Question

An ideal diode is connected in series with a capacitor. The free ends of the capacitor and the diode are connected across a 220 V ac source. Now the potential difference across the capacitor is:

• A. 110 V
• B. 311 V
• C. \( 2 \sqrt{110} \) V
• D. \( \sqrt{220} \) V

Hint:

In AC circuits with diodes, the peak voltage is given by \( V_{\text{peak}} = V_{\text{RMS}} \times \sqrt{2} \), and this voltage appears across the capacitor in a half-wave rectifier setup.

Answer 1

The Correct Option is B

When a diode is connected in series with a capacitor, the capacitor will only charge during the half cycle when the diode is forward biased. 
In an alternating current (AC) circuit, the diode allows current to pass during one half of the AC cycle, effectively acting as a half-wave rectifier. 
The voltage across the capacitor will then be equal to the peak value of the AC voltage because the capacitor charges during the positive half cycle. The root-mean-square (RMS) value of the applied voltage is given as 220 V. 
To find the peak value \( V_{\text{peak}} \), we use the relationship: \[ V_{\text{peak}} = V_{\text{RMS}} \times \sqrt{2} \] Substituting the given RMS value of 220 V: \[ V_{\text{peak}} = 220 \times \sqrt{2} \approx 311 \, \text{V} \] Thus, the potential difference across the capacitor will be 311 V.