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.

--

Roger Hayter