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

Published by Patrick Mutisya · 14 days ago

Cambridge A-Level Physics 9702 – Characteristics of Alternating Currents

Characteristics of Alternating Currents

Objective

Understand and use the terms period, frequency and peak value as applied to an alternating current (AC) or voltage.

Key Concepts

  • An alternating current or voltage varies periodically with time.

  • The shape of the variation is often sinusoidal, but the definitions of period, frequency and peak value apply to any periodic waveform.

Definitions and Relationships

TermSymbolDefinitionUnits
Period\$T\$The time taken for one complete cycle of the waveform.seconds (s)
Frequency\$f\$Number of cycles per unit time. It is the reciprocal of the period.hertz (Hz)
Peak (or Amplitude) Value\$I{0}\$ or \$V{0}\$Maximum instantaneous value of current or voltage in either direction.ampere (A) or volt (V)
Angular Frequency\$\omega\$Rate of change of phase, related to frequency by \$\omega = 2\pi f\$.radians per second (rad·s⁻¹)
Root‑Mean‑Square (RMS) Value\$I{\mathrm{rms}}\$, \$V{\mathrm{rms}}\$Effective value of AC, equal to the DC value that would produce the same heating effect. For a sinusoid, \$I{\mathrm{rms}} = \dfrac{I{0}}{\sqrt{2}}\$.A or V

Mathematical Description of a Sinusoidal AC

The instantaneous current (or voltage) in a sinusoidal AC source can be written as

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

where

  • \$I_{0}\$ is the peak current,

  • \$\omega = 2\pi f\$ is the angular frequency,

  • \$\phi\$ is the phase angle (often zero for a simple source).

Similarly, the instantaneous voltage is

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

Relationship Between Period and Frequency

The period \$T\$ and frequency \$f\$ are inverses of each other:

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

For example, a mains supply in the United Kingdom has \$f = 50\ \text{Hz}\$, giving a period of \$T = 0.020\ \text{s}\$ (20 ms).

Example Calculation

  1. Given a sinusoidal voltage with a peak value \$V_{0}=325\ \text{V}\$ and a frequency \$f=50\ \text{Hz}\$, find the RMS voltage.

  2. Use the RMS relation for a sinusoid:

    \$V{\mathrm{rms}} = \frac{V{0}}{\sqrt{2}}\$

  3. Substitute the numbers:

    \$V_{\mathrm{rms}} = \frac{325\ \text{V}}{\sqrt{2}} \approx 230\ \text{V}\$

  4. Thus the standard UK mains voltage of 230 V RMS corresponds to a peak of about 325 V.

Practical Points for Examination

  • Always state the units when giving a numerical answer.

  • Remember that \$f\$ is measured in hertz (cycles per second) and \$T\$ in seconds.

  • Peak values are often denoted by a subscript “0” (e.g., \$I{0}\$, \$V{0}\$).

  • RMS values are the quantities that appear on most electrical specifications.

Suggested diagram: A sinusoidal voltage versus time showing one complete period, the peak value \$V{0}\$, and the corresponding RMS value \$V{\mathrm{rms}}\$ marked on the vertical axis.