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To calculate the various values that have been presented above you need to define the question and then analyze the resulting wave form.
Some of the above comments relate to the peak voltage of the sine wave source. These all imply a capacitor input filter that as a first approximation, if the capacitance is large enough, captures the peak voltage of the waveform. In this case for any of the normal rectifier circuits the DC output voltage equals the peak voltage of the sine wave or Vrms/0.707 or Vrms*1.414 . So from a 120 V supply for any of 1/2 wave, full wave, 3 phase, and 6 phase the result is 120*sq-root of 2 = 169.7 V.
If instead, the question is what is the average DC voltage from the rectifier, then you need to look at the waveform and calculate the average value from this. The average is determined by the area under the curve divided by the averaging period. Working on a full cycle period the averaging time is 2*Pi.
The area under a sine wave can be mathematically determined with integral calculus. The integral of the sine is = -cos . To calculate the area under the curve the integral has to be evaluated at the beginning and ending points of interest.
The integral is evaluated by determining the value at the upper limit of the range and subtracting the value at the lower limit of the integration range.
For a half wave rectifier the area under the curve is -cos 180 - (-cos 0) = -(-1)-(-1) = 2 . This is the evaluation of the half sine wave for the period 0 to 180 deg. Pi radians = 180 deg. Note the integral over the second half of the waveform is zero because the voltage is zero. Since the area is 2 and the averaging period is 2*Pi the average value is 1/Pi = 0.318309886 . The average value relative to the peak voltage of the sine wave is 0.318309886*Vpeak. To get the value relative to the RMS voltage you divide by (1/2)*sq-root of 2. The result is 0.450158158*Vrms .
For the full wave rectifier it is obvious that the value is double that of the half wave because the area doubles. The results are 0.636619772*Vpeak and 0.900316316*Vrms. Thus, at 120 Vrms the average DC output voltage is 120*0.9 = 108 V DCaverage. If you connect an inductive load, such as a clutch, supplied from a bridge rectifier from 120 V single phase supply, then the DC voltage across the clutch coil would be 108 V minus the voltage drops in the diode rectifier. So about 106 to 107 V.
Now to the 3 phase 3 diode rectifier. This is one circuit of what may be described as an "n-phase half wave rectifier". Draw the composite rectified waveform. Conduction from any one diode is for 120 deg. Thus, the first diode conducts over the range 90+/-60 or from 30 to 150 deg. In radians this is from Pi/6 to 5*Pi/6. The area under this curve is -(-0.866)-(-0.866) = sq-root of 3 = 1.732050808, and the averaging time is 4*Pi/6 = 2.094395102. Thus, the average voltage is 0.826993343*Vpeak, and 1.169545202*Vrms.
From the information I provided you can try to figure out the 6 phase 6 diode half wave rectifier. This waveform is the same as a 3 phase full wave rectifier.
Also I make calculation and logic errors so see if there are any mistakes.
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