understand and use the terms period, frequency and peak value as applied to an alternating current or voltage

Characteristics of Alternating Currents (AC)

Objective

To understand and correctly use the terms period, frequency, peak (amplitude), peak‑to‑peak and RMS (root‑mean‑square) for an alternating current or voltage, and to relate these quantities to the mean (average) power of a sinusoidal AC.

Key Concepts

  • An AC quantity repeats its shape at regular intervals – it is periodic.

  • Definitions of period, frequency, peak, peak‑to‑peak and RMS apply to any periodic waveform, not only to sinusoids.

  • RMS values are the quantities that appear on specifications because they give the same heating effect as an equivalent DC value.

Definitions and Symbols

TermSymbolDefinitionUnits
PeriodTTime taken for one complete cycle of the waveform.s (seconds)
FrequencyfNumber of cycles per unit time; the reciprocal of the period.Hz (hertz)
Peak (Amplitude) ValueI0, V0Maximum instantaneous current or voltage (positive or negative) reached in a cycle.A or V
Peak‑to‑Peak ValueIpp, VppDifference between the maximum positive and maximum negative values; for a sinusoid, Vpp = 2 V0.A or V
Root‑Mean‑Square (RMS) ValueIrms, VrmsEffective value of AC; the DC current or voltage that would produce the same average heating in a resistor.A or V
Angular FrequencyωRate of change of phase; ω = 2πf.rad · s⁻¹

Note on non‑sinusoidal waveforms: the RMS‑to‑peak factor is waveform‑dependent. For example, a square wave has RMS = peak / √1 = peak, a triangular wave has RMS = peak / √3. The sinusoidal factor (1/√2) is therefore a special case.

Mathematical Description of a Sinusoidal AC

Instantaneous current and voltage are expressed as

\[

i(t)=I_{0}\sin(\omega t+\phi),\qquad

v(t)=V_{0}\sin(\omega t+\phi)

\]

  • I0, V0 – peak (amplitude) value.

  • ω = 2πf – angular frequency (rad · s⁻¹).

  • φ – phase angle (zero for a simple source).

Derivation of the RMS Value (sinusoid)

\[

I{\rm rms}= \sqrt{\frac{1}{T}\int{0}^{T} i^{2}(t)\,dt}

= \sqrt{\frac{1}{T}\int{0}^{T} I{0}^{2}\sin^{2}(\omega t+\phi)\,dt}

= I{0}\sqrt{\frac{1}{T}\int{0}^{T}\sin^{2}(\omega t+\phi)\,dt}

= I_{0}\sqrt{\frac{1}{2}}

= \frac{I_{0}}{\sqrt{2}}

\]

(Using \(\langle\sin^{2}\rangle = ½\) over one full cycle.)

Relationship Between Period and Frequency

\[

f=\frac{1}{T}\qquad\text{and}\qquad T=\frac{1}{f}

\]

Example: UK mains – f = 50 Hz ⇒ T = 1/50 = 0.020 s (20 ms).

RMS–Peak Relationship (Sinusoid)

\[

I{\rm rms}= \frac{I{0}}{\sqrt{2}},\qquad

V{\rm rms}= \frac{V{0}}{\sqrt{2}}

\]

Mean (Average) Power in a Sinusoidal AC

For a sinusoidal voltage of peak \(V_{0}\) applied to a resistor \(R\), the average power over one complete cycle is

\[

\bar P = \frac{1}{2}V{0}I{0}

= I_{\rm rms}^{2}R

= \frac{V_{\rm rms}^{2}}{R}

\]

  • The factor \(½\) originates from the time‑average of \(\sin^{2}\) (⟨sin²⟩ = ½).

  • Because RMS values appear in the last two forms, specifications of power‑rated equipment always use RMS quantities.

Example Calculations

  1. RMS voltage from a given peak.

    Given \(V{0}=325\;{\rm V}\) and \(f=50\;{\rm Hz}\), find \(V{\rm rms}\).

    \[

    V_{\rm rms}= \frac{325\ {\rm V}}{\sqrt{2}}\approx 230\ {\rm V}

    \]

    This is why UK mains is quoted as 230 V RMS.

  2. Mean power in a 230 V RMS domestic circuit.

    For a purely resistive load of \(R=100\;\Omega\):

    \[

    \bar P = \frac{V_{\rm rms}^{2}}{R}

    = \frac{(230\ {\rm V})^{2}}{100\ \Omega}

    = 529\ {\rm W}

    \]

    Using peak values gives the same result: \(V{0}\approx325\ {\rm V}\), \(I{0}=V{0}/R\approx3.25\ {\rm A}\), then \(\tfrac12V{0}I_{0}=529\ {\rm W}\).

  3. Peak‑to‑peak voltage.

    For the same sinusoid, \(V{pp}=2V{0}=2\times325\ {\rm V}=650\ {\rm V}\).

Practical Points for Examination

  • Always write the correct units (s, Hz, V, A, Ω, W).

  • Remember: f (Hz) = 1/T (s⁻¹) and ω = 2πf.

  • Peak values are denoted by subscript “0”; RMS values have subscript “rms”.

  • Peak‑to‑peak values are denoted by subscript “pp”.

  • State clearly that RMS is the DC‑equivalent heating value; it is not the same as the peak.

  • When a question asks for power, use the RMS form \(\bar P = V{\rm rms}^{2}/R\) or \(\bar P = I{\rm rms}^{2}R\) unless a peak value is explicitly required.

Suggested diagram: A sinusoidal voltage versus time showing one complete period, the peak value \(V{0}\), the RMS value \(V{\rm rms}\) (as a horizontal line), the peak‑to‑peak value \(V{pp}=2V{0}\), and the period \(T\) marked on the time axis.