2 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 2
A battery is connected to a lightbulb by copper wires. Which statement best explains why charge moves through the wires?
Answer and reasoning
AThe battery releases stored charge into wires that had no free charge. A student who thinks a battery is a store of charge picks this. The wires contain vast numbers of free electrons before the battery is connected; the battery's potential difference sets them moving, and the same charge circulates around the circuit.
BThe battery's emf is a force, in newtons, that pushes on each charge directly. A student who takes the name 'electromotive force' literally picks this. An emf is a potential difference, measured in volts; the forces on the charges come from the electric field that this potential difference sets up in the circuit.
CThe battery's potential difference makes charges already in the wires move.Correct The wires and filament are full of free electrons before the battery is connected. The battery provides a potential difference (its emf), which sets up an electric field in the circuit, and the charges already present move in response. The battery does not supply the charge.
DCharge flows out of both battery terminals and meets in the bulb filament. A student who pictures charge leaving both terminals picks this. Charge moves around the loop in one direction: conventional current leaves the positive terminal and returns to the negative terminal, and electrons move the opposite way.
The diagram shows positive and negative ions moving through a solution between ends X and Y. The labels give the magnitude of the charge that each kind of ion carries through the cross-section shown each second. What is the current through the cross-section?
Answer and reasoning
A3.0 A toward Y A student who takes the current to be in the direction the negative carriers move, and counts only them, picks this. Conventional current is the direction positive charge would move: the negative ions moving toward Y make a current toward X, and the positive ions add to it.
B1.0 A toward Y A student who thinks flows in opposite directions cancel picks this: 3.0 − 2.0 = 1.0. Negative charge moving toward Y and positive charge moving toward X both make conventional current toward X, so they add.
C5.0 A toward XCorrect Conventional current is in the direction positive charge moves. The positive ions carry 2.0 C each second toward X: 2.0 A toward X. The negative ions carry 3.0 C each second toward Y, which is equivalent to positive charge moving toward X: 3.0 A toward X. Both contributions point the same way, so the current is 5.0 A toward X.
D2.0 A toward X A student who thinks only positive charges count toward conventional current picks this. The negative ions carry charge too; their motion toward Y is a current of 3.0 A toward X, which adds to the 2.0 A.
Working Positive ions: 2.0 C/s toward X, a current of 2.0 A toward X. Negative ions: 3.0 C/s of negative charge toward Y, a current of 3.0 A toward X. Total I = 2.0 A + 3.0 A = 5.0 A toward X.
In preparation: 0 of 2 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
11.1.A.1 Electric current, I Fix
Electric current, I
The rate at which charge passes through a cross-sectional area of a wire, I = Δq/Δt. Unit: ampere (A), where 1 A = 1 C/s.
Charge carriers
The charged particles whose motion makes up a current: electrons in metal wires; positive and negative ions in a solution.
emf, ε
The potential difference that makes charge move around a circuit, provided, for example, by a battery. Although the name 'electromotive force' contains the word 'force', emf is a potential difference, measured in volts.
Zero current
When the current in a section of wire is zero, the net motion of the charge carriers through it is zero, although each carrier is still moving: in a metal, the free electrons move rapidly in random directions.
Students often think Current is the amount of charge in a wire, or the amount that has passed through it, so more charge means more current. In fact No. Current is the RATE at which charge passes through a cross-section of the wire, I = Δq/Δt, in coulombs per second (amperes). A large amount of charge can pass with a small current if it takes a long time, and a wire full of charge carriers can have no current at all.
Students often think The current at an instant is the area under the graph of charge passed against time, up to that instant. In fact No. Since I = Δq/Δt, the current is the slope of a graph of charge passed against time. The area under a q–t graph has units of C·s, which is not a current.
11.1.A.2 Direction of current Fix
Direction of current
Current is not a vector, but it has a direction along its conductor: the direction in which positive charge would move. That direction is not tied to any coordinate system: a current does not become negative because its wire points along −x.
Conventional current
The direction chosen for current: the direction in which positive charge would move. In a circuit driven by a single battery, conventional current runs from the battery's positive terminal through the external circuit to its negative terminal.
Electron current in metals
In common circuits the moving charge carriers are electrons, which move opposite to the conventional current. Negative charge moving one way is equivalent to positive charge moving the other way.
