AP Physics 2 · Topic 11.1

Topic 11.1: Electric Current

Unit 11: Electric Circuits15-18% of the multiple-choice section

Electric current is the rate at which charge passes through a cross-sectional area, in amperes: one ampere is one coulomb per second. Current is not a vector, but it does have a direction, and on the AP Physics 2 exam that direction is conventional current, the way positive charge would move.

AP Physics: Unit 11 (topics 11.1 Electric Current). AP Physics 2 Unit 11, Topic 11.1. The single learning objective, 11.1.A, asks students to describe the movement of electric charges through a medium. Two essential knowledge statements support it: 11.1.A.1 defines current as the rate at which charge passes through a cross-sectional area of a wire and gives the equation I = Δq/Δt, with sub-statements on potential difference and emf (11.1.A.1.i) and on zero current meaning zero net motion but not zero carrier speed (11.1.A.1.ii); 11.1.A.2 states that current is not a vector but does have a direction, tied to the motion of positive charge rather than to any coordinate system, with sub-statements defining conventional current (11.1.A.2.i) and noting that common circuits carry current by electron motion (11.1.A.2.ii). Topic 11.1 has no boundary statement of its own. The CED's suggested skills here are 1.A, 2.C, 3.B, and 3.C. Unit 11 carries 15 to 18 percent of the multiple-choice section and a suggested 12 to 20 class periods.

What Topic 11.1 requires

Topic 11.1 carries one learning objective, 11.1.A: describe the movement of electric charges through a medium. Two essential knowledge statements sit under it, and each of them adds two sub-statements.

  • 11.1.A.1 states that current is the rate at which charge passes through a cross-sectional area of a wire, and gives the relevant equation I=ΔqΔtI = \frac{\Delta q}{\Delta t}.
  • 11.1.A.1.i states that electric charge moves in a circuit in response to an electric potential difference, sometimes referred to as electromotive force, or emf (ε\varepsilon).
  • 11.1.A.1.ii states that if the current is zero in a section of wire, the net motion of charge carriers in the wire is also zero, although individual charge carriers will not have zero speed.
  • 11.1.A.2 states that although current is not a vector quantity, it does have a direction, and that the direction of current is associated with what the motion of positive charge would be but not with any coordinate system in space.
  • 11.1.A.2.i states that the direction of conventional current is chosen to be the direction in which positive charge would move.
  • 11.1.A.2.ii states that in common circuits, current is actually due to the movement of electrons (negative charge carriers).

Topic 11.1 has no boundary statement of its own. The unit's first one belongs to Topic 11.2, and it is worth reading here anyway: unless otherwise specified, all circuit schematic diagrams will be drawn using conventional current.

Read the learning objective again and notice the noun. It says medium, not wire. Essential knowledge 11.1.A.1 narrows to a wire because that is the usual case, but 11.1.A covers charge moving through anything that can carry it, which is why the third worked example below counts two kinds of carrier at once.

The CED lists four suggested skills for this topic: 1.A, create diagrams, tables, charts, or schematics to represent physical situations; 2.C, compare physical quantities between two or more scenarios or at different times and locations in a single scenario; 3.B, apply an appropriate law, definition, theoretical relationship, or model to make a claim; and 3.C, justify or support a claim using evidence from experimental data, physical representations, or physical principles or laws. Unit 11 is weighted at 15 to 18 percent of the multiple-choice section, and the CED suggests roughly 12 to 20 class periods for the whole unit.

Notice what is absent. Resistance, resistivity and Ohm's law are Topic 11.3; power is Topic 11.4. Topic 11.1 is the definition plus the direction convention that travels with it, placed first so every later equation in the unit has something well defined to move around.

Reading I = Δq/Δt term by term

I=ΔqΔtI = \frac{\Delta q}{\Delta t}

Read it as a rate. Δq\Delta q is the amount of charge that crosses a chosen cross-section, in coulombs. Δt\Delta t is the time it takes, in seconds. Divide and you get amperes: one ampere is one coulomb per second. Both the ampere (A) and the coulomb (C) appear in the unit-symbol table printed on the AP Physics 2 Table of Information.

