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How do RC circuits charge, discharge and filter signals, and how do we predict their behaviour over time and frequency?

The capacitor charge and discharge through a resistor, the time constant, and passive RC filters selecting signals by frequency.

An SQA Advanced Higher Engineering Science answer on analogue signal processing, covering capacitor charge and discharge through a resistor, the RC time constant, and passive low-pass and high-pass RC filters that select signals by frequency.

Generated by Claude Opus 4.815 min answer

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

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  1. What this key area is asking
  2. Charging and discharging through a resistor
  3. The time constant relationship
  4. Passive RC filters
  5. Examples in context
  6. Try this

What this key area is asking

The SQA wants you to describe how a capacitor charges and discharges through a resistor, to calculate and use the time constant τ=RC\tau = RC, and to explain how a passive RC filter selects signals by frequency, distinguishing low-pass from high-pass behaviour and finding the cut-off frequency. This is the analogue signal-processing core of the electronics area.

Charging and discharging through a resistor

When a capacitor charges through a resistor, the current is largest at the start (the capacitor is empty, so the full supply voltage is across the resistor) and falls as the capacitor voltage rises to oppose the supply. The voltage therefore climbs on an exponential curve, fast at first and then levelling off. Discharging is the mirror image: the voltage falls quickly at first and then more slowly towards zero.

The time constant relationship

The time constant fixes the speed of every RC process: charging, discharging, the settling of a smoothing capacitor in a power supply, and the timing interval that triggers a later circuit. Choosing RR and CC is how an engineer sets a delay or a filter frequency.

Passive RC filters

A capacitor's opposition to alternating current, its reactance, falls as frequency rises. An RC filter exploits this to separate signals by frequency, and which output you take decides the type:

  • Low-pass filter - output across the capacitor. At low frequency the capacitor's high reactance keeps most of the voltage across it, so low frequencies pass; at high frequency the capacitor's low reactance shorts the signal to earth, so high frequencies are attenuated.
  • High-pass filter - output across the resistor. High frequencies, which the capacitor passes easily, appear across the resistor; low frequencies are blocked by the capacitor.

Examples in context

A power-supply smoothing capacitor charges through the supply resistance and holds the rectified voltage between peaks; its time constant must be long compared with the mains period to keep the output steady. A camera flash stores energy in a capacitor charged slowly through a resistor, then dumps it quickly. A tone control on an amplifier is a pair of RC filters: the low-pass section sets the treble roll-off and the high-pass section sets the bass. An anti-aliasing filter in front of an analogue-to-digital converter is a low-pass RC stage removing frequencies the converter cannot sample correctly.

Try this

Q1. State the relationship for the time constant of an RC circuit. [1 mark]

  • Cue. τ=RC\tau = RC, in seconds.

Q2. A 220 μF220\ \mu\text{F} capacitor discharges through a 10 kΩ10\ \text{k}\Omega resistor. Find the time constant. [2 marks]

  • Cue. τ=RC=10×103×220×10−6=2.2 s\tau = RC = 10 \times 10^3 \times 220 \times 10^{-6} = 2.2\ \text{s}.

Q3. State which output (across R or across C) gives a high-pass filter. [1 mark]

  • Cue. Output across the resistor gives a high-pass filter.

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 style4 marksA 100 μF100\ \mu\text{F} capacitor charges through a 47 kΩ47\ \text{k}\Omega resistor from a 9 V9\ \text{V} supply. Calculate the time constant and state the voltage across the capacitor after one time constant.
Show worked answer →

The time constant is the product of resistance and capacitance.

Relationship: τ=RC\tau = RC.

Substitution: τ=47×103×100×10−6=4.7 s\tau = 47 \times 10^3 \times 100 \times 10^{-6} = 4.7\ \text{s}.

After one time constant a charging capacitor has reached about 63%63\% of the final voltage.

V=0.63×9=5.7 VV = 0.63 \times 9 = 5.7\ \text{V} (approximately).

Markers reward τ=RC=4.7 s\tau = RC = 4.7\ \text{s} with the correct powers of ten, and the value of about 5.7 V5.7\ \text{V} from the 63%63\% rule. Forgetting to convert microfarads is the usual error.

SQA AH style5 marksA passive RC circuit uses a resistor and capacitor in series with the output taken across the capacitor. State whether this is a low-pass or high-pass filter, calculate the cut-off frequency for R=1.6 kΩR = 1.6\ \text{k}\Omega and C=100 nFC = 100\ \text{nF}, and explain its action.
Show worked answer →

Output across the capacitor gives a low-pass filter, because the capacitor passes low frequencies (high reactance) to the output and shorts high frequencies to earth.

Relationship: fc=12πRCf_c = \dfrac{1}{2\pi RC}.

Substitution: fc=12π×1.6×103×100×10−9≈1.0 kHzf_c = \dfrac{1}{2\pi \times 1.6 \times 10^3 \times 100 \times 10^{-9}} \approx 1.0\ \text{kHz}.

Action: signals well below 1.0 kHz1.0\ \text{kHz} pass with little loss; signals well above it are increasingly attenuated.

Markers reward identifying it as low-pass, the cut-off relationship, the value of about 1.0 kHz1.0\ \text{kHz}, and a correct description of which frequencies pass.

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