Worked examples: mains voltage and a scope trace

Two problems — one going from RMS to peak (the mains case), one going from peak-to-peak to RMS (the oscilloscope case). Every AC calculation you do for the next nine topics will start with one of these two conversions.

Example 1 — UK mains

A UK mains outlet reads $V_{rms} = 230\,\text{V}$ at $f = 50\,\text{Hz}$. Find the peak voltage $V_p$, the peak-to-peak voltage $V_{pp}$, the period $T$, and the angular frequency $ω$.

1. Peak from RMS.
   $$V_p = V_{rms} \cdot \sqrt{2} = 230 \cdot 1.414 \approx 325\,\text{V}$$
2. Peak-to-peak. $V_{pp} = 2 V_p \approx 650\,\text{V}$. The mains waveform swings across 650 V from trough to crest — which is why a capacitor in a mains filter needs to survive at least 400 V, not just 230 V.
3. Period. $T = 1/f = 1/50 = 0.02\,\text{s} = 20\,\text{ms}$.
4. Angular frequency. $\omega = 2\pi f = 2\pi \cdot 50 \approx 314\,\text{rad/s}$. Every sinusoidal calculation from now on uses $ω$ rather than $f$, because the $\sin(\omega t)$ inside the formula is cleaner than $\sin(2\pi f t)$.

The plot shows two full 20 ms cycles of the mains waveform. The +Vp line at 325 V is where every crest sits; the ±V_rms dashed orange lines at ~230 V mark the DC-equivalent heating level.

Example 2 — an oscilloscope trace

You put your scope probe on a signal generator's output. The trace swings between +5 V and −5 V once every 1 ms. Find $V_{pp}$, $V_p$, $V_{rms}$, $T$, $f$, and $ω$.

1. Read peak-to-peak directly. $V_{pp} = 5 - (-5) = 10\,\text{V}$. Scope readings are almost always peak-to-peak — it's what you measure by eye from trough to crest.
2. Peak is half. $V_p = V_{pp} / 2 = 5\,\text{V}$.
3. RMS from peak. $V_{rms} = V_p / \sqrt{2} = 5/1.414 \approx 3.54\,\text{V}$. This is the equivalent DC voltage that would heat a resistor as much as our 10 Vpp signal does.
4. Period and frequency. $T = 1\,\text{ms} = 10^{-3}\,\text{s}$, so $f = 1/T = 1000\,\text{Hz} = 1\,\text{kHz}$.
5. Angular frequency. $\omega = 2\pi f \approx 6283\,\text{rad/s}$.