In a half-wave rectifier using a PN junction | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The peak current $I_m$ is given by $I_m = \frac{V_m}{R_f + R_L} = \frac{25}{10 + 1000} = \frac{25}{1010} \approx 0.02475 \ A = 24.75 \ mA$.The $DC

Vedclass · Accepted Answer

$1.21$

Vedclass · Answer

(C) The peak current $I_m$ is given by $I_m = \frac{V_m}{R_f + R_L} = \frac{25}{10 + 1000} = \frac{25}{1010} \approx 0.02475 \ A = 24.75 \ mA$.The $DC$ current $I_{dc}$ is $I_{dc} = \frac{I_m}{\pi} = \frac{24.75}{\pi} \approx 7.88 \ mA$.The $RMS$ current $I_{rms}$ for a half-wave rectifier is $I_{rms} = \frac{I_m}{2} = \frac{24.75}{2} = 12.375 \ mA$.The ripple factor $\gamma$ is defined as $\gamma = \sqrt{\left( \frac{I_{rms}}{I_{dc}} \right)^2 - 1}$.Substituting the values: $\gamma = \sqrt{\left( \frac{12.375}{7.88} \right)^2 - 1} = \sqrt{(1.57)^2 - 1} = \sqrt{2.46 - 1} = \sqrt{1.46} \approx 1.21$.

In a half-wave rectifier using a $PN$ junction diode,the input voltage has an amplitude of $25 \ V$ and a frequency of $50 \ Hz$. No filter is used,and the load resistance is $1000 \ \Omega$. The forward resistance of the diode is $10 \ \Omega$. The ripple factor is:

Similar Questions

Write the types of rectifier.

The input a.c. voltage of frequency $60 \,Hz$ is applied to a half-wave rectifier and also to a full-wave rectifier. The output frequency in the case of the half-wave rectifier and that in the case of the full-wave rectifier is respectively:

The output current versus time curve of a rectifier is shown in the figure. The average value of the output current in this case is

What is rectification and rectifier? Why is a $p-n$ junction diode used as a rectifier?

In a rectifier circuit,the filter circuit:

Vedclass Test Series

Exam Paper Generator

Online Exam Module