Series: one current, several drops
You will be able to: Find equivalent resistance, current and individual voltage drops in a series circuit.
How does a source voltage divide in one path?
A 2 Ω resistor and a 4 Ω resistor lie in a single path across 12 V. Every charge passing one must pass the other, so the current is the same. The larger resistance accounts for the larger energy transfer per coulomb.
A useful starting point: Compare bulbs using power →
Words and symbols before equations
- Equivalent resistance R_eq
- One resistance giving the same total current for a specified applied voltage.
- Series connection
- No branch between the elements through which current could divert.
- Voltage drop
- Decrease of electric potential moving through a resistor with conventional current.
What this picture assumes
Ideal wires and voltage source; positive, fixed resistances. Circuit geometry is schematic, not a physical length or speed scale.
Read the picture in three steps
- Locate the labeled sources, system boundary or graph axes. Read the units before comparing values.
- R_eq=6 Ω; I=2 A. Loop: +12−4−8=0 V. Graph: source rise at 1, resistor drops over 2–3 and 4–5.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the physics
Steady charge cannot build up indefinitely between series resistors, so the current in each is the same. Voltage drops add: V=IR₁+IR₂=I(R₁+R₂). Thus R_eq=R₁+R₂.
Calculate total current from the whole loop first, then apply V_i=IR_i to each resistor. A voltage-divider relation follows: V₁=V R₁/(R₁+R₂). Increasing one resistance changes the current throughout the loop.
Check both conservation laws: current is the same before and after each resistor, and voltage drops sum to the supply voltage. An element’s physical closeness to the battery does not give it a larger current.
A worked example, step by step
For a 12 V ideal source with 2 Ω and 4 Ω in series, find all currents and voltage drops.
- R_eq=2+4=6 Ω.
- I=12/6=2 A through each resistor.
- V₁=(2)(2)=4 V; V₂=(2)(4)=8 V.
- 4+8=12 V, verifying the loop energy balance.
Resistors divide voltage in a series path, not current.
Swap the positions of the two resistors. Does the current change?
Compare with an explanation
No. The series sum and source voltage are unchanged.
Predict. Change one thing. Explain.
Increase the second resistance at fixed supply. Predict the change in total current and both voltage drops; verify that the sum stays fixed.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
R_eq=6 Ω; I=2 A. Loop: +12−4−8=0 V. Graph: source rise at 1, resistor drops over 2–3 and 4–5.
Ideal wires and voltage source; positive, fixed resistances. Circuit geometry is schematic, not a physical length or speed scale.
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 9 V source drives series resistors 3 Ω and 6 Ω. (a) Find R_eq. (b) Find I. (c) Find each drop. (d) Check the loop.
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Compare with the answer and four-point rubric
- 1 point: R_eq=9 Ω.
- 1 point: I=1 A everywhere in the path.
- 1 point: Drops 3 V and 6 V.
- 1 point: 9−3−6=0 V.
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 1How do series resistances combine?
Add them.
RECALL 2Why is series current shared?
There is one path and no steady charge accumulation.
RECALL 3What sets a resistor’s fraction of the series voltage?
Its resistance divided by the total series resistance.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Series: one current, several drops
- R_eq=ΣR_i in series.
- All series elements share I.
- V_i=IR_i; ΣV_drops=source voltage for this single-source loop.
Remember: Resistors divide voltage in a series path, not current.
Conditions: Ideal wires and voltage source; positive, fixed resistances. Circuit geometry is schematic, not a physical length or speed scale.
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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