Calculate sine-wave peak current from RMS, peak, or peak-to-peak current, or from load voltage and resistance or power and voltage.
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Peak Current Formula
For a sinusoidal AC waveform, peak current relates to RMS current by a factor of √2 (about 1.4142).
Ipeak = √2 × Irms
Ipp = 2 × Ipeak
Irms = Vrms / R
Irms = P / Vrms
- Ipeak = peak current (amperes)
- Irms = RMS (root-mean-square) current (amperes)
- Ipp = peak-to-peak current (amperes)
- Vrms = RMS voltage (volts)
- R = resistance (ohms)
- P = real power for a resistive load (watts)
The Known Current tab converts between Irms, Ipeak, and Ipp. If you enter Ipeak, it divides by √2 to get Irms and multiplies by 2 to get Ipp. If you enter Ipp, it halves the value first, then divides by √2.
The From Load tab finds RMS current from circuit values, then scales by √2 to give peak current. Use voltage and resistance for a resistive load, or power and voltage when you know the wattage.
Reference Values
The conversion factors below apply to any sine wave. Multiply or divide by these constants to move between forms.
| Conversion | Factor | Example |
|---|---|---|
| Irms to Ipeak | × 1.4142 | 10 A → 14.14 A |
| Ipeak to Irms | ÷ 1.4142 | 14.14 A → 10 A |
| Ipeak to Ipp | × 2 | 14.14 A → 28.28 A |
| Irms to Ipp | × 2.8284 | 10 A → 28.28 A |
Typical peak currents for common single-phase resistive loads:
| Load | Vrms | Irms | Ipeak |
|---|---|---|---|
| 60 W lamp | 120 V | 0.50 A | 0.71 A |
| 1500 W heater | 120 V | 12.5 A | 17.68 A |
| 3000 W kettle | 230 V | 13.04 A | 18.45 A |
| 5500 W dryer | 240 V | 22.92 A | 32.41 A |
Example Problems
Example 1. A heater draws 12.5 A RMS from a 120 V supply. Find the peak current.
Ipeak = √2 × 12.5 = 1.4142 × 12.5 = 17.68 A.
Example 2. A 230 V RMS source feeds a 50 Ω resistor. Find Irms and Ipeak.
Irms = 230 / 50 = 4.6 A. Ipeak = √2 × 4.6 = 6.51 A.
FAQ
Does this work for non-sinusoidal waveforms? No. The √2 factor only holds for pure sine waves. Square, triangular, or distorted waveforms have different crest factors.
Does it apply to inductive or capacitive loads? The current relationships still hold for the current itself. The power formula Irms = P / Vrms only works when P is real power and the load is purely resistive, or when you use apparent power (VA) instead.
Why is peak current important? Component ratings, fuse selection, and insulation stress depend on peak values, not RMS. A 10 A RMS line actually swings to 14.14 A at the top of each cycle.
What is peak-to-peak current? It is the full swing from the negative peak to the positive peak, equal to twice the peak value for a symmetric sine wave.
