A half-wave rectifier provides a load current | vedclass

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(C) For a half-wave rectifier,the peak voltage $V_m$ is given by $V_m = \sqrt{2} 	imes V_{rms}$.Given $V_{rms} = 220 \ V$,so $V_m = 1.414 	imes 220

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$99 \ V, 99 \ mA$

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(C) For a half-wave rectifier,the peak voltage $V_m$ is given by $V_m = \sqrt{2} 	imes V_{rms}$.Given $V_{rms} = 220 \ V$,so $V_m = 1.414 	imes 220 \approx 311.13 \ V$.The average $DC$ voltage $V_{dc}$ is given by $V_{dc} = \frac{V_m}{\pi}$.$V_{dc} = \frac{311.13}{3.14} \approx 99.08 \ V \approx 99 \ V$.The average $DC$ current $I_{dc}$ is given by $I_{dc} = \frac{V_{dc}}{R_L}$.Given $R_L = 1 \ k\Omega = 1000 \ \Omega$.$I_{dc} = \frac{99 \ V}{1000 \ \Omega} = 0.099 \ A = 99 \ mA$.Thus,the average $DC$ voltage is $99 \ V$ and the average $DC$ current is $99 \ mA$.

$A$ half-wave rectifier provides a load current of $1 \ k\Omega$. The input voltage is $220 \ V$. Neglecting the diode resistance,the values of average $DC$ voltage and average $DC$ current are respectively:

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