Three things in that definition are easy to read past.

  • Current is counted at a surface, not along a length. You pick a cross-section, you stand there, and you count coulombs going by. "The current in the wire" means that count, taken at any point along the wire.
  • The equation gives an average over the interval Δt\Delta t. When the current is steady the average equals the instantaneous value and the distinction never bites. When it is not steady, as in the discharging capacitor of Topic 11.8, the average over an interval and the value at an instant are different numbers.
  • Charge goes in and the same charge comes out. Charge is conserved, which is Topic 10.2's statement, so nothing accumulates in a plain length of wire and nothing is consumed there.

Rearranged, the definition gives the form you use more often in practice: Δq=IΔt\Delta q = I \Delta t. A steady 1 A running for 1 s moves 1 C past every cross-section of the conductor.

It also helps to hear the equation as one member of a family of rate definitions. In fluids, volume flow rate counts cubic meters past a cross-section every second, which is what conservation applied to a flowing fluid is built on. Current counts coulombs instead of cubic meters: same grammar, different noun.

What makes charge move: potential difference and emf (11.1.A.1.i)

Essential knowledge 11.1.A.1.i is the causal half of the topic: electric charge moves in a circuit in response to an electric potential difference, sometimes referred to as electromotive force, or emf (ε\varepsilon).

Two readings of that sentence matter.

First, the driver is a potential difference, which is Topic 10.5's quantity rather than a new one. A battery does not hand charge to the circuit. It maintains a difference in electric potential between its terminals, and the charge already sitting in the conductor responds. The CED calls out the alternative belief by name: the unit's Preparing for the AP Exam note gives "students may believe that batteries store charge, or that current is used up in a circuit" as a misconception to challenge.

Second, emf is a potential difference, not a force, despite what the words spell out. The name is historical and the CED keeps it only as an alias, with the symbol ε\varepsilon. The AP Physics 2 equation sheet prints no equation for emf in its Electricity column at all; the symbol ε\varepsilon for emf shows up in the symbol list beside the Magnetism column. The quantitative treatment, a battery's emf set against the potential difference actually measured across its terminals, is Topic 11.5, where essential knowledge 11.5.B.1.iii defines the emf as the potential difference measured across the terminals when there is no current in the battery.

For Topic 11.1 the working statement is short: a loop with no source of potential difference in it carries no current. Read that at the level of the loop, not the element. An ideal wire carries current with no potential difference across it at all, and that is precisely what makes a short circuit a short circuit in Topic 11.2.

Zero current is not zero motion (11.1.A.1.ii)

Essential knowledge 11.1.A.1.ii: if the current is zero in a section of wire, the net motion of charge carriers in the wire is also zero, although individual charge carriers will not have zero speed.

This is the CED taking a convenient shortcut away from you. "No current" does not mean "the electrons have stopped". Carriers in a metal are always moving, quickly and in effectively random directions. What zero current says is that the counts balance: over the interval you watch, as much charge crosses your cross-section one way as the other, so Δq=0\Delta q = 0 and therefore I=0I = 0.

The distinction is between net motion and motion. Current is a statement about the net, and the net can be zero while every single carrier is moving. Two consequences follow.

  • An open switch does not freeze the electrons in the wire behind it. It removes the closed path, so the net transport goes to zero.
  • Switching a current on does not start the carriers moving. It biases motion that was already there, so that it stops cancelling.

This is also the statement that makes suggested skills 3.B and 3.C answerable in words. "There is no current, so nothing in the wire is moving" is exactly the sort of claim a justification question hands you to correct, and 11.1.A.1.ii is the evidence you cite.

Current has a direction, but it is not a vector (11.1.A.2)

Essential knowledge 11.1.A.2 is worded with unusual care, and the care is the point: although current is not a vector quantity, it does have a direction, and the direction of current is associated with what the motion of positive charge would be but not with any coordinate system in space.

