Charge balances at a junction
You will be able to: Apply the junction rule with signed directions and combine it with branch voltage relations.
Why must currents balance where wires meet?
If 5 coulombs enter a junction each second and 2 leave along one branch, 3 must leave along the other in steady operation. Otherwise charge would keep accumulating at that connection.
A useful starting point: Let a negative current correct your arrow →
Words and symbols before equations
- Incoming current
- Current directed toward the chosen junction.
- Outgoing current
- Current directed away from the chosen junction.
- Steady state
- Macroscopic values do not change with time; no continuing charge accumulation at an ideal junction.
- Signed branch current
- Value defined relative to an assumed branch arrow.
What this picture assumes
One ideal steady junction with no charge accumulation. Two specified currents determine the signed third. This local accounting diagram does not specify a complete driving circuit.
Read the picture in three steps
- Locate the labeled sources, system boundary or graph axes. Read the units before comparing values.
- 2=5+I₃, so I₃=-3 A. 3 A actually enters on the third branch.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the physics
Charge conservation gives ΣI_in=ΣI_out at a steady ideal junction. This does not assert equal branch currents. A negative solved value means that branch’s true direction is opposite your initial arrow.
Combine the junction rule with the loop rule and resistor relations to analyze a complete circuit. For two resistors across the same source, their shared voltage sets I₁=V/R₁ and I₂=V/R₂; the source current is then I₁+I₂.
The Explore panel isolates one junction to practice accounting. Its two specified currents may require the third arrow to reverse. It is not a complete source-and-resistor circuit, so no resistor values or energy supply are inferred from those current controls alone.
A worked example, step by step
At a junction 2 A enters, 5 A leaves on one branch, and I is assumed to leave on another. Find I.
- Use incoming=outgoing: 2=5+I.
- Solve I=−3 A.
- Thus 3 A actually enters along the branch whose arrow was drawn outward.
- Check physical directions: 2+3=5 A enters and leaves.
Current need not split equally; only the total incoming and outgoing charge rates must balance.
If incoming current exceeds one outgoing branch by 1 A, what does the other outgoing branch carry?
Compare with an explanation
1 A outward under the stated two-branch model.
Predict. Change one thing. Explain.
Change incoming current and the first outgoing current. Predict when the second branch must reverse direction, then check the signed readout.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
2=5+I₃, so I₃=-3 A. 3 A actually enters on the third branch.
One ideal steady junction with no charge accumulation. Two specified currents determine the signed third. This local accounting diagram does not specify a complete driving circuit.
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 questionA 12 V supply drives 6 Ω and 4 Ω parallel branches. (a) Find each current. (b) Write the junction equation. (c) Find source current. (d) Explain why equal splitting is wrong.
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Compare with the answer and four-point rubric
- 1 point: Branch currents are 2 A and 3 A.
- 1 point: I_source=I₁+I₂.
- 1 point: I_source=5 A.
- 1 point: Equal voltage across unequal resistances gives unequal V/R currents.
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 does the junction rule conserve?
Electric charge.
RECALL 2What does a negative branch solution indicate?
Actual current flows opposite the chosen arrow.
RECALL 3Does the junction rule alone fix every circuit current?
No. You also need source and component relations and suitable loop equations.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Charge balances at a junction
- ΣI_in=ΣI_out for a steady junction.
- Use one consistent arrow for each branch.
- Combine charge conservation with loop energy conservation.
Remember: Current need not split equally; only the total incoming and outgoing charge rates must balance.
Conditions: One ideal steady junction with no charge accumulation. Two specified currents determine the signed third. This local accounting diagram does not specify a complete driving circuit.
Refresh Kid · AP Physics 2 Unit 3 (official Unit 11) · Objectives 11.7.A · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 11.7, objectives 11.7.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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