Short on time? This page is the recap. For the derivations, the figures and the worked examples, use the full study guide.
Recap
- Capacitors in parallel have the same ΔV; in series they have the same Q, because the joined inner plates form an isolated, neutral conductor.
- Series C_eq is less than the smallest capacitance in the group.
- The loop rule plus I = ±dq/dt gives a first-order differential equation. Its solutions are exponentials with time constant τ = RC.
- Current is largest just after the switch moves and decays towards zero, whether charging or discharging.
- After times much longer than τ (about 5τ), use steady-state conditions: no current in the capacitor branch.
Key relationships
| Situation | Result |
|---|---|
| Parallel | C_eq = C₁ + C₂ + … |
| Series | 1/C_eq = 1/C₁ + 1/C₂ + … |
| Charging equation | R dq/dt + q/C = ℰ |
| Charging solution | q = Cℰ(1 − e^(−t/RC)); I = (ℰ/R)e^(−t/RC) |
| Discharging equation | R dq/dt + q/C = 0 (I = −dq/dt) |
| Discharging solution | q = Q₀e^(−t/RC); ΔV_C and I have the same factor e^(−t/RC) |
| Time constant | τ = RC (Ω·F = s); 1 − e⁻¹ ≈ 0.63, e⁻¹ ≈ 0.37 |
| Energy | U = ½CΔV²; when charging, half of Cℰ² is stored and half dissipated |
| Linear graph | ln ΔV_C = ln ΔV₀ − t/(RC), slope −1/(RC) |
Just after and long after
| Moment | Capacitor behaves like | Current in its branch | ΔV_C |
|---|---|---|---|
| Just after closing, uncharged | a wire | largest (set by the resistors) | 0 |
| Just after, already charged | a battery of its present ΔV | jumps at once (ΔV/R if it discharges through a single resistor R) | unchanged |
| Long after (t ≫ τ) | a break | 0 | its steady-state value |
Assumptions
- Ideal battery, wires and meters unless stated; capacitors start uncharged unless stated.
- The resistance and capacitance stay constant.
Mistakes to avoid
- Treating series capacitors like series resistors. Reciprocals for series capacitors.
- “Half charged after τ.” It is 63%; half takes 0.69τ.
- Blocking current at t = 0. An uncharged capacitor acts like a wire at first.
- Using the wrong R. τ uses the resistance the capacitor charges or discharges through.
- Energy falling like charge. U ∝ q², so U decays as e^(−2t/RC).
Quick self-check
- What is τ for R = 2.0 kΩ and C = 3.0 μF? (6.0 ms)
- What is C_eq for 3.0 μF and 6.0 μF in series? In parallel? (2.0 μF; 9.0 μF)
- What fraction of its charge does a discharging capacitor keep after 3τ? (e⁻³ ≈ 0.050, about 5%)
Next: practice questions.