Take the second clause first. The direction of a current is never "40 degrees above the x-axis". It is "this way around the loop", or "from a to b through the resistor". A current follows its conductor, and the conductor is free to bend, so the direction travels with the wire rather than with your axes. That is what the phrase about coordinate systems rules out.

Now the first clause. Current behaves as a scalar carrying a sign attached to a chosen sense. In practice:

  • You never resolve a current into xx and yy components.
  • You never combine two currents with a parallelogram or with the Pythagorean theorem. At a junction they add as signed numbers, which is what Kirchhoff's junction rule formalises in Topic 11.7.
  • A negative answer for a current is not a mistake to clean up. It means the current runs opposite to the sense you drew. Leave the sign and read it back as a direction at the end.

Set it beside a Unit 10 quantity that genuinely is a vector.

Electric currentElectric field
Vector or scalarNot a vector; has a direction (11.1.A.2)Vector
How the sheet writes itIIE\vec{E}, with the arrow printed
ComponentsNever resolved into componentsResolved into components routinely
Combining two of themSigned sum at a junctionVector sum at a point
What sets the directionThe path of the conductorDirections in space

The short version to say out loud: a current has a direction along a wire, not a direction in space. Topic 10.3 is where the vector in the right-hand column is developed.

Conventional current and electron flow (11.1.A.2.i and 11.1.A.2.ii)

Two sub-statements, and they sound like opposites on purpose.

  • 11.1.A.2.i The direction of conventional current is chosen to be the direction in which positive charge would move.
  • 11.1.A.2.ii In common circuits, current is actually due to the movement of electrons (negative charge carriers).

Both hold at once because conventional current is a choice of bookkeeping, made before anyone knew what the carriers were and kept because it costs nothing. Negative charge moving left transports exactly as much charge across your cross-section, in the rightward sense, as positive charge moving right does, so the convention never changes an answer.

What settles the ambiguity for you is printed. The AP Physics 2 Table of Information carries a list of nine conventions under the heading "The following conventions are used in this exam unless otherwise stated", and one of them reads "Current is conventional current." The Topic 11.2 boundary statement repeats it for diagrams: unless otherwise specified, all circuit schematic diagrams will be drawn using conventional current. In both places the escape clause is explicit, which is what leaves room for a question that hands you electron flow instead.

Conventional currentElectron flow
DirectionThe way positive charge would move (11.1.A.2.i)The way the electrons drift, in a metal
In a metal wireOut of the higher-potential terminal and around the external loopOpposite to the conventional current
Status on the AP Physics 2 examThe default, by printed conventionOnly when a question says so

Neither description is more correct than the other, and 11.1.A.2.ii is the CED making sure you know the everyday case is the one where the two disagree.

One habit follows. An unlabelled arrow on a wire in an AP Physics 2 diagram is conventional current. When a question instead describes electrons moving in some direction, the current is in the other direction, and writing that out in a response is the sort of vocabulary precision the unit's Preparing for the AP Exam note asks for by name.

What the equation sheet gives you for Topic 11.1

The Electricity group of the AP Physics 2 equation sheet holds 20 entries. Exactly one of them is Topic 11.1's:

I=ΔqΔtI = \frac{\Delta q}{\Delta t}

The other 19 belong to Unit 10 and to the later topics of Unit 11: Coulomb's law with the field, potential-energy and potential lines that follow it, the four capacitance and stored-energy lines, R=ρAR = \frac{\rho \ell}{A} and I=ΔVRI = \frac{\Delta V}{R} for Topic 11.3, P=IΔVP = I \Delta V for Topic 11.4, the four series and parallel combination rules for resistors and capacitors, and τ=ReqCeq\tau = R_{\text{eq}} C_{\text{eq}} for Topic 11.8.

Two of those deserve flagging here because they get misquoted.

