Let a negative current correct your arrow
You will be able to: Solve a single loop with opposing ideal emfs and interpret the sign of current.
What if two batteries oppose each other?
Two ideal batteries can push charge in opposing directions around one resistive loop. Choose an arrow first; the algebra will tell you whether the stronger source drives current with or against it.
A useful starting point: Walk a loop and account for every volt →
Words and symbols before equations
- Assumed current
- A reference arrow chosen before solving; a negative result reverses its physical direction.
- Opposing emfs
- One source gives a rise while the other gives a drop along the chosen traversal.
- Source absorption
- Current entering a source’s positive terminal transfers energy into that ideal source.
What this picture assumes
Two ideal opposing emfs in one series loop with positive resistance. Clockwise current is the signed reference; a negative result is counterclockwise. This is not unequal ideal sources in parallel.
Read the picture in three steps
- Locate the labeled sources, system boundary or graph axes. Read the units before comparing values.
- +6−10−I(2)=0. I=-2 A (counterclockwise actual); resistor power=8 W.
- 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 clockwise reference current through a total resistance R, with source rises ε₁ and opposing ε₂, the loop equation is ε₁−ε₂−IR=0. Thus I=(ε₁−ε₂)/R. Both source polarities must be shown.
If ε₂ exceeds ε₁, I is negative: actual conventional current is counterclockwise. Do not replace the result by zero or treat a negative current as impossible. Resistor heating remains I²R≥0.
When emfs match, this ideal resistive-loop model has zero current. These sources are in one series loop, not unequal ideal batteries directly connected in parallel. Real charging also requires suitable chemistry and control; the diagram is an energy-accounting model, not a battery-building instruction.
A worked example, step by step
Opposing sources ε₁=6 V and ε₂=10 V share a 2 Ω loop. Find current relative to the ε₁-driven arrow.
- Write +6−10−2I=0.
- I=−2 A relative to the chosen arrow.
- Actual current is 2 A in the opposite direction, driven by the 10 V source.
- The resistor dissipates I²R=8 W; net source delivery is (10−6)(2)=8 W.
The arrow is a reference choice. A negative solution reverses the current direction, not the resistance sign.
When both ideal emfs are 8 V, what is the current in this resistive loop?
Compare with an explanation
Zero: the opposing rises cancel.
Predict. Change one thing. Explain.
Increase the opposing emf through the first emf’s value. Predict where current is zero and how its direction changes.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
+6−10−I(2)=0. I=-2 A (counterclockwise actual); resistor power=8 W.
Two ideal opposing emfs in one series loop with positive resistance. Clockwise current is the signed reference; a negative result is counterclockwise. This is not unequal ideal sources in parallel.
Explain what you noticed: Which quantity changed? Which stayed fixed? Use the relevant current, voltage or energy 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.
Original written challenge
4 points · self-check · not an official AP questionIn one loop ε₁=12 V and opposing ε₂=8 V, with R=4 Ω. (a) State your reference arrow. (b) Write the loop equation. (c) Find I. (d) Find resistor power.
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Compare with the answer and four-point rubric
- 1 point: Choose current driven by ε₁ as positive.
- 1 point: 12−8−4I=0.
- 1 point: I=1 A in the chosen direction.
- 1 point: P=I²R=4 W, equal to net source delivery.
Accept equivalent correct methods and explanations. This is a Refresh Kid teaching rubric, not an official AP scoring guideline.
Retrieve it before you reveal it.
RECALL 1What should you do with a negative current result?
Reverse the physical direction while retaining the magnitude.
RECALL 2What fixes battery signs in a loop?
The polarity crossed along the chosen traversal.
RECALL 3Can a resistor dissipate negative power?
A passive resistor dissipates I²R, which is nonnegative.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Let a negative current correct your arrow
- One opposing-source loop: I=(ε₁−ε₂)/R.
- Negative I means opposite the assumed arrow.
- Resistor power I²R remains nonnegative.
Remember: The arrow is a reference choice. A negative solution reverses the current direction, not the resistance sign.
Conditions: Two ideal opposing emfs in one series loop with positive resistance. Clockwise current is the signed reference; a negative result is counterclockwise. This is not unequal ideal sources in parallel.
Refresh Kid · AP Physics 2 Unit 3 (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. Refresh Kid calls this the third AP Physics 2 unit; College Board numbers it Unit 11; the first unit in this course is official Unit 9. 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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