Capacitors

WPH14Topic 4.44 questions

Capacitance, energy stored in a capacitor, and exponential charging and discharging through a resistor.

Specification points

Understand that capacitance is defined as $C = \frac{Q}{V}$ and be able to use this equation

Previous topic4.3Electric fields Next topic 4.5Magnetic fields and forces

Be able to use the equations for the energy stored in a capacitor $W = \frac{1}{2}QV$ , $W = \frac{1}{2}CV^2$ and $W = \frac{Q^2}{2C}$

Be able to draw and interpret charge and discharge curves for resistor-capacitor circuits and understand the significance of the time constant $RC$

CORE PRACTICAL 11: Use an oscilloscope or data logger to display and analyse the potential difference (p.d.) across a capacitor as it charges and discharges through a resistor

Be able to use the equation $Q = Q_0 e^{-t/RC}$ and derive and use related equations for exponential discharge in a resistor-capacitor circuit, $I = I_0 e^{-t/RC}$ and $V = V_0 e^{-t/RC}$ , and the corresponding log equations, e.g. $\ln Q = \ln Q_0 - \frac{t}{RC}$

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Questions

Multiple choice: 2
Short answer: 1
Calculation: 1
Total: 4

Sample questions

Original practice questions written in the style of the exam.

Question 13 marks
A charged capacitor is discharged through a resistor. Explain why the current decreases exponentially with time.
Mark scheme
- The current is proportional to the pd across the capacitor ( $I = V/R$ ), and $V$ is proportional to the charge stored ( $V = Q/C$ ) (1)
- As charge flows off, the pd falls, so the current (rate of loss of charge) falls (1)
- The rate of decrease of charge is proportional to the charge remaining, which gives an exponential decay (1)
Question 24 marks
A $2200\,\mu\text{F}$ capacitor is charged to $12\,\text{V}$ and then discharged through a $1.5\,\text{k}\Omega$ resistor.

(a) Calculate the energy initially stored in the capacitor.

(b) Calculate the time taken for the potential difference across the capacitor to fall to $3.0\,\text{V}$ .

3.0

Mark scheme

$W = \tfrac{1}{2}CV^2 = \tfrac{1}{2} \times 2200 \times 10^{-6} \times 12^2 = 0.16\,\text{J}$ (0.158 J) (1)
$RC = 1.5 \times 10^{3} \times 2200 \times 10^{-6} = 3.3\,\text{s}$ (1)
Uses $V = V_0 e^{-t/RC}$ : $3.0 = 12e^{-t/3.3}$ , so (1)
$t = 4.6\,\text{s}$ (1)

3.3s

12e−t/3.3

t = 3.3\ln 4