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Average–Peak Power for a Resistive Sinusoid

Relate cycle-average power to instantaneous peak power for an ideal resistor driven by a sinusoidal voltage.

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How to use this calculator

  1. Enter the known values in the units shown. Results update as you type.
  2. Where results are editable, change one to solve backwards. Lock a value to hold it fixed.
  3. Use the worked example to check the method. Reset restores the starting fields.

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Formula and method

For an ideal resistor under a sinusoidal voltage, 50 W average dissipation has a 100 W instantaneous peak.

Instantaneous peak power = 2 × average power for in-phase sinusoidal voltage/current

Worked example

Enter these known values and leave the other values blank.

Cycle-average power
50 W
Instantaneous peak power
100 W

Assumptions and limitations

  • For sinusoidal voltage and current in phase in an ideal resistor, instantaneous power is proportional to sin²(t); its maximum is twice its cycle average.
  • The square-root-of-two RMS-to-peak factor applies to voltage/current amplitudes, not directly to average power in watts.
  • This relation does not convert arbitrary amplifier advertising ratings or non-sinusoidal/reactive waveforms. Both powers are nonnegative.
  • Arithmetic uses 64 significant decimal digits, with separate display rounding.

Common questions

Why is the factor two instead of √2?

Voltage or current amplitudes differ by √2 under a sine wave. Resistive power depends on their product, so instantaneous peak power is twice cycle-average power.

References

Bookify labels the model as cycle-average versus instantaneous peak power for a resistive sinusoid and uses the power factor of two, replacing the captured voltage-amplitude factor of √2 applied to watts.

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