Stored charge can drive a current
You will be able to: Interpret discharge graphs and distinguish charge decay from energy decay.
Where does capacitor energy go when it discharges?
Disconnect a charged capacitor from its battery, then connect a resistor across its plates. The capacitor now supplies the energy. Current flows from its initially positive plate through the external resistor toward its negative plate.
A useful starting point: Watch a capacitor charge →
Words and symbols before equations
- Initial voltage V₀
- Capacitor voltage at the start of discharge.
- Discharge current magnitude
- Positive size of current in the defined discharge direction.
- Exponential decay
- The same fraction remains after each equal time interval.
- Energy fraction
- Because U∝V², the energy fraction is the square of the voltage fraction.
What this picture assumes
Battery removed; capacitor initially at 10 V discharges only through R. Current readout is its magnitude in the discharge direction. Lost capacitor energy becomes resistor heat; no leakage or radiation.
Read the picture in three steps
- Locate the labeled sources, system boundary or graph axes. Read the units before comparing values.
- τ=1 s; t/τ=1. V_C=3.679 V, |I|=0.3679 mA, |Q|=367.9 μC. Stored energy=0.6767 mJ; cumulative resistor heat=4.323 mJ.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the physics
AP scope: explain RC trends and graphs qualitatively and calculate initial and final states. The simulation supplies intermediate values for exploration; deriving or memorizing time-dependent exponential equations is not required in AP Physics 2.
Optional model detail: with the battery removed, the simulation uses V_C=V₀exp(−t/RC), Q=CV_C and current magnitude is V_C/R. All three decrease toward zero. Current is opposite the charging direction in the external resistor path for the usual changeover arrangement.
After one τ=RC, about 37% of initial charge and voltage remain. Only about 14% of initial energy remains because (0.368)²≈0.135. The rest has become thermal energy in the resistor under this ideal model.
An experiment can estimate τ from the time when voltage falls to 0.37V₀, and then infer C=τ/R. Repeat trials and account for meter loading and component tolerances. The ideal curve is synthetic; real data would include uncertainty.
| Quantity | Charging from zero | Discharging from V₀ |
|---|---|---|
| Capacitor voltage | Rises toward source voltage | Falls toward zero |
| Current magnitude | Falls from ε/R toward zero | Falls from V₀/R toward zero |
| At one time constant | About 63% of final charge | About 37% of initial charge |
A worked example, step by step
A 100 μF capacitor initially at 10 V discharges through 10 kΩ. Find voltage and energy after 1 s.
- τ=(10,000)(100×10⁻⁶)=1 s, so this is one τ.
- V_C≈0.368×10=3.68 V.
- Initial energy=½(100×10⁻⁶)(10²)=0.005 J=5 mJ.
- Remaining energy≈0.135×5=0.677 mJ; about 4.32 mJ has become heat.
Charge and voltage decay by the same fraction; stored energy decays faster because it depends on voltage squared.
Doubling discharge resistance changes the initial stored energy by what factor?
Compare with an explanation
It does not change it when C and V₀ are fixed; it doubles the time constant and halves initial current.
Predict. Change one thing. Explain.
Move time to one τ, then two τ. Compare voltage, current magnitude, stored energy and cumulative heat. Predict how R changes timing without changing initial stored energy.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
τ=1 s; t/τ=1. V_C=3.679 V, |I|=0.3679 mA, |Q|=367.9 μC. Stored energy=0.6767 mJ; cumulative resistor heat=4.323 mJ.
Battery removed; capacitor initially at 10 V discharges only through R. Current readout is its magnitude in the discharge direction. Lost capacitor energy becomes resistor heat; no leakage or radiation.
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 capacitor discharges through 20 kΩ. Its voltage falls from 8 V to about 2.94 V in 2 s. (a) Identify τ. (b) Find C. (c) Find initial current magnitude. (d) State the energy fraction at 2 s.
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Compare with the answer and four-point rubric
- 1 point: 2.94/8≈0.368, so τ≈2 s.
- 1 point: C=τ/R=2/20,000=100 μF.
- 1 point: I₀=8/20,000=0.40 mA.
- 1 point: Remaining fraction≈0.135, or 13.5%.
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 1Where does discharge energy go in the ideal resistor model?
Into thermal energy.
RECALL 2How can a voltage curve estimate capacitance?
Read τ at 37% of initial voltage and use C=τ/R.
RECALL 3Does capacitor voltage jump at switching with finite current?
No. Its charge, and therefore voltage at fixed C, cannot change instantaneously.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Stored charge can drive a current
- τ=RC.
- At one discharge τ: Q/Q₀=V/V₀≈0.368.
- U/U₀=(V/V₀)²≈0.135 at τ.
- Battery-free ideal discharge: lost capacitor energy becomes resistor heat.
Remember: Charge and voltage decay by the same fraction; stored energy decays faster because it depends on voltage squared.
Conditions: Battery removed; capacitor initially at 10 V discharges only through R. Current readout is its magnitude in the discharge direction. Lost capacitor energy becomes resistor heat; no leakage or radiation.
Refresh Kid · AP Physics 2 Unit 3 (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. 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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