Calculate rms and peak alternating values
Convert between peak and root-mean-square values for sinusoidal current and voltage.
A Level; Cambridge International Physics 9702, 2025–2027 syllabus, Topic 21. These original explanations and calculations are study aids; full timed assessment and human subject-teacher review remain outstanding.
Before you begin
- Use period, frequency, sine graphs, current, voltage, power, diodes, capacitors and SI units.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What does the r.m.s. value of AC current represent for a resistor?
What you will learn
- Distinguish r.m.s. value from peak value
- Use Irms = I₀/√2 and Vrms = V₀/√2
- Interpret r.m.s. as the equivalent steady value for resistive heating
R.m.s. describes heating equivalence
The root-mean-square (r.m.s.) value of an alternating current is the value of a steady direct current that would produce the same mean heating power in a resistor. It is useful because an AC current reverses but still transfers energy in both half-cycles.
For a sinusoidal current, Irms = I₀/√2 ≈ 0.707 I₀. For a sinusoidal voltage, Vrms = V₀/√2 ≈ 0.707 V₀. The r.m.s. value is smaller than the peak magnitude; it is not the arithmetic mean of the signed waveform, which is zero for a symmetric cycle.
Rearrange carefully when the peak is required
To convert from peak to r.m.s., divide by √2. To convert back, multiply by √2: I₀ = √2 Irms and V₀ = √2 Vrms. Keep the same quantity and unit on both sides; do not divide a voltage by a current or confuse peak-to-peak with peak.
A mains supply stated as 230 V AC usually gives the r.m.s. voltage unless a question says otherwise. Its sinusoidal peak is about 325 V. The relation here assumes a sinusoidal waveform; rectified or distorted waveforms need their own r.m.s. analysis.
Pause and explain
A sinusoidal current has peak value 10 A. What is its r.m.s. value?
Worked example
A sinusoidal supply has r.m.s. voltage 230 V. Find its peak voltage.
Show the worked solution
- Use V₀ = √2 Vrms.
- Substitute 230 V.
- Round suitably and keep volts.
Answer V₀ = √2 × 230 ≈ 325 V.
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
A sinusoidal alternating current has peak value 8.0 A. Calculate its r.m.s. value.
Give me a hint
Divide the peak by √2.
Compare my reasoning
- Irms = I₀/√2.
- Irms = 8.0/√2.
- Irms ≈ 5.66 A.
Irms ≈ 5.7 A.
Look for these in your work
- Correct conversion used
- Calculation shown
- Current unit stated
A sinusoidal voltage is 120 V r.m.s. Find its peak and peak-to-peak values.
Give me a hint
Multiply by √2 to get peak; a symmetric waveform spans twice the peak.
Compare my reasoning
- V₀ = √2 × 120 ≈ 170 V.
- Peak-to-peak = 2V₀.
- Peak-to-peak ≈ 340 V.
Peak ≈ 170 V; peak-to-peak ≈ 340 V.
Look for these in your work
- Peak conversion applied
- Symmetry used
- Two values distinguished
A sinusoidal voltage and a steady d.c. voltage both read 12 V across the same resistor. Explain what extra comparison is needed before saying they produce the same mean heating power.
Give me a hint
Determine whether each 12 V is a peak, r.m.s. or steady value.
Compare my reasoning
- A stated AC voltage might be a peak or an r.m.s. value; identify which one.
- For a sinusoidal supply, the r.m.s. voltage determines mean resistor power.
- Only if the AC r.m.s. value equals 12 V will it match 12 V steady d.c. across the same resistor.
The AC reading must be identified as r.m.s.; equal r.m.s. and steady d.c. voltages across the same resistor give equal mean power.
Look for these in your work
- Voltage definition checked
- Heating link made
- Comparison condition stated
Common mistakes
- Using arithmetic mean as the r.m.s. value
- Dividing by √2 when converting r.m.s. back to peak
- Confusing peak with peak-to-peak
- Applying the sinusoidal conversion to a distorted waveform without justification
What can you explain now?
What does the r.m.s. value of AC current represent for a resistor?
Compare with the explanation
The equivalent steady current for the same mean heating power
R.m.s. expresses the heating effect of an alternating current.
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Make it stick
How do peak and r.m.s. values differ, and why does r.m.s. predict heating?
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.