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

(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

Vedclass · Accepted Answer

$40$

Vedclass · Answer

(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 $40 \ \%$.

Similar Questions

If in a $p-n$ junction diode, a square input signal of $5 \,V$ to $-5 \,V$ is applied as shown, then the output signal across $R_L$ will be:

$A$ diode is connected to $220 \, V$ $(rms)$ $ac$ in series with a capacitor as shown in the figure. The voltage across the capacitor is..... $V$

$A$ sinusoidal voltage of peak value $200\,V$ is connected to a diode and resistor $R$ in the circuit shown,so that half-wave rectification occurs. If the forward resistance of the diode is negligible compared to $R$,the rms value of voltage across $R$ is approximately (in volt):

In the circuit shown below, the voltage appearing across the diode $D$ will be of the form:

$A$ full-wave rectifier provides a load of $1 \ k\Omega$. The diode is supplied with an $a.c.$ voltage of $220 \ V_{rms}$. If the diode resistance is neglected,the average $d.c.$ voltage and average $d.c.$ current are respectively:

Vedclass Test Series

Exam Paper Generator

Online Exam Module