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How do capacitors store charge and behave in d.c. and a.c. circuits?

Capacitance and the energy stored, charging and discharging through a resistor with the time constant, and the behaviour of capacitors in d.c. and a.c. circuits.

An SQA Advanced Higher Physics answer on capacitors, covering capacitance and the energy stored, the charging and discharging of a capacitor through a resistor with the time constant, and the behaviour of capacitors in direct-current and alternating-current circuits.

Generated by Claude Opus 4.814 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

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  1. What this key area is asking
  2. Capacitance and energy stored
  3. Charging and discharging: the time constant
  4. Capacitors in d.c. and a.c. circuits
  5. Examples in context
  6. Try this

What this key area is asking

The SQA wants you to use capacitance C=QVC = \frac{Q}{V} and the energy stored E=12QV=12CV2E = \tfrac{1}{2}QV = \tfrac{1}{2}CV^2, analyse the charging and discharging of a capacitor through a resistor including the time constant τ=RC\tau = RC, and describe the behaviour of capacitors in d.c. and a.c. circuits.

Capacitance and energy stored

The energy stored is not simply QVQV, because the voltage grows from zero to its final value as the capacitor charges. Averaging, the energy is:

E=12QV=12CV2=Q22C.E = \tfrac{1}{2}QV = \tfrac{1}{2}CV^2 = \frac{Q^2}{2C}.

The factor of one half is the most commonly forgotten detail. A graph of charge against voltage is a straight line, and the energy stored is the area under it, a triangle of area 12QV\tfrac{1}{2}QV.

Charging and discharging: the time constant

A larger resistance or capacitance gives a longer time constant and slower charging. On discharge, the current is largest at the start (when the voltage is highest) and decays away; on charge, the current starts large and falls to zero as the capacitor fills. The exponential shape, and the fact that the capacitor is effectively fully charged after about five time constants, are standard exam content.

Capacitors in d.c. and a.c. circuits

Because a capacitor opposes a change in voltage, it lets through rapidly changing (high-frequency) signals while blocking steady (d.c.) ones. This is why capacitors are used to block d.c. while passing a.c. signals, and in filters that select frequencies. The opposition a capacitor offers to a.c. decreases as frequency rises, the opposite of an inductor.

Examples in context

Camera flashes charge a capacitor slowly and release the energy 12CV2\tfrac{1}{2}CV^2 in a brief, bright burst. Smoothing capacitors in power supplies charge during the peaks of rectified a.c. and discharge between them, flattening the output. Timing circuits use the RCRC time constant to set delays in flashing indicators and timers. Coupling capacitors in audio amplifiers block d.c. bias while passing the a.c. signal between stages.

Try this

Q1. Write the relationship for the energy stored on a capacitor in terms of CC and VV. [1 mark]

  • Cue. E=12CV2E = \tfrac{1}{2}CV^2.

Q2. Write the relationship for the time constant of an RC circuit. [1 mark]

  • Cue. τ=RC\tau = RC.

Q3. State what a fully charged capacitor does to a steady direct current. [1 mark]

  • Cue. It blocks it (no steady current flows once fully charged).

Exam-style practice questions

Practice questions written in the style of SQA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

SQA AH style5 marksA 470 μF470\ \mu\text{F} capacitor is charged to 12 V12\ \text{V}. Calculate the charge stored and the energy stored.
Show worked answer →

Charge stored: Q=CV=470×106×12=5.6×103 CQ = CV = 470 \times 10^{-6} \times 12 = 5.6 \times 10^{-3}\ \text{C}.

Energy stored: E=12CV2=12×470×106×122E = \tfrac{1}{2}CV^2 = \tfrac{1}{2} \times 470 \times 10^{-6} \times 12^2.

Evaluate: E=12×470×106×144=3.4×102 JE = \tfrac{1}{2} \times 470 \times 10^{-6} \times 144 = 3.4 \times 10^{-2}\ \text{J}.

Markers reward Q=CVQ = CV for the charge, E=12CV2E = \tfrac{1}{2}CV^2 for the energy, and both with units. The half is essential, because the average voltage during charging is half the final voltage.

SQA AH style4 marksA 2.0 μF2.0\ \mu\text{F} capacitor discharges through a 1.0 MΩ1.0\ \text{M}\Omega resistor. Calculate the time constant and state what fraction of the initial charge remains after one time constant.
Show worked answer →

The time constant is τ=RC\tau = RC.

Substitute: τ=1.0×106×2.0×106=2.0 s\tau = 1.0 \times 10^{6} \times 2.0 \times 10^{-6} = 2.0\ \text{s}.

After one time constant the charge falls to 1e\frac{1}{e} of its initial value, which is about 0.370.37, so about 37%37\% remains.

Markers reward the relationship τ=RC\tau = RC, the value in seconds, and that one time constant leaves 1/e1/e (about 37 per cent) of the charge.

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