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 :
To analyze the potential difference across the capacitor, let's first understand the behavior of the circuit elements involved. An ideal diode allows current to flow only in one direction, blocking current in the reverse direction. A capacitor, on the other hand, stores energy in the form of an electric field and can charge and discharge based on the applied voltage. The ac source has a voltage of 220 V . The given voltage is typically an RMS (Root Mean Square) value, which is a form of averaging used for ac voltages. The peak voltage ( V‌peak ‌ ) of the ac source can be calculated using the relationship:
V‌peak ‌=V‌RMS ‌×√2 Substituting the given RMS value, we have: V‌peak ‌‌=220V×√2 V‌peak ‌‌‌≈220V×1.414 V‌peak ‌‌‌≈311V Since the diode is ideal, during the positive half cycle of the ac voltage, the diode will conduct and allow current to flow, charging the capacitor to the peak voltage of the ac source, which is 311 V . During the negative half cycle, the diode will be in reverse bias and will not conduct, thus isolating the capacitor and maintaining its charge. Therefore, the potential difference across the capacitor will be equal to the peak voltage of the ac source. Hence, the potential difference across the capacitor is: