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RMS and peak voltage for a square waveform are the same. There is a small caveat, and that is that you'd have to have a "perfect" square wave with a rise time of zero. Let's have a look. If we have a perfect square wave, it has a positive peak and a negative peak (naturally). And if the transition from one peak to the other can be made in zero time, then the voltage of the waveform will always be at the positive or the negative peak. That means it will always be at its maximum, and the effective value (which is what RMS or root mean square is - it's the DC equivalent or the "area under the curve of the waveform") will be exactly what the peak value is. It's a slam dunk. If we have a (perfect) square wave of 100 volts peak, it will always be at positive or negative 100 volts. As RMS is the DC equivalent, or is the "heating value for a purely resistive load" on the voltage source, the voltage will always be 100 volts (either + or -), and the resistive load will always be driven by 100 volts. Piece of cake.

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15y ago
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13y ago

RMS means root mean square.If voltage is dependent on time,RMS can be calculated by integrating square of voltage with respect to time,and dividing it by time period(between the limits) and the taking square root of it.

These two quantities i.e.RMS voltage and Peak divided by root 2 are equal for sinusoidally varying voltage,but not in all cases.For example in case of half wave rectified voltage root mean square is equal to peak divided by 2.

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Q: What is the relationship between RMS and peak voltage for a square waveform?
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