Students often think Current is a vector, so currents in wires that meet at an angle combine by vector addition, and a current directed along −x is negative. In fact No. Current has a direction along its wire, but it is not a vector: it is the charge passing per second. Where wires meet, the charge arriving per second equals the charge leaving per second (charge is conserved), whatever the angles between the wires, so currents add as ordinary numbers; a current does not take a sign from a coordinate axis.
Students often think Current is the flow of electrons or other negative carriers, so its direction is the direction in which they move. In fact No. The direction of conventional current is defined as the direction in which positive charge would move. Negative carriers, such as electrons, moving one way make a current in the opposite direction.
9 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 9
The graph shows the total charge q that has passed through a cross-section of a wire as a function of time t. What is the current in the wire at t = 4.0 s?
Answer and reasoning
A6.0 A A student who takes the current to be the amount of charge that has passed picks this: the graph reads 6.0 C at 4.0 s. Current is the rate at which charge passes, the slope of the graph, not its height.
B3.0 ACorrect Current is the rate at which charge passes, I = Δq/Δt, which is the slope of the q–t graph. At t = 4.0 s the graph is on the straight section from (3.0 s, 3.0 C) to (5.0 s, 9.0 C), so I = (6.0 C)/(2.0 s) = 3.0 A.
C1.5 A A student who divides the total charge by the total time picks this: (6.0 C)/(4.0 s). That is the average current over the first 4.0 s, which includes the slower first 3.0 s; the current at 4.0 s is the slope of the graph there.
D9.0 A A student who uses the area under the graph from 0 to 4.0 s picks this: 4.5 + 4.5 = 9.0. That area has units of C·s, not amperes; the current is the slope of the graph.
Working At t = 4.0 s the graph is on the straight segment from (3.0 s, 3.0 C) to (5.0 s, 9.0 C): I = Δq/Δt = (9.0 C − 3.0 C)/(5.0 s − 3.0 s) = 3.0 A.
A copper wire has a thick section X joined to a thin section Y, and there is a steady current in the wire. How do the currents IX and IY in the two sections compare?
Answer and reasoning
AThey are equal: the same charge passes each cross-section per second.Correct In a steady current, charge does not pile up anywhere in the wire, so all the charge that passes a cross-section of X each second also passes a cross-section of Y. Current is the charge passing per second, so IX = IY.
BIY is greater, since the charges move faster through the narrow section Y. A student who takes the current to be the speed of the charge carriers picks this. The carriers do move faster in the thin section, but each length of it contains fewer carriers, so the charge passing per second is the same.
CIX is greater, since the thick section X holds more of the charge carriers. A student who takes the current to be the amount of charge present picks this. The thick section does contain more charge carriers, but current is the charge passing a cross-section per second, and that is the same in both sections.
DIt depends on the direction of the current, as current decreases along a wire. A student who thinks current is used up along a wire picks this, expecting the section the charge reaches first to carry more. Charge is conserved and does not pile up, so every cross-section of the wire carries the same current.
A length of copper wire is not connected to anything, so there is no current in it. Which statement correctly describes the free electrons in the wire?
Answer and reasoning
AThey are at rest, because a current is needed to set the carriers moving. A student who thinks no current means no moving charge picks this. The free electrons are moving, rapidly and randomly; there is no current because their motion has no net direction.
BThere are none; free electrons enter a wire from a battery connected to it. A student who thinks the charge in a circuit comes from the battery picks this. Copper contains an enormous number of free electrons whether or not it is connected to anything; a battery sets them drifting in one direction.
CThey are fixed in place; in a metal, the positive charges are what move. A student who thinks positive charges carry the current in metals picks this. In a metal the positive ions are held in place; the free electrons are the charges that move.
DThey move in random directions, so their net motion through the wire is zero.Correct Zero current means zero net motion of the charge carriers, not zero motion. The free electrons in the copper move rapidly in random directions, so as many cross any cross-section one way as the other, and no net charge is carried along the wire.
In the circuit shown, a battery is connected to a lightbulb by metal wires. In which direction do electrons move through the filament of the bulb?
Answer and reasoning
ADown through the filament, in the same direction as the current A student who takes the direction of the current to be the direction electrons move picks this. Conventional current is defined by the motion of positive charge; the electrons that carry the current in metal move the opposite way, up through the filament.
