Alternating current and voltage, measuring frequency and peak values from an oscilloscope, and the relationship between peak and root-mean-square values.

What this key area is asking

The SQA wants you to describe alternating current and voltage, read the frequency and peak value of an AC signal from an oscilloscope trace, and use the relationship between the peak and root-mean-square (rms) values of a sinusoidal supply.

Alternating current and voltage

The period $T$ is the time for one complete cycle, related to frequency by $f = \frac{1}{T}$. AC is used for mains distribution because its voltage can be stepped up and down efficiently by transformers, reducing transmission losses.

Reading an oscilloscope

An oscilloscope plots voltage (vertical) against time (horizontal) on a divided screen. Two settings let you turn divisions into physical quantities: the time-base (seconds per horizontal division) and the y-gain or volts-per-division (volts per vertical division).

Peak and root-mean-square values

The rms value of an alternating quantity is the value of the steady direct current (or voltage) that would dissipate the same average power in a resistor. Because the instantaneous power follows $V^2$, averaging over a cycle introduces a factor of one half, and taking the square root gives the factor of $\sqrt{2}$.

The quoted "$230\ \text{V}$ mains" is an rms value; its peak is $\sqrt{2} \times 230 \approx 325\ \text{V}$. Meters and power ratings use rms because that is the value equivalent to a steady DC supply.

Examples in context

Your phone charger and laptop brick take in mains AC and rectify it to DC; the input is quoted as an rms voltage. A multimeter set to AC reads rms directly, so a reading of $230\ \text{V}$ corresponds to a $325\ \text{V}$ peak that the insulation must withstand. Oscilloscopes in a school lab let you measure the $50\ \text{Hz}$ mains frequency by reading the period of one cycle off the screen. An audio amplifier rated for a certain rms power output is being described by the equivalent steady power, even though the music signal swings between positive and negative peaks far above that level.

Try this

Q1. An AC supply has a peak voltage of $10\ \text{V}$. Calculate its rms voltage. [2 marks]

• Cue. $V_{\text{rms}} = \frac{10}{\sqrt{2}} = 7.1$ V.

Q2. One cycle of an AC signal occupies $8.0\ \text{ms}$. Calculate its frequency. [2 marks]

• Cue. $f = \frac{1}{T} = \frac{1}{8.0 \times 10^{-3}} = 125$ Hz.

Q3. State why mains voltage is quoted as an rms value rather than a peak value. [1 mark]

• Cue. The rms value is the equivalent steady DC value that delivers the same average power.