On Thu, 8 Oct 2009 09:18:59 -0700, DaveC wrote:

peak = RMS * sqrt(2) = 110 * 1.414 ~ 155V

JF

This is for half-wave rect.?

No, it's for AC and for a full-wave rectified sine wave.

What the deal is is that if 120VDC is placed across, say, a 120 ohm
resistor then the resistor will dissipate 120 watts and generate heat.

In order to get the same amount of heat generated by an AC voltage
connected across the resistor, (called the root-mean-square, or 'RMS'
voltage) it'll have to go above the steady DC value, on its peaks,
because that's the only way it can make up for the valleys which fall
below the steady DC voltage.

For AC and full-wave rectified AC, the number of peaks and valleys are
the same, and to get either of those voltages to heat up the resistor
the same amount requires that the peaks rise to the steady DC voltage
multiplied by the square root of two, and that voltage is called the
'peak' voltage.

For half-wave rectification it's an entirely different story because
half of the half sine waves are missing.