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Flashcards · 9702 · A Level Topic 19
Capacitance (A Level) flashcards
15 cards on A Level Topic 19, subtopics 19.1 to 19.3. Pick a subtopic or study them all.
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19.1 Capacitors and capacitance
- Capacitance
- The charge stored per unit potential difference: C = Q ÷ V. Unit farad (1 F = 1 C V⁻¹).
- Capacitors in parallel and in series
- Parallel: C = C₁ + C₂ + … Series: 1/C = 1/C₁ + 1/C₂ + … The opposite of the resistor rules.
- 4 μF and 12 μF capacitors: find the combined capacitance in series and in parallel.
- Series: 1/C = 1/4 + 1/12 = 1/3, so 3 μF (less than the smallest). Parallel: 4 + 12 = 16 μF (more than the largest).
- Derive the formula for capacitors in parallel
- Each has the same p.d. V and the total charge is Q = Q₁ + Q₂. So CV = C₁V + C₂V, giving C = C₁ + C₂.
- Capacitance of an isolated conducting sphere of radius r
- Its potential is V = Q ÷ (4πε₀r), so C = Q ÷ V = 4πε₀r.
- What happens when a capacitor is connected to a d.c. supply?
- Charge flows briefly: one plate gains charge +Q and the other −Q. The current falls to zero when the p.d. across the capacitor equals the supply p.d.
19.2 Energy stored in a capacitor
- Energy stored in a capacitor
- W = ½QV = ½CV² = ½Q² ÷ C.
- Why is the energy stored ½QV and not QV?
- The p.d. rises from zero as charge is added, so the energy is the area under the potential–charge graph: a triangle, ½QV.
- A 100 μF capacitor is charged to 20 V. Find the energy stored.
- W = ½CV² = 0.5 × 100 × 10⁻⁶ × 20² = 0.020 J.
- The p.d. across a capacitor doubles. What happens to the energy stored?
- It quadruples, because W = ½CV² ∝ V².
19.3 Discharging a capacitor
- Equations for a capacitor discharging through a resistor
- x = x₀e^(−t/RC), where x is the charge Q, the p.d. V or the current I. All three decay with the same time constant.
- Time constant
- τ = RC (seconds, with R in Ω and C in F). After one time constant, Q, V and I have fallen to 1/e (about 37%) of their initial values.
- A 100 μF capacitor discharges through 10 kΩ. Find the time constant.
- τ = RC = 10 000 × 100 × 10⁻⁶ = 1.0 s.
- How can RC be found from discharge data?
- Plot ln Q (or ln V) against t. ln Q = ln Q₀ − t ÷ RC, so the line has gradient −1 ÷ RC.
- Half-life of a capacitor discharge
- t½ = RC ln 2 ≈ 0.69 RC.