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

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(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

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$1.21$

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(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: $\gamma = \sqrt{\left( \frac{12.375}{7.88} \right)^2 - 1} = \sqrt{(1.57)^2 - 1} = \sqrt{2.46 - 1} = \sqrt{1.46} \approx 1.21$.

List $I$	List $II$
$(a)$ Rectifier	$(i)$ Used either for stepping up or stepping down the $a.c.$ voltage
$(b)$ Stabilizer	$(ii)$ Used to convert $a.c.$ voltage into $d.c.$ voltage
$(c)$ Transformer	$(iii)$ Used to remove any ripple in the rectified output voltage
$(d)$ Filter	$(iv)$ Used for constant output voltage even when the input voltage or load current change

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:

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