A) \[220\,V\]
B) \[110\,V\]
C) \[311.1\,V\]
D) \[\frac{220}{\sqrt{2}}V\]
Correct Answer: D
Solution :
The diode D will conduct for positive half cycle of AC supply because this is forward biased. For negative half cycle of AC supply, this is reverse biased and does not conduct. So output would be half wave rectified and for half wave rectified output will \[{{V}_{rms}}=\frac{{{V}_{0}}}{2}=\frac{220\sqrt{2}}{2}=\frac{220}{\sqrt{2}}\]
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