The shape of the lesson
By the end of the lesson, learners can
- Watching the decay: Adjust R and C to change the time constant and watch the curve stretch or steepen.
- The time constant τ = RC: The product RC has units of seconds and sets the pace of the decay: a larger resistance or capacitance means a slower discharge.
- Capacitance: C = Q/V. In parallel C = C₁ + C₂; in series 1/C = 1/C₁ + 1/C₂.
- Energy: W = ½QV = ½CV² = ½Q²/C, the area under the V-Q graph.
- Discharge: x = x₀ e⁻ᵗּᵣᶜ; the time constant τ = RC is the time to fall to 1/e (about 37%).
One picture every idea returns to
When a capacitor discharges through a resistor, the charge, voltage and current all decay exponentially: x = x₀ e⁻ᵗּᵣᶜ. The time constant τ = RC is the time to fall to 1/e, about 37%, of the starting value.
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, The slow fade of an exponential.. Let a charged capacitor empty through a resistor and the charge does not fall steadily: it decays exponentially, fast at first and ever more slowly, set by a single time constant. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 17 min | Teach and check: Watching the decay | Adjust R and C to change the time constant and watch the curve stretch or steepen. Project tab 3 of the lab (Discharging) and let a student drive it while the class predicts. | Whole class, then pairs |
| 17 to 30 min | Teach and check: The time constant τ = RC | The product RC has units of seconds and sets the pace of the decay: a larger resistance or capacitance means a slower discharge. | Mix-Pair-Share |
| 30 to 43 min | Check and plenary | The quick-check questions from the student lesson, answered on mini whiteboards one at a time; re-teach the one most get wrong, then a random call on the last. | Individual, then whole class |
| 43 to 45 min | Exit and homework | Set the worksheet below (3 questions, 35 marks) from the topic practice set. Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in The time constant τ = RC, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Mix-Pair-Share
Students mix around the room until you call pair; they pair with the nearest classmate, discuss the prompt, then mix again. Three prompts of rising difficulty, three different partners.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
No examiner trap is recorded for this lesson. Use the quiz key below to spot the distractor most of the class picks.
Discharging a capacitor: topic worksheet
3 questions, 35 marks. Written in the style of the papers; no past paper question is reproduced. These questions cover the whole topic, so sibling lessons share them.
A 470 µF capacitor is charged to 6.0 V and then discharged through a 10 kΩ resistor. Fig. 3.1 shows how the voltage across the capacitor falls with time.
(a) Calculate the time constant of the circuit. [2]
(b) Write down the equation giving the voltage at time t. [1]
(c) Calculate the voltage across the capacitor after one time constant. [2]
(d) Calculate the time taken for the voltage to fall to 1.5 V. [3]
(e) State and explain the effect on the time constant of doubling the resistance. [2]
(f) Describe how the discharge current varies with time. [2]
A 220 μF capacitor charged to 10 V discharges through a 47 kΩ resistor, as in Fig. 9.1.
(a) Write the equation for the charge on the capacitor during the discharge. [1]
(b) Calculate the time constant. [2]
(c) State the physical meaning of the time constant. [2]
(d) Calculate the p.d. across the capacitor after 5.0 s. [3]
(e) Calculate the time for the p.d. to fall to 2.0 V. [3]
A 470 μF capacitor is charged to 6.0 V and then discharged through a 10 kΩ resistor. This question uses all three lessons of the topic: capacitance, energy, and discharge.
(a) Calculate the initial charge stored. [2]
(b) Calculate the initial energy stored. [2]
(c) Calculate the time constant of the discharge. [2]
(d) Calculate the p.d. across the capacitor after 3.0 s. [3]
(e) Calculate the energy still stored at that time. [3]
Answer key and marking notes
- (a)τ = RC = 10 000 × 470 × 10⁻⁶ ✓ = 4.7 s ✓
- (b)V = V₀ e⁻ᵗּᵣᶜ = 6.0 e⁻ᵗּ⁴.⁷ ✓
- (c)V = 6.0 × e⁻¹ = 6.0 × 0.368 ✓ = 2.2 V ✓
- (d)1.5 = 6.0 e⁻ᵗּ⁴.⁷ ⇒ e⁻ᵗּ⁴.⁷ = 0.25 ✓; −t/4.7 = ln 0.25 = −1.386 ✓; t = 6.5 s ✓
- (e)The time constant doubles ✓, since τ = RC is proportional to R; the capacitor discharges more slowly ✓
- (f)The current starts at V₀/R and decays exponentially with the same time constant ✓, approaching zero ✓
- (a)Q = Q₀ e^(−t/RC) ✓
- (b)τ = RC = 47 × 10³ × 220 × 10⁻⁶ ✓ = 10 s ✓
- (c)The time for the charge (or p.d.) to fall to 1/e, about 37%, of its initial value ✓✓
- (d)V = 10 e^(−5.0/10.34) = 10 × e^(−0.484) ✓✓ = 6.2 V ✓
- (e)2.0 = 10 e^(−t/τ), so t = τ ln(10/2) = 10.34 × 1.61 ✓✓ = 17 s ✓
- (a)Q = CV = 470 × 10⁻⁶ × 6.0 ✓ = 2.8 × 10⁻³ C ✓
- (b)W = ½CV² = ½ × 470 × 10⁻⁶ × 6.0² ✓ = 8.5 × 10⁻³ J ✓
- (c)τ = RC = 10 × 10³ × 470 × 10⁻⁶ ✓ = 4.7 s ✓
- (d)V = 6.0 e^(−3.0/4.7) = 6.0 × e^(−0.638) ✓✓ = 3.2 V ✓
- (e)W = ½CV² = ½ × 470 × 10⁻⁶ × 3.17² ✓✓ = 2.4 × 10⁻³ J ✓
The lesson's own checks as a quiz
The quick-check and exam-style questions from the student lesson, printable with the key.
During discharge through a resistor, the charge on a capacitor:
- Adecays exponentially
- Bdecreases at a constant rate
- Cstays constant
- Dincreases
The time constant of a resistor-capacitor discharge circuit is:
- ARC
- BR/C
- CC/R
- DRC²
After one time constant, the charge has fallen to about:
- A37% of its initial value
- B50% of its initial value
- C63% of its initial value
- Dzero
Doubling the resistance in the discharge circuit:
- Adoubles the time constant, slowing the discharge
- Bhalves the time constant
- Chas no effect
- Dstops the discharge
Quiz key
Charge, voltage and current all follow x = x₀ e⁻ᵗּᵣᶜ, an exponential decay.
The time constant is τ = RC, with units of seconds.
After t = τ the factor is e⁻¹ ≈ 0.37, so about 37% remains.
τ = RC is proportional to R, so doubling R doubles the time constant and slows the decay.
What to project
The student lesson embeds these tabs. Open them full screen on the projector; every slider has a drag handle on the stage.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.