Final answer:
The student is looking to find the output DC voltage for a 1-ph half-wave rectifier with an inductive load. Without specific load impedance and phase shift details, it is not possible to give a precise value for the output voltage. However, the output voltage is generally going to be less than the peak input voltage due to inductive effects.
Step-by-step explanation:
The student is asking to calculate the output DC voltage of a 1-ph half-wave rectifier supplying an inductive load (R+L) with the given Beta angle of 210 degrees and an AC voltage source of 100V. The output voltage of a rectifier, especially with an inductive load, can be complex to determine without the detailed waveforms or further information about the circuit dynamics. However, in general terms, a half-wave rectifier with an inductive load would give an average output lower than the peak voltage level of the source due to the inductive drop-out during the periods when the diode is not conducting.
Nonetheless, since we are lacking specific details about the impedance of the load, the exact phase shift of the current due to the inductor, and the circuit configuration, it is not possible to provide a precise calculation. Generally, the average power output for a resistive-inductive (RL) load in a rectified AC circuit is affected by the average value of the rectified voltage and the power factor, which in turn is influenced by the phase shift β caused by the inductance L.