΔVY with the current alone picks this. Y's resistance has tripled as well, so ΔVY = IR changes by (1/2)(3) = 3/2.
Working Before: I₁ = ε/(2R); ΔVY = I₁R = ε/2. After: I₂ = ε/(R + 3R) = ε/(4R); ΔVY = I₂(3R) = 3ε/4. Ratio (3ε/4)/(ε/2) = 3/2. (Errors: constant current ε/(2R) gives 3ε/2, ×3; equal shares give ε/2, ×1; ΔV proportional to the halved current gives ×1/2.)
An ideal battery of emf 9.0 V, a 3.0 Ω resistor R₁ and a second resistor R₂ are connected in series in a single loop by ideal wires. An ideal voltmeter connected across R₁ reads 6.0 V. What is the resistance of R₂?
Answer and reasoning
A4.5 Ω A student who thinks each element has the battery's full potential difference across it takes ΔV₂ = 9.0 V and picks this: 9.0 V/2.0 A. Around the loop the drops must add up to 9.0 V; R₁ already has 6.0 V, so R₂ has 3.0 V.
B7.5 Ω A student who adds the emf and the voltmeter reading as positive terms picks this: ΔV₂ = 9.0 V + 6.0 V = 15 V, and 15 V/2.0 A = 7.5 Ω. The battery is a rise and the resistors are drops, so ΔV₂ = 9.0 V − 6.0 V = 3.0 V.
C6.0 Ω A student who finds ΔV₂ = 3.0 V but thinks the resistor with the smaller potential difference has the larger resistance scales R₁ by 6.0 V/3.0 V and picks this. The same current passes through both, so ΔV = IR: the smaller potential difference belongs to the smaller resistance, 1.5 Ω.
D1.5 ΩCorrect The voltmeter reading gives the current: I = 6.0 V/3.0 Ω = 2.0 A, which is also the current in R₂ because the loop is a single path. By the loop rule the drops across R₁ and R₂ add up to the 9.0 V rise across the battery, so ΔV₂ = 3.0 V and R₂ = 3.0 V/2.0 A = 1.5 Ω.
Working Current: I = ΔV₁/R₁ = 6.0 V/3.0 Ω = 2.0 A, the same in R₂ (single loop). Loop rule: ε − ΔV₁ − ΔV₂ = 0, so ΔV₂ = 9.0 V − 6.0 V = 3.0 V. R₂ = ΔV₂/I = 3.0 V/2.0 A = 1.5 Ω. (m08, ΔV₂ = 9.0 V: 4.5 Ω; m09, ΔV₂ = 9.0 V + 6.0 V = 15 V: 7.5 Ω; m14, resistances in inverse ratio to the potential differences: R₂ = (3.0 Ω)(6.0 V/3.0 V) = 6.0 Ω.)
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
