COMEDK2024Morning ShiftPhysicsSemiconductorsActual
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 :
Options
- A311 ~V
- B220 ~V
- C110 ~V
- D2 110 ~V
Correct answer
A. 311 ~V
Step-by-step solution
The input voltage is an AC source given by V(t) = V₀ ( t) , where V_ rms = 220 V . The peak voltage V₀ is given by V₀ = V_ rms 2 = 220 2 311 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 V₀ = 311 V . During the negative half-cycle, the diode becomes reverse-biased and acts as an open circuit. Since the diode preve