Question

A sinusoidal voltage of amplitude $25$ volt and frequency $50Hz$ 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)

$I_m=\frac{V_m}{R_f+R_L}=\frac{25}{(10+1000)}=24.75mA$
$I_{dc}=\frac{I_m}{\pi }=\frac{24.75}{3.14}=7.87mA$
$I_{rms}=\frac{I_m}{2}=\frac{24.75}{2}=12.37mA$
$P_{dc}=I_{dc}^2\times R_L={\left(7.87\times 10^{-3}\right)}^2\times 10^3=61.9mW$
$P_{ac}=I_{rms}{}^2\left(R_f+R_L\right)={\left(12.37\times 10^{-3}\right)}^2\times (10+1000)$
$=154.54mW$
Rectifier efficiency
$\eta =\frac{P_{dc}}{P_{ac}}\times 100=\frac{61.9}{154.54}\times 100=40.05\%$