Capacitors flashcards

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Capacitance C
The charge stored (on one plate) per unit potential difference across the capacitor: C = Q/V.
The farad
1 F = 1 C V⁻¹: a capacitor that stores 1 C for each volt across it.
Capacitor
Two conductors separated by an insulator; charging leaves +Q on one plate and −Q on the other.
Time constant
τ = RC: the time for a discharging capacitor's charge, p.d. or current to fall to 1/e (37%) of its starting value. Unit: s.
Time to halve (capacitor discharge)
The time for Q, V or I to fall to half; t½ = RC ln 2, independent of the starting value.
Capacitance equation
C = Q/V (Yes)
Energy stored by a capacitor (three forms)
W = ½QV = ½CV² = ½Q²/C (Yes)
Discharge equations
Q = Q₀e^(−t/RC), I = I₀e^(−t/RC), V = V₀e^(−t/RC) (Yes)
Log form of the discharge equation
ln Q = ln Q₀ − t/RC (same for I and V). (Yes)
Time to halve
t½ = RC ln 2 (put Q = ½Q₀ in the discharge equation). (No — derive it)
P.d. across a charging capacitor
V_C = V₀(1 − e^(−t/RC)), because V_C = V₀ − V_R and V_R = V₀e^(−t/RC). (No — derive it)
Energy transferred by the supply while charging
W = QV (twice the energy stored). (No — derive it)
V against Q for a capacitor
Straight line through the origin, gradient 1/C; the area under it is the energy stored, ½QV.
Charging through a resistor: V_C and I against t
V_C rises steeply then levels off at V₀ (63% after RC); I starts at V₀/R and decays exponentially.
Discharge: Q, V or I against t
Exponential decay: falls by the same fraction in each equal time; 37% left after RC; initial tangent meets the t-axis at RC.
ln V against t for a discharge
Straight line, gradient −1/RC, intercept ln V₀.
Charge–discharge circuit
Supply → two-way switch → resistor → capacitor; switch position B joins resistor and capacitor in a loop without the supply; voltmeter or data logger across C.
Area under a current–time graph for a discharge
Equals the total charge that flowed, Q₀.
Why does the charging current decrease?
Charge builds up → V_C rises → p.d. across R (V₀ − V_C) falls → current falls.
Why is the energy stored ½QV, not QV?
Charge is added while the p.d. rises from 0 to V; on average it is moved through ½V — the triangle under the V–Q graph.
Where does the other half of the supply's energy go when charging?
It is dissipated as thermal energy in the circuit resistance; the fraction is ½ whatever the resistance.
Effect of a larger R on charging
Charging takes longer (larger RC) but the final charge and energy are unchanged.
Why is a discharge exponential?
The current, and so the rate of loss of charge, is proportional to the charge remaining.
Core Practical 11: why use a data logger or oscilloscope?
Many readings at precise, short intervals, including the fast early part; no reaction-time error.
Core Practical 11: why must the meter or logger have a very high resistance?
Otherwise it is a second discharge path in parallel with R, making the measured time constant too small.
Charge on a 470 µF capacitor at 9.0 V?
Q = CV = 470 × 10⁻⁶ × 9.0 = 4.2 × 10⁻³ C.
Energy stored by 150 µF at 330 V?
W = ½CV² = ½ × 150 × 10⁻⁶ × 330² = 8.2 J.
220 µF discharges through 47 kΩ from 8.0 V. V after 15 s?
RC = 10.3 s; V = 8.0e^(−15/10.34) = 1.9 V.
RC = 2500 s. Time for V to fall from 5.0 V to 2.0 V?
t = RC ln(V₀/V) = 2500 × ln 2.5 = 2.3 × 10³ s.
Time to halve for 6.8 µF discharging through 1.5 MΩ?
RC = 10.2 s; t½ = RC ln 2 = 7.1 s.
ln V–t gradient −0.025 s⁻¹, R = 100 kΩ. Capacitance?
RC = 1/0.025 = 40 s; C = 40 ÷ 100 × 10³ = 4.0 × 10⁻⁴ F (400 µF).
Charging from 9.0 V with RC = 4.7 s. V_C after 3.0 s?
V_C = 9.0(1 − e^(−3.0/4.7)) = 4.2 V.
Using V = V₀e^(−t/RC) for a charging capacitor
Wrong: that is the p.d. across the resistor. V_C = V₀ − V_R.
Forgetting µF and kΩ prefixes
Convert first: 470 µF = 470 × 10⁻⁶ F, 68 kΩ = 68 × 10³ Ω; RC then comes out in seconds.
Fraction of energy left after one time constant
Not 37%: W ∝ V², so e⁻² ≈ 14%.
Gradient of a ln V–t graph
It is −1/RC (unit s⁻¹), so RC = −1/gradient — not the gradient itself.
Uncertainty in C = RC ÷ R
Add the percentage uncertainties in RC (or the gradient) and in R.

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