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

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(D) 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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$40$

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(D) 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 is $I_{dc} = \frac{I_m}{\pi} = \frac{24.75}{\pi} \approx 7.88 \ mA$.The $RMS$ current is $I_{rms} = \frac{I_m}{2} = \frac{24.75}{2} = 12.375 \ mA$.The $DC$ output power is $P_{dc} = I_{dc}^2 	imes R_L = (7.88 	imes 10^{-3})^2 	imes 1000 \approx 0.0621 \ W = 62.1 \ mW$.The $AC$ input power is $P_{ac} = I_{rms}^2 	imes (R_f + R_L) = (12.375 	imes 10^{-3})^2 	imes (10 + 1000) \approx 0.1547 \ W = 154.7 \ mW$.The efficiency $\eta$ is given by $\eta = \frac{P_{dc}}{P_{ac}} 	imes 100 = \frac{62.1}{154.7} 	imes 100 \approx 40.14 \ \%$.Thus, the efficiency is approximately $40 \ \%$.

List-$I$	List-$II$
$(a)$ Rectifier	$(i)$ Used for stepping up or stepping down the $\text{A.C.}$ voltage
$(b)$ Stabilizer	$(ii)$ Used to convert $\text{A.C.}$ voltage into $\text{D.C.}$ voltage
$(c)$ Transformer	$(iii)$ Used to remove any ripple in the rectified output voltage
$(d)$ Filter	$(iv)$ Used for constant output voltage

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