On 08/06/2019 11:16, Roger Hayter wrote:
Steve Walker wrote:

On 08/06/2019 09:45, Harry Bloomfield wrote:
Steve Walker expressed precisely :
Of a sinewave, not a half-wave rectified sine wave.

Sinewave: 339V / sqrt(2) = 240V RMS

Half-wave rectified sinewave: 339V / 2 = 169.5V RMS

SteveW

169.5v x 0.707= 119.83v RMS..

Peak voltage is 339V for both full and half-wave rectified sinewaves.
The equations for RMS voltage differ in each case. A quick google will
show you them and they are as I quoted above. You have taken the RMS
(not peak) voltage of a half-wave rectified sinewave and multiplied it
by the factor for converting a full-wave sinewave's peak voltage into an
RMS one!

SteveW

You can confirm your result by simple inspection. The mean square
voltage of a half wave rectified sine wave must be half that of a full
wave sine wave because half the readings are replaced by 0^2. But for
RMS we take the square root of mean square voltage. Square root of a
half is 1/1.414 which is 0.707. So half wave rectification halves the
power into a resistive load. Any other result would be surprising
because it is applying the same waveform for half the time, ignoring
polarity which we can for a resistive load.

Different ways of getting to the same result, unlike Harry's oddball
guess. Yours of course does have the advantage of showing how it gets to
the result, whereas mine just applies the different equations.

SteveW