  • Ohm's law prints as I=ΔVRI = \frac{\Delta V}{R}, with the delta. The numerator is a potential difference, not a potential. Write it the way the sheet writes it, and if you want the routine for using it, the Ohm's law guide and the Ohm's law calculator have it.
  • I=ΔqΔtI = \frac{\Delta q}{\Delta t} and I=ΔVRI = \frac{\Delta V}{R} are different kinds of statement. The first defines what current is. The second says how much current a particular ohmic element carries under a given potential difference. Only the first is Topic 11.1, and only the first still applies when the element is not ohmic.

Things the sheet does not hand you for this topic: no equation for drift speed, none for emf, and none for a count of charge carriers. When a question asks how many electrons a current corresponds to, you take the elementary charge e=1.60×1019e = 1.60 \times 10^{-19} C from the Constants and Conversion Factors table and divide. Every line of the sheet is laid out on the AP Physics 2 formula page.

How Topic 11.1 is tested, and where it leads

The four suggested skills point at four shapes of question.

  • 1.A, create diagrams, tables, charts, or schematics. Turn a description into a labelled diagram with a current arrow on it, drawn as conventional current.
  • 2.C, compare physical quantities between two or more scenarios or at different times and locations in a single scenario. Two conductors and two currents, or one conductor at two different times. The comparison is the answer, and the second worked example below is exactly this shape.
  • 3.B, apply an appropriate law, definition, theoretical relationship, or model to make a claim. Usually the definition I=Δq/ΔtI = \Delta q / \Delta t used to support a claim rather than to produce a number.
  • 3.C, justify or support a claim using evidence from experimental data, physical representations, or physical principles or laws. The written half. 11.1.A.1.ii and 11.1.A.2 are the two statements you will cite most.

The unit's Preparing for the AP Exam note is unusually specific about language. Students should know the differences in meaning between "current", "potential difference", "resistance", "resistivity", and "capacitance", it says, and it warns that students can inadvertently miscommunicate their answers by using words incorrectly or not fully understanding their nuances. It adds that on the free-response section, written expression and justification represent many of the available points. A sentence that swaps current for potential difference costs you there.

Where the topic leads:

  • Topic 11.2, Simple Circuits puts this current into a loop and gives you the schematic to read it off.
  • Topic 11.3 relates current to potential difference and resistance.
  • Topic 11.4 turns current and potential difference into a rate of energy transfer.
  • Topic 11.7 makes the signed arithmetic of the section above into a rule at a junction.

The unit opener also sets up what comes after Unit 11: in Unit 12, students will expand their investigations of the similarities and differences between electric and magnetic fields.

Charge delivered by a steady current, and how many electrons that is

A steady current of 0.75 A runs in a copper wire for 4.0 minutes. Find (a) the charge that passes one cross-section of the wire in that time, (b) the number of electrons that corresponds to, and (c) the direction those electrons move relative to the current arrow drawn on the wire.

  1. Convert the time before touching anything else. Δt=4.0×60=240s\Delta t = 4.0 \times 60 = 240 \, \mathrm{s}. The definition is written in coulombs and seconds, so the minutes have to go first.

  2. Rearrange the definition rather than the numbers. I=Δq/ΔtI = \Delta q / \Delta t gives Δq=IΔt\Delta q = I \Delta t.

  3. (a) Δq=(0.75A)(240s)=180C\Delta q = (0.75 \, \mathrm{A})(240 \, \mathrm{s}) = 180 \, \mathrm{C}. That is charge crossing a cross-section, not charge stored anywhere. The same 180 C leaves the far end of the wire.

  4. (b) Each electron carries charge of magnitude e=1.60×1019Ce = 1.60 \times 10^{-19} \, \mathrm{C}, taken from the Constants and Conversion Factors table on the AP Physics 2 sheet. So N=Δq/e=180/(1.60×1019)=1.125×1021N = \Delta q / e = 180 / (1.60 \times 10^{-19}) = 1.125 \times 10^{21}.

