COMEDK · Physics · 30. Semiconductors
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 :
- A \(311 \mathrm{~V}\)
- B \(\sqrt{220} \mathrm{~V}\)
- C \(110 \mathrm{~V}\)
- D \(2 \sqrt{110} \mathrm{~V}\)
Answer & Solution
Correct Answer
(A) \(311 \mathrm{~V}\)
Step-by-step Solution
Detailed explanation
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.…
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