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LESSON 14 / 22 · TOPIC 11.6

A loop returns to its starting potential

You will be able to: Use consistent traversal signs and interpret a potential profile around a loop.

Calculus-based circuit analysisFree study resourceReview editionTeacher review pending

What does a negative calculated current tell you?

Two batteries in one loop can oppose each other. Choose a current direction before calculating; the sign of the answer tells you whether that guess matched the actual flow.

A useful starting point: A measuring instrument becomes part of the circuit →

Words and symbols before equations

Traversal
Chosen direction for walking around the circuit.
Source rise
Crossing a battery from − to + changes potential by +ε.
Resistor drop
Traversing a resistor along its assigned current changes potential by −IR.
Opposing sources: signed clockwise reference+6 V1 Ω3 Ω+10 VReference I →Schematic geometry; ideal wires unless labeled otherwise
Read this model snapshot. Signed I = -1 A. Loop changes: +6, 1, 3, −10 V; sum = 0. Profile indices: 0 start, 1 after source 1, 2 after R₁, 3 after R₂, 4 after source 2. Distance is not physical wire length.
What this picture assumes

Single loop: reference traversal rises through source 1, follows both resistors and drops through source 2. Positive source terminals are at the top. Negative I means actual counterclockwise flow. Lumped circuit with no changing linked magnetic flux.

Read the picture in three steps

  1. Locate the labeled sources, system boundary or graph axes. Read the units before comparing values.
  2. Signed I = -1 A. Loop changes: +6, 1, 3, −10 V; sum = 0. Profile indices: 0 start, 1 after source 1, 2 after R₁, 3 after R₂, 4 after source 2. Distance is not physical wire length.
  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

For the drawn loop, traverse source ε₁ as a rise, then R₁ and R₂, then source ε₂ as a drop. The equation is ε₁−IR₁−IR₂−ε₂ = 0.

Thus I = (ε₁−ε₂)/(R₁+R₂). A negative I reverses actual flow relative to the reference arrows; keep the algebraic sign when evaluating voltage changes.

The potential profile starts at zero, rises across ε₁, changes across both resistors and returns through ε₂ to zero. Kirchhoff’s loop rule expresses energy conservation in this lumped circuit with no time-varying magnetic flux.

A worked example, step by step

Let ε₁ = 6 V, ε₂ = 10 V, R₁ = 1 Ω and R₂ = 3 Ω. Find signed I.

  1. Choose the reference traversal shown.
  2. Write 6−I(1)−I(3)−10 = 0.
  3. I = −1 A, so actual flow opposes the reference.
  4. The signed changes are +6, +1, +3, −10 V; their sum is zero.
Common mix-up

A negative answer is not automatically an error. It can correct an initially chosen direction.

CHECK THE IDEA

If you traverse the same loop backward, does the physical current change?

Compare with an explanation

No. All traversal signs reverse consistently; the physical solution is unchanged.

Now investigate one change Explore →

Predict. Change one thing. Explain.

Raise ε₂ above ε₁ and watch the signed current reverse. Follow each voltage step and check that the loop still ends at zero.

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

Opposing sources: signed clockwise reference+6 V1 Ω3 Ω+10 VReference I →Schematic geometry; ideal wires unless labeled otherwise

Signed I = -1 A. Loop changes: +6, 1, 3, −10 V; sum = 0. Profile indices: 0 start, 1 after source 1, 2 after R₁, 3 after R₂, 4 after source 2. Distance is not physical wire length.

Potential after each element in the reference traversalPotential (V)Traversal boundary index (see readout)0-1122538411

Single loop: reference traversal rises through source 1, follows both resistors and drops through source 2. Positive source terminals are at the top. Negative I means actual counterclockwise flow. Lumped circuit with no changing linked magnetic flux.

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 loop rule follows conservation of…

Show answer and reasoning

energy. Net energy change per charge around the closed loop is zero.

2. With opposing 12 V and 4 V sources and total 4 Ω, I is…

Show answer and reasoning

2 A. The net driving emf is 8 V.

Original written challenge

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

For ε₁ = 9 V, ε₂ = 3 V, R₁ = 2 Ω and R₂ = 4 Ω, calculate current, each resistor drop and the complete potential sequence.

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

Compare with the answer and four-point rubric
  1. 1 point: I = (9−3)/(2+4) = 1 A.
  2. 1 point: Drops are 2 V and 4 V.
  3. 1 point: Potential sequence is 0, 9, 7, 3, 0 V.
  4. 1 point: The changes +9−2−4−3 sum to zero.

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 fixes a source sign?

Which terminal you cross first during traversal.

RECALL 2What does negative I indicate?

Actual current opposes the chosen reference.

RECALL 3Why are wire segments flat on the profile?

They are modeled as ideal zero-resistance wires.

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

A loop returns to its starting potential

  • ΣΔV_loop = 0.
  • Battery −→+: +ε; +→−: −ε.
  • Resistor along assigned I: −IR.

Remember: A negative answer is not automatically an error. It can correct an initially chosen direction.

Conditions: Single loop: reference traversal rises through source 1, follows both resistors and drops through source 2. Positive source terminals are at the top. Negative I means actual counterclockwise flow. Lumped circuit with no changing linked magnetic flux.

Refresh Kid · AP Physics C: Electricity and Magnetism Unit 4 (official Unit 11) · Objectives 11.6.A · Review edition

Framework, scope and review status

Mapped to College Board CED, Topic 11.6, objectives 11.6.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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