  5. Round to the two significant figures the data supports: 1.1×10211.1 \times 10^{21} electrons. Check it back: (1.125×1021)(1.60×1019)=180C(1.125 \times 10^{21})(1.60 \times 10^{-19}) = 180 \, \mathrm{C}.

  6. (c) The arrow on the wire is conventional current, the direction positive charge would move (11.1.A.2.i). The carriers in copper are electrons, which are negative (11.1.A.2.ii), so they drift opposite to the arrow.

(a) Δq=180C\Delta q = 180 \, \mathrm{C}. (b) About 1.1×10211.1 \times 10^{21} electrons. (c) Opposite to the current arrow: the arrow is conventional current and the carriers are negative.

Comparing the current at two different times from charge data

A meter records the total charge that has passed one cross-section of a conductor since t=0t = 0. The readings are 0 C at 0 s, 1.6 C at 2.0 s, 3.2 C at 4.0 s, 4.4 C at 6.0 s, and 5.6 C at 8.0 s. Find the current during the first 4.0 s and during the last 4.0 s, find the average current over the whole 8.0 s, and say what the three numbers together tell you.

  1. Each pair of readings gives an average current over the interval between them, because I=Δq/ΔtI = \Delta q / \Delta t uses a change in charge, not a total.

  2. First 4.0 s: Δq=3.20=3.2C\Delta q = 3.2 - 0 = 3.2 \, \mathrm{C} over Δt=4.0s\Delta t = 4.0 \, \mathrm{s}, so I=3.2/4.0=0.80AI = 3.2/4.0 = 0.80 \, \mathrm{A}.

  3. Last 4.0 s: Δq=5.63.2=2.4C\Delta q = 5.6 - 3.2 = 2.4 \, \mathrm{C} over Δt=4.0s\Delta t = 4.0 \, \mathrm{s}, so I=2.4/4.0=0.60AI = 2.4/4.0 = 0.60 \, \mathrm{A}.

  4. Check the 2.0 s sub-intervals before calling either half a constant current. They give 1.6/2.0=0.801.6/2.0 = 0.80, (3.21.6)/2.0=0.80(3.2-1.6)/2.0 = 0.80, (4.43.2)/2.0=0.60(4.4-3.2)/2.0 = 0.60 and (5.64.4)/2.0=0.60(5.6-4.4)/2.0 = 0.60, all in amperes. Steady within each half, at two different values.

  5. Whole interval: Iavg=5.6/8.0=0.70AI_{\text{avg}} = 5.6/8.0 = 0.70 \, \mathrm{A}.

  6. Compare, which is what suggested skill 2.C asks for. The current dropped by a factor of 0.80/0.60=4/30.80/0.60 = 4/3 between the two halves, and the 0.70 A average matches neither half. An average current over an interval is a real number, but it is not the current at any instant unless the current is steady throughout.

  7. One more reading of the same data: on a graph of charge against time, these currents are slopes. A steeper line is a larger current, and a horizontal stretch would be zero current.

First 4.0 s: I=0.80AI = 0.80 \, \mathrm{A}. Last 4.0 s: I=0.60AI = 0.60 \, \mathrm{A}. Over the full 8.0 s: Iavg=0.70AI_{\text{avg}} = 0.70 \, \mathrm{A}. The current is steady within each half but not across the interval, and the overall average equals neither half's value.

Two kinds of charge carrier in one medium

Learning objective 11.1.A is about charge moving through a medium, and some media carry charge both ways at once. Across a fixed plane in an ionic solution, over an interval of 5.0 s, 3.0 C of positive ions cross moving to the right while 2.0 C of negative charge crosses the same plane moving to the left. Find the current through the plane, with its direction.

  1. Fix the positive sense first and keep it for the whole problem: take rightward to be the positive direction for current.

  2. The positive ions move rightward and carry +3.0C+3.0 \, \mathrm{C} with them. Their contribution to the rightward transport of charge is +3.0C+3.0 \, \mathrm{C}.

