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 a.c. supply because this is forward biased. For negative half cycle of a.c. supply, this is reverse biased and does not conduct. So output would be half wave rectified and for half wave rectified out put \[{{V}_{rms}}=\frac{{{V}_{0}}}{2}=\frac{200\sqrt{2}}{2}=\frac{200}{\sqrt{2}}\]
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