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18 April, 20:28

A certain full-wave rectifier has a peak output voltage of 30 V. A 50 mF capacitor-input filter is connected to the rectifier. Calculate the peak-to-peak ripple and the dc output voltage devel-oped across a 600 V load resistance.

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  1. 18 April, 20:33
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    Peak to peak Ripple voltage 8.33mV

    DC output voltage = 19.11 V

    Explanation:

    Peak voltage (Vp) = 30v

    Load resistance = 600 ohms

    Capacitor filter = 50mF

    Frequency of supply = 120Hz

    The peak to peak Ripple is not only dependent on the capacitor value but also on the frequency and load current.

    To calculate the,

    Peak to peak ripple = I (load) / f*c

    I (load) = loadd current = 30/600 = 0.05 A

    Peak to peak ripple = 0.05/6

    = 8.33mV

    The average Dc output voltage for a full wave rectifier is double that of a half wave rectifier.

    The DC output voltage is equal to 0.637Vp assuming no losses.

    Vdc = 0.637 * 30

    Vdc = 19.11V
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Home » Engineering » A certain full-wave rectifier has a peak output voltage of 30 V. A 50 mF capacitor-input filter is connected to the rectifier. Calculate the peak-to-peak ripple and the dc output voltage devel-oped across a 600 V load resistance.
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