131026-1745 EDT
There may be no single answer to my original post question, but with the suggested items it is likely there is only a single answer.
This was not intended to be any sort of trick question, but just one to stimulate thought, and work from basics.
My problem was I did a calculation on the RMS current or voltage of a sine wave turned on at the 90 degree point, and I wanted a way to verify that my calculation was probably correct.
The method I used was to connect a diode between the chopped sine wave and a DC voltmeter. Thus, measure the average voltage of a sine wave over a full cycle period.
The average value of a full wave rectified sine wave is 0.636 of the peak value. The average of the waveform of interest is 0.5*0.5*0.636*Peak. If the input sine wave is 120 V RMS, then the average voltage of the half wave rectified 90 degree chopped sine wave is 27.0 V DC. The RMS value of the waveform is the square root of ( (1/2)-(1/4) )*0.707)*Peak = 0.707*RMS = 0.707*120 = 84.8 V RMS.
Experimentally I got results sufficiently close to check my calculation.
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