A diode is connected to 220 , V (rms) ac in s | vedclass

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(C) The diode $D$ conducts during the positive half-cycle of the $ac$ supply because it is forward-biased. During the negative half-cycle,it is revers

Vedclass · Accepted Answer

$311.1$

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(C) The diode $D$ conducts during the positive half-cycle of the $ac$ supply because it is forward-biased. During the negative half-cycle,it is reverse-biased and does not conduct. The capacitor charges to the peak value of the input voltage during the positive half-cycle. The peak voltage $V_0$ is given by $V_0 = V_{rms} 	imes \sqrt{2} = 220\sqrt{2} \, V$. Since there is no load resistor to discharge the capacitor,it remains charged to the peak voltage $V_0 = 220\sqrt{2} \, V$. However,if we consider the average $DC$ output voltage for a half-wave rectifier,$V_{dc} = \frac{V_0}{\pi} = \frac{220\sqrt{2}}{\pi} \approx 99 \, V$. Given the options provided,the question likely refers to the peak voltage or a specific $RMS$ interpretation. Re-evaluating the standard interpretation for this specific textbook problem,the capacitor charges to the peak voltage $V_0 = 220\sqrt{2} \approx 311.1 \, V$. Thus,option $(C)$ is the correct answer.

$A$ diode is connected to $220 \, V$ $(rms)$ $ac$ in series with a capacitor as shown in the figure. The voltage across the capacitor is..... $V$

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