In a full-wave P-N junction diode rectifier,a | vedclass

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

Question

(B) The peak current $I_m$ is given by $I_m = \frac{V_m}{R_f + R_L}$.Given $V_m = 30 \ V$,$R_f = 10 \ \Omega$,and $R_L = 1500 \ \Omega$:$I_m = \frac{3

Vedclass · Accepted Answer

$19.9 \ mA, 12.66 \ mA, 14.08 \ mA$

Vedclass · Answer

(B) The peak current $I_m$ is given by $I_m = \frac{V_m}{R_f + R_L}$.Given $V_m = 30 \ V$,$R_f = 10 \ \Omega$,and $R_L = 1500 \ \Omega$:$I_m = \frac{30}{10 + 1500} = \frac{30}{1510} \approx 0.019868 \ A \approx 19.9 \ mA$.For a full-wave rectifier,the average $(DC)$ current $I_{dc}$ is given by $I_{dc} = \frac{2 I_m}{\pi}$.$I_{dc} = \frac{2 	imes 19.9 \ mA}{3.14159} \approx 12.66 \ mA$.The $RMS$ current $I_{rms}$ for a full-wave rectifier is given by $I_{rms} = \frac{I_m}{\sqrt{2}}$.$I_{rms} = \frac{19.9 \ mA}{1.414} \approx 14.08 \ mA$.Thus,the values are $19.9 \ mA, 12.66 \ mA, 14.08 \ mA$.

Similar Questions

Which of the following depicts the output of the full-wave rectifier with a capacitor filter for the given $AC$ input?

If a full wave rectifier circuit is operating from $50 \,Hz$ mains, the fundamental frequency in the ripple will be (in $\,Hz$)

If the frequency of the input voltage is $50 \,Hz$, applied to a $(a)$ half-wave rectifier and $(b)$ full-wave rectifier, what is the output frequency in both cases, respectively?

As shown in the figure,the input is applied between $A$ and $C$,and the output is taken between $B$ and $D$. What should the output be?

If in a $p-n$ junction,a sinusoidal input signal is applied at $A$ as shown,then the output voltage $V_0$ will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module