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

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

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$95.43$

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(D) For a full-wave rectifier,the peak voltage $V_m$ is given by $V_m = V_{rms} 	imes \sqrt{2}$.Given $V_{rms} = 220 \ V$,so $V_m = 220 	imes 1.414 = 311.13 \ V$.The average $d.c.$ voltage is $V_{dc} = \frac{2V_m}{\pi} = 0.6366 	imes 311.13 \approx 198.07 \ V$.The ripple factor $r$ for a full-wave rectifier is $0.482$.The ripple voltage $(rms)$ is given by $V_{r(rms)} = r 	imes V_{dc}$.$V_{r(rms)} = 0.482 	imes 198.07 \approx 95.47 \ V$.Rounding to the nearest option,the value is $95.43 \ V$.

$A$ full-wave rectifier supplies a load of $1 \ k\Omega$. The $a.c.$ voltage applied to the diode is $220 \ V_{rms}$. If the diode resistance is neglected,the value of the ripple voltage $(rms)$ is .... $V$.

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$A$ half-wave rectifier provides a load current of $1 \ k\Omega$. The input voltage is $220 \ V$. Neglecting the diode resistance,the value of the ripple voltage $(rms)$ is .... $V$.

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