AeroLab Reading room

Physics / A2 level

Alternating currents

Use rms values to compare varying current with steady heating.

Learning objectives

The big idea

Alternating current reverses direction periodically. For a sinusoid through a fixed resistor, rms current gives the same average heating power as a direct current of that value. The signed mean over a full cycle is zero.

A diode can rectify a signal so current has one direction. A smoothing capacitor reduces ripple rather than making a perfect steady source. Transformers require changing magnetic flux and have losses in real operation.

1. Read a sinusoid precisely

A sinusoidal current can be written I = I₀ sin(ωt + φ). The peak I₀, angular frequency ω and phase φ describe different features. Period is 2π/ω. A graph crossing zero does not imply that the source has permanently stopped: current reverses through the next half-cycle. The signed mean over a complete symmetric cycle is zero, while the mean square current is positive.

2. Rms compares heating effects

For a sinusoid, Irms = I₀/√2 and Vrms = V₀/√2. In a constant resistor, average power is Irms²R, equal to the heating effect of a direct current of the same rms value. If peak current is 4 A through 5 Ω, mean power is (4²/2) × 5 = 40 W. The peak power is larger, so do not substitute peak values into an average-power formula.

3. Rectification changes directionality

A diode allows much greater conduction in one direction than the other. An ideal half-wave rectifier removes one half of an input cycle; a full-wave arrangement routes both halves into the same output-current direction. The result still varies with time. Full-wave rectification gives more frequent peaks, but it does not automatically produce a constant voltage. Real diodes also have forward-voltage drops and other non-ideal behaviour.

4. Smoothing depends on the load

A capacitor across a load charges near suitable waveform peaks and discharges into the load between them, reducing ripple. Greater capacitance or a larger load resistance generally slows this discharge and reduces ripple in the simple model. More load current can increase the voltage fall between peaks. The remaining waveform is not a pure sinusoid, so the sinusoidal peak-to-rms conversion should not be applied automatically to every rectified or smoothed signal.

A method you can reuse

  1. Read peak, period and phase from the waveform.
  2. Convert between period, frequency and angular frequency.
  3. Use sinusoidal rms relations only for sinusoidal signals.
  4. Apply average-power expressions to the specified load.
  5. Explain diode paths and capacitor discharge when discussing rectification and ripple.
Why can alternating current have zero mean yet still heat a resistor?

Positive and negative current cancel in a signed average, but resistive heating depends on current squared. Both directions contribute positive energy transfer to the resistor.

Equations and relationships

Irms=I02I_{\text{rms}}=\frac{I_0}{\sqrt2}

For a sinusoid; I₀ is peak current.

Paverage=Irms2RP_{\text{average}}=I_{\text{rms}}^2R

Average heating power in a fixed resistor.

I=I0sin⁡(ωt)I=I_0\sin(\omega t)

Sinusoidal current with the chosen zero phase.

Worked example

A sinusoidal current has peak 10 A. What is its rms value to three significant figures?

Follow the solution

Irms = 10/√2 = 7.07 A.

Investigate and explain

Use a simulated sinusoidal source and resistor to compare peak readings with rms predictions.

Inspect simulated waveforms and resistive heating before considering rectification. Component availability differs between models; use circuit diagrams where a diode bridge is not provided, and never practise on mains supplies.

Open the interactive investigation

Check your understanding

1. Why is rms current useful?

  1. It always equals maximum current.
  2. It predicts equivalent resistive heating.
  3. It eliminates resistance from the power calculation.
Check the answer and reasoning

It predicts equivalent resistive heating.

Mean power depends on mean square current.

2. A sinusoidal current has peak 10 A. What is its rms value to three significant figures?

  1. 5.00 A
  2. 14.1 A
  3. 7.07 A
Check the answer and reasoning

7.07 A

Irms = 10/√2 = 7.07 A.

3. What is the signed mean of a full symmetric sinusoidal current cycle?

  1. Zero
  2. Its rms value
  3. Its positive peak
Check the answer and reasoning

Zero

Positive and negative half-cycles cancel in an arithmetic mean.