Answer :
Final answer:
The potential difference V across the capacitor when connected to a 220 V A.C. supply with an ideal diode in the circuit will be approximately 220 V in terms of RMS voltage. This does not depend on the capacitance of the capacitor.So, the correct answer is option B.
Explanation:
When a 220 V A.C. supply is connected between points A and B in a circuit containing an ideal diode and a capacitor, the behavior of the circuit depends on the orientation of the diode. However, if the diode is oriented such that it allows current to flow (forward-biased), the capacitor will eventually charge up to the peak voltage of the A.C. supply.
During the charging period, since it's an ideal diode, it won't allow the reverse flow of charges once the capacitor is charged. Thus, the potential difference V across the capacitor will oscillate between 0 and the peak voltage of the A.C. supply. The peak voltage is √2 times the RMS voltage (which is 220 V here), so it would be approximately 311 V. But since the question is about the RMS voltage which is generally considered when dealing with A.C., the correct answer in the context of an RMS value would be B. 220 V.
It's important to note that once the capacitor is fully charged, it will block direct current as it acts like an open circuit. The potential difference across the capacitor in a DC steady state (ignoring the transient charging phase) would in fact be 220 V, matching the supply potential, while the A.C. characteristics would determine the waveform of the voltage across the capacitor.
The potential difference does not depend on the capacitance in the case of an ideal diode with an A.C. supply because capacitance would only determine how quickly the capacitor charges to that voltage, not what the final voltage across it will be.