Question

A sinusoidal voltage of amplitude $25$ volt and frequency $50\, Hz$ is applied to a half wave rectifier using P-n junction diode. No filter is used and the load resistor is $1000\, \Omega$. The forward resistance $R _{ f }$ of ideal diode is $10\, \Omega$. The percentage rectifier efficiency is

• A. $40 \%$
• B. $20 \%$
• C. $30 \%$
• D. $15 \%$

Answer 1

Answer: (A)

$I _{ m }=\frac{ V _{ m }}{ R _{ f }+ R _{ L }}=\frac{25}{(10+1000)}=24.75\, mA$
$I _{ dc }=\frac{ I _{ m }}{\pi}=\frac{24.75}{3.14}=7.87\, mA$
$ I _{ rms }=\frac{ I _{ m }}{2}=\frac{24.75}{2}=12.37\, mA $
$P _{ dc } = I _{ dc }^2 \times R _{ L }=\left(7.87 \times 10^{-3}\right)^2 \times 10^3=61.9\, mW $
$P _{ ac }= I _{ rms }{ }^2\left( R _{ f }+ R _{ L }\right)=\left(12.37 \times 10^{-3}\right)^2 \times(10+1000) $
$=154.54 \, mW$
Rectifier efficiency
$\eta=\frac{P_{d c}}{P_{a c}} \times 100=\frac{61.9}{154.54} \times 100=40.05 \%$