A full-wave rectifier provides a load of 1 k | vedclass

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(C) For a full-wave rectifier,the average $d.c.$ voltage is given by $V_{dc} = \frac{2V_m}{\pi} \approx 0.636 V_m$,where $V_m$ is the peak voltage of

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$198 \ V, 198 \ mA$

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(C) For a full-wave rectifier,the average $d.c.$ voltage is given by $V_{dc} = \frac{2V_m}{\pi} \approx 0.636 V_m$,where $V_m$ is the peak voltage of each secondary half-cycle.Given the $rms$ voltage $V = 220 \ V$,the peak voltage is $V_m = V\sqrt{2} = 220\sqrt{2} \approx 311.13 \ V$.Substituting this into the formula: $V_{dc} = 0.636 	imes 311.13 \approx 198 \ V$.Alternatively,$V_{dc} = \frac{2 	imes (V\sqrt{2})}{\pi} = \frac{2\sqrt{2}}{\pi} V \approx 0.9 	imes 220 = 198 \ V$.The average $d.c.$ current is $I_{dc} = \frac{V_{dc}}{R_L} = \frac{198 \ V}{1000 \ \Omega} = 0.198 \ A = 198 \ mA$.

$A$ 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:

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