Estimate effective current for circuits, heaters, and motors. Choose a method and enter measured values. Download clean reports to share with teams today easily.
| Case | Given | Value | Computed Irms | Notes |
|---|---|---|---|---|
| 1 | Sinusoidal Ipeak | 5 A | 3.535 A | Irms = Ipeak / √2 |
| 2 | Sinusoidal Ipp | 10 A | 3.535 A | Ipeak = Ipp/2 |
| 3 | Power and resistance | P = 60 W, R = 10 Ω | 2.449 A | Irms = √(P/R) |
| 4 | Vrms and |Z| | Vrms = 120 V, |Z| = 24 Ω | 5.000 A | Irms = Vrms/|Z| |
| 5 | Samples | 0, 2, -2, 2 | 1.732 A | Irms = √(mean(i²)) |
Tip: For non-sinusoidal waveforms, the samples method is usually the safest.
RMS current is the “heating‑equivalent” value of a varying current. If a resistor carries a time‑varying current i(t), the average thermal power is proportional to the average of i(t)². RMS compresses that square‑average into a single current that would produce the same power as steady DC.
Wire sizing, fuse selection, transformer loading, and motor heating are primarily RMS‑driven. In many standards, RMS defines thermal current ratings. For example, a 10 Ω heater dissipates 60 W when Irms = √(P/R) ≈ 2.449 A. Using peak current alone can overstate heating by about 41% for a sine wave.
For a pure sine, Irms = Ipeak/√2 and Irms = Ipp/(2√2). That means a 5 A peak sinusoid corresponds to 3.535 A RMS. These relationships are widely used for 50/60 Hz mains and many inverter outputs when distortion is low.
Crest factor is Ipeak/Irms. A sine has 1.414, while pulsed or rectifier loads can exceed 2–3. Higher crest factor increases conductor and component stress even if RMS stays similar, so measuring or sampling the waveform is important in power electronics.
If you know real power and resistance, RMS current follows directly from heating: Irms = √(P/R). This is ideal for resistive loads such as heaters and lamps. For non‑resistive loads, real power depends on power factor, so voltage‑impedance methods may be better.
For AC steady state, Irms = Vrms/|Z|. If resistance and reactance are known, |Z| = √(R² + X²). Example: with 120 V RMS and |Z| = 24 Ω, RMS current is 5 A. This is useful for coils, capacitors, and mixed loads.
When waveforms are distorted, compute RMS from samples: Irms = √(mean(i²)). More samples improve accuracy. If your instrument provides current readings over time (CSV from a scope or logger), paste them here, keep units consistent, and let the calculator handle the square‑average.
Ensure sensors are rated for the expected peak and bandwidth. For switching converters, sample fast enough to capture ripple; otherwise RMS will be underestimated. If you only have peak‑to‑peak ripple and a DC bias, use samples or a dedicated RMS meter. Document results using the export buttons.
No. Average can be near zero for AC, while RMS is positive and tracks heating. A sine wave has zero average over a cycle but nonzero RMS.
Use it for pulsed, chopped, rectified, or distorted currents. It works for any waveform as long as the samples represent the full operating cycle.
It is the DC current that produces the same resistive heating as the AC waveform. For a resistor, equivalent DC equals the RMS current.
It works for linear steady‑state conditions using RMS magnitudes. If impedance changes with time or the waveform is highly non‑sinusoidal, samples or direct measurement is safer.
More is better. Aim for at least one full cycle and dozens of points. For switching supplies, capture multiple switching periods to include ripple energy.
High peaks can stress semiconductors, fuses, and magnetic components even when RMS looks acceptable. It can also increase electromagnetic interference and nuisance trips.
Yes. Pick the unit that matches your measurement range. The calculator converts internally to amps, then formats the output in your chosen unit for readability.
Important Note: All the Calculators listed in this site are for educational purpose only and we do not guarentee the accuracy of results. Please do consult with other sources as well.