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LESSON 01 / 22 · TOPIC 11.1

Current is the rate at which charge crosses a surface

You will be able to: Differentiate transferred charge and integrate signed current.

Calculus-based circuit analysisFree study resourceReview editionTeacher review pending

How do charge and current graphs describe the same flow?

Imagine counting charge that crosses a marked section of wire. If 6 coulombs cross in 3 seconds, the average current is 2 amperes. The instantaneous rate may still change during that interval.

A useful starting point: Capacitance and charge →

Words and symbols before equations

Current I
Signed charge-transfer rate, in amperes: 1 A = 1 C/s.
Transferred charge q
Cumulative signed charge crossing the chosen surface, in coulombs.
Positive direction
An assigned direction for conventional positive-charge flow.
Signed current over timeI (A)Time (s)0-111.7524.537.25410
Read this model snapshot. At t = 3 s: I = 7 A; net transferred q = 12 C. Positive transfer follows the chosen conventional-current direction.
What this picture assumes

Prescribed current I = a + bt; q(0) = 0. Positive direction is the selected surface orientation. Graph area is signed charge, not energy.

Read the picture in three steps

  1. Locate the labeled sources, system boundary or graph axes. Read the units before comparing values.
  2. At t = 3 s: I = 7 A; net transferred q = 12 C. Positive transfer follows the chosen conventional-current direction.
  3. Check what the picture assumes below. Use the Explore task to predict one change before moving a control.

Connect the picture to the physics

Current is I = dq/dt: the slope of the transferred-charge graph. Average current is Δq/Δt, which need not equal every instantaneous value.

Conversely, Δq = ∫I dt is the signed area under a current-time graph. Negative area represents transfer opposite the chosen positive direction; it is not negative elapsed time.

Conventional current follows positive-charge motion. In metal, electrons drift opposite that direction. Current is a signed flow rate through a cross-section, not a spatial vector to resolve into components.

A worked example, step by step

For I(t) = 1 + 2t A with t in seconds, find charge transferred from 0 to 3 s and current at 3 s.

  1. Choose the stated direction as positive.
  2. Integrate Δq = ∫₀³(1+2t)dt = [t+t²]₀³.
  3. Δq = 12 C; the average current is 12/3 = 4 A.
  4. Instantaneous current at 3 s is 1+2(3) = 7 A.
Common mix-up

Graph height gives current; area gives transferred charge. Average and instantaneous current differ.

CHECK THE IDEA

Can net transferred charge be zero with nonzero current earlier?

Compare with an explanation

Yes. Opposite signed areas can cancel.

Now investigate one change Explore →

Predict. Change one thing. Explain.

Change the initial current and its rate of change. Compare the selected-time current with accumulated charge, including an interval where the current reverses.

On narrow screens, swipe or scroll diagrams sideways to read all labels.

Signed current over timeI (A)Time (s)0-111.7524.537.25410

At t = 3 s: I = 7 A; net transferred q = 12 C. Positive transfer follows the chosen conventional-current direction.

Accumulated signed transferq (C)Time (s)0-114.5210315.5421

Prescribed current I = a + bt; q(0) = 0. Positive direction is the selected surface orientation. Graph area is signed charge, not energy.

Explain what you noticed: Which quantity changed? Which stayed fixed? Use the relevant current, circuit topology, charge or energy conservation, or RC relationship to justify your prediction.

Apply the idea to a fresh problem Practice →

Show what you understand.

Two original questions are a starting check, not proof of mastery. Explain your choice before revealing the answer.

1. The slope of q(t) represents…

Show answer and reasoning

current. Differentiating charge gives charge per time.

2. A steady −2 A for 4 s transfers…

Show answer and reasoning

−8 C. Δq = IΔt = −8 C in the chosen convention.

Original written challenge

4 points · self-check · not an official AP question

For I = 4−2t A between 0 and 4 s, find the reversal time, net charge transfer and total magnitude of charge crossing.

This response is not submitted or saved. Copy it before leaving.

Compare with the answer and four-point rubric
  1. 1 point: Set I = 0 to find t = 2 s.
  2. 1 point: Integrate [4t−t²]₀⁴ = 0 C net.
  3. 1 point: The first triangular area is +4 C and the second is −4 C.
  4. 1 point: The sum of magnitudes is 8 C, despite zero net transfer.

Accept equivalent correct methods and explanations. This is a Refresh Kid teaching rubric, not an official AP scoring guideline.

Recall the ideas without notes Review →

Retrieve it before you reveal it.

RECALL 1What does the sign of I mean?

Direction relative to the chosen surface orientation.

RECALL 2Do electrons follow conventional current?

No, electron drift is opposite conventional current.

RECALL 3What does current-time area measure?

Signed transferred charge.

Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.

Current is the rate at which charge crosses a surface

  • I = dq/dt.
  • Δq = ∫I dt.
  • 1 A = 1 C/s.

Remember: Graph height gives current; area gives transferred charge. Average and instantaneous current differ.

Conditions: Prescribed current I = a + bt; q(0) = 0. Positive direction is the selected surface orientation. Graph area is signed charge, not energy.

Refresh Kid · AP Physics C: Electricity and Magnetism Unit 4 (official Unit 11) · Objectives 11.1.A · Review edition

Framework, scope and review status

Mapped to College Board CED, Topic 11.1, objectives 11.1.A. CED effective Fall 2024, current PDF ©2026; checked September 16, 2026. This is E&M Unit 4: Electric Circuits, numbered Unit 11 in the official combined Physics C sequence. Topics 11.1–11.8 retain their official identifiers. Models include signed charge flow, prescribed current-density and resistivity integrals, DC resistor networks, nonideal batteries and meters, capacitor combinations and finite-resistance RC transients. Circuit schematics use conventional-current references and explicit node connectivity. Initial capacitor voltage and positive time constants are stated. Unequal ideal sources are never directly wired in parallel. The optional spatial wire view supplements a complete 2D current-density explanation. Checked with the Fall 2026 clarifications. The lesson breakdown and questions are original Refresh Kid work, not official topic subdivisions.

Implementation and automated checks are separate from independent teacher review and observation of students. Both human review stages remain pending. This is a review edition, not a certified or validated assessment.

Optional further resource: College Board’s released questions and scoring guides. Papers can combine units; this link is an archive, not an assignment of every question to this lesson.

Our learn, explore, practice and recall sequence is informed by the IES learning guide. The exact Refresh Kid implementation has not been evaluated for learning effectiveness.

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