BUp through the filament, opposite to the conventional currentCorrect Conventional current leaves the battery's positive terminal, at the top, passes along the top wire to the bulb, goes down through the filament and returns along the bottom wire. In metal wires the current is carried by electrons, which move opposite to the conventional current: up through the filament.
CNowhere; positive charges move down through the filament A student who thinks positive charges carry the current in metal wires picks this. In metals the positive ions are held in place; the electrons move, opposite to the conventional current.
DInto the filament from both of its ends, where the two flows meet A student who pictures charge flowing from both battery terminals into the bulb picks this. The electrons move around the loop in one direction: from the negative terminal along the bottom wire, up through the filament, and back to the positive terminal.
A battery drives a steady current of 2.0 A around a circuit whose wires form a rectangle. In the top wire the current is directed in the +x direction, and in the bottom wire in the −x direction. A student claims that the current in the bottom wire is −2.0 A. Which response to the student's claim is correct?
Answer and reasoning
AThe claim is right, since a current directed along −x is a negative current. A student who treats current as a vector component picks this. A current has a direction along its wire, but it is not a vector, so it does not take a sign from the x axis; the bottom wire carries the same 2.0 A as the top wire.
BThe two wires carry opposite currents, so the net current of the circuit is zero. A student who thinks currents in opposite directions cancel picks this. There is one current, 2.0 A, around a single loop; the charge moving along the bottom wire is the same charge that moved along the top wire, and nothing cancels.
CThe bottom wire carries less than 2.0 A, since the charge reaches it last. A student who thinks current is used up around a circuit picks this. Charge is conserved and does not pile up, so every cross-section of a single loop carries the same 2.0 A, whatever order the charge reaches the wires in.
DBoth wires carry 2.0 A; a current's direction is along its wire, not a sign on an axis.Correct Current is not a vector. Its direction is the direction in which positive charge would move along the wire, and it is not tied to a coordinate axis. The same 2.0 A passes every cross-section of the single loop, in the top wire and in the bottom wire alike.
Working Single loop, steady current: the same charge per second passes every cross-section, so both wires carry 2.0 A. Current is a scalar; its direction along the wire does not give it a sign on the x axis.
The current in a wire increases at a steady rate from zero at time t = 0 to I₀ at time t = T. Which expression gives the charge Δq that passes through a cross-section of the wire between t = 0 and t = T?
Answer and reasoning
AΔq = I₀T/2Correct Because the current rises steadily from 0 to I₀, its average value is I₀/2, and I = Δq/Δt gives Δq = (I₀/2)T. This is the area of the triangle under the current–time graph, (1/2)(T)(I₀).
BΔq = I₀T A student who treats the current at one instant as the total charge divided by the total time, I₀ = Δq/T, picks this. That would hold only if the current had been I₀ throughout; here it was smaller at every earlier moment, and the average current is I₀/2.
CΔq = I₀/T A student who takes the charge from the slope of the current–time graph picks this. The slope, I₀/T, is the rate at which the current increases, in A/s; the charge is the area under the graph, in A·s = C.
DΔq = I₀ A student who thinks the current is the amount of charge that has passed takes the final current as the charge and picks this. Current is a rate, charge per second; the charge that passes also depends on how long the current flows.
Working The current rises linearly from 0 to I₀, so the average current over the interval is (0 + I₀)/2 = I₀/2. From I = Δq/Δt, Δq = Iavg·Δt = (I₀/2)T = I₀T/2. Equivalently, Δq is the area under the I–t graph, a triangle of base T and height I₀: (1/2)I₀T.
An ion source produces a steady beam in vacuum. Each second it emits N ions, each with charge +q, moving at speed v, and every ion reaches the target. The beam current is I₀. The source is changed so that each second it still emits N ions, but each now has charge +2q and moves at speed v/2. Because the ions are slower, they are now twice as closely spaced along the beam. What is the new beam current?
Answer and reasoning
A0.50I₀ A student who takes the current to be set by how fast the ions move picks this: the speed is halved, so they halve the current. Current is the charge crossing a cross-section per second, and N ions still cross it each second.
B4.00I₀ A student who takes the current to be the amount of charge present in the beam picks this: the ions are twice as close together and each carries twice the charge, so the beam holds four times as much charge. The current is the charge passing per second, which only doubles.
