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Learn7408Paper 2 · Paper 2Fields and their consequencesCharacteristics of alternating currents
Fields and their consequences · Subtopic 4 of 13

Characteristics of Alternating Currents

An alternating current reverses direction periodically: I=I0sin⁡ωtI=I_0\sin\omega t, with peak value I0I_0, period TT and frequency f=1Tf=\dfrac1T. The power in a resistor, I2RI^2R, varies between 00 and I02RI_0^2R, so the mean power is 12I02R\tfrac12I_0^2R. The r.m.s. value is the steady d.c. value giving the same mean power: Irms=I02I_{\text{rms}}=\dfrac{I_0}{\sqrt2} and Vrms=V02V_{\text{rms}}=\dfrac{V_0}{\sqrt2}. Pakistan's mains is 230230 V r.m.s., peaking at about 325325 V.

Cambridge 9702 Paper 4AQA 7408 Paper 2≈ 30 minRevision 7 · updated 2026-10-27
Section 1 of 14

Learning objectives

  • Use period, frequency and peak value for a.c.
  • Use x=x0sin⁡ωtx=x_0\sin\omega t for a.c. currents and voltages.
  • Explain why mean power is half the peak power.
  • Define and use r.m.s. values.
Pause and think

Mains is "230230 V", but an oscilloscope shows it swinging between +325+325 V and −325-325 V. Which is the real voltage?

Section 2 of 14

Core idea

Start hereWhat does "230 V" mean?

Mains voltage swings between +325 V and −325 V fifty times a second. Its average is zero, yet a heater clearly gets hot. "230 V" is the steady voltage that would heat it just as well: the r.m.s. value.

The big idea

V=V0sin⁡ωtV=V_0\sin\omega t. Vrms=V02V_{\text{rms}}=\frac{V_0}{\sqrt2}; mean power = half the peak power.

1. r.m.s. values

Voltage and power in a resistor on a.c.
24681012−1−0.50.510timeVpower ∝ V²mean power = ½ peakV rms = V₀/√2
Power is always positive (heating happens in both directions) and averages half its peak. That is why Vrms=V0/2V_{\text{rms}}=V_0/\sqrt2.
Why V02\frac{V_0}{\sqrt2}
  1. P=V2R=V02Rsin⁡2ωtP=\frac{V^2}R=\frac{V_0^2}R\sin^2\omega t
    Instantaneous power.
  2. ⟨sin⁡2ωt⟩=12\langle\sin^2\omega t\rangle=\tfrac12
    sin⁡2=12(1−cos⁡2ωt)\sin^2=\frac12(1-\cos2\omega t); the cosine averages to zero.
  3. ⟨P⟩=V022R=(V0/2)2R\langle P\rangle=\frac{V_0^2}{2R}=\frac{(V_0/\sqrt2)^2}R
    Same as a d.c. voltage V0/2V_0/\sqrt2.
Worked exampleUK-style mains

Peak 325 V, 50 Hz; 100 Ω heater.

Goal
r.m.s.; TT, ω\omega; mean and peak power.
Strategy
Formulas.
  1. 3252=230 V\frac{325}{\sqrt2}=230\text{ V}
    r.m.s.
  2. T=20 ms, ω=2π(50)=314 rad s−1T=20\text{ ms},\ \omega=2\pi(50)=314\text{ rad s}^{-1}
    Timing.
  3. ⟨P⟩=2302100=529 W, P0=3252100=1056 W\langle P\rangle=\frac{230^2}{100}=529\text{ W},\ P_0=\frac{325^2}{100}=1056\text{ W}
    Mean is half the peak.
Answer230230 V; 2020 ms, 314314 rad s−1^{-1}; 529529 W mean, 10561056 W peak.
✓ Check1056/2=5281056/2=528 ✓.
Common misconceptionMean voltage The mean of a sine wave over a cycle is zero: use r.m.s.

Summary

  • Vrms=V0/2V_{\text{rms}}=V_0/\sqrt2.
  • Mean power half of peak.
Key words
r.m.s.
Root-mean-square: equivalent d.c. value.
HistoryThe war of the currents. In the 18801880s Thomas Edison championed direct current, while Nikola Tesla and George Westinghouse pushed alternating current. A.c. won because transformers can step its voltage up for efficient long-distance transmission and down again for homes, something d.c. couldn't do easily then.
Fun factWhy lights don't flicker. On a 5050 Hz supply, the current passes through zero 100100 times a second, so a filament lamp's power dips 100100 times a second. The filament stays hot between dips, and anyway our eyes can't follow flicker that fast.
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