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

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(C) The input voltage is an $ac$ source with $V_{rms} = 220 \, V$. The peak voltage $V_0$ is given by $V_0 = V_{rms} \sqrt{2} = 220 \sqrt{2} \approx 3

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$311.1 \, V$

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(C) The input voltage is an $ac$ source with $V_{rms} = 220 \, V$. The peak voltage $V_0$ is given by $V_0 = V_{rms} \sqrt{2} = 220 \sqrt{2} \approx 311.1 \, V$.When the diode is forward-biased (during the positive half-cycle),it conducts and charges the capacitor to the peak value of the input voltage,$V_0$.Once the capacitor is charged to the peak voltage $V_0$,the diode becomes reverse-biased during the negative half-cycle and prevents the capacitor from discharging (assuming an ideal diode and no load).Therefore,the voltage across the capacitor will be equal to the peak voltage of the $ac$ source,which is $V_0 = 220 \sqrt{2} \approx 311.1 \, V$.

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

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