Battery work divides between stored energy and resistor heat
You will be able to: Integrate RC power and verify the energy balance during charging.
Where does the other half of the battery’s charging energy go?
A battery charges an initially empty capacitor through a resistor. The battery supplies energy as charge moves; some becomes electric-field energy and some is dissipated in the resistor.
A useful starting point: A logarithmic plot reveals the RC timescale →
Words and symbols before equations
- Battery work W_b
- Energy supplied by a constant-emf source: ε times transferred charge.
- Stored U
- Capacitor energy ½CV_C².
- Dissipated H
- Integral ∫I²R dt; thermal energy generated in the resistor.
What this picture assumes
Initially uncharged capacitor, ideal constant emf, positive series R. All work values are accumulated from t = 0. R = 0 is excluded. Energy units are mJ when C is entered in mF.
Read the picture in three steps
- Locate the labeled sources, system boundary or graph axes. Read the units before comparing values.
- At 2 s: τ = 2 s; battery 63.21 mJ = stored 19.98 + resistor heat 43.23 mJ. V_C = 6.321 V; I = 1.839 mA.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the physics
With x = e^(−t/RC), q = Cε(1−x). Battery work is W_b = εq = Cε²(1−x).
Stored energy is U = ½Cε²(1−x)². Integrating resistor power gives H = ½Cε²(1−x²), and W_b = U+H at every time.
At long times, W_b→Cε² while both U and H approach ½Cε². Changing positive R changes the charging rate, not these final totals. The exact R=0 ideal-source connection is a singular transient and is not modeled by setting R to zero.
A worked example, step by step
A 10 V battery charges 1 mF through a positive resistor. Find the long-time battery work, stored energy and heat.
- Final charge is Cε = 10 mC.
- Battery work is εq = 100 mJ.
- Stored energy is ½Cε² = 50 mJ.
- The remaining 50 mJ is dissipated in the resistor, regardless of its positive resistance value.
The capacitor’s ½CV² is not the entire battery work for ordinary resistive charging from zero.
Does a larger R dissipate more total energy after complete charging?
Compare with an explanation
No. It dissipates more slowly, but the final heat is still ½Cε² under these assumptions.
Predict. Change one thing. Explain.
Move through time and verify the energy ledger. Change R at the same time, then compare long-time limits. All terms are positive for this charging-from-zero model.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
At 2 s: τ = 2 s; battery 63.21 mJ = stored 19.98 + resistor heat 43.23 mJ. V_C = 6.321 V; I = 1.839 mA.
Initially uncharged capacitor, ideal constant emf, positive series R. All work values are accumulated from t = 0. R = 0 is excluded. Energy units are mJ when C is entered in mF.
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 questionAt t = RC for C = 1 mF and ε = 10 V, calculate battery work, stored energy and heat and check their sum.
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Compare with the answer and four-point rubric
- 1 point: x = e⁻¹ ≈ 0.3679, so W_b = 100(1−x) ≈ 63.212 mJ.
- 1 point: U = 50(1−x)² ≈ 19.979 mJ.
- 1 point: H = 50(1−x²) ≈ 43.233 mJ.
- 1 point: U+H ≈ 63.212 mJ equals battery work.
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 integrating I²R give?
Energy dissipated in the resistor.
RECALL 2Why is R required to be positive here?
The finite RC transient assumes finite current and a resistive path.
RECALL 3What changes when R changes?
The timescale and instantaneous powers, not final energy totals.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Battery work divides between stored energy and resistor heat
- W_b = εq.
- U = ½CV_C².
- H = ∫I²R dt = ½Cε²(1−e^(−2t/RC)).
- W_b = U+H.
Remember: The capacitor’s ½CV² is not the entire battery work for ordinary resistive charging from zero.
Conditions: Initially uncharged capacitor, ideal constant emf, positive series R. All work values are accumulated from t = 0. R = 0 is excluded. Energy units are mJ when C is entered in mF.
Refresh Kid · AP Physics C: Electricity and Magnetism Unit 4 (official Unit 11) · Objectives 11.8.B · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 11.8, objectives 11.8.B. 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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