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Electrical
Peak and RMS values (AC)
Peak voltage, current and power of a sinusoidal AC supply from its RMS voltage and the load resistance.
- Ref
- ELE-06
- Rev
- v1
- Field
- Electrical
- Result
- P0 [W]
Subscript 0 is the peak value and rms the RMS value. These factors hold for a sine wave. The calculator finds the peak power in a resistive load, P0 = 2Vrms²/R, and shows the peak voltage and current as derived values.
An AC supply swings up to a peak and back, so its useful heating value is lower than its peak. The RMS value is the steady DC value that would heat a resistor equally. For a sine wave the peak is √2 times the RMS value, and the peak power is twice the mean power.
Calculator
- Peak voltage V0 = √2 Vrms325.3 V
- RMS current Irms = Vrms/R2.3 A
- Peak current I03.253 A
- Mean power Prms529 W
What drives the result
Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.
- Peak values are √2 ≈ 1.414 times the RMS values for voltage and current, and 2 times the mean value for power. A 230 V RMS supply swings to about ±325 V.
- Power goes as the square of the voltage, so peak power rises 21% for a 10% rise in Vrms. Components must withstand the peak voltage even though heating follows the RMS value.
- Mean power is Vrms²/R, not V0²/R. Using the peak voltage there overstates the mean power by a factor of 2.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (W) | Local step | Change in result |
|---|---|---|---|---|---|---|
| R Load resistance | 10 to 1,000 Ω | 105.8 | 10,580 | 10,470 | +10% | −9.09 % |
| Vrms RMS voltage | 12 to 400 V | 2.88 | 3,200 | 3,197 | +10% | +21 % |
Sweep of RMS voltage
| RMS voltage (V) | Peak power (W) |
|---|---|
| 12 | 2.88 |
| 18.6 | 6.92 |
| 28.83 | 16.63 |
| 44.7 | 39.95 |
| 69.28 | 96 |
| 107.4 | 230.7 |
| 166.5 | 554.3 |
| 258 | 1,332 |
| 400 | 3,200 |
Sweep one variable
Variables and how each one contributes
Peak power
What the equation gives you. Everything else feeds it.
RMS voltage
The RMS value of the supply voltage, as quoted for mains (for example 230 V). Peak power goes as Vrms², so a 10% higher supply voltage gives about 21% more peak power.
Load resistance
A purely resistive load connected across the supply. Peak power falls as 1/R: a smaller resistance draws more current and so more power.
Worked example with the values above
- Peak voltage: V0 = √2 Vrms = 1.4142 × 230 = 325.3 V.
- RMS current: Irms = Vrms/R = 230 / 100 = 2.3 A. Peak current: I0 = √2 Irms = 3.253 A.
- Mean power: Prms = Vrms Irms = 230 × 2.3 = 529 W.
- Peak power: P0 = 2 Prms = 2 × 529 = 1,058 W. Check: V0 I0 = 325.3 × 3.253 = 1,058 W.
Assumptions and limits
- Pure sinusoidal waveform. The √2 and 2 factors are specific to a sine wave.
- Purely resistive load, so current is in phase with voltage (power factor 1).
- Steady RMS value.
Common mistakes
- Using peak values in P = V²/R and getting twice the mean power.
- Quoting mains as its peak value. 230 V is the RMS value.
- Applying √2 to square or distorted waveforms.
Statistics and uncertainty
RMS is itself a statistical quantity: the square root of the mean of the squared waveform. The factor √2 holds only for a pure sine wave. Distorted waveforms have a different peak-to-RMS ratio (the crest factor), and meters that are not true-RMS read wrongly on them.
Where it is used
- Insulation and component voltage ratings, which must cover the peak.
- Mean heating power of AC loads.
- Converting between oscilloscope readings (peak) and meter readings (RMS).
Maths you need
- Square roots: √2 ≈ 1.414
- Sine waves: amplitude, and why the mean of the squared value gives a factor ½
- Ohm’s law: I = V/R at each instant