AP · AP C:E&M · extended response
A discharging capacitor transfers finite energy to a resistor
AP Physics C: Electricity and Magnetism · Original GioPhysics question with a detailed, mark-by-mark answer guide.
- Demand
- discriminating
- Marks
- 6
- Topics
- 1
- Answer
- Complete
Capacitor discharge voltage
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Voltage / V
Horizontal axis (x): Time / s
Read the exact data as a table
| Time / s | Voltage / V |
|---|---|
| 0 | 10 |
| 1.386 | 5 |
| 2.772 | 2.5 |
| 4.158 | 1.25 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Choose your AP course · School year 2026–27 · May 2027 exam
Choose a course to check this question’s reviewed parts and required depth. Its original AP label and topic codes do not establish suitability for the current syllabus.
Choose reviewed AP practice →A 100 μF capacitor initially at 10.0 V discharges through a 20.0 kΩ resistor. The supplied voltages follow V(t) = V0 exp(−t/RC). There is no power supply during discharge. Use ln2 = 0.693 and neglect leakage and meter loading. Supplied data (Time / s; Voltage / V): (0, 10), (1.386, 5), (2.772, 2.5), (4.158, 1.25).
- (a)
Find the time constant, initial current, initial stored energy and time for voltage to reach half its initial value.
3 marks - (b)
Find energy remaining and energy dissipated when voltage halves. Derive the energy decay law and explain why current also halves at that time.
3 marks
Ready to self-mark?Reveal the detailed answer guide
(a)
Find the time constant, initial current, initial stored energy and time for voltage to reach half its initial value.
Build a chain of claim, evidence and physics reasoning. Address more than one relevant factor, identify a limitation or assumption, and finish with a conclusion that is conditional on the evidence rather than absolute.
- 1
The time constant is RC = (2.0×10⁴)(1.0×10⁻⁴) = 2.0 s; initial current magnitude is V0/R = 10/20000 = 0.50 mA.
- 2
Initial energy is ½CV0² = ½(1.0×10⁻⁴)(100) = 0.0050 J.
- 3
Setting V/V0 = 1/2 gives t = RC ln2 = 1.386 s, consistent with the supplied data. The time constant is not the voltage half-time.
(b)
Find energy remaining and energy dissipated when voltage halves. Derive the energy decay law and explain why current also halves at that time.
Build a chain of claim, evidence and physics reasoning. Address more than one relevant factor, identify a limitation or assumption, and finish with a conclusion that is conditional on the evidence rather than absolute.
- 1
Since U = ½CV² and C is fixed, halving voltage quarters stored energy, leaving 5.0/4 = 1.25 mJ.
- 2
The resistor receives the missing 3.75 mJ as thermal energy. Substitution of V(t) gives U(t) = U0 exp(−2t/RC), so energy decays with a time constant RC/2.
- 3
Ohm's law gives I(t) = V(t)/R, so current halves when voltage halves. Integrating I²R over the full discharge returns the finite initial capacitor energy.
Complete marking guidance
Use these worked steps alongside the part-by-part answers. They explain the method and mark allocation; they do not add extra marks.
- The time constant is RC = (2.0×10⁴)(1.0×10⁻⁴) = 2.0 s; initial current magnitude is V0/R = 10/20000 = 0.50 mA.
- Initial energy is ½CV0² = ½(1.0×10⁻⁴)(100) = 0.0050 J.
- Setting V/V0 = 1/2 gives t = RC ln2 = 1.386 s, consistent with the supplied data. The time constant is not the voltage half-time.
- Since U = ½CV² and C is fixed, halving voltage quarters stored energy, leaving 5.0/4 = 1.25 mJ.
- The resistor receives the missing 3.75 mJ as thermal energy. Substitution of V(t) gives U(t) = U0 exp(−2t/RC), so energy decays with a time constant RC/2.
- Ohm's law gives I(t) = V(t)/R, so current halves when voltage halves. Integrating I²R over the full discharge returns the finite initial capacitor energy.
Private on this device