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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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adedayo9

Answer:

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

lhmarianateixeira

In this exercise we have to use electronic knowledge to calculate the voltage value from the resistance. In this way we can conclude that:

  • Ripple voltage: [tex]8.33mV[/tex]
  • DC output voltage: [tex]19.11 V[/tex]

So from the data given in the text, we have that:

  • Voltage: [tex]30v[/tex]
  • Load resistance: [tex]600 ohms[/tex]
  • Capacitor filter: [tex]50mF[/tex]
  • Frequency of supply: [tex]120Hz[/tex]

So with the formula given below, we can calculate what is being asked by it:

[tex]Peak \ to \ peak \ ripple = I (load)/(f)(c)[/tex]

They are using the data previously informed and putting it in the given formula, we would be with:

[tex]I (load) = load \ current = 30/600 = 0.05 A\\Peak \ to \ peak \ ripple = 0.05/6\\= 8.33mV[/tex]

The average Dc something produced power for a complete wave exist double :

[tex]Vd_c= 0.637 * 30\\Vd_c =19.11V[/tex]

See more about electronic at brainly.com/question/1255220

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