Reduce a network, then work backward
You will be able to: Reduce a simple mixed resistor network and reconstruct its branch values.
How do you solve a series–parallel combination?
A 2 Ω resistor comes before a junction feeding 6 Ω and 3 Ω branches. The first resistor carries the total current. The two branches share only the voltage left after that first resistor’s drop.
A useful starting point: Parallel: one voltage, several currents →
Words and symbols before equations
- Reduction
- Replace a recognizable group with an equivalent element.
- Reconstruction
- Use the reduced circuit’s results to recover individual branch values.
- Series with a group
- The total group current must pass the series element first.
What this picture assumes
Ideal wires and voltage source; positive, fixed resistances. Circuit geometry is schematic, not a physical length or speed scale. Fixed source 12 V. The upstream resistor is in series with the two-branch parallel group.
Read the picture in three steps
- Locate the labeled sources, system boundary or graph axes. Read the units before comparing values.
- R_parallel=2 Ω; R_total=4 Ω. I_total=3 A; upstream drop=6 V; branch voltage=6 V. I₁+I₂=1+2=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
Identify nodes before calculating. Reduce the 6 Ω and 3 Ω parallel pair to 2 Ω, then add the first 2 Ω series resistor. The total is 4 Ω, not 11 Ω.
For 12 V, total current is 3 A. The first resistor drops 6 V, leaving 6 V across the branch group. Branch currents are then 1 A and 2 A; they recombine to the total 3 A.
Adding a parallel branch within a mixed network changes the group resistance and therefore the total current and upstream voltage drop. The voltage across an existing branch may change even when the source is ideal, because that branch is not directly across the source.
A worked example, step by step
Find currents and drops for 12 V across a 2 Ω resistor in series with (6 Ω parallel 3 Ω).
- R_parallel=(6×3)/(6+3)=2 Ω; R_total=2+2=4 Ω.
- I_total=12/4=3 A; upstream drop=(3)(2)=6 V.
- Parallel voltage=12−6=6 V; branch currents 1 A and 2 A.
- Check 1+2=3 A and 6+6=12 V.
A branch in a mixed circuit may not receive the entire battery voltage.
Can the two branch resistors be added directly to the upstream resistor?
Compare with an explanation
No. Reduce the parallel pair first, then add that equivalent in series.
Predict. Change one thing. Explain.
Change the upstream resistance. Watch how both branch currents change together as the group’s shared voltage changes.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
R_parallel=2 Ω; R_total=4 Ω. I_total=3 A; upstream drop=6 V; branch voltage=6 V. I₁+I₂=1+2=3 A.
Ideal wires and voltage source; positive, fixed resistances. Circuit geometry is schematic, not a physical length or speed scale. Fixed source 12 V. The upstream resistor is in series with the two-branch parallel group.
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 source feeds 3 Ω in series with two 6 Ω parallel branches. (a) Reduce the pair. (b) Find total current. (c) Find branch voltage. (d) Find each branch current.
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Compare with the answer and four-point rubric
- 1 point: Parallel equivalent is 3 Ω.
- 1 point: Total resistance 6 Ω; current 2 A.
- 1 point: Upstream drop 6 V; branch voltage 6 V.
- 1 point: Each branch current 1 A; sum 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 the first step in a mixed network?
Identify the connected nodes and reducible groups.
RECALL 2Why reconstruct after reducing?
An equivalent resistor hides individual branch values.
RECALL 3Which two checks catch many errors?
Junction current sums and loop voltage sums.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Reduce a network, then work backward
- Reduce known parallel or series groups in order.
- Reconstruct individual currents and drops afterward.
- Check each junction and complete loop.
Remember: A branch in a mixed circuit may not receive the entire battery voltage.
Conditions: Ideal wires and voltage source; positive, fixed resistances. Circuit geometry is schematic, not a physical length or speed scale. Fixed source 12 V. The upstream resistor is in series with the two-branch parallel group.
Refresh Kid · AP Physics 2 Unit 3 (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. 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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