Series resistors share current and divide voltage
You will be able to: Find series equivalent resistance and individual voltage drops.
What stays the same along a single unbranched path?
Two resistors placed one after another have no alternate route between them. In steady state, the same charge per second must pass both; otherwise charge would accumulate at their connection.
A useful starting point: Current transfers energy without being used up →
Words and symbols before equations
- Series connection
- One unbranched route through each element.
- Equivalent resistance
- Single resistance with the same terminal voltage-current relation.
- Voltage division
- Sharing the total drop among series elements according to resistance.
What this picture assumes
One unbranched loop with ideal source and wires. Both resistors have positive constant resistance.
Read the picture in three steps
- Locate the labeled sources, system boundary or graph axes. Read the units before comparing values.
- R_eq = 9 Ω; I = 1.333 A; drops 8 and 4 V; resistor power total = 16 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
The same steady I flows through R₁ and R₂. Their drops are IR₁ and IR₂.
Loop energy balance gives V = I(R₁+R₂), so R_eq = R₁+R₂. A larger series resistance receives a larger voltage drop at this shared current.
The resistor power ratio is R₁/R₂ because I is common. The diagram uses ideal wires and a fixed ideal source; physical spacing does not determine the series relationship.
A worked example, step by step
A 12 V source feeds 2 Ω and 4 Ω in series. Find current and both drops.
- R_eq = 2+4 = 6 Ω.
- I = 12/6 = 2 A through both.
- V₁ = 4 V and V₂ = 8 V.
- The drops sum to 12 V, checking loop balance.
Series elements share current, not necessarily voltage.
Does adding a series resistor to an ideal-voltage circuit increase I?
Compare with an explanation
No. The total resistance rises, so the current falls.
Predict. Change one thing. Explain.
Increase R₂ at fixed source voltage. Track total current and both voltage drops rather than assuming current stays fixed.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
R_eq = 9 Ω; I = 1.333 A; drops 8 and 4 V; resistor power total = 16 W.
One unbranched loop with ideal source and wires. Both resistors have positive constant resistance.
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 questionA 15 V source supplies 5 Ω and 10 Ω in series. Calculate current, voltage drops and total dissipated power.
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Compare with the answer and four-point rubric
- 1 point: R_eq = 15 Ω gives I = 1 A.
- 1 point: Drops are 5 V and 10 V.
- 1 point: P₁ = 5 W and P₂ = 10 W.
- 1 point: Total 15 W equals source VI = 15 W.
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 1Why is current equal?
Steady charge cannot accumulate between series elements.
RECALL 2Which resistor gets more voltage?
The larger R at common I.
RECALL 3How can you check the result?
Sum drops and compare total dissipated power with source power.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Series resistors share current and divide voltage
- R_eq = R₁+R₂.
- I₁ = I₂.
- V_i = IR_i.
Remember: Series elements share current, not necessarily voltage.
Conditions: One unbranched loop with ideal source and wires. Both resistors have positive constant resistance.
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.
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