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A2 · Practice questions · Characteristics of alternating current

The steady current that heats equally.

Six original Cambridge-style questions on peak and r.m.s. values, i = I₀ sin ωt, mean and peak power in a resistor, and why the factor 1/√2 belongs only to sinusoids.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.
Keep these straight

R.m.s. is defined through power.

01
Calculation
[2 marks]

An alternating current is described by i = 4.2 sin(314t), with i in amperes and t in seconds. Calculate the frequency and period of the current, and its r.m.s. value.

  • ω = 314 rad s⁻¹, so f = ω / 2π = 314 / 6.28 = 50 Hz and T = 1/f = 20 ms. ✓
  • Irms = I₀ / √2 = 4.2 / 1.414 = 3.0 A. ✓
02
Analysis
[3 marks]

Define the root-mean-square value of an alternating current. Explain why the simple average of a sinusoidal current is not useful, and why the mean power dissipated by a sinusoidal current in a resistor is half the peak power.

  • The r.m.s. value is the value of the steady direct current that would dissipate the same mean power in the same resistor (equivalently, the square root of the mean of the square of the current). ✓
  • A sinusoidal current is positive for half of each cycle and equally negative for the other half, so its simple average over a cycle is zero, which says nothing about the heating it produces. ✓
  • The power i²R = I₀²R sin²ωt is always positive, and the mean of sin² over a cycle is ½, so the mean power is ½I₀²R, half the peak power. ✓
03
Calculation
[3 marks]

A filament lamp is rated 60 W at 230 V r.m.s. Calculate the r.m.s. current in the lamp, the peak current, and the peak instantaneous power dissipated.

  • Irms = P / Vrms = 60 / 230 = 0.26 A. ✓
  • I₀ = √2 × Irms = 1.414 × 0.261 = 0.37 A. ✓
  • Peak power = I₀V₀ = 2 × mean power = 120 W (the lamp's resistance is 230² / 60 = 880 Ω). ✓
04
Calculation
[3 marks]

A signal generator produces a sinusoidal p.d. of peak value 20 V at a frequency of 400 Hz across a 50 Ω resistor. Calculate the angular frequency, the instantaneous p.d. at t = 0.30 ms after the p.d. passes through zero going positive, and the mean power dissipated in the resistor.

  • ω = 2πf = 2π × 400 = 2.5 × 10³ rad s⁻¹ (2513). ✓
  • v = V₀ sin ωt = 20 sin(2513 × 3.0 × 10⁻⁴) = 20 sin(0.754 rad) = 14 V. ✓
  • Mean power = Vrms² / R = (20 / √2)² / 50 = 200 / 50 = 4.0 W. ✓
05
Analysis
[2 marks]

A square-wave current alternates between +2.0 A and −2.0 A, spending equal times at each value. State its r.m.s. value and explain why the factor 1/√2 must not be used here.

  • The square of the current is 4.0 A² at every instant, so the mean square is 4.0 A² and Irms = 2.0 A, equal to the peak value. ✓
  • The factor 1/√2 comes from the mean of sin² over a cycle being ½, which is true only for a sinusoidal waveform; for other shapes the r.m.s. value must be found from the mean of the square directly. ✓
06
Calculation
[4 marks]

An electric heater is rated 2.0 kW when connected to the 230 V r.m.s., 50 Hz mains.

(a) Calculate the resistance of the heater. (b) Calculate the r.m.s. current and the peak current. (c) Calculate the peak power and the energy dissipated in one cycle of the supply.

  • (a) R = Vrms² / P = 230² / 2000 = 26 Ω. ✓
  • (b) Irms = P / Vrms = 2000 / 230 = 8.7 A. ✓
  • I₀ = √2 × 8.70 = 12 A. ✓
  • (c) Peak power = 2 × 2.0 kW = 4.0 kW; energy per cycle = mean power × T = 2000 × 0.020 = 40 J. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.