Q180 / 180 · COMEDK 2024 (Morning Shift) · Physics
An ideal diode is connected in series with a capacitor. The free ends of the capacitor and the diode are connected across a $220 \mathrm{~V}$ ac source. Now the potential difference across the capacitor is :
- Correct answerA$311 \mathrm{~V}$
- B$\sqrt{220} \mathrm{~V}$
- C$110 \mathrm{~V}$
- D$2 \sqrt{110} \mathrm{~V}$
Solution
The input voltage is an AC source given by $V(t) = V_0 \sin(\omega t)$, where $V_{rms} = 220 \text{ V}$. The peak voltage $V_0$ is given by $V_0 = V_{rms} \sqrt{2} = 220 \sqrt{2} \approx 311 \text{ V}$.
The circuit consists of an ideal diode in series with a capacitor. During the positive half-cycle of the AC source, the diode is forward-biased and acts as a short circuit. The capacitor charges up to the peak value of the input voltage, which is $V_0 = 311 \text{ V}$.
During the negative half-cycle, the diode becomes reverse-biased and acts as an open circuit. Since the diode prevents the discharge of the capacitor, the capacitor retains its charge and remains at the peak potential difference of $311 \text{ V}$.
Therefore, the potential difference across the capacitor is $311 \text{ V}$.
Answer: $311 \mathrm{~V}$
The circuit consists of an ideal diode in series with a capacitor. During the positive half-cycle of the AC source, the diode is forward-biased and acts as a short circuit. The capacitor charges up to the peak value of the input voltage, which is $V_0 = 311 \text{ V}$.
During the negative half-cycle, the diode becomes reverse-biased and acts as an open circuit. Since the diode prevents the discharge of the capacitor, the capacitor retains its charge and remains at the peak potential difference of $311 \text{ V}$.
Therefore, the potential difference across the capacitor is $311 \text{ V}$.
Answer: $311 \mathrm{~V}$
Hence, the correct answer is option: 'A'.