Reduce a network one valid group at a time
You will be able to: Reduce a series-plus-parallel network and reconstruct branch quantities.
How can a series resistor affect both parallel branches?
A source feeds a resistor before the path splits into two branches. The first resistor carries all source current; the branches receive only the remaining voltage.
A useful starting point: Parallel branches share voltage and split current →
Words and symbols before equations
- Series feeder R_s
- Resistor before the junction carrying total current.
- Parallel block R_p
- Equivalent resistance of the two downstream branches.
- Reconstruction
- Working backward from total current to voltages and branch currents.
What this picture assumes
Series feeder before two parallel branches. Reduce the parallel group first, then reconstruct its voltage and each branch current.
Read the picture in three steps
- Locate the labeled sources, system boundary or graph axes. Read the units before comparing values.
- R_eq = 4 Ω; branch voltage = 6 V; I₁ = 1, I₂ = 2, source I = 3 A.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the physics
Combine the true parallel group first: R_p = R₁R₂/(R₁+R₂). Then R_total = R_s+R_p.
Find I_total = V_source/R_total. The parallel block voltage is V_p = I_total R_p, not automatically the full battery voltage.
Recover each branch current from V_p/R_i and check their sum. Equivalent resistance preserves terminal behavior; it does not imply equal current through every original component.
A worked example, step by step
A 12 V source feeds R_s = 2 Ω followed by parallel 6 Ω and 3 Ω. Find currents.
- R_p = 6(3)/(6+3) = 2 Ω.
- R_total = 4 Ω and I_total = 3 A.
- V_p = 3(2) = 6 V.
- Branch currents are 1 A and 2 A; their sum is 3 A.
Do not apply the full source voltage to branches behind a series voltage drop.
Can two resistors touching at a node always be combined in series?
Compare with an explanation
No. If a third conducting branch leaves that node, their currents need not be equal.
Predict. Change one thing. Explain.
Increase the feeder resistance at fixed source voltage and branch resistances. Explain why both branch currents fall.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
R_eq = 4 Ω; branch voltage = 6 V; I₁ = 1, I₂ = 2, source I = 3 A.
Series feeder before two parallel branches. Reduce the parallel group first, then reconstruct its voltage and each branch current.
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.
Original written challenge
4 points · self-check · not an official AP questionFor a 10 V source, R_s = 3 Ω and two parallel 4 Ω resistors, find source current, branch voltage and each branch current.
This response is not submitted or saved. Copy it before leaving.
Compare with the answer and four-point rubric
- 1 point: R_p = 2 Ω.
- 1 point: R_total = 5 Ω, so I_total = 2 A.
- 1 point: V_p = 4 V.
- 1 point: Each branch carries 1 A; 1+1 = 2 A.
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 is reduced first here?
The two resistors sharing both nodes.
RECALL 2What must be reconstructed afterward?
Original branch voltages and currents.
RECALL 3What check catches many errors?
Branch-current sum and total loop voltage.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Reduce a network one valid group at a time
- R_p = R₁R₂/(R₁+R₂).
- R_total = R_s+R_p.
- V_p = I_total R_p.
Remember: Do not apply the full source voltage to branches behind a series voltage drop.
Conditions: Series feeder before two parallel branches. Reduce the parallel group first, then reconstruct its voltage and each branch current.
Refresh Kid · AP Physics C: Electricity and Magnetism Unit 4 (official Unit 11) · Objectives 11.5.A · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 11.5, objectives 11.5.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.
Want to work through this with a tutor?
Bring your question about Reduce a network one valid group at a time. Your explanation and answers remain free to access.