C1.00I₀ A student who takes the current to be the number of ions passing each second picks this: the source still emits N ions per second, so the current seems unchanged. Each ion now carries twice the charge, so twice as much charge passes each second.
D2.00I₀Correct The beam is steady and no ions are lost, so N ions still cross each cross-section every second. Each now carries 2q, so the charge passing per second, the current, doubles: 2.00I₀. The slower ions are closer together along the beam, but the number passing each second is unchanged.
Working In a steady beam in which no ions are lost, N ions cross every cross-section each second. I = Δq/Δt = N × (charge per ion). Before: I₀ = Nq. After: I = N(2q) = 2Nq = 2.00I₀. The speed does not appear.
A particle with charge +q moves at constant speed v in a circle of radius r. Its motion is equivalent to a current around the circle, equal to the rate at which charge passes any fixed point on the circle. Which expression gives this current?
Answer and reasoning
AI = qv A student who multiplies the charge by its speed, as if more charge moving faster were all that sets the current, picks this. qv has units C·m/s, not amperes. Current is the charge passing a fixed point per unit time: charge q passes once per revolution, which takes 2πr/v, so I = qv/(2πr).
BI = qv/(2πr)Correct Charge q passes a fixed point once in every revolution. One revolution is a distance 2πr, which takes T = 2πr/v, so I = Δq/Δt = q/T = qv/(2πr).
CI = 2πrq/v A student who multiplies the charge by the time instead of dividing picks this: q = I/T gives I = qT = 2πrq/v, which has units of C·s. Current is charge per unit time, I = q/T = qv/(2πr).
DI = qv/r A student who takes v/r as the number of revolutions per second picks this. v/r is the angular speed in radians per second; one revolution is 2π radians, a distance 2πr, so the particle goes around v/(2πr) times per second and I = qv/(2πr).
Working The particle passes a fixed point on the circle once per revolution, so Δq = q in each time Δt = T. One revolution is a distance 2πr at speed v, so T = 2πr/v. I = Δq/Δt = q/T = qv/(2πr). Errors: charge multiplied by its speed (m22): I = qv; charge multiplied by the time instead of divided, I = qT (m21): I = 2πrq/v; ω = v/r taken as revolutions per second (new): I = qv/r.
In an electron beam in a vacuum tube, 1.5 × 10¹⁵ electrons pass through a cross-section of the beam in 8.0 s at a steady rate. What is the magnitude of the current in the beam? Use e = 1.60 × 10⁻¹⁹ C.
Answer and reasoning
A2.4 × 10⁻⁴ A A student who takes the current to be the amount of charge that has passed picks this, stopping at 2.4 × 10⁻⁴ C. Current is a rate: the same charge passing in a longer time is a smaller current, I = Δq/Δt = 3.0 × 10⁻⁵ A.
B1.9 × 10⁻³ A A student who multiplies the charge by the time instead of dividing picks this: I = Δq × Δt. That product has units of C·s, not amperes. I = Δq/Δt = 3.0 × 10⁻⁵ A.
C3.0 × 10⁻⁵ ACorrect The charge that passes is Δq = Ne = (1.5 × 10¹⁵)(1.60 × 10⁻¹⁹ C) = 2.4 × 10⁻⁴ C. Current is the rate at which charge passes: I = Δq/Δt = (2.4 × 10⁻⁴ C)/(8.0 s) = 3.0 × 10⁻⁵ A.
D1.9 × 10¹⁴ A A student who takes the current to be the number of electrons passing each second picks this: 1.5 × 10¹⁵/8.0 s. Each electron carries charge e, so the charge per second is (1.9 × 10¹⁴ s⁻¹)(1.60 × 10⁻¹⁹ C) = 3.0 × 10⁻⁵ A.
Working Charge passing: Δq = Ne = (1.5 × 10¹⁵)(1.60 × 10⁻¹⁹ C) = 2.4 × 10⁻⁴ C. I = Δq/Δt = (2.4 × 10⁻⁴ C)/(8.0 s) = 3.0 × 10⁻⁵ A. Errors: current taken as the charge passed (m01): 2.4 × 10⁻⁴; charge multiplied by the time instead of divided, I = Δq·Δt (m21): 1.9 × 10⁻³; current taken as electrons per second (m18): 1.5 × 10¹⁵/8.0 = 1.9 × 10¹⁴.
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
