A sinusoidal voltage of rms value 200 V is connected to the diode and capacitor C in the circuit shown so that half-wave rectification occurs. The final potential difference in volt, across C, is

Question

A sinusoidal voltage of rms value 200 V is connected to the diode and capacitor C in the circuit shown so that half-wave rectification occurs. The final potential difference in volt, across C, is (A) 500 (B) 200 (C) 283 (D) 141

Tardigrade Admin · Accepted Answer

A junction diode conducts during alternate half-cycles of ac input supply. During the half-cycle of conduction, the capacitor will charge itself to peak value of supply voltage.

∴

Voltage across capacitor

= E_{rms} 2 = 200 \times 2 V = 282.8 V \approx 283 V

Q. A sinusoidal voltage of rms value 200V is connected to the diode and capacitor C in the circuit shown so that half-wave rectification occurs. The final potential difference in volt, across C, is

500

200

283

141

A junction diode conducts during alternate half-cycles of ac input supply. During the half-cycle of conduction, the capacitor will charge itself to peak value of supply voltage. ∴ Voltage across capacitor =Erms ​2​=200×2​V=282.8V≈283V

Q. A sinusoidal voltage of rms value $200 V$ is connected to the diode and capacitor $C$ in the circuit shown so that half-wave rectification occurs. The final potential difference in volt, across $C$ , is

A junction diode conducts during alternate half-cycles of ac input supply. During the half-cycle of conduction, the capacitor will charge itself to peak value of supply voltage.
$∴$ Voltage across capacitor
$= E_{rms} 2 = 200 \times 2 V = 282.8 V \approx 283 V$