  3. The negative charge moves leftward. Carrying 2.0C-2.0 \, \mathrm{C} to the left is the same transfer as carrying (2.0)=+2.0C-(-2.0) = +2.0 \, \mathrm{C} to the right, so its contribution to the rightward transport is also +2.0C+2.0 \, \mathrm{C}. This is 11.1.A.2.i doing real work: the current direction is defined by what positive charge would do.

  4. Add the contributions, do not subtract them: Δq=3.0+2.0=5.0C\Delta q = 3.0 + 2.0 = 5.0 \, \mathrm{C} transported rightward in 5.0 s.

  5. I=Δq/Δt=5.0/5.0=1.0AI = \Delta q / \Delta t = 5.0/5.0 = 1.0 \, \mathrm{A}, directed to the right.

  6. Name the trap so you do not fall into it later. Subtracting to get 1.0 C and then 0.20 A treats the two carrier types as opposing currents. They are not opposing. Opposite charges moving in opposite directions produce current in the same direction.

I=1.0AI = 1.0 \, \mathrm{A}, directed to the right. The two carrier types add rather than cancel, because negative charge moving left is equivalent, as far as current is concerned, to positive charge moving right.

Frequently asked questions

What is electric current in AP Physics 2?

Electric current is the rate at which charge passes through a cross-sectional area, which is essential knowledge 11.1.A.1 of AP Physics 2 Topic 11.1. The equation is I = Δq/Δt: the charge in coulombs that crosses a chosen cross-section, divided by the time in seconds it takes to cross. The unit is the ampere, and one ampere is one coulomb per second. The learning objective, 11.1.A, describes charge moving through a medium, so it is not restricted to metal wires.

What is the formula for electric current?

I = Δq/Δt, printed in the Electricity group of the AP Physics 2 equation sheet. Δq is the charge in coulombs that passes a cross-section of the conductor and Δt is the time interval in seconds, so the current comes out in amperes. Rearranged as Δq = IΔt, it gives the charge a steady current delivers in a given time. Ohm's law, which the same sheet prints as I = ΔV/R, is a different statement and belongs to Topic 11.3.

Is electric current a vector or a scalar?

Current is not a vector, but it does have a direction. Essential knowledge 11.1.A.2 states that although current is not a vector quantity, it does have a direction, and that the direction is associated with what the motion of positive charge would be but not with any coordinate system in space. In practice that means you never resolve a current into x and y components and never combine two currents with a parallelogram. Currents add as signed numbers, which is what Kirchhoff's junction rule does in Topic 11.7.

What is the difference between conventional current and electron flow?

Conventional current points the way positive charge would move, which is essential knowledge 11.1.A.2.i. Electron flow points the way the electrons actually drift, and in a common metal circuit that is the opposite direction, because the carriers are negative (11.1.A.2.ii). Both describe the same transport of charge, so neither is wrong. The AP Physics 2 Table of Information settles which one the exam means: "Current is conventional current" is one of the conventions used unless a question states otherwise.

If the current is zero, are the electrons in the wire standing still?

No. Essential knowledge 11.1.A.1.ii states that if the current is zero in a section of wire, the net motion of charge carriers in the wire is also zero, although individual charge carriers will not have zero speed. The carriers keep moving in effectively random directions. Zero current means the counts balance: as much charge crosses a given cross-section one way as the other, so the net charge transported over the interval is zero.

What unit is electric current measured in?

The ampere, written A, which is listed in the unit-symbol table on the AP Physics 2 Table of Information. One ampere is one coulomb per second, straight from I = Δq/Δt. Circuit problems often quote currents in milliamps, and 1 mA is 0.001 A, using the milli prefix from the prefix table printed on the same page.

Is current used up as it goes around a circuit?

No. The AP Physics 2 course description names that belief as a misconception to challenge in Unit 11, alongside the belief that batteries store charge. Charge is conserved, so the same amount of charge that enters a length of wire leaves it, and the current is the same at every point along a single unbranched path. What the circuit transfers out to its surroundings is energy, not charge.