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

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(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

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

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(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 $19.9 \ mA, 12.66 \ mA, 14.08 \ mA$.

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