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AS & A-Level Physics 21 — Alternating Currents

AS & A-Level Physics 21 — Alternating Currents

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Independent Deckloop revision aligned with the Cambridge International AS & A Level Physics (9702) syllabus, 2025–2027. Not affiliated with or endorsed by Cambridge International Education. Chapter 21 of 26: Alternating currents. Concepts, worked applications and misconception checks.

Physics EN A-Level
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Sinusoidal AC Equations

For a sinusoidally varying alternating current or voltage, its instantaneous value can be described by an equation similar to that used for simple harmonic motion. The general form is x=x0sin⁡ωtx = x_0 \sin \omega t, where xx represents the instantaneous current (II) or voltage (VV), and x0x_0 is the peak value (I0I_0 or V0V_0). The term ω\omega is the angular frequency, measured in radians per second (rad s−1^{-1}). It is related to the linear frequency ff and period TT by the relations ω=2πf\omega = 2\pi f and ω=2π/T\omega = 2\pi/T. The argument of the sine function, ωt\omega t, is in radians. This equation assumes the waveform starts at zero and is increasing at t=0t=0. If it started at a different point in its cycle, a phase constant would be included, but for basic AC analysis, this simplified form is standard.

Key points

  • Instantaneous AC current: I=I0sin⁡ωtI = I_0 \sin \omega t.
  • Instantaneous AC voltage: V=V0sin⁡ωtV = V_0 \sin \omega t.
  • Angular frequency ω=2πf=2π/T\omega = 2\pi f = 2\pi/T.
  • ωt\omega t must be in radians for the sine function.

Worked example

Question

An AC voltage supply has a peak voltage of 340 V340 \text{ V} and a frequency of 60 Hz60 \text{ Hz}. Write the equation for the instantaneous voltage VV in terms of time tt. Then, calculate the instantaneous voltage at t=5.0 mst = 5.0 \text{ ms}. Choose t = 0 when the voltage crosses zero in the positive direction; quote the calculated voltage to two significant figures.

Solution

1. Calculate the angular frequency: ω=2πf=2π×60 Hz=120π rad s−1≈377 rad s−1\omega = 2\pi f = 2\pi \times 60 \text{ Hz} = 120\pi \text{ rad s}^{-1} \approx 377 \text{ rad s}^{-1}.
2. Write the equation for instantaneous voltage: V=V0sin⁡ωt=340sin⁡(120πt)V = V_0 \sin \omega t = 340 \sin(120\pi t).
3. Substitute t=5.0 ms=5.0×10−3 st = 5.0 \text{ ms} = 5.0 \times 10^{-3} \text{ s} into the equation:
4. V=340sin⁡(120π×5.0×10−3)V = 340 \sin(120\pi \times 5.0 \times 10^{-3})
5. V=340sin⁡(0.6π)V = 340 \sin(0.6\pi)
6. V=340sin⁡(1.885 rad)V = 340 \sin(1.885 \text{ rad})
7. V≈340×0.951=323.34 VV \approx 340 \times 0.951 = 323.34 \text{ V}.
8. The calculator gives about 323.36 V. To two significant figures, this is 320 V.

With this time origin, V=340sin⁡(120πt)V=340\sin(120\pi t) V, with t in seconds. At 5.0 ms, V = 320 V (two significant figures).

Common pitfalls

  • Using degrees instead of radians for ωt\omega t in the sine function. Always ensure your calculator is in radian mode for these calculations.
  • Confusing peak value (x0x_0) with instantaneous value (xx). The equation gives the value at a specific moment, not the maximum.

Prerequisites

  • Understanding of peak value, frequency, and period is essential for defining the parameters in the sinusoidal equation.
  • Familiarity with sine functions and radian measure.
  • Understanding of angular frequency and its relation to linear frequency from wave motion.
Further resources