Question

An AC waveform has an RMS voltage of 60 VAC. Whats the waveforms peak voltage?

Answer 1

60√2 ≈ 84.9 V

Step-by-step explanation:

You want the peak voltage of an AC waveform that has an RMS value of 60 VAC.

RMS

The square root of the average of the square of a sine wave is ...

`\displaystyle\sqrt{(1)/(2\pi)\int_0^(2\pi){\sin^2(x)}\,dx}=\sqrt{(1)/(2\pi)\left[(1)/(2)x-(\sin(2x))/(4)\right]_0^(2\pi)}=\sqrt{(1)/(2)}=(1)/(√(2))`

The sine wave has a peak value of 1, which is √2 times its RMS value.

The peak voltage of a 60 Vrms sine wave is 60√2 ≈ 84.9 V.

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Additional comment

The RMS value for any given waveform depends on the shape of the waveform. Here, we assumed the description "AC waveform" means the waveform is a sinusoid. If it is a pulse, square wave, triangle, sawtooth, or other waveform, the peak value may be different.